An introduction to Critical Point Theory

Maria Joao Borges

March 18, 2019

1 Brief History

It is common to look for maximum or minimum values of functions using the critical points.Nevertheless it is widely known that not all critical points turn out to correspond to relativeextrema.

Example 1. For f : R2 → R, defined by

f(x, y) = x2 − y2,

we have that ∇f = (fx, fy) = (0, 0) at (0, 0) and yet f(0, 0) is neither a maximum nor aminimum of f .

The typical sort of problem we will study is to find among all functions with prescribedboundary conditions, those which minimize a given functional. More precisely, we will try tofind

minI(u) : u ∈ X and u = u0 on ∂Ωwhere

I(u) =

∫Ω

F (x, u(x), Du(x)) dx,

and Ω ⊂ Rn is a bounded open set, u : Ω→ Rm, Du ∈ Rnm, F ∈ C(Ω×Rm ×Rmn → R, u0 isa given function and X is a real Banach space.

The history of the calculus of variations is older than the greek civilization; it is widelyknonw the interest of ancient civilizations in solving problems related to “optimal forms”.The development of calculus implied a development of the calculus of variations. The firsttextbook on the area is due to Euler: “A method for finding curves enjoying certain maximumor minimum properties”, 1744. Euler expresses his belief that every effect in nature follows amaximum or minimum rule.

The calculus of variations has generated a great deal of mathematical work since then.During the 19th century, new methods were developed and important contributions were madeby, among many others, Lagrange, Riemann, Weierstrass, Jacobi, Hamilton. These are the

1

so called classical methods. During the 20th century, different techniques were introduced byHilbert and Lebesgue in connection with the study of the integral. These methods are nowknown as the direct methods.

We will list some classical examples:

- Dido’s Problem: given a certain closed curve, choose its shape so that it encloses thelargest possible area.

- Brachistochrome Problem: Given two points (x0, y0), (x1, y1) in the plane, choose thecurve that joins them such that a particle under gravity slides from one to the other inthe least amount of time possible. This was solved in 1696 by Bernoulli. The solution isobtained by finding the minimizer of the functional

I(u) =

∫ x1

x0

√1 + (u′(x))2√

2gu(x)dx,

with u(x0) = y0, u(x1) = y1. 1

- Fermat’s Principle of Least Time: Given two points (x0, y0), (x1, y1) in the plane, a lightray from one to the other, chooses the path between the points which requires the leasttime to transverse. Once again the solution corresponds to the minimizer of

I(u) =

∫ x1

x0

√1 + (u′(x))2

γ(t, u(x))dx,

where γ is the speed of light in the medium.

- Shortest Distance: Given two points (x0, y0), (x1, y1) in the plane, we want to find thecurve joining them with smallest length. We obtain it, by minimizing the functional

I(u) =

∫ x1

x0

√1 + (u′(x))2 dx.

If we work in a more general setting, we are given a Riemannian metric, and

L(u) = I(u) =

∫ 1

0

(∑aij(u(t))uiuj)

)1/2

dt

Usually we try to minimize the length of the curve, i.e., find p such that

I1(p) ≤ I1(u)

for all admissible u. p is called a minimal geodesic

1Notice that√

1 + (u′(x))2 dx = dS and√

2gu(x) = v. Hence the integrand function is dSv .

2

- Hamilton’s Principle: Consider the system formed with n particles of mass mi, i =1, . . . , n. We will denote by xi(t) the position of the ith particle at time t and xi(t) itsvelocity. The kinetic energy associated to this system is

K(t) =n∑i=1

mi

2x2i ,

and the potential energy is a function V = V (x1, . . . , xn). Then Hamilton’s Principlestates that the motion of the system, for to ≤ t ≤ t1 is such that

I(x) =

∫ t1

t0

( n∑i=1

mi

2x2i − V (x1, . . . , xn)

)dt

is extremized.

- Dirichlet’s Principle: Let D ⊂ Rn be any bounded domain with smooth boundary, andconsider the boundary value problem

∆u = 0 for x ∈ Du = φ on ∂D

where φ is any prescribed continuous function defined on ∂D. The solutions of thisproblem minimize the functional ∫

D|∇u|2dx,

over the class of functions satisfying the boundary condition u = φ.

- Plateau’s Problem: Given γ a smooth, Jordan curve in R3, find the surface of minimalarea bounded by γ, i.e., find u : B1(0) ⊂ R2 → R3 such that u|∂B1(0) = γ and

A(u) =

∫B1(0)

det(|∇u|T |∇u|)1/2dx

is minimal.

The critical points of the functional I, which are smooth functions will satisfy a differentialequation, the so called Euler-Lagrange equations, together with some auxiliary conditions.

The classical approach to minimization problems consists in constructing the associatedEuler equation, solve it and go back to prove that the solution is in fact a minimum value of I.On the other hand, direct methods work directly with the functionals.

In Rn, suppose that we have a function with 2 relative minimum, satisfying a coerciveproperty (f(x) → ∞ when |x| → ∞). If n = 1 then f must also have a relative maximum.If n > 1, we can’t guarantee the existence of a maximum, but f must have at least one morecritical point. Our goal is to extend these sort of considerations to functionals defined otherthan in Rn.

3

Another type of application is to prove the existence of special type of solutions of an ODE.Consider

x = f(x) , x ∈ [a, b] and f(a) = f(b) = 0.

Then x(t) ≡ a and x(t) ≡ b are equilibrium solutions of the equation, i.e. they are timeindependent solutions of the boundary value problem. One possible thing to do is to prove theexistence of heteroclinic solutions, asymptotic to the equilibrium, that is we seek for solutionsof the ordinary differential equation satisfying

x(t)→ a as t→∞,

andx(t)→ b as t→ −∞.

Another class of problems is to solve variational problems associated to functionals exhibitingsome sort of symmetry proprieties. Consider f ∈ C1(Rn,R) such that f(−x) = f(x) forall x ∈ Rn. Let’s look at f |Sn−1 . The critical points occur at antipodal points, and if onecorresponds to a maximum value of f its antipodal will correspond to a minimum value of f .In this case it is easy to prove that f has at least n pairs of critical points.

Example 2. For the special case

f(x) =1

2(Mx, x),

where M is a n×n symmetric matrix, with the constraint |x| = 1. Using Lagrange multipliers,we can show that the critical points of f must satisfy the equation

Mx = λx

and we notice that this is just the eigenvalue equation. Since a symmetric matrix has n distincteigenvalues, f will have at least n distinct critical points.

2 Outline of the Course

1. Classical Methods: Euler equations and applications;

2. Topics in the calculus in Banach Spaces: function spaces, properties of functionals, Frechetdifferentiability.

3. Direct methods: Minima of functionals.

4. Applications:

Solutions of BVP, Periodic solutions, heteroclinic and homoclinic solutions.

5. Geodesics: Riemannian metric, curves of shortest length;

4

6. Minimax Theorems: Mountain Pass Theorem and Saddle point Theorem;

7. Applications.

References:

- Courant and Hilbert, Volume I;

- Schwarz, Nonlinear Functional Analysis;

- M. Struwe, ”Variational Methods”;

- P. Rabinowitz, ”Minimax Methods and Critical Point Theory and Applications to Differ-ential Equations”;

- J.Mawhin, M.Willem, Critical Point Theory and Hamiltonian Systems;

3 Euler Equation and Applications

The classical approach of the calculus of variations consists of, as remarked earlier, to find thecritical points of a given functional. As in the finite dimensional case, one way of studying theproblem is by finding the zeros of the “derivative” of I, I ′(u) = 0, which are known as theEuler-Lagrange equations. We will derive the Euler-Lagrange equations in some specific cases.

3.1 The One-dimensional Case

Consider

I(u) =

∫ b

a

F (x, u, u′) dx,

where F = F (x, u, p) ∈ C2([a, b]× R2,R). The natural set of admissible functions is

Γ = u ∈ C1([a, b],R) : u(a) = α , u(b) = β,

for some given real numbers α, β. Our goal is to find

infu∈Γ

I(u)

Assume that there exists v ∈ Γ such that I(v) = infu∈Γ

I(u). We will try to obtain some necessary

conditions for this to hold..

Pick any function ϕ ∈ C2([a, b],R) such that ϕ(a) = ϕ(b) = 0, and let δ ∈ R. It is easy tocheck that the function v + δϕ is still in C1([a, b],R) and by construction verifies

(v + δϕ)(a) = α , (v + δϕ)(b) = β

5

Hence v + δϕ ∈ Γ. Fixing ϕ, define

ψ(δ) = I(v + δϕ) =

∫ b

a

F (x, v + δϕ, v′ + δϕ′) dx.

Since v is the minimizer of Iψ(0) ≤ ψ(δ) , ∀δ ∈ R.

Then ψ has a global minimum at δ = 0. On the other hand, the regularity of F and the differ-entiability of v+ δϕ imply that ψ ∈ C1(R,R). It is knonw the the minimum of a differentiablereal function is attained at the zero of the derivative. Thus

ψ′(0) = 0

and we can derive the equation∫ b

a

(Fu(x, v, v

′)ϕ+ Fp(x, v, v′)ϕ′)dx = 0,

denoted as the weak form of the Euler equation. Notice that, although we have derived theequation for fixed ϕ, it will hold for any function ϕ in the conditions we assumed.

If we further assume that v ∈ C2([a, b],R), we can integrate by parts the second term of theequation, and obtain∫ b

a

Fu(x, v, v′)ϕ dx+ Fp(x, v, v

′)ϕ|ba −∫ b

a

d

dx

(Fp(x, v, v

′))ϕ dx = 0

As a consequence of the boundary conditions on ϕ

Fp(x, v, v′)ϕ|ba = 0,

and therefore ∫ b

a

(Fu(x, v, v

′)− d

dxFp(x, v, v

′))ϕ dx = 0,

for all ϕ ∈ C2([a, b],R), ϕ(a) = ϕ(b) = 0.

Homework 1. Prove that, if for h ∈ C([a, b])∫ b

a

h(x)ϕ(x) dx = 0,

for all ϕ ∈ C([a, b]), ϕ(a) = ϕ(b) = 0, then h(x) ≡ 0 on [a, b].

Using the above result, we can finally derive the Euler equation

Fu(x, v, v′)− d

dxFp(x, v, v

′) = 0,

6

or computing the derivative, we obtain the expanded form of the Euler equation

Fpp(x, v, v′)v′′ + Fpu(x, v, v

′)v′ + Fpx(x, v, v′)− Fu(x, v, v′) = 0,

with v(a) = α, v(b) = β. Minimizing I over Γ leads to this boundary value problem. It is anecessary condition for the existence of solution.

Example 3. Consider

I(u) =

∫ b

a

√1 + (u′)2 dx , u(a) = α , u(b) = β.

As we remarked before, the minimizer of I is the curve of shortest length joining (a, α) to (b, β).Then,

F (x, u, u′) =√

1 + (u′)2

and the Euler equation isd

dx

u′√1 + (u′)2

= 0,

which implies thatu′(x) = c

√1 + (u′)2 for |c| < 1.

Solving for u′ we obtain

(u′(x))2 =c2

1− c2,

and therefore we can conclude that u′ is constant. Hence the solution of the Euler equation isthe linear function through the points (a, α) to (b, β),

u(x) =x− ab− a

β +b− xb− a

α.

Nevertheless we still have to prove that u is exactly the solution of the minimization problem -it might exist a minimizer of I not in C2.

Example 4. Consider

I(u) =

∫ b

a

F (u, u′) dx,

where F is as before, but independent of x. Then

d

dx

(F (u, u′)− u′Fp(u, u′)

)= Fuu

′ + Fpu′′ − u′′Fp − u′

d

dxFp = u′

(Fu −

d

dxFp

)If u is a solution of the Euler equation

d

dx

(F (u, u′)− u′Fp(u, u′)

)= 0,

7

which implies that the solutions of the Euler equation satisfy the first order differential equation

F (u, u′)− u′Fp(u, u′) = const. (1)

which is a simpler equation to solve. As an application of this, consider the brachistochromeproblem,

F (x, u, u′) =

√1 + (u′)2

√2gu

,

which is independent of t. Using equation (1) evaluated at a solution of the Euler equation, weobtain √

1 + (v′)2

√2gv

− (v′)2

√2gv√

1 + (v′)2= c1

which implies1

√2gv√

1 + (v′)2= c1.

Take c =√

2gc1, and solving for v′

(v′)2 =1

c2v− 1.

Assume that v′ > 0, which is a reasonable assumption since we are trying to reach the point(x1, y1) in the least possible time. Then

dv

dx=

√1− c2v

c2v⇐⇒ dx

dv=

√c2v

1− c2v.

We will make some changes of variable, in order to obtain a parametrization of the solution.Take w = c2v. Then the equation becomes

c2 dx

dw=

√w

1− w.

Making the trigonometric substitution w = sin2 θ,

dx

dw=dx

dθ

dθ

dw=dx

dθ

1

2 sin θ cos θ

Replacing in the differential equation

dx

dθ=

1

c2

1

(2 sin θ cos θ)−1

dx

dw=

2

c2sin2 θ

Solving the differential equation

x(θ) = α +1

c2(θ − 1

2sin(2θ))

8

For simplicity, assume that the initial point is at (0, 0), implying α = 0. For R = 12c2

, theparametric solutions of the problem are

x(θ) = R(2θ − sin(2θ)) and v(θ) =sin2 θ

c2= R(1− cos(2θ))

Recalling that the parametric equations for a cycloid are

X(ϕ) = R(ϕ− sinϕ) , Y (ϕ) = R(1− cosϕ)

the solution for our problem is a cycloid.

3.2 The Vectorial Case

We will now derive the Euler equation associated to the some minimization problem, for x ∈ Rand u(x) = (u1(x), . . . , un(x)). We will take

Γ = u ∈ C1([a, b],Rn) : u(a) = α , u(b) = β , α, β ∈ Rn,

as the class of admissible functions. Once again we suppose that there exists v ∈ Γ minimizingI. Our goal is to derive a necessary condition satisfied by v. Pick ϕ ∈ C1([a, b],Rn) withϕ(a) = ϕ(b) = 0, and let δ ∈ R be arbitrary. Define

ψ(δ) = I(v + δϕ)

and for fixed ϕ, ψ is a differentiable function in δ. Moreover, since v is the minimum of I overΓ, and v + δϕ ∈ Γ

ψ(0) ≤ ψ(δ) , ∀δ ∈ R,which will imply, as in the previous case, that ψ′(0) = 0. Hence we can write weak form of theEuler equation∫ b

a

(Fu1(x, v, v

′)ϕ1 + . . .+ Fun(x, v, v′)ϕn + Fp1(x, v, v′)ϕ′1 + . . .+ Fpn(x, v, v′)ϕ′n

)dx = 0.

If, furthermore v ∈ C2([a, b],Rn), we can integrate by parts and conclude∫ b

a

(Fu(x, v, v

′) · ϕ− d

dxFp(x, v, v

′) · ϕ)dx = 0 , ∀ϕ.

Using a slight variation of the result of Homework 1 2

Fu(x, v, v′)− d

dxFp(x, v, v

′) = 0,

and this system of ordinary differential equations represents the Euler equation for this problem.

2Take ϕ = (ϕ1, 0, . . . , 0) and use Homework 1 to conclude that the first component of h ≡ 0. Use this processinductively.

9

Example 5. Consider the functional

I(p, q) =

∫ b

a

(p · q −H(p, q)) dt , p, q ∈ Rn.

The quantity p · q − H(p, q) is usually denoted by Lagrangian,∫ bap · q dt the action integral

and H(p, q) the Hamiltonian. By Hamilton’s Principle, the motion of a system of n particleswith no friction extremize I(p, q). Let’s derive the Hamilton’s equation: the necessary conditionthat any minimizer of I(p, q) must satisfy. If (p, q) is a minimizer of I, for ϕ, ψ ∈ C1 withϕ(a) = ϕ(b) = 0 and ψ(a) = ψ(b) = 0, we define

Ψ(δ) =

∫ b

a

((p+ δϕ) · ( ˙q + δψ)−H(p+ δϕ, q + δψ)

)dt.

As in the previous cases, Ψ is a differentiable function of δ attaining its minimum value atδ = 0, and therefore

0 = Ψ′(0) =

∫ b

a

(ϕ · q + p · ψ −Hp(p, q) · ϕ−Hq(p, q) · ψ

)dt.

Assuming that p is twice differentiable, we can integrate by parts and as a consequence of theboundary conditions

0 =

∫ b

a

((q −Hp(p, q)) · ϕ− (p+Hq(p, q)) · ψ

)dt , ∀ϕ, ψ

Then we can write Hamilton’s equation (which is a 2n dimensional system)q = Hp(p, q)p = −Hq(p, q)

For the simplest case

H(p, q) =1

2|p|2 + V (q),

that is, in the case where the Hamiltonian is the total energy of the system,q = pp = −∇V (q)

which can be written in the formq +∇V (q) = 0

corresponding to Newton’s second law of motion.

10

3.3 Case F = F (x, u, u′, ..., u(k))

Let us study next the case

I(u) =

∫ b

a

F (x, u, u′, . . . , u(k))dx.

The minimization of such functional is connected to problems of elasticity. We will minimize Iover the natural class of functions, Γ, of the k differentiable functions u verifying the boundaryconditions

(u, u′, . . . , u(k−1))(a) = (α, α1, . . . , αk−1) and (u, u′, . . . , u(k−1))(b) = (β, β1, . . . , βk−1).

for some given (α, α1, . . . , αk−1), (β, β1, . . . , βk−1) ∈ Rn. Suppose that there is a minimizer ofI, v ∈ Γ.

Homework 2. Derive the correspondent Euler equations (You might need to make some furtherassumptions).

3.4 The Scalar Case

Another problem is to minimize

I(u) =

∫Ω

F (x, u,Du) dx,

where Ω ⊂ Rn is a bounded, open set with smooth boundary. The set of admissible functionsis

Γ = u ∈ C1(Ω,R) : u|∂Ω = f(x)

where f is a given smooth function. Suppose it exists v ∈ Γ such that I(v) = infΓI. To find

the Euler equations we will proceed as in the previous examples. Take ϕ ∈ C1(Ω,R) such thatϕ|∂Ω = 0. Then v + δϕ ∈ Γ and we can define

ψ(δ) = I(v + δϕ) =

∫Ω

F (x, v + δϕ, (vx1 + δϕx1 , . . . , vxn + δϕxn))dx.

Although it is not as simple as the previous cases, we can prove that ψ is a differentiablefunction of δ, and

ψ(0) ≤ ψ(δ) , ∀δ ∈ R

Hence

0 = ψ′(δ) =

∫Ω

(Fu(x, v,Dv)ϕ+ Fp1(x, v,Dv)ϕx1 + . . .+ Fpn(x, v,Dv)ϕxn

)dx

=

∫Ω

(Fu(x, v,Dv)ϕ+ Fp(x, v,Dv) · ∇ϕ

)dx

11

Assuming that v ∈ C2, we can use the Divergence Theorem. We recall that∫Ω

∇ · v dx =

∫∂Ω

v · ν dx,

where ν(x) is the outward pointing unit normal to ∂Ω. Look at

∇ · (ϕFp) =∑i

∂

∂xi(ϕFpi) =

∑i

∂ϕ

∂xiFpi + ϕ

∑i

∂Fpi∂xi

= ∇ϕ · Fp + ϕ∇ · Fp

which implies∇ϕ · Fp = ∇ · (ϕFp)− ϕ∇ · Fp

Using this equality, we can write the weak Euler equation as

0 =

∫Ω

(Fu(x, v,Dv)−∇ · Fp

)ϕ+

∫Ω

∇ · (ϕFp)

and applying the Divergence Theorem

0 =

∫Ω

(Fu(x, v,Dv)−∇ · Fp

)ϕ+

∫∂Ω

Fpϕ · νdS

Since ϕ = 0 on ∂Ω ∫Ω

(Fu(x, v,Dv)−∇ · Fp

)ϕ dx = 0,

for all such ϕ. Using the usual arguments, one can show that this implies

Fu(x, v,Dv)−∇ · Fp = 0

or

Fu(x, v,Dv)−∑i

∂

∂xiFpi(x, v,Dv) = 0 (2)

Example 6. Take F (x, u,Du) = 12|∇u|2−G(x, u). Using (2), the correspondent Euler equation

is

−Gu(x, u)−∑i

∂

∂xiuxi = 0

which can be written as Gu(x, u) + ∆u = 0 , x ∈ Ωu = f on ∂Ω

12

3.5 Free Boundary Problem

We will consider next the case, where no boundary conditions are assumed. We will minimize

I(u) =

∫ b

a

F (x, u, u′)dx

over the class of admissible functions

Γ = C1([a, b],R).

This problem is usually denoted by free boundary problem. Proceeding as in the previous cases,for arbitrary ϕ ∈ Γ and v the minimizer of I over Γ, the weak Euler equation is verified, i.e.,∫ b

a

(Fu(x, v, v

′)ϕ+ Fp(x, v, v′)ϕ′)dx = 0 , ∀ϕ ∈ Γ

Further assuming that v ∈ C2, we can integrate by parts and obtain∫ b

a

(Fu(x, v, v

′)ϕ− d

dxFp(x, v, v

′)ϕ)dx+ Fp(x, v, v

′)ϕ|ba = 0 (3)

Choose ϕ ∈ Γ such that ϕ(a) = ϕ(b) = 0. For this particular case equation (3) becomes∫ b

a

(Fu(x, v, v

′)− d

dxFp(x, v, v

′))ϕ dx = 0

and, as before, this implies that v is a solution of the Euler equation

Fu(x, v, v′)− d

dxFp(x, v, v

′) = 0 (4)

For an arbitrary ϕ ∈ Γ, (3) and (4)yields

0 = Fp(b, v(b), v′(b))ϕ(b)− Fp(a, v(a), v′(a))ϕ(a).

Choose ϕ such that ϕ(a) = 0 and ϕ(b) = 1. Then

Fp(b, v(b), v′(b)) = 0. (5)

Reversing the roles of a and b, we also obtain

Fp(a, v(a), v′(a)) = 0. (6)

We conclude that although our problem has no boundary conditions, the minimizer must verifysome sort of boundary conditions. (5) and (6) are denoted as natural boundary conditions.

13

3.6 Constrained Variational Problem

We want to minimize

I(u) =

∫ b

a

F (x, u, u′)dx

subject to the integral constraint

J(u) =

∫ b

a

G(x, u, u′)dx = c

for some given constant c and G ∈ C2([a, b]× R2,R) 3. Define the class

Γ = u ∈ C1 : J(u) = c.

If v is a minimizer of I over the Γ, consider ϕ ∈ C1([a, b],R) and δ ∈ R such that v + δϕ ∈ Γ.Define ψ(δ) = I(v+ δϕ) subject to χ(δ) ≡ J(v+ δϕ) = c. By construction, the minimum valueof ψ verifying χ(δ) = c is achieved at δ = 0. Then using Lagrange multipliers

d

dδ

(ψ(δ) + λχ(δ)

)= 0 at δ = 0. (7)

Example 7. Minimize∫ π

0(u′(t))2dt subject to u(0) = u(π) = 0 and

∫ π0u2(t)dt = 1. Notice that

the constraint excludes the obvious candidate u ≡ 0. By (7) we may write the equation∫ π

0

(u′ϕ′ + λuϕ

)dt = 0

If u ∈ C2([a, b],R)

d

dx

(Fp(x, u, u

′) + λGp(x, u, u′))− (Fu(x, u, u

′) + λGu(x, u, u′)) = 0,

which is equivalent to u′′ − λu = 0u(0) = u(π) = 0

This is an eigenvalue problem for ordinary differential equations: we are looking for the nontriv-ial solutions of the boundary value problem. It is known that the eigenfunctions of this problemare uk(t) = sin(kt), corresponding to the eigenvalues λk = −k2.

Homework 3. Find the curve with endpoints u(a) = α, u(b) = β of shortest length which hasarea A below it.

To derive the Euler equation, we need to assume that the minimizer is a C2 function.Solving the Euler equation yields a C2 candidate for the minimum of I. But since the classof admissible functions is, roughly speaking, C1, the minimum value of I may be achieved inC1 \C2. We would like to derive a procedure, which ables us to show when the solution of theEuler equation is in fact the minimizer of the functional.

3Other type of constraints may be |u(x)| = 1, or more generally that u(x) lies on a given manifold.

14

Example 8. In the case of the curve with shortest length, we saw that the solution of the Eulerequation is the straight line through (a, α) and (b, β). We will prove that in fact is the minimizerof I. Writing the solution explicitly

v(x) =x− ab− a

β +b− xb− a

α

we can compute

I(v) =

∫ b

a

√1 + (v′)2dx =

∫ b

a

√1 + (

β − αb− a

)2dx =√

(b− a)2 + (β − α)2

which is the formula for the distance between two points.

Notice that minimizing I, depends on our choices of the endpoints A(a, α), B(b, β). Fix Aat (0, 0) and let B = (x, y) change. For each (x, y) we obtain a function v(t) and

I(v) = (x2 + y2)1/2 ≡ S(x, y).

Defineϕ(x, y, p) = Sx(x, y) + Sy(x, y)p,

For arbitraryγ ∈ Γ = u ∈ C1([0, x]) : u(0) = 0 and u(x) = y.

we will have

ϕ(t, γ(t), γ′(t)) = Sx(t, γ(t)) + Sy(t, γ(t))γ′(t) =d

dtS(t, γ(t))

Then ∫ x

0

ϕ(t, γ, γ′)dt =

∫ x

0

d

dtS(t, γ)dt = S(x, y) = I(v) (8)

On the other hand, ifϕ(x, y, p) ≤

√1 + p2 , ∀x, y, p (9)

then ∫ x

0

ϕ(t, γ, γ′)dt ≤∫ x

0

√1 + (γ′)2dt = I(γ) (10)

It is easy to see that (8) and (10) imply

I(v) ≤ I(γ) , ∀γ ∈ Γ

and therefore v is the minimizer of I over Γ. All we have to do is prove that (9) holds. Noticethat

ϕ(x, y, p) =x√

x2 + y2+

yp√x2 + y2

=(x, y) · (1, p)√

x2 + y2

Using Schwarz inequality(x, y) · (1, p) ≤

√x2 + y2

√1 + p2

and (9) is verified.

15

Let’s try to generalize this procedure, in order to prove that the solution of the Eulerequation is in fact the minimizer of the functional I(u) =

∫ baF (x, u, u′)dt over the class

Γ = u ∈ C1([a, b]) : u(a) = α and u(b) = β.

Fixing A at (0, 0) and leting B(x, y) change, for v the minimizer of I over Γ

I(v) = S(x, y).

Definingϕ(x, y, p) = Sx(x, y) + Sy(x, y)p.

for anyγ ∈ Γ = u ∈ C1([0, x]) : u(0) = 0 and u(x) = y,

as above

ϕ(t, γ(t), γ′(t)) =d

dtS(t, γ(t))

and therefore

I(v) =

∫ x

0

ϕ(t, γ(t), γ′(t)) dt

If we can prove

ϕ(x, γ(t), γ′(t)) ≤ d

dtS(t, γ(t))

doing the same analysis we can prove that the solution of the Euler equation is in fact theminimizer of I over Γ.

To summarize, we consider the functional I(u) =∫ baF (x, u, u′)dx, where F is a smooth

function in its arguments. The procedure we have been following is

1. Choose a class of admissible functions, Γ.

2. Existence of Minimizer. Assume that there exists v ∈ Γ such that I(v) = infΓI. This

implies that v is a solution of the weak Euler equation∫ b

a

(Fu(x, v, v′)ϕ+ Fp(x, v, v

′)ϕ′)dx = 0

for all ϕ ∈ C1([a, b]) such that ϕ(a) = α and ϕ(b) = β.

3. Regularity. Assume that v ∈ C2([a, b]) and the Euler equation is satisfied.

So far we have been working with the class

Γ = u ∈ C1([a, b],R) : u(a) = α , u(b) = β

but notice that we can choose to work with other classes of functions. The bigger the class Γis, the smaller the infimum might be. For example if we take the set of piecewise C1 functions,instead of C1,

infPC1

I ≤ infC1I.

16

Example 9. Consider I(u) =∫ 1

−1(1 − (u′)2)2dx with u(−1) = u(1) = 0. It is obvious that

I(u) ≥ 0, which implies that inf I ≥ 0. But notice that I(v) = 0 for v(x) = |x|. Hence inf I = 0and therefore |x| is one minimizer of I which is not C1([−1, 1]).

Homework 4. Find all the minimizers of I as in Example 9, that lie in PC1([−1, 1]). Noticethat none of it is in C1([−1, 1]).

Example 10. Consider I(u) =∫ 1

−1|1 − u|2u2dx with u(−1) = 0 and u(1) = 1. Once again

inf I ≥ 0 and for v(x) = max0, x, I(v) = 0. Hence inf I = 0. Check that v is the onlyminimizer of I.

Example 11. Let I(u) =∫ 1

−1(2x− u′)2u2dx with u(−1) = 0 and u(1) = 1. Then I(v) ≥ 0 and

for

v(x) =

0 if x ≤ 0x2 if x > 0

I(v) = 0. Note that v ∈ C1 but still not in C2 (meaning that it is not solution of the Eulerequation).

4 Topics in Calculus in Banach Spaces

In order to solve our minimization problems in a more general setting, we need some notionsin an abstract set. Consist of

- working on appropriate spaces,

- continuity properties,

- notions of differentiability,

Let E be a real Banach space.

. C([a, b],R) with norm ‖u‖ = maxx∈[a,b]

|u(x)|;

. C([a, b],Rn) with norm ‖u‖ = maxx∈[a,b]

|u(x)|;

. C(Ω,R), Ω ⊂ Rn compact, with norm ‖u‖ = maxx∈Ω|u(x)|;

. C1([a, b]) with norm ‖u‖C1([a,b]) = ‖u‖+ ‖u′‖;

. Lp([a, b]) with norm ‖u‖p =( ∫ b

a|u(x)|pdx

)1/p

.

17

The particular case p = 2, L2([a, b]) is a Hilbert space: the norm comes from the inner product

(u, v)L2([a,b]) =

∫ b

a

u(x)v(x) dx

In an integral setting, L2 is the analogue of continuous functions. The analogues of the spacesCp are the Sobolev spaces.

. W 1,2([a, b]) is the completion of C1([a, b]) under the norm

‖u‖W 1,2([a,b]) =(∫ b

a

(u2(x) + (u′(x))2

)dx)1/2

.

In the case [a, b] = [0, 2π], a function in L2([0, 2π]) may be seen as a Fourier series

u(x) =∑k∈Z

akeikx,

and by Parseval’s identity ∫ 2π

0

|u(x)|2dx =∑k∈Z

|ak|2.

If u ∈ W 1,2([0, 2π]), the same idea can be applied to u′ and therefore

‖u‖W 1,2([0,2π]) =∑k∈Z

(1 + k2)a2k.

. W 1,2T (R) is the set of T -periodic functions in W 1,2. Is the closure of C1

T (R) under thenorm of W 1,2([0, T ]).

. W 1,20 ([a, b]) is the set of W 1,2 functions vanishing at a and b, is the closure of C1

0([a, b])under the W 1,2([a, b]) norm.

Lemma 1. (Sobolev Embedding Theorem) If u ∈ W 1,2([a, b]) then u ∈ C([a, b]), and

‖u‖C([a,b]) ≡ ‖u‖L∞([a,b]) ≤ α‖u‖W 1,2([a,b]).

for some positive constant α.

Proof:

Suppose that u ∈ C1([a, b]) and consider a ≤ y < x ≤ b. Then

u(x)− u(y) =

∫ x

y

u′(t) dt,

18

which implies that

|u(x)− u(y)| ≤∫ x

y

|u′(t)| dt ≤ |b− a|1/2(∫ b

a

|u′(t)|2dt)1/2

On the other hand, since |u(x)− u(y)| ≥ ||u(x)| − |u(y)||

|u(x)| ≤ |u(y)|+ |b− a|1/2(∫ b

a

|u′(t)|2dt)1/2

Integrating in y over [a, b],

|u(x)| ≤ 1

|b− a|

∫ b

a

|u(y)|dy +1

|b− a|1/2(∫ b

a

|u′(t)|2dt)1/2

≤ 1

|b− a|(b− a)1/2

(∫ b

a

|u(t)|2dt)1/2

+1

|b− a|1/2(∫ b

a

|u′(t)|2dt)1/2

≤ α‖u‖w1,2

Taking the supremum over x ∈ [a, b] the inequality is proved. Consider now the case u ∈W 1,2([a, b]). By definition, there exist a sequence (um) ⊂ C1([a, b]) such that

‖um − u‖W 1,2([a,b]) → 0 as m→∞

Since um is a convergent sequence in W 1,2([a, b]), it will be a Cauchy sequence in W 1,2([a, b]).By definition um ∈ C1([a, b]) and as a consequence , for m, j ∈ N

‖um − uj‖C([a,b]) ≤ α‖um − uj‖W 1,2([a,b])

Hence um is a Cauchy sequence in C([a, b]). Then um converges uniformly in C([a, b]) to somefunction v ∈ C([a, b]). But notice that if um → v in C([a, b])∫ b

a

|um − v|2dx ≤ maxx∈[a,b]

|um(x)− v(x)||b− a| → 0 as m→∞

implying um → v in L2([a, b]). By uniqueness of limit u ≡ v. In particular this implies thatu ∈ C([a, b]).

♦

Definition 1. A function is Holder continuous of order α, if

|u(x)− u(y)| ≤ K|x− y|α , 0 < α < 1.

Corollary 1. If u ∈ W 1,2([a, b]) then u is Holder continuous of order 1/2.

Homework 5. Prove the result.

19

As a reference we state de general Sobolev Theorem, which is of extreme importance in theapplications to PDE’s.

Theorem 1. (Sobolev Imbedding Theorem) Let Ω ⊂ Rn be bounded with ∂Ω smooth.Then

u ∈ W 1,20 (Ω) ⇒ u ∈ L

2nn−2 (Ω)

for n ≥ 3, and there exists c = c(Ω, n) such that

‖u‖Ls(Ω) ≤ c‖u‖W 1,20 (Ω) , ∀s ∈ [1,

2n

n− 2]

This inequality implies that

W 1,20 (Ω) ⊂ Ls(Ω) , ∀s ∈ [1,

2n

n− 2]

and this imbedding is compact if s < 2nn−2

.

For the proof see

Adams - Sobolev Spaces;

Friedman - PDE.

4.1 Continuity Properties of Functionals

Let E be a real Banach space, and consider I : E → R.

Definition 2. I is continuous at u ∈ E if

∀ε > 0 ∃δ = δ(ε, u) : ‖v − u‖E ≤ δ ⇒ |I(v)− I(u)| ≤ ε.

I is continuous on E, I ∈ C(E,R), if it is continuous on every point on E.

Definition 3. I is sequentially continuous at u ∈ E if

∀um → u in E I(um)→ I(u).

Homework 6. Prove that I is continuous at u ∈ E iff I is sequentially continuous at u.

Example 12. Consider

I(u) =

∫ b

a

F (x, u, u′) dx

where F is continuous in its arguments (F ∈ C([a, b] × R2,R)), and E = C1([a, b]). For anyu ∈ E, let (um) ⊂ E such that ‖um − u‖E → 0 as m → ∞. Since F (x, um, u

′m) → F (x, u, u′)

uniformly for x ∈ [a, b]. Hence I is sequentially continuous and as a consequence continuous.

20

Example 13. Consider

I(u) =

∫ b

a

F (x, u, u′) dx

defined in E = W 1,2([a, b]). Notice that if F is just continuous, I may not be well defined in E.As an example of this, consider F (x, u, u′) = (u′(x))4, [a, b] = [0, 1] and let u(x) = x3/4. Thenu ∈ W 1,2([0, 1]) and yet I(u) = ∞. Therefore we need some further conditions on F . We willsuppose that F verifies the Growth Condition

|F (x, y, z)| ≤ γ(x, y)z2 + β(x, y)

where γ, β are continuous in its arguments. This condition means that F can’t grow more thanquadratically in z (since u′ ∈ L2 this should be enough for I to be well defined). Let us start toprove that I is well defined. For u ∈ W 1,2([a, b]) let ‖u‖W 1,2([a,b]) = M . By Sobolev EmbeddingTheorem ‖u‖C([a,b]) ≤ αM , and

|F (x, u, u′)| ≤ γ(x, u(x))|u′(x)|2 + β(x, u(x))

≤ maxx∈[a,b],|u|≤αM

|γ(x, u(x))||u′(x)|2 + maxx∈[a,b],|u|≤αM

|β(x, u(x))|

≤ M1|u′(x)|2 +M2

for some positive constants M1, M2. Hence

I(u) ≤M1‖u‖W 1,2([a,b]) + (b− a)M2

and thus I is well defined in W 1,2([a, b]). To prove that I is continuous we need some moreassumptions. The ideal situation is that F ∈ C2 in its arguments, but we don’t need so much.We will assume that F ∈ C1([a, b]× R2,R) and that

|Fy(x, y, z)| ≤ γ1(x, y)|z|2 + β1(x, y)

and|Fz(x, y, z)| ≤ γ2(x, y)|z|+ β2(x, y)

for γi, βi, i = 1, 2, continuous in its arguments. Notice that the second condition makes sense,since we assumed that F grows at most quadratically in z, Fz grows at most linearly in z. Then

I(v)− I(u) =

∫ b

a

(F (x, u, u′)− F (x, v, v′))dx

=

∫ b

a

∫ 1

0

d

dsF (x, u+ s(v − u), u′ + s(v′ − u′))dsdx

=

∫ b

a

∫ 1

0

(Fu(x, u+ s(v − u), u′ + s(v′ − u′))(v − u) +

+Fp(x, u+ s(v − u), u′ + s(v′ − u′))(v′ − u′))dsdx

21

Using the growth conditions on the derivatives

|I(v)− I(u)| ≤∫ b

a

∫ 1

0

(γ1(x, u+ s(v − u))|u′ + s(v′ − u′)|2|v − u|

+β1(x, u+ s(v − u))|v − u|++γ2(x, u+ s(v − u))|u′ + s(v′ − u′)||v′ − u′|

+β1(x, u+ s(v − u))|v′ − u′|)dsdx

Let M = max‖u‖W 1,2([a,b]), ‖v‖W 1,2([a,b]). Then both ‖u‖W 1,2([a,b]) and ‖v‖W 1,2([a,b]) are boundedby M , and by Sobolev inequality, both ‖u‖C([a,b]) and ‖v‖C([a,b]) are bounded by αM . Notice thatwe can make a rough estimate

|u+ s(v − u)| ≤ |u|+ s(|u|+ |v|) ≤ 3αM

Thus ∫ b

a

∫ 1

0

γ1(x, u+ s(v − u))|u′ + s(v′ − u′)|2|v − u| dsdx

≤ maxx∈[a,b],|u+s(v−u)|≤3αM

γ1 maxx∈[a,b]

|v(x)− u(x)|∫ b

a

∫ 1

0

|u′ + s(v′ − u′)|2dsdx

≤ Γ1(M)‖u− v‖C([a,b])

and similarly ∫ b

a

∫ 1

0

β1(x, u+ s(v − u))|v − u| dsdx ≤ B(M)‖u− v‖C([a,b])

On the other hand using Schwarz inequality∫ b

a

∫ 1

0

γ2(x, u+ s(v − u))|u′ + s(v′ − u′)||v′ − u′| dsdx ≤ Γ2(M)‖u′ − v′‖L2([a,b])

and ∫ b

a

∫ 1

0

β1(x, u+ s(v − u))|v′ − u′| dsdx ≤ B2(M)‖u′ − v′‖L2([a,b])

Finally, it is easy to check that

|I(v)− I(u)| ≤ K(M)‖u− v‖W 1,2([a,b])

and this proves that I is continuous on E.

Homework 7. Prove that I(u) =∫ baG(x, u(x))dx defined on E = W 1,2([a, b]), is continuous,

where G ∈ C([a, b]× R,R).

22

We will now review some of the most important results in Functional Analysis. The nextproposition states that the distance to a closed, convex subset of a Hilbert space is achieved.

Proposition 1. Let K be a closed, convex set contained in E. Then for all points u ∈ E, thereis a unique v ∈ K such that

‖u− v‖ = infw∈K‖u− w‖

Proof:

Let d = infw∈K‖u−w‖. By definition of infimum there exists (vm) ⊂ K such that ‖u−vm‖ d

as m→∞. By the parallelogram law

2(‖u− vn‖2 + ‖u− vm‖2) = ‖vm − vn‖2 + ‖2u− (vn + vm)‖2 = ‖vm − vn‖2 + 4‖u− vn + vm2‖2

K is convex and therefore vn+vm2∈ K, since is the midpoint of the segment line joining vm and

vn. Moreover d is the infimum, hence

‖u− vn + vm2‖2 ≥ d2

Thus4d2 ≥ ‖vm − vn‖2 + 4d2

which implies that ‖vm − vn‖ ≤ 0. Then (vm) is a Cauchy sequence, and since K is closedvm → v as m → ∞ for some v ∈ K. To prove the uniqueness, let’s suppose that there aretwo such elements v, w. Take the sequence v, w, v, w, . . .. By definition this is a minimizingsequence. But we just proved that any minimizing sequence is a Cauchy sequence. Henceforthit must have a unique limit, which is a contradiction unless v = w.

♦

Theorem 2. (Projection Theorem) Let F be a closed linear subspace of E. Then, for allu ∈ E there exists unique v ∈ F and w ∈ F⊥ such that u = v + w.

Proof:

Since a closed subspace is a closed convex set, by the previous proposition, there existsv ∈ F such that

‖u− v‖ = infw∈F‖u− w‖

Choose any f ∈ F . By definition of v and the fact that v + αf ∈ F for all α ∈ R

‖u− v‖2 ≤ ‖u− v − αf‖2 = ‖u− v‖2 − 2α(u− v, f) + α2‖f‖2.

If α ∈ R+,

(u− v, f) ≤ α

2‖f‖2 , ∀f ∈ F , α ∈ R+

23

and letting α→ 0(u− v, f) ≤ 0 , ∀f ∈ F

On the other hand, for α ∈ R− converging to 0, we will obtain

(u− v, f) ≥ 0 , ∀f ∈ F

Then (u− v, f) = 0 for all f ∈ F , and therefore u− v ∈ F⊥. Since u = v + (u− v), the resultis proved. To show uniqueness, suppose that u = v1 + w1 = v2 + w2 with v1, v2 ∈ F and w1,w2 ∈ F⊥. Then v1 − v2 = w1 − w2 and since F ∩ F⊥ = 0 we must necessarily have v1 = v2

and w1 = w2.

♦

Corollary 2. If F is a closed subspace of E and F 6= E, there exists w ∈ F⊥ \ 0.

Proof:

By hypothesis F 6= E, and therefore ∃u ∈ E \ F . By the Projection Theorem

u = v + w , v ∈ F w ∈ F⊥

and w 6= 0 otherwise u = v ∈ F .♦

Definition 4. Let E be a real Banach space. l : E → R is a linear functional on E, if

∀α, β ∈ R ∀u, v ∈ E l(αu+ βv) = αl(u) + βl(v)

l is bounded if∃M ∈ R+ : |l(ϕ)| ≤M‖ϕ‖E ∀ϕ ∈ E

The classE ′ = bounded linear functional over E

is known as the dual space of E. We define

‖l‖ = infE

|l(ϕ)|‖ϕ‖

= sup‖ϕ‖=1

|l(ϕ)|

which is a norm in E ′. Moreover, E ′ is a Banach space relatively to ‖ · ‖E′.

Notice that as a consequence of the definition, if a linear functional l is bounded, then

|l(ϕ)| ≤ ‖l‖E′‖ϕ‖E , ∀ϕ ∈ E

Homework 8. If l is a bounded linear functional on E, prove that l is bounded iff l is contin-uous.

24

Consider the set of all bounded linear functionals on E. In the case of E being a Hilbertspace its dual is known.

Theorem 3. (Riesz Representation Theorem) If E is a real Hilbert space and l ∈ E ′,there exists a unique u ∈ E such that

l(e) = (u, e)E , ∀e ∈ E.

Moreover‖l‖E′ = ‖u‖E.

Proof:

Let N be the null space of l, i.e.

N = n ∈ E : l(n) = 0.

It is known that N is a closed linear subspace of E. If E = N then u = 0 and we are done.Otherwise, we are in the conditions of Corollary 2, and there exists v ∈ N⊥ \ 0, and inparticular l(v) can not be 0. Choose any e ∈ E. By linearity

l(e− l(e)

l(v)v)

= l(e)− l(e) = 0

which implies that e− l(e)

l(v)v ∈ N . Then

0 = (v, e− l(e)

l(v)v) = (v, e)− l(e)

l(v)‖v‖2

implying that

l(e) =(e, v)

‖v‖2l(v) = (

l(v)v

‖v‖2, e) , ∀e ∈ E.

Sincel(v)v

‖v‖2∈ E the existence result is proved. To prove the uniqueness, suppose that for u 6= w

l(e) = (u, e) = (w, e) , ∀e ∈ E.

Then

‖u− w‖2 = (u− w, u− w) = (u, u− w)− (w, u− w) = l(u− w)− l(u− w) = 0

which implies that u = w. To prove the norm equality, notice that

‖u‖2E = |(u, u)E| = |l(u)| ≤ ‖l‖E′‖u‖E,

and therefore‖u‖E ≤ ‖l‖E′ . (11)

25

On the other hand, by definition

‖l‖E′ = sup‖ϕ‖E=1

|l(ϕ)|

which means that∀ε > 0 ∃e ∈ E : ‖e‖ = 1 and |l(e)| ≥ ‖l‖E′ − ε.

But also|l(e)| = (u, e) ≤ ‖u‖E‖e‖E = ‖u‖E

Then‖u‖E ≥ |l(e)| ≥ ‖l‖E′ − ε.

Letting ε→ 0‖u‖E ≥ ‖l‖E′ . (12)

(11) and (12) imply ‖u‖E = ‖l‖E′ .♦

4.2 Weak Convergence of Sequences

Definition 5. If E is a real Banach space, um converges weakly to u, um u in E, ifl(um)→ l(u) for all continuous linear functional l on E.

In the case of a Hilbert space, by Riesz Representation Theorem

um u if ∀ϕ ∈ E (ϕ, um)→ (ϕ, u) as m→∞.

In the particular case E = L2([a, b])

um u if ∀ϕ ∈ L2([a, b])

∫ b

a

ϕ(x)um(x) dx→∫ b

a

ϕ(x)u(x) dx as m→∞.

Example 14. In E = L2([0, 2π]) take um(x) = cos(mx). We will determine the weak limit ofum in E. Since any ϕ ∈ E can be written as a Fourier series,

ϕ(x) =∑i∈N

ai cos(ix) + bi sin(ix)

then

< ϕ, um >=

∫ 2π

0

ϕ(x)um(x)dx =

∫ 2π

0

am(cos(mx))2dx = πam

By Parseval’s identity

∞ > ‖ϕ‖2E =

∑i∈N

(a2i + b2

i )

implying that both ai, bi → 0 as i → ∞. In particular, this implies that < ϕ, um >→ 0 asm→∞, and therefore um 0 as m→∞. We remark that um 6→ 0 as m→∞.

26

Theorem 4. Let E be a Hilbert space, and (um) ⊂ E be bounded. Then (um) has a weaklyconvergent subsequence, i.e., there exists u ∈ E such that (umi , ϕ)→ (u, ϕ) for all ϕ ∈ E.

Proof:

Suppose we can find a subsequence (umi) of (um) such that (umi , ϕ) is a Cauchy sequencefor all ϕ ∈ E. Then, we can define

l(ϕ) = limi→∞

(umi , ϕ).

Since (um) is bounded in E, there exists M > 0 such that ‖um‖E ≤ M for all m ∈ M , and asa consequence

|(umi , ϕ)| ≤ ‖umi‖‖ϕ‖ ≤M‖ϕ‖

Then|l(ϕ)| ≤M‖ϕ‖,

and therefore l ∈ E ′. By the Riesz Representation Theorem, there exists u ∈ E such that

l(ϕ) = (u, ϕ) , ∀ϕ ∈ E

By constructionlimi→∞

(umi , ϕ) = l(ϕ) = (u, ϕ) ∀ϕ ∈ E

as wanted. It remains to show that there exists such subsequence (umi). We can simplify theproblem a little further. If

F = span(um),

we know that F is a closed linear subspace of E. By the Projection Theorem

∀ϕ ∈ E ϕ = f + g for f ∈ F and g ∈ F⊥.

Therefore(um, ϕ) = (um, f) + (um, g) = (um, f)

since (um) ⊂ F and g ∈ F⊥. So it’s enough to prove that (umi , ϕ) is a Cauchy sequence for allϕ ∈ F . Consider (um, u1). By assumption

|(um, u1)| ≤ ‖um‖‖u1‖ ≤M2.

Hence |(um, u1)| is a bounded set of R which implies that we can extract a convergent subse-quence, which we will denote as

(u1m, u1)

Next, consider (u1m, u2). Once again this is a bounded sequence in R and therefore admits aconvergent subsequence

(u2m, u2)

27

Proceeding inductively we construct the convergent sequence (ukm, uk). With a diagonalizationprocess we can take umj = ujj. To show that (ujj, ϕ) is a Cauchy sequence, for all ϕ ∈ F , noticethat the result is obvious for any ϕ which is a linear combination of uj’s, since by construction(ujj, ϕ) is convergent for any such ϕ. It remains to show that the same thing holds for anelement on the closure. Take any ψ ∈ F and any ε > 0, we can find a finite linear combinationof elements of (um), ε close to ψ, i.e., we can find

f =k∑i=1

αiui : ‖f − ψ‖ ≤ ε.

Then|(uii − ujj, ψ)| ≤ (uii − ujj, ψ − f) + (uii − ujj, f).

On one hand|(uii − ujj, ψ − f)| ≤ ‖uii − ujj‖‖ψ − f‖ ≤ 2Mε,

on the other hand, since f ∈ F as we remarked earlier, by construction

(uii − ujj, f)→ 0 as i, j →∞.

Hence, for i, j sufficiently large |(uii − ujj, ψ)| is as small as we want.

♦

4.3 Lower Semicontinuity of Functionals

Definition 6. I : E → R is lower semicontinuous at x if for all sequences xm → x on E

I(x) ≤ lim infm→∞

I(xm).

Example 15. Let I : R→ R defined by

I(x) =

|x|+ 1 if x 6= 00 if x = 0

The definition of lower semicontinuity allows the function to jump down: note that I(0) = 0 ≤limx→0

I(x) = 1.

Definition 7. If I : E → R, with E a Banach space, then I is weakly lower semi-continuousat u ∈ E, if for all sequences um u

I(u) ≤ lim infm→∞

I(um).

28

Example 16. Let I(u) = ‖u‖2 defined on a Hilbert space E. Recall that in Example 14, weproved that a sequence of unit vectors, converges weakly to 0, so I can’t be continuous in areasonable way. For um u

0 ≤ ‖um − u‖2 = ‖um‖2 − 2(um, u) + ‖u‖2,

which implies that2(um, u)− ‖u‖2 ≤ ‖um‖2. (13)

Choose a subsequence (umi) verifying

limi→∞‖umi‖ = lim inf

m→∞‖u‖.

By (13) applied to (umi)2(umi , u)− ‖u‖2 ≤ ‖umi‖2.

Letting i→∞‖u‖2 ≤ lim inf

m→∞‖um‖2,

and therefore ‖ · ‖2 is weakly lower semicontinuous.

Definition 8. If E is a Banach space, I : E → R is said to be weakly sequentially continuousat u ∈ E, if whenever um u, I(um)→ I(u).

Notice that functionals which are weakly sequentially continuous have some kind of com-pactness properties.

Remarks:

1 - If I is weakly sequentially continuous then I is weakly lower semicontinuous.

2 - If I1 is weakly lower semicontinuous and I2 is weakly sequentially continuous, then I1 +I2

is weakly lower semicontinuous.

Example 17. Let E = W 1,2([a, b]) and

I2(u) =

∫ b

a

g(x, u) dx

where g ∈ C([a, b] × R,R). First notice that I2 is well defined in E (if u ∈ E, then u iscontinuous and therefore g(·, u) is continuous). We claim that that I2 is weakly sequentiallycontinuous. Let um u on E. By definition

(um, ϕ)→ (u, ϕ) , ∀ϕ ∈ W 1,2([a, b])

Then (um, ϕ) is a convergent sequence in R and therefore is bounded.

29

Theorem 5. Banach-Steinhaus Theorem Suppose that E is a Hilbert space, and considerS ⊂ E. If

supx∈S|(x, ϕ)| ≤ Aϕ , ∀ϕ ∈ E

then S is a bounded subset of E.

By direct application of the Theorem, we can conclude that (um) is bounded in E. Hence,there exists M ≥ 0 such that

‖um‖E ≤M , ∀m ∈ N

By the Sobolev inequality‖um‖L∞ ≤ αM

and therefore (um) is bounded in L∞([a, b]). As we remarked before, if u ∈ W 1,2([a, b]) then uis Holder continuous of order 1/2. In particular, for all m ∈ N and x, y ∈ [a, b]

|um(x)− um(y)||x− y|1/2

≤ ‖u′m‖2L2([a,b]) ≤M2

and therefore (um) is an equicontinuous family of functions.

Theorem 6. (Ascolli-Arzella Theorem) If F ⊂ C(K,R), where K ⊂ Rn is compact, andF is a uniformly bounded and equicontinuous set, then F is pre-compact (any bounded sequencehas a convergent subsequence).

Then, along some subsequence, there exists v ∈ C([a, b]) such that umi → v uniformly, i.e.,

maxx∈[a,b]

|umi(x)− v(x)| → 0 as i→∞

This implies ∫ b

a

umi(x)ϕ(x)dx→∫ b

a

v(x)ϕ(x)dx , ∀ϕ ∈ E

On the other hand

|∫ b

a

um(x)ϕ(x)dx| ≤ ‖um‖L2‖ϕ‖L2 ≤ ‖um‖E‖ϕ‖E ≤M‖ϕ‖E

and therefore we can conclude that∫ baum(x)ϕ(x)dx is a bounded linear functional on E. Since

um u, ∫ b

a

umi(x)ϕ(x) dx→∫ b

a

u(x)ϕ(x) dx , ∀ϕ ∈ E

By uniqueness of limit ∫ b

a

u(x)ϕ(x) dx =→∫ b

a

v(x)ϕ(x)dx , ∀ϕ ∈ E

30

Since C1([a, b]) is dense in E, we will also have∫ b

a

(u− v)(x)ϕ(x)dx = 0 , ∀ϕ ∈ C1([a, b])

which as we have seen before, this implies that u = v. Hence umi → u in L∞([a, b]), asi → ∞. The next thing to prove is that all the sequence converges. If not, we could findanother subsequence (umi) such that ‖umi −u‖ ≥ ε for some positive ε. But as above (umi) is auniformly bounded sequence and equicontinuous. Ascolli-Arzella Theorem implies the existenceof v ∈ C([a, b]) such that umi → v in L∞([a, b]), as i→∞. Using the same arguments we canprove that u = v. Then

um u in E ⇒ um → u in L∞([a, b])

which implies that I2(um)→ I2(u) as we wanted.

Example 18. Let E = W 1,2([a, b],R) and

I(u) =

∫ b

a

1

2|u′|2dx

We want to prove that I is weakly lower semicontinuous on E. First notice that we can writeI in the form

I(u) =1

2

∫ b

a

(|u|2 + |u′|2) dx− 1

2

∫ b

a

|u|′dx ≡ I1(u) + I2(u)

In Example 16 we proved that I1 is weakly lower semicontinuous, and in Example 17 we provedthat I2 is weakly sequentially continuous. Hence I is weakly lower semicontinuous.

4.4 Minima of Functionals

Theorem 7. Let E be a Hilbert space, K ⊂ E bounded and closed under weak convergentsequences. Let I : E → R be a weakly lower semicontinuous functional. Then there existsv ∈ K such that

I(v) = infKI > −∞

Proof:

Choose (um) ⊂ K a minimizing sequence, i.e., I(um) infKI. Since K is a bounded set,

(um) is bounded in E. Then there exists a subsequence umi ⊂ K such that umi v, and v ∈ K(since K is closed). By the weak lower semicontinuity of I

I(v) ≤ lim infi→∞

I(umi) = infKI

Then I(v) = infKI > −∞.

♦

31

Corollary 3. If E is a Hilbert space, I : E → R weakly lower semicontinuous and I(u)→∞as ‖u‖ → ∞, then I attains its infimum over E, i.e.,

∃v ∈ E : I(v) = infEI

Proof:

Choose z ∈ E such that I(z) <∞. By the coercivity hypothesis, for ‖u‖ > R, I(u) > I(z)for R sufficiently large. Therefore

infEI = inf

‖u‖≤RI(u)

But then we are in the conditions of Theorem 7 and the result follows.

♦

Corollary 4. If I is weakly sequentially continuous on K, where K is as in the Theorem 7,then I achieves its infimum and supremum on K.

Proof:

Since I is weakly sequentially continuous, both I and −I are weakly lower semicontinuous.By Theorem 7, I and −I achieve the infimum on K. Moreover − inf

k(−I) = sup

KI.

♦

Proposition 2. Let I : K → R with I weakly sequentially continuous, K ⊂ E bounded, and Eis an infinite dimensional Banach space. Then

I(∂K) ⊃ I(intK)

Proof:

Since E is an infinite dimensional Banach space, we can consider a sequence of unit vectors(en)n∈N verifying ‖em‖ = 1 and em 0 as m→∞. Let u ∈ intK (then I(u) ⊂ I(intK)), andchoose tm such that

um = u+ tmem ∈ ∂K (14)

Notice that |tm| = ‖um−u‖, and since K is bounded, tm is a bounded sequence of real numbers.Letting m→∞ in (14), by construction em 0 and tm is bounded. Then the right hand side

u+ tmem u

and as a consequence um u. Since I is weakly sequentially continuous, I(um) → I(u). Butnotice that I(um) ∈ I(∂K). Thus

I(u) ∈ I(∂K)

as wanted.♦

As an immediate consequence we can state the following result.

32

Corollary 5. (Maximum/Minimum Modulus Theorem) In the same hypothesis of Propo-sition 2

supKI = sup

∂KI , inf

KI = inf

∂KI

Example 19. Let V ∈ C2([a, b]× Rn,R), and consider the corresponding Hamiltonian system

(HS) q + Vq(t, q) = 0

As we have seen before, (HS) is the Euler equation for the minimization of the functional

I(q) =

∫ b

a

L(q) dt

where L(q) = 12|q|2 − V (t, q) is the associated Lagrangian.

As a simple Special Case, consider n = 1 and V (q) = 1 − cos q. Then the Euler equationbecome

q + sin q = 0 , q(a) = α q(b) = β

which is the equation of motion of a simple pendulum.

To prove that I has a minimum value, we can’t work in such a generality. So we furtherassume that

V (t, x) ≤ 0 , ∀t ∈ [a, b] and x ∈ Rn (V 1)

(V1) implies L(q) ≥ 0 (and therefore I is bounded by below), and I(q) is well defined forq ∈ E = W 1,2([a, b],Rn). By the Sobolev Theorem we can assume that q is continuous and thusit makes sense to consider the boundary conditions in E.

Consider the problem of minimizing I over

Γ = q ∈ W 1,2([a, b],Rn) : q satisfies BC

and letc = inf

ΓI(q)

As an immediate consequence of (V1), c ≥ 0. Let (qm) ⊂ Γ be a minimizing sequence, i.e.,I(qm) c as m → ∞. Then I(qm) is a bounded sequence in R and therefore we can assumethat ∃M ≥ 0 such that I(qm) ≤M . Using (V1)

M ≥ I(qm) ≥∫ b

a

1

2|qm|2dt =

1

2‖qm‖L2([a,b]) (15)

and therefore (qm) is a bounded sequence in L2([a, b]). On the other hand

|qm(t)| ≤ |qm(a)|+∫ b

a

|qm|dt

≤ α + |b− a|1/2‖qm‖L2([a,b])

≤ α + |b− a|1/2√

2M ≡M1

33

Taking the supremum over t ∈ [a, b], we obtain ‖qm‖L∞ ≤M1 and as a consequence

‖qm‖2L2([a,b]) =

∫ b

a

|qm|2dt ≤ ‖qm‖2L∞|b− a| = M2

1 |b− a| (16)

(15) and (16) imply that (qm) is bounded in W 1,2([a, b]). Then, there exists v ∈ E such thatqm v in E (possibly along a subsequence that we will denote again by qm). Also, notice that

I(q) =1

2

∫ b

a

(|q|2 + |q|2)dt−∫ b

a

(1

2|q|2 + V (t, q)) dt =

1

2‖q‖2

W 1,2 − I2(q)

and as we have seen before ‖ · ‖ is weakly lower semicontinuous and I2 is weakly continuous(since 1

2|q|2 + V (t, q) is continuous). Hence I is weakly lower semicontinuous, which implies

thatI(v) ≤ lim inf I(qm) = c (17)

Next we will prove that v ∈ Γ. By its definition v ∈ E, so all we have to show is that it satisfiesthe boundary conditions. But note that since qm converges weakly to v in E, by Sobolev Theoremqm → v uniformly and therefore v(a) = lim qm(a) = α and v(b) = lim qm(b) = β. Thus v ∈ Γand as a consequence

I(v) ≥ c (18)

(17) and (18) imply I(v) = c. We still have to show that v ∈ C2([a, b]) and that it satisfies(HS). We will do this later.

Example 20. In Example 19, we used the condition V (t, x) ≤ 0 to obtain estimates on aminimizing sequence in E. In this example we will assume a growth condition instead

|V (t, x)| ≤ A+B|x|γ , with 0 < γ < 2 (V 1)

where A and B are positive real numbers. Condition (V 1) forces V not to grow fast, in factit forces V to grow less than quadratically. Let (qm) be a minimizing sequence, our goal is toobtain estimates for (qm) in E. Once we do that, the rest follows as in Example 19. SinceI(qm) converges, there exists M ≥ 0 such that I(qm) ≤M for all m ∈ N. Then∫ b

a

1

2|qm|2dt ≤M +

∫ b

a

|V (t, qm)| dt

and by (V 1)1

2‖qm‖2

L2([a,b]) ≤M + (b− a)A+B

∫ b

a

|qm|γdt

As noticed, our goal is to get an estimate on∫ ba|qm|γdt in terms of ‖qm‖L2. Since qm is Holder

continuous of order 1/2, for t ∈ [a, b]

|q(t)| ≤ |q(a)|+ (t− a)1/2(∫ t

a

|q|2dt)1/2

34

and therefore

1

2‖qm‖2

L2([a,b]) ≤M1 +B

∫ b

a

(α + (t− a)1/2

(∫ t

a

|q|2dt)1/2)γ

dt (19)

Notice that ifz > w then (z + w)γ ≤ (2z)γ

and ifz < w then (z + w)γ ≤ (2w)γ

In either case(z + w)γ ≤ (2z)γ + (2w)γ

Using this fact in (19)

1

2‖qm‖2

L2([a,b]) ≤ M1 + 2γB

∫ b

a

(αγ + (t− a)γ/2

(∫ b

a

|qm|2dt)γ/2)

dt

≡ M2 + 2γB

∫ b

a

(t− a)γ/2(∫ b

a

|qm|2dt)γ/2

dt

≤ M2 +M3

(∫ b

a

|qm|2dt)γ/2

Lemma 2. (Young’s Inequality) If x, y ∈ R+ and σ, τ ≥ 1 satisfying 1σ

+ 1τ

= 1, then

xy ≤ xσ

σ+yτ

τ

Homework 9. Prove the result. Hint: Use the sign of the second derivative of the logarithimfunction.

Choosing x = M3

εand y = ε

( ∫ ba|qm|2dt

)γ/2, the inequality implies

M3

(∫ b

a

|qm|2dt)γ/2

≤ Mσ3

σ+ ετ

γ

2

(∫ b

a

|qm|2dt) γ

2τ

Take τ γ2

= 1 (notice that since γ < 2, τ > 1), and σ such that the inequality holds. Then

M3

(∫ b

a

|qm|2dt)γ/2

≤ Mσ3

σ+γ

2ε2/γ‖qm‖2

L2([a,b])

Finally1

2‖qm‖2

L2([a,b]) ≤M2 +Mσ

3

σ+γ

2ε2/γ‖qm‖2

L2([a,b])

Choose ε so small so that ετ γ2

= 14, and conclude

1

4‖qm‖2

L2([a,b]) ≤M4

and from here we proceed as in Example 19.

35

4.5 Differentiability in Banach Spaces

Let E, F be real Banach spaces and L : E → F a linear map.

Example 21. 1. Let E = F = Rn and Lu = Au where A is a (n× n) matrix.

2. Let E = C1([a, b]), F = C([a, b]) and Lu = u′.

3. Let E = C2([a, b]), F = C([a, b]) and Lu = u′′ + a(x)u where a is a given continuousfunction.

Definition 9. L is a bounded map if there exists M > 0 such that

supx 6=0,x∈E

‖Lx‖F‖x‖E

≤M or sup‖x‖=1

‖Lx‖F ≤M

We can easily prove, like in the case of linear functionals, that a linear map is bounded iffis continuous.

DefineL(E,F ) = Continuous Linear maps

which is a Banach space under the norm

‖L‖L(E,F ) = sup‖x‖E=1

‖Lx‖F

Definition 10. Let O ⊂ E be open and suppose T : O → F . Then T is Frechet Differentiableat u ∈ O if there exists L = L(u) ∈ L(E,F ) such that

∀ε > 0 ∃δ = δ(ε, u) : ‖v‖E ≤ δ ⇒ ‖T (u+ v)− T (u)− L(v)‖F ≤ ε‖v‖E

The linear map L = L(u), is called the Frechet derivative of T at u, denoted by T ′(u).

Example 22. 1. Let E = Rn, F = Rm and T : Rn → Rm. Then the Frechet derivative ofT at u ∈ E is T ′(u) = ( ∂Ti

∂uj), which is the Jacobian matrix of T .

2. T ∈ L(E,F ). Then, by linearity

‖T (u+ v)− T (u)− T ′(u)v‖ = ‖T (v)− T ′(u)v‖

and therefore choosing T ′(u)v = T (v) the definition is verified independently of u.

3. For E = C([0, 1]), consider T : E → E defined by T (u) = u3. Then

T (u+ v)− T (u) = (u+ v)3 − u3 = 3uv2 + 3u2v + v3

Looking at the linear part in v, we guess that T ′(u) = 3u2. In fact

‖T (u+ v)− T (u)− T ′(u)v‖ = ‖3uv2 + v3‖ = o(‖v‖)

as wanted.

36

Remarks:

1. The Frechet derivative T ′(u) is unique. If there were two linear maps S ′(u) and T ′(u),verifying the definition,

‖T (u+ v)− T (u)− T ′(u)v‖ = o(‖v‖)

and‖T (u+ v)− T (u)− S ′(u)v‖ = o(‖v‖)

for ‖v‖ ≤ δ. Then

‖T ′(u)− S ′(u)‖L(E,F ) = sup‖v‖E=δ

‖(T ′(u)− S ′(u))v‖

≤ ‖T ′(u)v − T (u+ v) + T (u)‖+ ‖S ′(u)v − T (u+ v) + T (u)‖≤ o(‖v‖) = ε‖v‖

for all ε > 0, implying S ′(u) = T ′(u).

2. If T is Frechet differentiable at u, then T is continuous at u, since

‖T (u+ v)− T (u)‖ ≤ ‖T ′(u)v‖+ o(‖v‖) ≤M‖v‖+ o(‖v‖)

which implies the continuity at u since M‖v‖+ o(‖v‖) is small whenever ‖v‖ is small.

3. If S and T are Frechet differentiable at u then S + T is Frechet differentiable at u, and

(S + T )′(u) = S ′(u) + T ′(u)

Example 23. Let E = C1([0, 1]), F = R and consider

I(u) =

∫ 1

0

G(x, u, u′) dx

where G = G(x, u, p) ∈ C1([0, 1]× R2,R). Let us study the Frechet differentiability of I.

I(u+ v)− I(u) =

∫ 1

0

(G(x, u+ v, u′ + v′)−G(x, u, u′)

)dx

=

∫ 1

0

∫ 1

0

d

dsG(x, u+ sv, u′ + sv′) ds dx

=

∫ 1

0

∫ 1

0

(Gu(x, u+ sv, u′ + sv′)v +Gp(x, u+ sv, u′ + sv′)v′

)ds dx

Looking at the “linear part” in v, we guess

I ′(u)v =

∫ 1

0

(Gu(x, u, u

′)v +Gp(x, u, u′)v′)dx

37

To show the claim we need to prove that

|I(u+ v)− I(u)−∫ 1

0

(Gu(x, u, u′)v +Gp(x, u, u

′)v′)dx| = o(‖v‖)C1([0,1])

By the above computations

I(u+ v)− I(u)−∫ 1

0

(Gu(x, u, u′)v +Gp(x, u, u

′)v′) dx

=

∫ 1

0

∫ 1

0

[(Gu(x, u+ sv, u′ + sv′)v +Gp(x, u+ sv, u′ + sv′)v′)ds

−(Gu(x, u, u′)v −Gp(x, u, u

′)v′]dx

=

∫ 1

0

∫ 1

0

[(Gu(x, u+ sv, u′ + sv′)−Gu(x, u, u

′))v

+(Gp(x, u+ sv, u′ + sv′)−Gp(x, u, u′))v′

]ds dx

Since G ∈ C1([0, 1]× R2), both Gu and Gu′ are continuous. Hence, for ‖v‖E ≤ δ

|Gu(x, u+ sv, u′ + sv′)−Gu(x, u, u′)| ≤ max

x∈[0,1],‖v‖E≤δ|Gu|ε ≡M1ε

and similarly

|Gp(x, u+ sv, u′ + sv′)−Gp(x, u, u′)| ≤ max

x∈[0,1],‖v‖E≤δ|Gu′ |ε ≡M2ε

Then

|I(u+ v)− I(u)− I ′(u)v| ≤∫ 1

0

(M1|v|+M2|v′|) dx

≤ M3 maxx∈[0,1]

(|v(x)|+ |v′(x)|) = M3‖v‖E

Definition 11. If T is Frechet differentiable for all u ∈ O and T ′(u) is continuous (in thetopology of L(E,F ), we write that T ∈ C1(O, F ).

We have seen in the examples, that somehow we have to guess the Frechet derivative. This maynot always be as simple as it has been so far. We will introduce a weaker notion of derivative,the Gateaux derivative, that in particular will help us to find the Frechet derivative.

Definition 12. Let T : O → F , where O ⊂ E is open. Then T is Gateaux differentiable atu ∈ O if

T ′G(u)v = limh→0

T (u+ hv)− T (u)

h

exists for all v ∈ E.

38

The difference between the two notions of derivative is that the Gateaux dervative involvesonly ‖ · ‖F , while the Frechet derivative involves both ‖ · ‖E and ‖ · ‖F . In the Euclidean space,the Gateaux derivative corresponds to the directional derivative.

If T is Frechet differentiable then is Gateaux differentiable and T ′(u) = T ′G(u). The conversemay not be true.

We are interested in studying maps I ∈ C1(E,R), where E is either a Banach or a Hilbertspace. Then I ′(u) ∈ L(E,R), and via the Riesz Representation Theorem I ′(u) will be associatedto some w ∈ E.

Consider I ≡ I1 + I2, where

I1(q) =

∫ b

a

1

2|q|2dt

and for V ∈ C1([a, b]× R,Rn)

I2(q) =

∫ b

a

−V (t, q) dt

defined in E = W 1,2([a, b],Rn). We will prove that I ∈ C1(E,R).

Lemma 3. I1 ∈ C1(E,R).

Proof:

By its definition

I1(q + v) =

∫ b

a

1

2|q + v|2dt =

∫ b

a

1

2|q|2 + (q, v) +

1

2|v|2 dt

= I1(q) + I ′1(q)v + o(‖v‖)

Hence, it is obvious that

I ′1(q)v =

∫ b

a

(q, v) dt

We must prove that I ′1 is continuous in the appropriate topology. We will show that if qm → qin E then I ′1(qm)→ I ′1(q) in L(E,F ).

|I ′1(qm)ϕ− I ′1(q)ϕ| =∣∣∣ ∫ b

a

( ˙qm − q) · ϕ dt∣∣∣ ≤ ‖qm − q‖L2([a,b])‖ϕ‖L2([a,b])

≤ ‖qm − q‖E‖ϕ‖E

Then

sup‖ϕ‖6=0

|I ′1(qm)ϕ− I ′1(q)ϕ|‖ϕ‖E

≤ ‖qm − q‖E

and therefore‖I ′1(qm)− I ′1(q)‖E′ ≤ ‖qm − q‖E

So I1 ∈ C1(E.R).♦

39

Lemma 4.

I2(q) =

∫ b

a

V (t, q) dt ∈ C1(E,R)

for all V ∈ C1([a, b]× Rn,R).

Proof:

To figure out what the Frechet derivative might be, we will start by computing the Gateauxderivative. For any ϕ ∈ E

I2(q + hϕ)− I2(q)

h=

1

h

∫ b

a

(V (t, q + hϕ)− V (t, q)

)dt

=1

h

∫ b

a

∫ 1

0

d

dsV (t, q + hsϕ) dt

=1

h

∫ b

a

∫ 1

0

Vq(t, q + hsϕ)hϕ dt

which implies

(I ′2(q))Gϕ = limh→0

I2(q + hϕ)− I2(q)

h=

∫ b

a

Vq(t, q)ϕ dt

We now have to verify that this is the Frechet derivative, i.e., if∣∣∣I2(q + ϕ)− I2(q)−∫ b

a

Vq(t, q)ϕ dt∣∣∣ =

∫ b

a

[V (t, q + ϕ)− V (t, q)− Vq(t, q)ϕ

]dt = o(‖ϕ‖E)

Pick any ε > 0. Since V ∈ C1([a, b],Rn), for all (t, z) ∈ [a, b] × Rn there exists δ = δ(ε, t, z)such that

|w| < δ ⇒ |V (t, z + w)− V (t, z)− Vq(t, z)w| ≤ ε|w| (20)

since t ∈ [a, b] and this is a compact interval, δ = δ(ε, z). If furthermore we restrict to |z| ≤M ,then δ = δ(ε,M). Using Sobolev inequality, if q ∈ E, then ‖q‖L∞([a,b]) ≤ α‖q‖E. ChooseM = α‖q‖E + 1 and ϕ ∈ E such that ‖ϕ‖E ≤ δ. Without loss of generality we can assume thatαδ ≤ 1. Then

‖q + ϕ‖L∞([a,b]) ≤ α(‖q‖E + ‖ϕ‖E) ≤ α‖q‖E + 1 = M

Hence we are in the conditions of (20) and therefore

|V (t, q + ϕ)− V (t, q)− Vq(t, q)ϕ| ≤ ε|ϕ(t)|

which implies

|I2(q + ϕ)− I2(q)−∫ b

a

Vq(t, q) · ϕ dt| ≤∫ b

a

ε|ϕ(t)|dt ≤ ε(b− a)1/2‖ϕ‖E

40

Thus I2 is Frechet differentiable. We still have to prove that I ′2(q) is continuous in E ′. Letqm → q in E.

‖I ′2(qm)− I ′2(q)‖L(E,R) = sup‖ϕ‖E=1

|I ′2(qm)ϕ− I ′2(q)ϕ|

= sup‖ϕ‖E=1

∣∣∣ ∫ b

a

(Vq(t, qm)− Vq(t, q)) · ϕ dt∣∣∣

≤ sup‖ϕ‖E=1

‖Vq(t, qm)− Vq(t, q)‖L2([a,b])‖ϕ‖L2([a,b])

≤ ‖Vq(t, qm)− Vq(t, q)‖L2([a,b])

Since, by Sobolev Theorem, qm is converging uniformly to q, and V ∈ C1([a, b]× Rn)

‖Vq(t, qm)− Vq(t, q)‖L2([a,b]) → 0

as m→∞ and therefore I ′2(q) is continuous.

♦

By Lemma 3 and Lemma 4, we can conclude that I = I1 − I2 ∈ C1(E,R).

Definition 13. If I ∈ C1(E,R), E is a real Banach space, u is a critical point of I if I ′(u) ≡ 0,that is I ′(u)v = 0 for all v ∈ E.

Lemma 5. If I ∈ C1(E,R), a local maximum or minimum of I is a critical point of I.

Proof:

Let v be a local minimum of I. As a consequence, for any ϕ ∈ E and δ ∈ R+, I(v + δϕ) ≥I(v), implying

I(v + δϕ)− I(v)

δ≥ 0

Letting δ → 0 we obtain I ′G(v)ϕ ≥ 0, but replacing ϕ by −ϕ, it is obvious that I ′G(v)ϕ ≤ 0.Hence I ′G(v)ϕ = 0 for all ϕ ∈ E.

♦

Summarizing a litle bit, we are proving the existence of a solution of (HS), by minimizing thefunctional

I(q) =

∫ b

a

(1

2|q|2− V (t, q)) dt

over the classΓ = q ∈ E : q(a) = α , q(b) = β

So far, we proved the existence of v ∈ Γ such that infΓI = I(v), and that I ∈ C1(E,R). To

finish our problem we still have to prove a couple of things: that v is a critical point of I, andthat v ∈ C2([a, b],Rn) and that way, as we have seen, it will satisfy the Euler equation.

41

Lemma 6. If v is the minimizer of I over Γ, then v is a critical point of I.

Proof:

Choose ϕ ∈ E and arbitrary δ ∈ R. If v is the minimizer of I over Γ, in order to havev+δϕ ∈ Γ, we must restrict to the case when ϕ(a) = ϕ(b) = 0. Then, as in the proof of Lemma5,

I ′(v)ϕ = 0 , ∀ϕ ∈ E

with ϕ(a) = ϕ(b) = 0. This means that we still have some work to do to show that v is a criticalpoint of I in Γ. If we consider the special case, α = β = 0, we can replace E = W 1,2([a, b],Rn)by E0 = W 1,2

0 ([a, b],Rn) which is a proper subspace of E. We showed that I ∈ C1(E,R), andit is easy to understand that with similar arguments one can show that I ∈ C1(E0,R). If thisis the case, v is a critical point since I ′(v)ϕ = 0 for all ϕ ∈ E0.

To prove the general case, we have to use a trick to reduce the problem to E0. Choose anyfunction p ∈ E such that p(a) = α and p(b) = β. In particular p can be the linear function

p(t) =t− ab− a

β +b− tb− a

α

DefineJ(q) = I(q + p)

where q ∈ E0. Notice that we proved that if Q is a minimizer of J over E0, then J ′(Q) ≡ 0.But, if v is a minimizer of I over E, by its definition, v − p ∈ E0 and is a minimizer of J overE0. Then

I ′(v) = J ′(v − p) = 0

meaning that v is a critical point of I.♦

Lemma 7. If v ∈ Γ is such that infΓI = I(v), then v ∈ C2([a, b],Rn).

Proof:

Consider the linear boundary value problem

q + Vq(t, v) = 0

q(a) = α

q(b) = β

Homework 10. Solve this Boundary Value Problem (write the solution explicitely).

There is w ∈ C2([a, b],Rn) solution to this problem. Then

w + Vq(t, v) = 0 , w(a) = α , w(b) = β

42

Take any ϕ ∈ E0, multiply both sides of the equation and integrate over [a, b], yields∫ b

a

(w + Vq(t, v)) · ϕdt = 0 , ∀ϕ ∈ E0

Integrate by parts the first term of the integral

wϕ|ba −∫ b

a

w · ϕdt+

∫ b

a

Vq(t, v) · ϕdt = 0 , ∀ϕ ∈ E0

and since ϕ(a) = ϕ(b) = 0∫ b

a

(w · ϕ− Vq(t, v)

)· ϕdt = 0 , ∀ϕ ∈ E0 (21)

On the other hand, by Lemma 6∫ b

a

(v · ϕ− Vq(t, v)

)· ϕdt = 0 , ∀ϕ ∈ E0 (22)

Subtracting equations (21) and (22) ∫ b

a

( ˙w − v) · ϕdt = 0

for all ϕ ∈ E0. In particular, since w − v ∈ E0, we can take ϕ = w − v, and∫ b

a

|w − v|2dt = 0

which implies |w − v| = 0 a.e. in [a, b].

Lemma 8. If ϕ ∈ W 1,2([a, b],Rn) and ϕ′ = 0 a.e., then ϕ = constant.

Proof:

Suppose that ϕ ∈ C1([a, b]). Then we can write

ϕ(x) = ϕ(a) +

∫ x

a

ϕ′(t)dt

and since ϕ′ = 0, we conclude ϕ(x) = ϕ(a) for all x ∈ [a, b]. By definition, every function ofE is the limit (in the E norm) of C1([a, b]) functions. Then the result holds for an arbitraryϕ ∈ E.

♦If w − v is constant, using the boundary conditions we conclude that w − v = 0 which impliesin particular that v ∈ C2([a, b],Rn).

♦

43

We have shown that any minimizer of I is in C2([a, b]) and therefore is a solution of theEuler equation. Let’s look now at a more general setting. Consider

I(u) =

∫ b

a

F (x, u, u′)dx

and suppose that there exists a critical point v. We want to indicate a method in how to checkthat the critical point has more regularity that the one it seems. All we have to start withis the weak form form of the Euler equation I ′(v)ϕ = 0, for all ϕ in an appropriate class offunctions, ∫ b

a

(Fu(x, v, v′)ϕ+ Fp(x, v, v

′)ϕ′)dx = 0 (23)

and suppose for the moment that v ∈ W 3,2 (which corresponds to v ∈ C3), and take ϕ = v′′ .Then (23) is equivalent to ∫ b

a

(Fu(x, v, v′)v′′ + Fp(x, v, v

′)v′′′)dx = 0

Integrate by parts∫ b

a

Fp(x, v, v′)v′′′dx = Fp(x, v, v

′)v′′|ba −∫ b

a

(Fpx(x, v, v′) + Fpu(x, v, v

′)v′ + Fpp(x, v, v′)v′′)v′′dx

and assume that ϕ ≡ 0 at the boundary points. Since v′′ = ϕ, that will imply v′′(a) = v′′(b) = 0and therefore∫ b

a

Fp(x, v, v′)v′′′dx = −

∫ b

a

(Fpx(x, v, v′) + Fpux, v, v

′)v′ + Fpp(x, v, v′)v′′)v′′dx

Hence∫ b

a

Fpp(x, v, v′)(v′′)2dx = −

∫ b

a

Fp(x, v, v′)v′′′dx−

∫ b

a

(Fpx(x, v, v′) + Fpu(x, v, v

′)v′)v′′dx

Notice that the left hand side is quadratic in v′′, while the right hand side is linear in v′′.Further assume that

Fpp(x, y, z) ≥ γ > 0

Then

γ‖v′′‖2L2 ≤

∫ b

a

Fpp(x, v, v′)(v′′)2dx

and we can use Schwarz inequality to obtain estimates on ‖v′′‖L2 .

How can we get around the assumptions we made? In the general case we have to localize.Consider

ϕ(t) = (χ(t)vh)−h

44

where χ is a cut off function and

vh(x) =v(x+ h)− v(x)

h

ϕ is an approximation of the second derivative. Then instead of integrating by parts, we candifference by parts and conclude that v ∈ W 2,2 (v′ ∈ C1) which is enough to write the Eulerequation (a.e. sense). Then using the same arguments we can show that v′′ exists.

5 Applications to Differential Equations

5.1 Existence of Periodic Solutions of (HS)

After having proved the existence of solution for the boundary value problem, we are interstedin finding some special solutions of (HS): we will start by proving the existence of periodicsolutions. Suppose we are in a simplified setting, i.e., we are solving

q + V ′(q) = 0

where V ∈ C1(R,R). If q is a solution, multiplying both sides of the equation by q

q · q + V ′(q) · q = 0

which is equivalent tod

dt

(1

2|q|2 + V (q)

)= 0

Then1

2|q|2 + V (q) = constant

meaning that the total energy of the system is conserved. Since we are working in R, we canmake the analysis in the phase plane: the solutions of (HS) will be the level curves of the energyfuncyion, i.e., will correspond to the curves

1

2p2 + V (q) = c

Suppose that we can prove the following

if V (q) = H and V ′(q) = 0 then x(t) = q is an equilibrium solution of (HS);

Then, in any other type of solution, V ′(q) 6= 0, and we can interpret the condition

1

2p2 + V (q) = c

as a C1 manifold (the gradient of the expression is everywhere nonozero - the curve has noedges). Consider any initial point on such manifold and turn on time:

45

. since energy is constant it will remain on the curve

. V ′(q) 6= 0 means that it can’t stop

If the curve is closed, the two conditions correspond to a periodic solution (this is the qualitativeway to obtain a periodic solution).

If n is arbitrary, things are not as intuitive. We will use the calculus of variations.

Let T > 0 arbitrary, and consider

q + Vq(t, q) = 0 (HS)

where V ∈ C2(R× Rn,R) is T -periodic in t.

Question: Is there a periodic solution to this problem?

Finding periodic solutions is equivalent to find solutions of (HS) with boundary conditionsq(0) = q(T ) and q′(0) = q′(T ).

Example 24. 1. q+a(t) sin q = 0, where a is continuous and periodic. If a(t) = 1, we havethe model of the simple pendulum.

2. q + qR2(t)+q2(t)

= 0 where R is a periodic function, is the equation for the interaction

between earth/sun.

3. n-body problem: Let q1,..., qn be the position function of n bodies. By the second law ofNewton

miqi =∂V

∂qi

where V is the n body potential

V (q) =∑i 6=j

mimj

|qi − qj|

which is a singular potential.

We will seek the periodic solutions of (HS), by minimizing the corresponding functional

I(q) =

∫ T

0

(1

2|q|2 − V (t, q))dt

over ET = W 1,2T (R,Rn). The following example show us that it is not enough to consider

V (t, x) ≤ 0 , t ∈ [0, T ] , x ∈ Rn

If (qm) is a minimizing sequence, I(qm) ≤ M implies (qm) bounded in L2([0, T ]), but it is notenough to prove that (qm) is bounded in ET .

46

Example 25. Consider

I(q) =

∫ T

0

(1

2|q|2 +

1

1 + |q|2)dt

Take qm = Rm →∞ as m→∞, where Rm is a constant sequence (which means in particularthat is periodic). Then

I(qm) =

∫ T

0

1

1 + |Rm|2dt→ 0 as m→∞

and therefore infETI(q) = 0. But it is easy to see that there is no T -periodic function for which

0 is achieved.

Theorem 8. If V (t, x) ∈ C2(R× Rn,R) is T -periodic, for some T > 0, and verifies

(V 1) −V (t, x) ≥ α|x|β − γ , t ∈ [0, T ] x ∈ Rn

and α, β, γ fixed positive constants, then (HS) admists at least one T -periodic solution.

Proof:

As mentioned, we will prove the Theorem using a variational argument. For (qm) ⊂ ET , aminimizing sequence, there exists M > 0 such that

M ≥∫ T

0

(1

2|qm|2 − V (t, qm)) dt

Using condition (V1) ∫ T

0

(1

2|qm|2 + α|qm|β )dt ≤M + γT

This implies directly that‖qm‖L2([0.T ]) ≤ 2(M + γT )

and ∫ T

0

|qm|β) dt ≤ 1

α(M + γt) (24)

Then (qm) is bounded in L2([0, T ]). To prove that (qm) is also bounded L2([0, T ]), we have toconsider two distinct cases:

Case 1. β > 2

Using Holder’s inequality

‖qm‖2L2([0,T ]) =

∫ T

0

|qm|2dt ≤(∫ T

0

1pdt)1/p(∫ T

0

(|qm|2)qdt)

1/q

47

Choose q such that 2q = β, and p such that 1p

+ 1q

= 1 (notice that β > 2 implies q > 1 and

therefore p is well defined). Then

‖qm‖2L2([0,T ]) ≤ T 1/p

(∫ T

0

|qm|βdt)2/β

As a consequence of (24)

‖qm‖2L2([0,T ]) ≤ T 1/p

( 1

α(M + γT )

)2/β

and we can conclude that (qm) is bounded in ET .

Case 2. 0 < β ≤ 2

As noticed, in this case we can’t use Holder’s inequality. Any q ∈ ET can be writen as a Fourierseries

q(t) =∑j∈Z

ajei2πjTt = a0 + remaining terms

The above decomposition q(t) = a0 +R(t), corresponds to the decomposition of our space in 2orthogonal subspaces, i.e., a0 ∈ Rn and R(t) ∈ (Rn)⊥. By definition

a0 =1

T

∫ T

0

q(t) dt ≡ [q]

which is the mean value or average of q. Define

q = [q] + q

Computing the derivative

q(t) =∑j 6=0

aji2πj

Tei2πjTt

and by Parseval’s identity, we can easily verify that

‖q‖2L2([0.T ]) = ‖q − [q]‖2

L2([0,T ]) =∑j 6=0

a2j

On the other hand

‖ ˙q‖2L2([0,T ]) = ‖q‖2

L2([0,T ]) =∑j 6=0

(aj

2πj

T

)2

≥ (2π

T)2∑j 6=0

a2j = (

2π

T)2‖q‖2

L2([0,T ])

i.e.‖q‖2

L2([0,T ]) ≤ c(T )‖ ˙q‖2L2([0,T ])

48

We will now proceed in finding bounds for (qm) in ET . Since

‖qm‖2L2([0,T ]) ≤ 2M1

we can write‖qm‖2

L2([0,T ]) ≤ C(T )‖ ˙qm‖2L2([0,T ]) = C1(T )‖qm‖2

L2([0,T ])

and therefore qm is bounded in L2([0, T ]). It remains to show that [qm] is also bounded, andsince it is a constant we have to find a bound for |[qm]|. Start noticing that

|[qm]| = |qm − qm| ≤ |qm|+ |qm|

Raising to β

|[qm]|β ≤(|qm|+ |qm|

)β≤ 2β

(|qm|β + |qm|β

)as used before. Integrating t over [0, T ]

T |[qm]|β ≤ 2β∫ T

0

(|qm|β + |qm|β

)dt

Once again, by (24) ∫ T

0

|qm|βdt ≤M1

α

and therefore

|[qm]|β ≤ 2β

T

(M1

α+ T‖qm‖βL∞([0,T ])

)Using Sobolev Inequality

|[qm]|β ≤ 2β

T

(M1

α+ cT‖qm‖βET

)Since we proved that (qm) is bounded in ET , we can finally conclude that (qm) is bounded inET and therefore (qm) admits a weakly convergent subsequence (which we will denote again by(qm)), i.e., there exists v ∈ ET such that qm v in ET and qm → v in L∞([0, T ]). As before Iis weakly lower semicontinuous, implying

I(v) ≤ lim infm→∞

I(qm)

Since v ∈ ET ,I(v) = inf

q∈ETI(q)

We still have to show that v ∈ C2([0, T ]) (the variational characterization implies that it isa T -periodic solution of (HS)). Let us start proving that v is a critical point of I, i.e., thatI ′(v)ϕ = 0, for all ϕ ∈ ET . But in this case this is straight forward, since for any ϕ ∈ ET andδ ∈ R, v + δϕ ∈ ET (so this case is easier than the BVP). Thus

0 = I ′(v)ϕ =

∫ T

0

(v · ϕ− Vq(t, v)ϕ) dt , ∀ϕ ∈ ET

49

We want to show next that this implies v ∈ C2([0, T ],Rn). We will use a variation of theregularity argument used for the BVP. Given that v is the minimizer of I over ET , we want tosolve the linear boundary value problem

w′′ = −Vq(t, v)w(0) = w(T )w′(0) = w′(T )

Nevertheless, we must point out that the solution is not unique: if w is such a solution, thenw + kT is also a solution. Hence we must impose an additional condition

[w] = [v]

Homework 11. Show that there exists a unique solution to this problem. (Hint: Notice thatas a consequence of the periodicity, [Vq] = 0 and then use calculus or Fourier series).

Let w ∈ C2([0, T ]) be the solution of the considered problem. For an arbitrary ϕ ∈ ET ,multiply both sides of the equation and integrate over t ∈ [0, T ]. Then∫ T

0

ϕ · w′′ dt = −∫ T

0

ϕ · Vq(t, v) dt

Integrating by parts the left hand side

w′ · ϕ|T0 −∫ T

0

ϕ · w dt = −∫ T

0

ϕ · Vq(t, v) dt

Since ϕ and w are T -periodic, ∫ T

0

ϕ · w dt =

∫ T

0

ϕ · Vq(t, v) dt

Subtracting the weak Euler equation∫ T

0

ϕ · (v − w) dt = 0

Take ϕ = w− v, we can conclude that v− w = 0 and henceforth w− v =constant. But by ournormalization [w] = [v] which implies that w = v, in particular v ∈ C2([0, T ]).

♦

When seeking for periodic solutions of (HS), by minimizing I over ET , some problems mayarise:

50

1. If V is independent of t then is obviously T -periodic, for all T , which implies by ourresults, that there exists a T periodic solution for all T > 0. But notice, that since weassumed

−V (q) ≥ α|q|β − γ , α, β, γ > 0

−V is very large for large values of |q| and therefore it must have a global minimum ξ.Consider q(t) ≡ ξ. By construction V ′(ξ) = 0 and hence ξ is a solution of (HS) (it is anequilibrium solution). If this is the case there may not exist any other periodic solution.

2. If V is T periodic in t, then V is kT periodic in t, for all k > 0. By our results, thereexists a solution of (HS) which is kT periodic, but it may be k copies of the T periodicsolution. To show the existence of subharmonic solutions we need more work.

Homework 12. In the same setting as in Theorem 8, for V = V (q), what (“a priori” conditionsmust we impose, so that, for v1, v2 verifying

I(v1) = infETI ≡ cT

andI(v2) = inf

E2T

I ≡ c2T

exist and are geometrically distinct.

Example 26. (Multiple Pendulum Type of Solutions) We consider the system

q + Vq(t, q) = 0 (HS)

where V = V (t, q) ∈ C2(R×Rn,R), T periodic in t and Ti periodic in qi. The simplest exampleis the simple pendulum

q + α(t) sin q = 0

where α is a periodic function. In the multiple pendulum, the variables qi correspond to theseveral angles. We will consider the case

V (t, q) = W (t, q)− f(t) · q

where f is continuous, T periodic and satisfies [f ] = 0 (to ensure the existence of periodicsolutions) and W satisfies the same conditions as V . For this V , the equation takes the form

(HS) q +Wq(t, q) = f(t)

Theorem 9. Under the stated conditions, there exists a T -periodic solution of (HS).

Proof:

We will prove the Theorem, by minimizing

I(q) =

∫ T

0

(1

2|q|2 − V (t, q)) dt

51

over ET = W 1,2T (R,Rn). Let

cT = infETI

and (qm) ⊂ ET a minimizing sequence. Using the periodicity of W and the condition [f ] = 0,it is obvious that

I(q + (k1T1, ..., knTn)) = I(q)

for all k1, ..., kn ∈ Z. This invariance property is important, since with it we will control thenorm of a minimizing sequence. As before, we can decompose

qm = [qm] + qm

and by the invariance property we can replace (qm) by any qm = qm + (k1T1, ..., knTn), with(k1, ..., k : n) ∈ Zn. We will normalize by choosing (k1, ..., kn) so that

0 ≤ [(qm)i] < Ti

Without loss of generality, assume that (qm) verifies

0 ≤ [(qm)i] < Ti

which is a very strong control on the mean value of the minimizing sequence.

Let’s prove that the minimizing sequence is bounded in ET . We can start to assume thatI(qm) ≤M . We will consider two cases f = 0 and f 6= 0. If f = 0,

1

2

∫ T

0

|qm|2dt ≤M +

∫ T

0

V (t, qm) dt

Since V is periodic and C2([0, T ]), it is uniformly bounded in C([0, T ]), and therefore

1

2

∫ T

0

|qm|2dt ≤M + ( maxt∈[0,T ]

V (t, qm(t)))T ≤M1

uniformly in m. From here we proceed as in the previous examples. If f 6= 0,

I(qm) =

∫ T

0

(1

2|qm|2 −W (t, qm) + f(t) · qm) dt ≤M

then ∫ T

0

1

2|qm|2dt =

∫ T

0

1

2| ˙qm|2dt ≤M +

∫ T

0

W (t, qm) dt+

∫ T

0

|f · qm| dt

Since W is C2([0, T ]) and periodic, is uniformly bounded, and therefore∫ T

0

W (t, qm) dt ≤M1 ∀m

52

On the other hand∫ T

0

f · qm dt =

∫ T

0

f · ([qm] + qm) dt =

∫ T

0

f · [qm] dt+

∫ T

0

f · qm dt

Since by hypothesis [f ] = 0,∫ T

0

f · qm dt =

∫ T

0

f · qm dt ≤ ‖f‖L2([0,T ])‖qm‖L2([0,T ])

On one hand, as we have seen previously

‖qm‖L2([0,T ]) ≤ C(T )‖ ˙qm‖L2([0,T ])

on the other hand‖f‖L2([0,T ]) ≤ T 1/2‖f‖L∞([0,T ])

As a consequence ∫ T

0

f · qm dt ≤ C(T )‖f‖L∞([0,T ]‖ ˙qm‖L2([0,T ])

Then1

2‖ ˙qm‖2

L2 =

∫ T

0

1

2|qm|2dt ≤M1 +M2‖ ˙qm‖L2([0,T ])

Notice that the left hand side is quadratic in ‖ ˙qm‖L2 while the right hand side is linear. Solvingthe quadratic inequality, we conclude that ‖ ˙qm‖L2([0,T ]) is bounded uniformly in m.

♦

Remarks:

1. As a consequence of the invariance property, there is no uniqueness of minimizer. Insteadthere is a lattice of minimizers. This suggests the existence of more critical points (wewill do this later, using the Mountain Pass Theorem).

2. This was a simplified version of the multiple pendulum. In the more general case, thekinetic energy

K(t) =

∫ T

0

∑i,j

ai,j(q)qi · qj dt

where (ai,j(q)) is a symmetric, positive definite matrix, Ti periodic in qi. The harder partis to prove the weak lower semicontinuity.

Example 27. (Singular Systems) Our next example deals with potentials, having a singu-larity. For example

V (q) = − 1

|q|β, β > 0

53

These are potentials arising from the 2-body problem. Our goal is to find a periodic solutionwinding around the singularity. We will start by formulating the problem. Consider

(HS) q + V ′(q) = 0

where V satisfies

(V1) V ∈ C2(R2 \ (0, 0));

(V2) V (x) ≤ 0 for all x ∈ R2 \ (0, 0);

(V3) V (x)→ −∞ as |x| → 0.

To prove the existence of such solution we will minimize the functional

I(q) =

∫ T

0

(1

2|q|2 − V (q))dt

over the classΓT = q ∈ W 1,2

T (R,R2) : WN0(q) 6= 0

We have to explicit what we mean by WN0(q) 6= 0. For q ∈ ΓT we can consider its characteri-zation in polar coordinates with respect to 0:

q(t) = R(t)eiθ(t)

where R(0) is in the positive x-axis and θ(0) = 0. Since q(t) is continuous, θ is continuous.We will define

WN0q|T0 = Θ(T )−Θ(0)

As a consequence of the periodicity of the functions in ET , we will have

WN0q|T0 = 2kπ

for some k ∈ Z. k will denote the number of times q winds around 0 in the interval [0, T ].Before we proceed, we can make a couple of simplifications. Since V does not depend explicitelyin t, for q ∈ ΓT , q(−t) is the same curve running in the opposite direction, and

I(q(−t)) = I(q(t))

Hence if v ∈ ΓT is a minimizer of I, so it will v(−t). Consider

ΓT = Γ+T ∪ Γ−T

and by our considerations, it is enough to minimize over Γ+T . Define

cT = infΓ+T

I

54

and consider (qm) ⊂ Γ+T a minimizing sequence. Then, there exists M such that I(qm) ≤ M

and together with condition (V2) implies ‖qm‖L2([0,T ]) is uniformly bounded. As we have seenpreviously, to find bounds on ‖qm‖L2([0,T ]) is equivalent to find bounds for [qm], where qm =

[qm] + qm. As before, we can prove that ‖ ˙qm‖L2([0,T ]) is bounded uniformly in m ∈ N. If themean value piece, [qm], is unbounded, then along a subsequence, [qmi → ∞ as i → ∞. But, ifthis is the case, there is no winding, and therefore for i sufficiently large, qmi 6 inΓ+

T , which isa contradiction. Thus we can conclude that [qm] is uniformly bounded, and as a consequence(qm) is bounded in W 1,2

T (R,R2). Then, there exists v ∈ W 1,2T (R,R2), such that, possibly along a

subsequenceqm v in W 1,2

T (R,R2)

andqm → v in L∞T (R,R2)

The only thing left to prove is that v ∈ Γ+T , and since v ∈ W 1,2

T (R,R2) by construction, we haveto show that WN0v|T0 > 0. But this doesn’t happen necessarily: the limit may pass through theorigin, in which case the winding number does not make sense anymore.

Example 28. Let V (q) ≈ − 1|q| (the gravitational potential), and q(t) ≈ tα near the origin.

Then we need two things to happen:

tα ∈ W 1,2([0, T ]) and I(tα) <∞

For the first condition to occur

q ∈ W 1,2([0, T ]) ⇔ α >1

2

since if q(t) ≈ tα then q(t) ≈ tα−1, and∫t2α−2dt <∞ ⇔ 2α− 2 > −1

On the other hand

−∫V (tα) dt = −

∫1

tαdt <∞ ⇔ α < 1

Then for 12< α < 1 everything is finite. So it is possible that the minimizer approaches the

origin.

Such a minimizer is called a collision orbit. To avoid the existence of such orbits we have toassume an extra condition on the potential, which as we will see, keeps candidates to minimizersaway from the origin.

(V4) Strong Force Condition: There exists a neighborhood of 0, N , and a function U ∈C1(N \ 0,R2) such that

|U ′(q)|2 ≤ −V (q) for all q ∈ N \ 0

55

|U(x)| → ∞ as x→ 0.

Homework 13. Show that there exists V satisfying such condition. Consider V (q) = − 1|q|α for

an appropriate α.

Lemma 9. Let V verify (V1)-(V4). If for q ∈ ET , I(q) ≤ M for some M > 0, there existsβ = β(M) such that

|q(t)| > β

for all t ∈ [0, T ]

Proof:

Let’s suppose that q(t) ∈ N \ 0, for t ∈ [σ, s], where N is given by (V4). Then

U(q(s))− U(q(σ)) =

∫ σ

s

d

dtU(q(t)) dt =

∫ σ

s

U ′(q(t) · q(t) dt

≤(∫ σ

s

|U ′(q(t))|2dt)1/2(∫ σ

s

|q|2dt)1/2

≤(∫ σ

s

−V (q)dt)1/2(∫ σ

s

|q|2dt)1/2

≤√

2I(q)

Since I(q) ≤MU(q(s))− U(q(σ)) ≤

√2M

Further suppose that q(s) ∈ ∂N . Based on what we have shown

|U(q(σ))| ≤ C

for some constant C. But since |U | blows up as we approach the origin, q(σ) can’t get too closeto it and therefore

|q(σ)| ≥ β(M)

for all σ.♦

Homework 14. Using Lemma , finish the proof of the existence of a periodic solution for thesingular hamiltonian system.

5.2 Existence of Heteroclinic Solutions of (HS)

Considerq + Vq(t, q) = 0 (HS)

where V satisfies the following conditions

56

(V1) V = V (t, q) ∈ C2(R× Rn,R) is 1-periodic in t and qi, 1 ≤ i ≤ n,

(V2) V (t, 0) = 0 > V (t, x) for all x ∈ Rn \ Zn and all t ∈ R.

Condition (V2) implies that the potential V has a global maximum at q = 0. The periodicityof V implies that the global maximum, 0, is achieved at all integer points (lattice of maximum).Then,

Vq(t, 0) = Vq(t, x) = 0 , ∀x ∈ Zn

and as a consequence all points in Zn are equilibrium solutions of (HS).

Question: Are there solutions of (HS) connecting 0 with other equilibrium solutions? Thatis, we want to find solutions of (HS) having the following asymptotic behavior

Q(−∞) = limt→−∞

Q(t) = 0

andQ(∞) = lim

t→∞Q(t) = η ∈ Zn \ 0

These type of solutions (connecting equilibrium solutions) are known as heteroclinic solutions.

Example 29. (Simple Pendulum)

Considerq = sin(2πq)

which as mentioned earlier is the equation of motion of the simple pendulum, and correspondsto the potential

V (q) =1

2π(cos(2πq)− 1) , q ∈ R

It is easy to see, that for n ∈ Z, q(t) ≡ n is an equilibrium solution of the differential equation.Furthermore, it is known that near the minimum values of V , the curves

1

2p2 + V (q) = C

represent the periodic solutions of (HS).

Homework 15. Sketch the phase plane for this equation.

Theorem 10. If V satisfies (V1)-(V2), for each ξ ∈ Zn there is η ∈ Zn \ ξ and a solutionof (HS) heteroclinic from ξ to η.

Without loss of generality we will assume ξ = 0. Notice that we have to change our previousarguments, since in this case we are working in an unbounded interval. Our goal is to find aminimizer for

I(q) =

∫R(1

2|q|2 − V (t, q)) dt

over an appropriate class of functions: the elements of this class must verify the following:

57

- have the right asymptotic behavior;

- give sense to I(q).

We start by remarking that we can’t expect the functions of our class to be square integrable:in particular we want lim

t→∞(q(t) = η 6= 0. This means W 1,2(R,Rn) is not a reasonable choice.

We will chooseE = W 1,2

loc (R,Rn) = q ∈ W 1,2([a, b],Rn) : ∀a < b ∈ Rwhich is a Hilbert space when endowed with the norm

‖q‖2E =

∫R|q|2dt+ |q(0)|2

and define the class

Γ = q ∈ E : q(−∞) = 0 and q(∞) ∈ Zn \ 0

Forc = inf

q∈EI(q),

as an immediate consequence of condition (V2), c ≥ 0. The Theorem follows from the followingpropositions:

Proposition 3. There exists Q ∈ Γ such that I(Q) = c.

Proposition 4. Q is a solution of (HS) satisfying |Q(t)| → 0 as |t| → ∞.

The definition of Γ and the asymptotic behavior of Q, will imply that Q is an heteroclinicsolution of (HS) as wanted. Let’s assume Proposition 3 for the moment, and prove Proposition4.

Proof: (of Proposition 4)

Choose a pair of real numbers r < s and a function ϕ ∈ C2(R,R) such that ϕ(t) = 0 fort 6∈ (r, s). If δ ∈ R is arbitrary, the function (Q+ δϕ)(t) behaves like Q(t) for large values of t,since ϕ has compact support. This implies, in particular, that Q + δϕ ∈ Γ, and by definitionof Q

ψ(δ) ≡ I(Q)− I(Q+ δϕ) ≤ 0 = ψ(0)

Note that ψ is a continuously differentiable function of δ, implying

ψ′(0) =

∫ s

r

(Q · ϕ− Vq(t, Q) · ϕ

)dt = 0 (25)

Let x ∈ C2([r, s]) denote the unique solution of the linear problem

x = −V q(t, Q)

x(s) = Q(s)

x(r) = Q(r)

58

and consider ϕ as above. Multiplying both sides of the equation by ϕ, and integrate in t ∈ [r, s]∫ s

r

(x · ϕ− Vq(t, Q) · ϕ

)dt = 0 (26)

Subtracting (25) of (26) ∫ s

r

˙(x−Q) · ϕdt = 0

In the previous examples we made the choice ϕ = x−Q. We can’t do it in this case, since wedon’t know if Q ∈ C1([r, s]). Consider ϕ such that ϕ = x − Q in (r, s) and ϕ(r) = ϕ(s) = 0.Integrating by parts

0 =

∫ s

r

˙(x−Q) · ϕdt = −∫ s

r

(x−Q) · ϕdt = −∫ s

r

|x−Q|2dt

which implies that x = Q and therefore Q ∈ C2((r, s)). Since r and s were chosen arbitrarily,Q ∈ C2(R,Rn), and as a consequence Q is a classical solution of (HS). It remains to show that

|Q(t)| → 0 as |t| → ∞

By Proposition 3,Q(−∞) = 0 and Q(∞) = η ∈ Zn \ 0

and by the first part of the proof,Q verifies (HS). Hence

lim|t|→∞

(Q+ Vq(t, Q)) = 0

Since Vq(t, Q)→ 0 as |t| → ∞, we conclude

Q(±∞) = 0

To conclude the desired asymptotic behavior on Q we need an interpolation result.

Lemma 10. Let ϕ ∈ C2([a, b],Rn). Then, for all ε > 0, there exists k = k(ε, b− a) such that

‖ϕ‖L∞([a,b]) ≤ ε‖ϕ‖L∞([a,b]) + k‖ϕ‖L∞([a,b])

If we assume the Lemma,

‖Q‖L∞([t−1,t+1]) ≤ ε‖Q‖L∞([t−1,t+1]) + k‖Q‖L∞([t−1,t+1])

Letting t→ −∞Q(t)→ 0 as t→ −∞

For the other side, apply Lemma 10 to Q(t) = Q(t)− η.

♦

59

Proof: (of Lemma 10)

Let ε > 0 be arbitrary. Then there exists n ∈ N such that

ε

n<|b− a|

2

Without loss of generality, take n = 1. For x ∈ [a, b] choose z ∈ [a, b] such that |x− z| = ε. Bythe Mean value Theorem, there exists y ∈ (x, z) (or y ∈ (z, x)) such that

ϕ(x)− ϕ(z)

x− z= ϕ′(y)

Then

ϕ′(x) = ϕ′(y) +

∫ x

y

ϕ′′(t)dt =ϕ(x)− ϕ(z)

x− z+

∫ x

y

ϕ′′(t)dt

Hence, for all x

|ϕ′(x)| ≤ 2

ε‖ϕ‖L∞[a,b] + ε‖ϕ′′‖L∞[a,b]

Taking the supremum over x ∈ [a, b] we obtain the result.

♦

Homework 16. Prove the analogous result for L2([a, b]).

Before we prove Proposition 3, let us make some remarks. For k ∈ Z denote τk(q) = q(t−k).Then, as a consequence of the periodicity of V

I(τk(q)) = I(q)

i.e., I is invariant under time phase shifts. This has consequences for our problem.

Suppose that Q is a critical point of I. Then qm = τm(Q) is a minimizing sequence,since I(qm) = I(Q) = c for all m ∈ Z. But notice that the sequence τm(Q) = Q(t −m)converges weakly to 0 as m → ∞ and 0 6∈ Γ, so we need to be very careful - we need tocontrol the sequence so that this doesn’t happen.

With the invariance property we can normalize a minimizing sequence in an “good” form.If Q ∈ Γ, then Q(−∞) = 0 and Q(∞) = η ∈ Zn \ 0. For 0 < ρ < 1

3, Q will intersect

∂Bρ(0) at some t(Q). We will use this fact to normalize a minimizing sequence, whichwill imply, in particular, that the limit will not be the trivial solution.

Proof: (Proposition 3)

Let (qm) ⊂ Γ be a minimizing sequence. For 0 < ρ < 13, denote by t(qm) the first time qm

hits the boundary of Bρ(0). Choose km ∈ Z, so that

t(τkm(qm)) ∈ [0, 1) ∀m

60

Without loss of generality, assume km = 0, for all m, i.e., assume that for all m, qm verifies

|qm(t)| < ρ , ∀t ∈ (−∞, t(qm))

and|qm(t(qm))| = ρ for t(qm) ∈ [0, 1)

As usual, I(qm) c implies the existence of M > 0 for which I(qm) ≤M . Thus

M ≥ I(qm) ≥∫R(1

2|qm|2 − V (t, qm)) dt

and as a consequence of (V2)‖qm‖2

L2 ≤ 2M

Then (qm) is bounded in L2(R,Rn). On the other hand

|qm(0)| ≤ |qm(t(qm))|+ |∫ 0

t(qm)

qm(t) dt| ≤ ρ+ |t(qm)|1/2‖qm‖L2

Since, by our normalization, |t(qm)| < 1, we conclude

|qm(0)| ≤ ρ+√

2M

Hence‖qm‖2

E ≤ 2√

2M + ρ

as wanted. Then, there exists Q ∈ E such that, possibly along a subsequence (again de-noted by qm), qm converges weakly to Q in E. For L > 0, by definition, (qm) is bounded inW 1,2([−L,L],Rn). Then, as a consequence of the normalization

qm(t)− qm(0) ≤∫ t

0

|qm|dt ≤ ρ+ L1/2√

2M

This proves that (qm) is an equicontinuous family, which implies that qm → Q in L∞([−L,L],Rn).Since L is arbitray, we can conclude that qm → Q in L∞loc(R,Rn). The strong convergence im-plies that there exists t = t(Q) ∈ [0, 1] such that Q(t) ∈ ∂Bρ(0). Hence Q(t) can not be thetrivial solution. We will start by proving that I(Q) <∞.

For L ∈ R+, since qm Q in W 1,2([−L,L]), and

I(q)|L−L =

∫ L

−LL(q)dt

is weakly lower semicontinuous,∫ L

−LL(Q)dt ≤ lim inf

m→∞

∫ L

−LL(qm)dt ≤ lim inf

m→∞

∫RL(qm)dt ≤M

61

Taking the limit as L → ∞, we conclude I(Q) ≤ M < ∞ as wanted. Next we want to showthat Q ∈ Γ, i.e., that

Q(−∞) = 0 and Q(∞) = η ∈ Zn \ 0

For ρ as in the normalization, choose 0 < ε < ρ, and set

β(ε) = infx 6∈Bε(Zn),t∈R

(−V (t, x))

As an immediate consequence of condition (V2), β(ε) is strictly positive.

Lemma 11. If w ∈ E, consider real numbers σ < s such that w(t) 6∈ Bε(Zn) for t ∈ [σ, s].Then ∫ s

σ

L(w) dt ≥√

2β(ε)|w(σ)− w(s)|

Proof:

Let l ≡ |w(σ)− w(s)|. Then

l = |∫ s

σ

w(t) dt| ≤ |s− σ|1/2(∫ s

σ

|w(t)|2 dt)1/2

implying ∫ s

σ

L(w) dt =

∫ s

σ

(1

2|w(t)|2 − V (t, w)) dt ≥ l2

2|s− σ|+ β(ε)|σ − s|

For z = |σ − s|, ∫ s

σ

L(w) dt ≥ l2

2z+ β(ε)z ≡ ϕ(z)

It is easy to prove that ϕ has a minimum value achieved at at z = l√2β(ε)

. Then

∫ s

σ

L(w) dt ≥ ϕ(l√

2β(ε)) =

√2β(ε)l

as wanted.♦

Corollary 6. If I(w) <∞, then w ∈ L∞(R,Rn).

Homework 17. Prove Corollary 6.

Lemma 12. If w ∈ E and I(w) <∞, then w(±∞) ∈ Zn.

Proof:

We will prove w(−∞) ∈ Zn, that w(∞) ∈ Zn follows in a similar form. By Corollary 6,w ∈ L∞(R,Rn). Then both lim sup

t→−∞w(t) and lim inf

t→−∞w(t) exist, and as a consequence, there

62

exists a sequence tm → −∞ and ξ ∈ Rn such that w(tm) → ξ as m → ∞. Suppose for themoment that we can prove

t→ −∞ ⇒ w(t)→ ξ (27)

Then ξ must necessarily belong to Zn. If not, for large values of t, V (t, w) is srtictly positive,i.e., there are strictly positive constants L and γ such that

−V (t, w) > γ , ∀t ≤ −L

As a consequence

I(w) >

∫R−V (t, w) dt >

∫ −L−∞

γ dt =∞

contradicting the hypothesis. All we have to prove is that (27) holds. Once again arguing bycontradiction, if not, there exists δ > 0 and sequences (tm), (sm) ⊂ R, converging to −∞ asm→∞, and such that

w(tm) ∈ ∂Bδ(ξ) and w(sm) ∈ ∂B2δ(ξ)

Without loss of generality, we may assume that 0 < δ < ρ3, so that

B2δ(ξ) \Bδ(ξ) ∩ Zn = ∅

This is obviously true for ξ ∈ Zn, and for ξ 6∈ Zn we can take δ as small as necessary.Furthermore, we may also assume that sm ∈ (tm, tm+1), and for t ∈ (tm, sm) require that

w(t) ∈ B2δ(ξ) \Bδ(ξ)

(refine the sequence if necessary), i.e., for t ∈ (tm, sm), w(t) does not belong to Bδ(ξ). Thus,we are in the conditions of Lemma 12, and therefore

∞ > I(w) ≥∑m

∫ sm

tm

L(w)dt ≥∑m

√2β(δ)δ =

√2β(δ)δ(#terms) =∞

which is a contradiction.♦

We will proceed the proof of the Theorem. By the previous Lemma, Q(−∞) ∈ Zn. On theother hand by the normalization

Q(t) ∈ Bρ(0) , ∀t < 0

ThenQ(−∞) ∈ Zn ∩Bρ(0) = 0

and by our choice of ρ, we must necessarily have Q(−∞) = 0. Also as a consequence of Lemma12 , Q(∞) ∈ Zn. But we have to be careful, since Q(∞) = 0 (which may happen if the

63

minimizing sequence breaks down in the limit). Consider 0 < δ << ρ and suppose Q(∞) = 0.Then, there exists T = T (δ) > 0 such that Q(t) ∈ Bδ(0) for all t ≥ T . Since qm → Q inL∞loc(R,Rn), for m sufficiently large qm(t) ∈ B2δ(0) for t ∈ [T, T + 1]. We will prove that thisyields a contradiction (the intuitive idea is that we can cut the extra part qm|T−∞ in (qm) andconstruct a new minimizing sequence vm with I(vm) < I(qm).) Define

vm(t) =

0 if t ≤ T(t− T )qm(T + 1) if t ∈ [T, T + 1]qm(t) if t ≥ T + 1

It is easy to check that vm ∈ Γ and

I(vm)− I(qm) =

∫ T+1

T

L(vm) dt−∫ T+1

−∞L(qm) dt

=

∫ T+1

T

(1

2|qm(T + 1)|2 − V (t, vm)

)dt−

∫ T+1

−∞L(qm) dt

By our assumptions ∫ T+1

T

1

2|qm(T + 1)|2dt = 2δ2

and for t ∈ [T, T + 1]|vm(t)| ≤ δ

which implies, by the properties of V ,∫ T+1

T

−V (t, vm) dt ≤ ε(δ)

where ε(δ) → 0 as δ → 0. On the other hand, for t ∈ (−∞, T + 1], qm(t) starts at 0 fort = −∞, hits the boundary of Bρ(0) for some t ∈ [0, 1), gets back to B2δ(0) for t ∈ [T, T + 1]and goes back to another point in the lattice. So it must cross at least twice, the boundary ofBρ(0) \Bρ/2(0) (remember that δ << ρ so we may assume that 2δ < ρ/2). By Lemma 11∫ 0

−∞L(qm) dt ≥ 2

√2β(ρ/2)ρ/2 = ρ

√2β(ρ/2)

FinallyI(vm)− I(qm) ≤ 2δ2 + ε(δ)− ρ

√2β(ρ/2)

and we can choose δ sufficiently small, so that

2δ2 + ε(δ) <ρ

2

√2β(ρ/2)

and thereforeI(vm)− I(qm) < 0

64

contradicting the fact that qm is a minimizing sequence of c = infΓI. Therefore Q(∞) 6= 0, and

as a consequence there exists η ∈ Zn \0 such that Q(∞) = 0. Then Q ∈ Γ. If this is the case

I(Q) ≥ c (28)

For ε > 0 and m large, using the definition of infimum

I(qm) ≤ c+ ε

Then

c+ ε ≥ I(qm) ≥∫ L

−LL(qm) dt

for all L > 0. Letting m→∞, and as a consequence of the weakly lower semicontinuity

c+ ε ≥∫ L

−LL(Q) dt

for all L ∈ R+ and ε > 0. Finally, letting L→∞ and ε→ 0

I(Q) ≤ c (29)

Conditions (28) and (29) imply I(Q) = c.♦

Using the same type of arguments, we can prove that this is not the only heteroclinic solutionof (HS) leaving 0.

Theorem 11. If V satisfies (V1)-(V2), for any ξ ∈ Zn, there are distinct η0, η1, ..., ηn ∈Zn \ 0, and at least Q0, ...., Qn solutions of (HS), where Qi is heteroclinic from ξ to ηi,0 ≤ i ≤ n.

In other words, the Theorem states that there are at least n+1 distinct heteroclinics leavingeach point in Zn.

Proof:

Once again we will assume that ξ = 0. By Theorem 10, there exists Q0 and η0 as wanted.Define the new class

Γ1 = q ∈ Γ : q(∞) 6∈ spanNη0i.e., the class of curves in W 1,2

loc (R,Rn), with q(−∞) = 0 and such that q(∞) is not any multipleof η0. Define

c1 = infΓ1

I(q)

We claim that there is a solution of (HS) which minimizes I over Γ1. Arguing as before, if(qm) is a minimizing sequence, (qm) is bounded in E (after normalization). Then there existsQ1 ∈ E such that qm → Q1 weakly in W 1,2

loc (R,Rn) and strongly in L∞loc(R,Rn). To show thatin fact Q1 ∈ Γ1, as last time it is easy to show that Q1(−∞) = 0. The possible dangers are

65

- Q1(∞) = 0 which will be excluded as in last time;

- Q1(∞) ∈ spanNη0. For simplicity assume that Q1(∞) = η0. For 0 < δ << ρ, there isT = T (δ) for which |Q1(t) − η0)| ≤ δ for all t ≥ T (δ). As a consequence of the strongconvergence, |qm(t)−η| ≤ 2δ for t ∈ [T, T + 1] and m sufficiently large. If this is the case,define

pm(t) =

0 if t ≤ T(T − t)(qm(T + 1)− η0) if T ≤ t ≤ T + 1qm(t)− η0 if t ≥ T + 1

i.e., we are considering the piece of qm, from η to qm(∞) = η1 translated by η1. Thenpm : 0→ η1 − η0 6∈ spanNη0, which implies that pm ∈ Γ2, and as before we can take δ sosmall so that I(pm) < I(qm) contradicting the fact that qm is a minimizing sequence.

So we produced another heteroclinic solution from 0 to η1 6∈ spanNη0. Why did we cut out allspanNη0 and not only η0? If we just cut out η0 the limit of the minimizing sequence may bejust a heteroclinic chain.

We proceed inductively. Define

Γ2 = q ∈ Γ1 : q(∞) 6∈ spanNη1

andc2 = inf

Γ2

I(q)

and we can continue this process until we span Zn, so we obtain n+ 1 distinct heteroclinics.

Remark: Choose a pair of points 0 and ξ. Can we find an heteroclinic solution such thatQ(−∞) = 0 and Q(∞) = ξ? We can’t prove the existence of such heteroclinic, but we canprove that there is at least an heteroclinic chain - the infimum may not be achieved.

5.3 Existence of Homoclinic Solutions for a Singular HamiltonianSystem

Considerq + Vq(t, q) = 0 (HS)

where V satisfies the following

(V1) There exists ξ ∈ R2 \0 such that V = V (t, q) ∈ C2(R×R2 \ξ,R), and V is 1-periodicin t;

(V2) limx→ξ

V (t, x) = −∞ uniformly in t ∈ [0, 1);

(V3) Strong Force Potential: There exists a neighborhood of ξ, N , and U ∈ C1(N \ ξ,R)such that

66

|U(x)| → ∞ as x→ ξ

|U ′(x)|2 ≤ −V (t, x) uniformly in t ∈ [0, 1) and x ∈ N \ ξ.

(V4) V (t, 0) = 0 > V (t, x) for all x 6= 0

(V5) There exists V0 < 0 such that lim sup|x|→∞

V (t, x) ≤ V0.

Condition (V5) implies that the potential is strictly negative at infinity, and therefore 0 is aglobal maximum of V .

Theorem 12. If V satisfies (V1)-(V5), there exists a pair of solutions of (HS), q+ and q−,which are homoclinic to 0 and wind, respectively positively and negatively around ξ.

We remark that, if V is independent of t, q+ and q− represent the same curve runned inopposite directions.

We will prove the Theorem by minimizing

I(q) =

∫R

(1

2|q|2 − V (t, q)

)dt ≥ 0

in an appropriate subclass of E = W 1,2loc (R,R2). Consider

Λ = q ∈ E : q(±∞) = 0 and q(t) 6= ξ ∀t ∈ R

Proposition 5. If I(q) <∞, there exists γ > 0 such that

|q(t)− ξ| > γ , ∀t ∈ R

Notice that γ depends only on the bounds for I(q).

Proposition 6. If I(q) < M then q ∈ L∞(R,R2) and q(t)→ 0 as |t| → ∞.

Homework 18. Prove Proposition 6.

Choose q ∈ Λ, such that I(q) <∞. By Proposition 6 the Range of q is a closed curve, andby Proposition 5 it avoids ξ. Then it makes sense in considering the winding number of q withrespect to ξ. By definition

WNξ(q) =1

2π(Θ(∞)−Θ(−∞))

where we are consideringq(t) = R(t)eiΘ(t)

a representation of q in polar coordinates with respect to ξ. Let

Γ± = q ∈ Λ : ±WNξ(q) > 0

andc± = inf

q∈Γ±I(q)

The proof of the Theorem follows from the next two Propositions.

67

Proposition 7. c± > 0 and there exists Q± ∈ Γ± such that I(Q±) = c±.

Proposition 8. Q± is a solution of (HS), verifying |Q±(±∞)| = 0.

The proof of Proposition 8 follows like the analogous results in our previous examples, sowe will ommit it. The variational characterization of Q and Proposition 8 prove Theorem 12.

Proof:(Proposition 7)

We will prove the existence of Q+ ∈ Γ+ with I(Q+) = c+. The proof for the existence ofQ− is similar. Let (qm) be a minimizing sequence for Γ+. Then there exists M > 0 for whichI(qm) ≤ M , which give us bounds for ‖qm‖L2(R,R2) uniformly for m ∈ N. As before we cannormalize the minimizing sequence, since we are working with an unbounded domain. Let

S = θξ : θ ≥ 1

i.e., S is the ray emanating from ξ and running to ∞. If q ∈ Γ+, it must intersect S, at leastonce. Replacing qm by τkm(qm) = qm(t− km), km ∈ Z, we can choose

qm((−∞, 0) ∩ S = ∅qm([0, 1)) ∩ S 6= ∅

that is, we can choose the minimizing sequence, so that, for every m ∈ N, the first time qmintersects S is in the unit interval. Moreover, Proposition 6 implies that ‖qm‖L∞(R,R2) ≤ k(M).Then

‖qm‖2E = ‖qm‖2

L2(R,R2) + |qm(0)|2 ≤ ‖qm‖L2(R,R2) + ‖qm‖L∞(R,R2) ≤M1

and as a consequence, there exists Q in E such that qm → Q weakly in W 1,2loc (R,R2) and strongly

in L∞loc(R,R2). The strong convergence of qm to Q in bounded intervals, imply

Q([0, 1]) ∩ S 6= ∅

Using the weakly lower semicontinuity, it is easy to show that, for all L ∈ R+,∫ L

−LL(Q) dt ≤ lim inf

m→∞I(qM) ≤M

Letting L → ∞, we conclude that I(Q) is finite. Proposition 5 implies that Q(t) 6= ξ, for allt ∈ R, and by Proposition 6, Q(±∞) = 0. Then Q ∈ Λ. To show that Q ∈ Γ+, it remains toprove that WNξ(Q) > 0. Once this is proved, it is easy to show that I(Q) = c+, which willfinish the proof. Arguing indirectly, suppose that WNξ(Q) ≤ 0. Since Q(±∞) = 0, for δ > 0,there exists T = T (δ) > 0, such that

|Q(t)| ≤ δ

2, ∀|t| ≥ T − 1

Then, for ε = ε(δ)→ 0 as δ → 0,

− ε8≤WNξ(Q)|−T−∞ ≤

ε

8

68

and− ε

8≤WNξ(Q)|∞T ≤

ε

8As a consequence

WNξ(Q) = WNξ(Q)|−T∞ + WNξ(Q)|T−T + WNξ(Q)|∞T ≤ 0

implies

WNξ(Q)|T−T ≤ε

4

Since qm → Q in L∞loc(R,R2), qm converges uniformly to Q in [−T, T ], and therefore, for msufficiently large

WNξ(qm)|T−T ≤ε

2

On the other hand, since by the normalization, qm((−∞,−T )) ∩ S = ∅, we may also assume(since in this interval there is not much change in the angle)

WNξ(qm)−T−∞ <ε

8.

Finally, by construction qm ∈ Γ+, for all m ∈ N, and therefore

WNξ(qm) ≥ 1

for all m ∈ N. Putting all this information together, we can conclude

WNξ(qm)|∞T ≥ 1− ε

2− ε

8≥ 1− 5ε

8≈ 1

We will prove that this yields a contradiction (intuitively, (qm|∞T does the job of a minimizingsequence by itself, so we can cut the extra part qm|T−∞ and obtain (pm) with I(pm) < I(qm)).Consider the comparison function

pm(t) =

0 if t ≤ T − 1(t− (T − 1))qm(T ) if T − 1 ≤ t ≤ Tqm(t) if t ≥ T

By construction, WNξ(qm)|∞T ≥ 1, which implies that pm ∈ Γ+, for all m ∈ N. On the otherhand

I(pm)− I(qm) =

∫ T

T−1

(1

2|qm(T )|2 − V (t, pm))dt−

∫ T

∞L(qm) dt

For t ∈ (−∞, T ), qm(t) starts at 0 for t = −∞, touches S for some t ∈ [0, 1) and returns toBδ(0) at time T . Thus

qm(t) ∈ B|ξ|(0) \B|ξ|/2(0)

for t ∈ [σ, τ ] ⊂ (−∞, T ). This implies∫ T

−∞L(qm) dt >

∫ τ

σ

L(qm) dt ≥√

2β(|ξ|/2)|ξ|/2 ≡ γ > 0

69

On the other hand ∫ T

T−1

(1

2|qm(T )|2dt =

δ2

2

and there exists ε = ε(δ)→ 0 as δ → 0, such that∫ T

T−1

−V (t, pm)dt ≤ ε

Finally

I(pm)− I(qm) ≤ δ2

2+ ε− γ

and we can take δ sufficiently small and m large, so that

δ2

2+ ε <

γ

2

ThenI(pm)− I(qm) ≤ −γ

2< 0

contradicting the definition of qm. Hence WNξ(Q) > 0 as wanted, and we finished the proof.

5.4 Existence of Heteroclinic Solutions to Periodics

Considerq + Vq(t, q) = 0 (HS)

where

(V1) V = V (t, q) ∈ C2(R× Rn,R) is 1-periodic in t and in q1, ..., qn.

We proved that, for such a potential, there exists at least a solution of (HS) which is 1-periodic.We obtained it, by minimizing

I1(q) =

∫ 1

0

L(q) dt

over E1 = W 1,21 (R,Rn). Let

c1 = infE1

I1

and defineK1 = q ∈ E1 : I1(q) = c1

As mentioned, K1 is nonempty, and as a consequence of (V1), if v ∈ K1 then v+ j ∈ K1 for allj ∈ Zn, i.e. K1 contains a lattice of 1-periodic solutions. We are intersted in finding solutionsof (HS) which are heteroclinic to these 1-periodic solutions. We need some extra assumptionsto prove their existence.

70

(V2) V is time reversible, i.e.,

V (−t, x) = V (t, x) , ∀t ∈ R , x ∈ Rn

We will start by showing some technical results, which are consequence of condition (V2).

Lemma 13. If (V1) and (V2) are satisfied then

c1 = 2 infq∈W 1,2([0,1],Rn)

∫ 1/2

0

L(q) dt

Moreover, if v ∈ K1 thenv(−t) = v(t) , ∀t ∈ R

Proof:

Choose q ∈ E1/2 ≡ W 1,2([0, 12],Rn), and define w by

w(t) =

q(t) t ∈ [0, 1

2]

q(1− t) t ∈ [12, 1]

Extending w periodically to R, and denoting the extension again by w, it is obvious thatw ∈ E1. Furthermore∫ 1

1/2

L(w) dt =

∫ 1

1/2

(1

2|q(1− t)|2 − V (t, q(1− t)) dt

=

∫ 1/2

0

(1

2|q(s)|2 − V (1− s, q)) ds

As a consequence of (V1) and (V2), we can conclude∫ 1

1/2

L(w) dt =

∫ 1/2

0

L(q) dt

and therefore

I1(w) = 2

∫ 1/2

0

L(q) dt

Thus, for all q ∈ E1/2 there exists w ∈ E1 for which

I1(w) = 2

∫ 1/2

0

L(q) dt

Taking the infimum over the appropriate sets

2 infq∈E1/2

∫ 1/2

0

L(q) dt =

∫ 1

0

L(w) dt ≥ infE1

∫ 1

0

L(w) dt = c1 (30)

71

On the other hand, for any v ∈ K1

c1 = I1(v) =

∫ 1/2

0

L(v) dt+

∫ 1

1/2

L(v) dt ≡ a+ b

If a ≤ b (implying c12≤ a, we can define

w(t) =

v(t) if t ∈ [0, 1

2]

v(1− t) if t ∈ [12, 1]

i.e., we are taking the restriction of v to the interval where mina, b is achieved and extend itto the other interval as an even function. As a consequence

2I1(w|120 ) ≤ c1

Taking the infimum over E1/2, equality is achieved in (30) as wanted. It remains to prove thatany element of K1 is an even function. For v, w as above, since they both minimize I1, theyare solutions of (HS). Moreover, by construction, they agree in [0, 1

2] or in [1

2, 1]. In either case,

by the uniqueness of solution for the associated IVP, they must equal in [0, 1]. Since w is aneven function about 1

2, also

v(1− t) = v(t)

But v is 1-periodic, which implies

v(−t) = v(t) , ∀t ∈ R

as wanted.♦

Corollary 7. If (V1) and (V2) are satisfied

c1 ≡ infEI1 = c1

Proof:

Since E is a bigger space, we must necessarily have c1 ≤ c1. We want to show that equalityholds, so let us assume that c < c1 and reach a contradiction. If that is the case, there existsq ∈ E such that I1(q) < c1. We can write I1(q) as

I1(q) =

∫ 1/2

0

L(q) dt+

∫ 1

1/2

L(q) dt ≡ a+ b

Defining w as the restriction to the interval where mina, b is achieved, extend it to the otherinterval as an even function about 1/2, and further extend it periodically to R, we constructw ∈ E1. But

I1(w) ≤ I1(q) < c1 ≡ infE1

I1

72

which contradicts the definition of infimum.

♦

We will now make another assumption, in order to restrict the set K1- it may be a very largeset, containing a continuous of solutions.

(*) K1 consists of isolated points.

How reasonable is this condition? Suppose v ∈ K1 and consider (HS), where V is replaced byW (t, x) = V (t, x)− ε|x− v(t)|2. Then

I1(q) =

∫ 1

0

(1

2|q(t)|2 − V (t, q) + ε|q − v(t)|2) dt

Notice that the extra term ε|q − v|2 will make the distinction between the elements of K1, butwe lost periodicity.

Homework 19. Use this idea to show that there is a function V verifying (V1), (V2) and (*).

We can normalize any element v ∈ K1, in a way that

0 ≤ [vi] < 1 , 1 ≤ i ≤ n

Lemma 14. If (V1), (V2) and (*) are verified, there is only a finite number of normalizedfunctions in K1.

Proof:

If not, there would be infinitely many, and therefore we could find a sequence of distinctelements (wm) ⊂ K1, verifying

0 ≤ [(wm)i] < 1 , 1 ≤ i ≤ n

Since I1(wm) = c1, there exists M = M(c1,maxV ), such that

‖wm‖L2([0,1],Rn) ≤M

Then, we can conclude that (wm) is bounded in W 1,2([0, 1],Rn), and by Sobolev inequality,there exists M1 = M1(M) such that

‖wm‖L∞[0,1]≤M1

On the other hand, for all m ∈ N, wm is a solution of (HS), and therefore

wm = −Vq(t, wm)

73

By (V1), Vq is continuous, and since (wm) is uniformly bounded, there exists M2 such that

‖wm‖L∞([0,1],Rn) ≤M2

Finally, by interpolation, there exists M3 such that

‖wm‖L∞([0,1],Rn) ≤M3

Then (wm) is bounded uniformly in C2([0, 1],Rn). By Ascoli-Arzela Theorem, it must convergein C1([0, 1],Rn), possibly along a subsequence, to another point in K1 contradicting (*).

♦

As an immediate consequence we can state the following result.

Lemma 15. If (V1)-(V2) and (*) holds, there exists b > 0 such that

|v(0)− w(0)| > b

for normalized v, w ∈ K1 with v 6= w

Proof:

Notice that we can have v(0) − w(0) = 0. If that was the case, since they are both evenfunctions, their derivatives will be odd functions, and as a consequence v(0) = w(0). But then,both v and w are solutions of the IVP

q + Vq(t, q) = 0q(0) = 0 = q(0)

and by uniqueness of solution, v ≡ w which contradicts the fact that v 6= w. So, the onlypossible situation where we can’t find such b is if we have a sequence (vm) ⊂ K1, such that

|vm(0)− vn(0)| → 0 as m,n→∞

Using familiar arguments, we can prove that vm converges for some v ∈ K1, violating condition(*).

♦

Corollary 8. Ifγ = inf‖v − w‖L∞ , v 6= w ∈ K1

then γ ≥ b.

This is obvious, since

γ = inf‖v − w‖L∞ , v 6= w ∈ K1= inf

v,wsupt|v(t)− w(t)| ≥ inf

v,w|v(0)− w(0)| ≥ b

For v ∈ K1 and r > 0, denote

Br(v) = q ∈ E : ‖q − v‖E1 < r

74

Proposition 9. If (V1)-(V2) and (*) hold, for any 0 < ρ < γ4, there exists α = α(ρ) > 0, such

that

(1) Bρ(v) ∩Bρ(w) = ∅ if v 6= w in K1.

(2) I1(u) > c1 for u ∈ Bρ(K1) \ K1.

(3) I1(u) ≥ c1 + α if u ∈ E \Bρ(K1).

Proof:

(1) Is just a consequence of Corollary 8.

(2) By definition of K1 and Corollary 8.

(3) Suppose this is not the case. Then, for all α ≥ 0,

I1(u) ≤ c1 + α

and as a consequence, there exists (um) ⊂ E \Bρ(K1) such that

I1(um)→ c1 , m→∞

As before, we can assume that, for all m ∈ N

0 ≤ [umi ] < 1 , 1 ≤ i ≤ n

and, we can conclude that (um) is a bounded sequence in E, and therefore, possibly along asubsequence, it converges weakly in E and strongly in L∞loc(R,Rn) to some u ∈ E. Since I1 isweakly lower semicontinuous

I1(u) ≤ lim infm→∞

I1(um) = c1

By definition of c1, I1(u) = c1, and thus u ∈ K1. For wm = um − u,

I1(um) = I1(u+ wm) =

∫ 1

0

(1

2|u+ wm|2 − V (t, u+ wm)) dt

=

∫ 1

0

(1

2|u|2 +

1

2|wm|2 + u · wm − V (t, u+ wm)) dt

=1

2‖wm‖2

E1+ I1(u) +

∫ 1

0

(u · wm −1

2|wm|2 − V (t, u+ wm) + V (t, u))dt

Notice that,

. wm ∈ E \Bρ(K1), as a consequence of condition (*);

. ‖wm‖E ≥ ρ since we considered (um) ∈ E \Bρ(K1) and v ∈ K1;

75

. since um u in E1, ∫ 1

0

(u · wm)dt→ 0 , m→∞

. since um → u in L∞([0, 1],Rn)∫ 1

0

|wm|2dt→ 0 , m→∞

and by the continuity of V , also

V (t, u+ wm)− V (t, u)→ 0 , m→∞

Then, taking the limit as m→∞,

c1 = limm→∞

I1(um) ≥ c1 +1

2ρ2

which is a contradiction.♦

Theorem 13. If (V1)-(V2) and (*) hold, for any v ∈ K1, there exists w ∈ K1 \ v and asolution Q of (HS) such that Q(t)− v(t)→ 0 as t→ −∞ and Q(t)− w(t)→ 0 as t→∞. Byits definition, Q is a heteroclinic solution from v to w.

Unfortunately, we can’t prove the Theorem by minimization of

I(q) =

∫RL(q)dt

Notice that, since we wantQ(t)− v(t)→ 0 as t→ −∞

there exists T ∈ N such thatQ(t) ≈ v(t) , ∀t ≤ −T

As a consequence∫ −T−∞L(Q) dt ≈

∫ −T−∞L(v) dt =

−T∑k=−∞

∫ k

k−1

L(v) dt =−T∑

k=−∞

c1 =∞

Therefore I is not well defined for the functions we want to find, and we have to formulate thevariational problem in a different way.

For q ∈ W 1,2loc (R,Rn) and p ∈ Z, define

ap(q) =

∫ p+1

p

L(q) dt− c1

Notice that W 1,2([p, p + 1],Rn) can be identified with W 1,2([0, 1],Rn) by a simple change ofcoordinates.

76

Homework 20. Show that ap(q) ≥ 0, for all p ∈ Z.

We define the renormalized functional

J(q) =∑p∈Z

ap(q)

We will minimize J over the set of admissible curves

Γ = q ∈ W 1,2loc (R,Rn) : q(−∞)− v(−∞) = 0 and ∃w ∈ K1 \ v : q(∞)− w(∞) = 0

It is easy to check that Γ is nonempty: define the function that coincides with v in (−∞, 1)runs to w for t ∈ [0, 1] and coincides with w for (1,∞). Further define

c = infΓJ(q)

The proof of the Theorem follows from two propositions.

Proposition 10. There exists Q ∈ Γ such that J(Q) = c.

Proposition 11. Q is a solution of (HS) and Q(t)− v(t)→ 0 as t→ −∞ and Q(t)− w(t)→ 0as t→∞.

The variational characterization of Q, and the asymptotic behavior of Q, imply that Q isin fact a heteroclinic solution between two periodic solutions of (HS). The proof of Proposition11 follows the some ideas as previous proofs, so we will ommit it.

Proof: (Proposition 10)

Let (qm) ⊂ Γ be a minimizing sequence for J over Γ. The periodicity of V implies that

J(τk(qm)) = J(qm(t− k)) = J(qm)

for all k ∈ Z. Thus, we can replace qm by an appropriate choice of τkqm verifying

|τkmqm(t)− v(t)| < ρ for all t < 0 and

|τkmqm(t)− v(t)| = ρ for some t ∈ [0, 1).

Without loss of generality we can assume that km = 0 for all m. Our goal is to obtain boundsfor qm in W 1,2

loc (R,Rn). Since J(qm)→ c, as m→∞, there exists M ≥ 0 such that J(qm) ≤Mfor all m ∈ N. For fixed L ∈ R+, we can find k ∈ N such that k − 1 ≤ L < k. Then

M ≥ J(qm) =∑p∈Z

ap(qm) ≥k−1∑p=−k

ap(qm)

=k−1∑p=−k

(∫ p+1

p

L(qm) dt− c1

)=

∫ L

−LL(qm) dt− 2kc1 ≥

∫ L

−LL(qm) dt− 2Lc1

77

This implies ∫ L

−L

1

2|qm|2 dt ≤M + 2Lc1 +

∫ L

−LV (t, qm) dt

Since V is continuous and periodic,∫ L−L V (t, qm) dt is bounded independently of m, and there-

fore ∫ L

−L

1

2|qm|2 dt ≤ K(L)

for all m ∈ N and all L ∈ R+. On the other hand, as a consequence of the normalization

|qm(0)| ≤ |v(0)|+ ρ

Then qm is bounded in W 1,2([−L,L],Rn) for every L > 0. Since W 1,2([−L,L],Rn) is a Hilbertspace, for each L ∈ R+, qm admits a convergent subsequence. By a diagonalization process wecan find Q ∈ W 1,2

loc (R,Rn) such that qm → Q weakly in W 1,2loc (R,Rn) and strongly in L∞loc(R,Rn).

We have to show that Q ∈ Γ and that J(Q) = c. Let’s start by proving that J(Q) < ∞.Consider L ∈ N arbitrary, then

L−1∑p=−L

ap(Q) =L−1∑p=−L

(∫ p+1

p

L(Q) dt− c1

)=

∫ L

−LL(Q) dt− 2Lc1

Since,∫ baL(q)dt is weakly lower semicontinuous

L−1∑p=−L

ap(Q) ≤ lim infm→∞

∫ L

−LL(qm) dt− 2Lc1 = lim inf

m→∞

L−1∑p=−L

ap(qm) ≤ lim infm→∞

J(qm) ≤M

for all L ∈ N. Letting L → ∞ we conclude J(Q) ≤ M < ∞, as wanted. As an immediateconsequence

ap(Q)→ 0 as |p| → ∞

For Qp = Q|p+1p ( which we can view as an element of W 1,2([0, 1],Rn)), we can choose p ∈ Z so

that

ap(Q) ≤ 1

2α(ρ)

where α is as defined in Proposition 9. Then

I1(Qp) ≤1

2α(ρ) + c1 (31)

and we conclude thatQp ∈ Bρ(K1) \ K1

If not we would haveI1(Qp) ≥ c1 + α(ρ)

78

as a consequence of Proposition 9, contradicting (31). Then, there exists vp ∈ K1 such that

‖Qp − vp‖W 1,2([p,p+1],Rn) ≤ ρ

and by Sobolev inequality‖Qp − vp‖L∞([p,p+1],Rn) ≤ 2ρ

On the other hand, we normalized (qm) so that

|qm(t)− v(t)| ≤ ρ , ∀t ≤ 0

For fixed t, letting m→∞ we will also obtain

|Q(t)− v(t)| ≤ ρ , ∀t ≤ 0

Then, for large negative p

‖vp − v‖L∞([p,p+1],Rn) ≤ ‖vp −Qp‖L∞([p,p+1],Rn) + ‖Qp − v‖L∞([p,p+1],Rn) ≤ 3ρ

Recall that we made the choice of

ρ <γ

4=

1

4infz 6=w‖z − w‖L∞([0,1],Rn)

and therefore

‖vp − v‖L∞([p,p+1],Rn) ≤ ρ <3

4γ < γ

which implies v = vp in [p, p+ 1] for all large negative p. Thus

‖Qp − v‖W 1,2([p,p+1],Rn) ≤ 2ρ

for all large negative p, and as a consequence (Qp) is a bounded sequence in W 1,2([0, 1]). Thus,there exists Q∗ ∈ W 1,2([0, 1]) such that Qp → Q∗ weakly in W 1,2([0, 1],Rn) and strongly inL∞([0, 1],Rn). By the weak lower semicontinuity

I1(Q∗) ≤ lim infp→−∞

I1(Qp)

But, since ap(Q) = ap(Qp)→ 0 as p→ −∞ we will have

lim infp→−∞

I1(Qp) = c1

and therefore I1(Q∗) ≤ c1. This together with the fact Q∗ ∈ W 1,2([0, 1]), implies I1(Q∗) = c1,i.e., Q∗ ∈ K1. On the other hand

|Qp(t)− v(t)| ≤ ρ

letting p→ −∞,|Q∗(t)− v(t)| ≤ ρ

79

and by choice of ρ, we must necessarily have Q∗(t) = v(t). We showed that a subsequence ofQp converges uniformly to v. The uniqueness of limit implies that all the sequence convergesto v. This gives the right asymptotic behavior as t→ −∞.

We will now prove thatlimt→∞|Q(t)− w(t)| = 0

for some w ∈ K1 \ v. Our previous reasoning is valid, up to the step

‖Qp − vp‖L∞([0,1],Rn) < 2ρ (32)

In the following arguments we can’t use the normalization, so we will have to work harder. (32)implies, in particular, that

|Qp(1)− vp(1)| ≤ 2ρ

and|Qp+1(0)− vp+1(0)| ≤ 2ρ

But since vp ∈ K1, we will also have

|Qp+1(0)− v(0)| ≤ 2ρ

Then|vp(0)− vp+1(0)| ≤ 4ρ

and since ρ < γ4,

|vp(0)− vp+1(0)| ≤ γ

By Corollary 8, we can conclude that vp = vp+1 for p large. If we denote the common functionby w, it is easy to prove that, (using the arguments we used to prove Q(−∞)− v(−∞) = 0),as t → ∞, |Q(t) − w(t)| → 0. Nevertheless, we still have to check that w(t) 6= v(t). For thatpurpose, we will use a comparison argument). Let’s suppose that w = v. For 0 < δ << α(ρ

2),

(for α(ρ/2) as in Proposition 9), since Q(t) → v(t) as t → ∞, there exists T = T (δ) > 1 suchthat

|Q(t)− v(t)| ≤ δ , ∀t ≥ T

Since qm → Q in L∞loc(R,Rn), for m sufficiently large

|qm(t)− v(t)| ≤ 2δ

for t ∈ [T − 1, T ]. Define the comparison function

vm(t) =

v(t) if t ≤ T − δ(t−(T−δ))

δqm(T ) + t−T

δv(T − δ) if T − δ ≤ t ≤ T

qm(t) if t ≥ T

It is easy to check that vm ∈ Γ. Moreover

J(vm)− J(qm) =∑p∈Z

(ap(vm)− ap(qm))

80

On one hand, vm(t) = v(t) for all t ≤ T − δ, and therefore

T−1∑−∞

ai(vm) = 0

On the other hand, vm(t) = qm(t) for t ≥ T , implying

∞∑T

ai(qm) =∞∑T

ai(vm)

Hence

J(vm)− J(qm) = aT−1(vm)−T−1∑p=−∞

ap(qm)

In the interval [0, 1) qm hits the boundary of Bρ(v) and returns to Bδ(v) for t ∈ [T − 1, T ]. Inparticular, we can find a negative integer p0 ≤ T such that

‖qm − v‖W 1,2([p0,p0+1],Rn ≥≥ρ

2

By Proposition 9, c) ∫ p0+1

p0

L(qm) dt ≥ c1 + α(ρ/2)

which is equivalent toap0(qm) ≥ α(ρ/2)

ThenJ(vm)− J(qm) ≤ aT−1(vm)− ap0(qm) ≤ aT−1(vm)− α(ρ/2)

Finally, let us now analyse aT−1(vm).

aT−1(vm) =

∫ T

T−1

L(vm) dt− c1 =

∫ T

T−1

(L(vm)− L(v)) dt

By construction, vm = v for t ∈ [T − δ, T ], and as a consequence

aT−1(vm) =

∫ T

T−δ(L(vm)− L(v)) dt

We proved before that v ∈ C2(R,Rn), therefore∣∣∣ ∫ T

T−δL(v) dt

∣∣∣ ≤ maxt∈[T−δ,T ]

(1

2|v|2 + V (t, v)

)≤ δM1

for some positive constant M1. On the other hand, for t ∈ [T − δ, T ], also by construction,|vm(t)! ≤ ε(δ)→ 0 as δ → 0, and thus∣∣∣ ∫ T

T−δV (t, vm) dt

∣∣∣ ≤M2δ

81

for some positive constant M2. Lastly, by its definition∣∣∣ ∫ T

T−δ

1

2|vm|2dt

∣∣∣ =1

2

∫ T

T−δ

∣∣∣qm(T )− v(T − δ)δ

∣∣∣2dtand

|qm(T )− v(T − δ)| ≤ |qm(T )− v(T )|+ |v(T )− v(T − δ)| ≤ 2δ + ‖v‖L∞[T−δ,T ]δ

Using, once again, the fact that v ∈ C2(R,Rn), we can conclude that there exists M3 ≥ 0 suchthat

|qm(T )− v(T − δ)| ≤M3δ

Hence ∣∣∣ ∫ T

T−δ

1

2|vm|2dt

∣∣∣ ≤M3δ

Finally, for some positive M

J(vm)− J(qm) ≤Mδ − α(ρ

2)

and we can choose δ so small so that Mδ <α( ρ

2)

2, implying

J(vm) < J(qm)

which is a contradiction to the fact that (qm) is a minimizing sequence for J over Γ. To finishthe proof it remains to show that J(Q) = c. Since J(Q) <∞

L∑−L

ap(Q) ≥ J(Q)− ε

Hence

c = lim infm→∞

J(qm) ≥ lim infm→∞

(∫ L

−LL(qm) dt− c1

)≥

(∫ L

−LL(Q) dt− c1

)=

L∑−L

ap(Q)

Letting L→∞ we obtain J(Q) ≤ c. Since Q ∈ Γ, by definition of c, J(Q) = c.

♦

82

6 Geodesics

Let M be a metric space, i.e., a set equipped with a map d :M×M→ R, verifying

(M1) d(x, y) = 0 iff x = y;

(M2) d(x, y) = d(y, x);

(M3) d(x, z) ≤ d(x, y) + d(y, z).

Intuitively, we can interpret d(x, y) as the distance from x to y.

Example 30. 1. M = Rn with d(x, y) = |x− y|.

2. M = C([0, 1],R) with

d(x, y) =

∫ 1

0

|x(t)− y(t)| dt

A function f defined in I ⊂ R, a closed interval, verifying f ∈ C(I,M) is a curve or pathin M. Consider P as the set of all partitions of the interval I = [a, b], i.e.

P = P = (t0, ..., tk) : a = t0 < t1 < ... < tk = b

For any P ∈ P , we can define the length of f relatively to P as

L(Pf) =k∑i=1

d(f(ti+1), f(ti))

Definition 14. We define Lenght of f as

L(f) = supP∈P

L(Pf)

Moreover, f is rectifiable if L(f) <∞.

Homework 21. Let f ∈ C(I,M) be rectifiable. Show that:

(a) For all τ ≤ s ≤ t in I = [a, b]

L(f |tτ ) = L(f |sτ ) + L(f |ts)

.

(b) For any ψ ∈ C([c, d], [a, b]) strictly monotone, with ψ(c) = a and ψ(d) = b, and h =f o ψ ∈ C([c, d],M), then

L(h) = L(f)

that is the length is independent of the parametrization.

83

Example 31. Let M = Rn, with d(x, y) = |x − y| and f ∈ C1([a, b],Rn). Then, using theMean Value Theorem, for any P ∈ P

L(Pf) =k∑i=1

|f(ti+1)− f(ti)| =k∑i=1

|f ′(θi)||ti+1 − ti|

for some θi ∈ [ti, ti+1]. Taking the supremum over all partitions, we conclude

L(f) = supP∈P

L(Pf) =

∫ b

a

|f ′(t)| dt

DefineR = f ∈ C(I,M) : L(f) <∞

i.e., R is the set of rectifiable curves onM. Then R can be made a metric space, by introducingthe metric

ρ(f, g) = maxt∈I

d(f(t), g(t)) ; f, g ∈ R

Example 32. If M = Rn, then ρ corresponds to the maximum norm:

ρ(f, g) = maxt∈[0,1]

|f(t)− g(t)|

The next example shows that the length map, L : R → R, is not continuous, in the topologydefined by ρ.

Example 33. Let f : [0, 1],R) be defined by f(t) = 1 − |1 − 2t|, and consider F : R → R the

periodic extension of f to R. Define the sequence fm ∈ R bfm(t) = F (mt)m

. It is easy to check

that, for all m ∈ N, L(fm) =√

5. But fm → 0 as m → ∞ and L(limit curve) = 1. HenceL(fm) 6→ L(limit curve) as m→∞.

Theorem 14. L is lower semicotinuous on R.

Proof:

We want to prove thatL(f) ≤ lim inf

m→∞L(fm)

for any (fm) ⊂ R, f ∈ R such that ρ(fm, f) → 0 as m → ∞. Choose s ≤ L(f) = supP∈P

L(Pf).

By definition of supremum, we can find a P0 ∈ P , such that

s < L(P0f) ≤ L(f)

and

L(P0f) =k∑i=1

d(f(ti−1), f(ti))

84

Pick ε > 0 and g ∈ R, close to f in the sense

ρ(f, g) <ε

k

Then, for each i ∈ 1, ...k

d(f(ti−1), f(ti)) ≤ d(f(ti−1), g(ti−1)) + d(g(ti−1), g(ti)) + d(g(ti−1), f(ti))

≤ 2ε

k+ d(g(ti−1), g(ti))

ThusL(P0f) ≤ 2ε+ L(P0g)

which is equivalent toL(P0g) ≥ L(P0f)− 2ε > s− 2ε

Taking the supremum over all P ∈ P , we conclude

L(g) ≥ s− 2ε

Since ρ(fm, f)→ 0, as m→∞, for m large, so that ρ(fm, f) < εk, also

L(fm) ≥ s− 2ε

Taking the limit as m→∞lim infm→∞

L(fm) ≥ s− 2ε

for all ε > 0, which implies

lim infm→∞

L(fm) ≥ s , ∀s < L(f)

and thereforelim infm→∞

L(fm) ≥ L(f)

♦

We want to prove the existence of a curve of shortest length joining two arbitrary pointsof a compact metric space. Before, we will prove a technic result which states that we canparametrize any curve in R in a way that a strong continuity condition is verified.

Proposition 12. For f ∈ R, there exists F ∈ R such that

(1) F (0) = f(0) and F (1) = f(1);

(2) L(F ) = L(f);

(3) ∀s, t ∈ [0, 1] d(F (s), F (t)) ≤ L(f)|s− t|.

85

Proof:

For t ∈ [0, 1], define γ(t) = L(f |t0). Then, it is obvious that γ(0) = L(f |00) = 0, γ(1) =L(f |10) = L(f) and γ is monotone non-decreasing. We are going to prove that γ is continuous.For any t ∈ [0, 1] and ε > 0, since f is continuous on a compact set, there exists δ = δ(ε) > 0,such that

|t− τ | ≤ δ ⇒ d(f(t), f(τ)) ≤ ε

2

Choose a partition P = (t0, ..., tk) of [0, 1], verifying

L(Pf) > L(f)− ε

We will assume thatti − ti−1 ≤ δ , 1 ≤ i ≤ k

and choose τ so that|t− τ | < min(ti − ti−1)

By our choices, either t, τ are in the same partition interval, or they belong to adjacent intervals.We claim that, if this is the case, then

|γ(τ)− γ(t)| ≤ 3ε (33)

which will prove that γ is continuous in t, for all t ∈ [0, 1]. Suppose (33) does not hold, i.e.,

|γ(τ)− γ(t)| > 3ε

Using the additivity property of the length, also

3ε < |γ(t)− γ(τ)| = L(f |τt )

We will prove that this yields a contradiction if the worst case ti−1 < t < ti < τ < ti+1 occurs.On one hand

L(Pf) + ε > L(f |ti−1

0 ) + L(f |ti+1

ti−1) + L(f 1

ti+1)

On the other hand

L(Pf) =∑

j 6=i,i+1

d(f(tj−1), f(tj)) + d(f(ti−1), f(ti)) + d(f(ti), f(ti+1)) + ε

and, by definition

L(f |ti−1

0 ) + L(f 1ti+1

) >∑

j 6=i,i+1

d(f(tj−1), f(tj))

Also, by our assumptionL(f |ti+1

ti−1) ≥ L(f |τt ) > 3ε

86

Then

d(f(ti−1), f(ti)) + d(f(ti), f(ti+1))

= L(Pf)−∑

j 6=i,i+1

d(f(tj−1), f(tj))

≥ L(Pf)− L(f |ti−1

0 )− L(f |1ti+1)

= L(Pf)− L(f) + L(f |ti+1

ti−1) < −ε+ 3ε = 2ε

But, the continuity of f implies

d(f(ti−1), f(ti)) + d(f(ti), f(ti+1) ≤ ε

which is a contradiction, and we conclude that γ is continuous.

Our goal is to define the inverse function of γ, but we have a problem since γ may not bestrictly monotone. Define ϕ : [0, L(f)]→ [0, 1], by

ϕ(r) = inft ∈ [0, 1] : γ(t) = r

and, our first approximation to F , f ∗ : [0, L(f)]→M, defined by

f ∗(r) = f(ϕ(r))

Notice that, f ∗ is well defined, since ϕ is well defined. Moreover, if γ(t) = r for t minimal, thenf ∗(r) = t. Also,

f ∗(0) = f(ϕ(0)) = inft ∈ [0, 1] : L(f |t0) = 0 = f(0);

f ∗(L(f)) = f(inft ∈ [0, 1] : L(f |t0) = L(f) ≡ f(t∗). If t∗ = 1 then f ∗(L(f)) = f(1)and we are done. Otherwise t∗ < 1 and therefore, by construction, f(t) ≡constant fort ≥ t∗, implying that f(t) = f(1) for all t ≥ t∗. In both cases f ∗(L(f)) = f(1).

Moreoverd(f ∗(s), f ∗(t)) = d(f(s), f(t)) (34)

which can be proved, using the same minimality argument we used above.

We will prove next, some properties of f ∗.

(1) f ∗ is continuous: by definition, f ∗ is the composition of a continuous function with ϕwhich can be discontinuous at a jump point. But, notice that, if we approach by belowthere is no problem, since ϕ is continuous. If we approach by above, there is also noproblem, since f ≡ constant in the “jump”.

87

(2) For all t, r such that γ(t) = rL(f |t0) = l(f ∗|r0)

Consider Pr = (r0, ..., rk) a partition of the interval [0, r]. Then, using (34)

L(Prf∗) =

k∑i=1

d(f ∗(ri−1), f ∗(ri)) =k∑i=1

d(f(ri−1)), f(ri)) ≤ L(f |t0)

As a consequenceL(Prf

∗) ≤ L(f |t0)

and taking the supremum over all partitions Pr,

L(f ∗|r0) ≤ L(f |t0)

Play the other way around, we will obtain

L(f ∗|r0) ≥ L(f |t0)

which implies the equality.

As a consequence of property (2)

r = γ(t) = L(f |t0) = L(f ∗|r0) ∀r ∈ [0, L(f)]

which will also be true for r + s, i.e.

r + s = L(f ∗|r+s0 ) = L(f ∗|r0) + L(f ∗|r+sr )

Subtracting the above equalitiess = L(f ∗|r+sr )

which is equivalent to (making r + s = σ)

L(f ∗|σs ) = σ − s

We denote f ∗ by the reparametrization of f by arclength. In particular, by definition of L

d(f ∗(r)f ∗(s)) ≤ L(f ∗|sr) ≤ |s− r| (35)

Finally, for t ∈ [0, 1], defineF (t) = f ∗(L(f)t)

Then

F (0) = f ∗(0) = f(0) and F (1) = f ∗(L(f)) = f(1).

L(F ) = L(f ∗) = L(f), since F is just a reparametrization of f ∗.

88

As a consequence of (35)

d(F (t), F (τ)) = d(f ∗(L(f)t), f ∗(L(f)τ)) ≤ |L(f)t− L(f)τ | = L(f)|t− τ |

♦

Theorem 15. Let M be a compact metric space and x, y ∈M. Define

S = f ∈ R : f(0) = x and f(1) = y

If S 6= ∅, there exists g ∈ S such that L(g) ≤ L(f) for all f ∈ S.

Proof:

Take a minimizing sequence in S, (fm) such that

M ≥ L(fm) infSL

By Proposition 12, we can replace (fm) by (Fm) where (Fm) has the additional property (3).Then

d(Fm(s), Fm(t)) ≤ L(fm)|s− t| ≤M |s− t|

meaning that (Fm) is an equicontinuous sequence in C([0, 1],M). Moreover for x = Fm(0)

d(x, Fm(t)) ≤M |t| ∀t ∈ [0, 1]

implying that it is uniformly bounded for t ∈ [0, 1]. By the Ascoli-Arzela Theorem somesubsequence of Fm will converge uniformly to some function g, and g(0) = x, g(1) = y. Also

L(g) ≤ lim inf L(Fm) = infSL ≤M (36)

and therefore g is rectifiable (g ∈ R). Hence g ∈ S. This fact and (36)imply

L(g) = infSL

♦

6.1 Curves defined on a Manifold

Definition 15. f is absolutely continuous on I = [a, b] ⊂ R if for all ε > 0 there existsδ = δ(ε) > 0, such that, whenever [a1, b1], ..., [ak, bk], closed, disjoint subintervals of [a, b], then

k∑i=1

|f(bi)− f(ai)| < ε

Note that, if f is absolutely continuous then f is uniformly continuous (just take k = 1).

89

Definition 16. f defined on [a, b] is of Bounded Variation (f ∈ BV ([a, b]), if for all P =(x0, ..., x1) ∈ P partition of the interval [a, b],

supP∈P

∑|f(xi−1)− f(xi)| <∞

Notice that in R the notion of bounded variation is equivalent to the notion of rectifiable.Also, it is easy to check that if f ∈ AC([a, b] then f ∈ BV ([a, b]). But, the same thing maynot be true, if we consider f ∈ AC(R).

Example 34. Consider f(x) = x and f(x) = sin x.

Theorem 16. f ∈ AC([a, b]) iff f(x)− f(a) =∫ xaf ′(t) dt

You can find the proof in Ch 8 of Rudin’s Real and Complex Analysis.

We will now generalize the notion of length to a curve defined on a manifold.

Let M ⊂ Rn be a k dimensional manifold, and consider c ∈ AC([0, T ],M), i.e., c(t) ∈ Mfor all t ∈ [0, T ]).

Definition 17. We define the Length of c, L(c), by

L(c) =

∫ T

0

|c(t)| dt

Since c ∈ AC([0, T ]), the definition makes sense, as a consequence of Theorem 16. Aswe have seen in a previous example, if c is a smooth curve, L(c) coincides with our previousdefinition.

Definition 18. We define the energy of c, E(c), by

E(c) =1

2

∫ T

0

|c(t)|2dt

We observe, that E(c) may not be well defined, if c 6∈ L2([0, T ]). So we have to make theextra assumption

c ∈ W 1,2([0, T ],M)

Lemma 16. (Invariance Properties)

1. For i : Rn → Rn, defined by i(y) = Ay + b where A is an orthogonal matrix (i is theeuclidean isometry), then

L(ic) = L(c) and E(i(c)) = E(c)

2. If ϕ ∈ C1([0, s], [0, T ]) is a diffeomorphism, then

L(c ϕ) = L(c)

90

Proof: In order to prove (2), notice that, by definition

L(c ϕ) =

∫ s

0

| ddsc(ϕ(s))| ds =

∫ s

0

|c(ϕ)||ϕ(s)| ds

=

∫ T

0

|c(t)| dt = L(c)

♦

Unfortunately, E is not invariant under this type of transformations. But we can observe thefollowing relationship. As a consequence of Schwarz inequality.

L(c) =

∫ T

0

|c(t)| dt ≤ T 1/2(∫ T

0

|c(t)|2 dt)1/2

=√

2TE(c)

Recall that ∫fg <

(∫f 2)1/2(∫

f 2)1/2

unless f = αg and equality holds. Then, if |c| = is constant, equality holds, and therefore

L(c) =√

2TE(c) (37)

If this is the case, c is said to be parametrized porportionally to the arclength. If |c| = 1, then cis said to be parametrized by the arclength.

Lemma 17. Let c : [0, L(c)]→ Rn be a curve parametrized on [0, L(c)]. Then the parametriza-tion of c by arclength (or the parametrization of E proportional to arclength), minimizes E overthe class of reparametrizations ϕ : [0, L(c)]→ [0, L(c)].

Proof:

Denote by c∗ the parametrization of c by arclength. Then L(c) = L(c∗) and by (37)

L(c) = L(c∗) =√

2L(c)E(c∗)

Squaring both sides, L(c) = 2E(c∗). As remarked above, if c is any other parametrization, then

L(c) = L(c) <√

2L(c)E(c)

Squaring both sides, and comparing

E(c∗) < E(c)

which concludes the proof.♦

91

Definition 19. M ⊆ Rn is a k-dimensional manifold, if for all p ∈M there are neighborhoodsW = W (p) ⊆ Rn, U ⊆ Rk and a diffeomorphism f : U → W , such that f(U) = M ∩W . Thetriple (f, U,W ) is denoted by local chart for M .

Suppose that c([0, T ]) ⊂ f(U) ⊂M (i.e, the image of the curve lies in a single local chart).By definition, we can find a curve γ(t) ∈ U such that

c(t) = f(γ(t))

and we can express L(c) in terms of f and γ. We will get more complicated expressions butwe no longer have to worry about the constraint (since U is an open set).

Homework 22. If f is smooth, c is absolutely continous in [0, T ] and γ(t) ∈ U , prove thatγ ∈ AC([0, T ], U).

We want to express L and E in terms of f and γ. For c = (c1, ..., cn), by definition

L(c) =

∫ T

0

|c(t)|2dt =

∫ T

0

( n∑m=1

(cm(t))2)1/2

dt

Sincec(t) = f(γ(t)) = (f1(γ1, ..., γk), ..., fn(γ1, ..., γk))

using the chain rule

cm =k∑i=1

∂fm∂xi

(γ) · γi

and, as a consequence

L(c) =

∫ T

0

( k∑i,j=1

n∑m=1

(∂fm∂xi

(γ) · γi)( ∂fl

∂xj(γ) · γj

))1/2

dt

In a similar way

E(c) =

∫ T

0

1

2|c|2dt =

1

2

∫ T

0

k∑i,j=1

n∑m=1

(∂fm∂xi

(γ) · γi)(∂fm

∂xj(γ) · γj

)dt

If we set

gij(x) =n∑

m=1

∂fm∂xi

(x)∂fm∂xj

(x)

[gi,j(x)]ni,j=1 defines a matrix G(x), which is denoted the metric tensor with respect to the chartf . Then

L(c) =

∫ T

0

( k∑i,j=1

gij(γ)γiγj

)1/2

dt

92

and

E(c) =

∫ T

0

1

2

k∑i,j=1

gij(γ)γiγj dt

Homework 23. Show that G is a symmetric and positive definite matrix, i.e,

a) gi,j = gji

b) Gy · y ≥ 0 and Gy · y = 0 iff y = 0.

The expressionk∑

i,j=1

gij γiγj

is called the riemannian metric. Our initial interest is to find curves defined in M of smallestlength or smallest energy.

If we look at the two expressions we derived for L(c):

L(c) =

∫ T

0

|c(t)| dt =

∫ T

0

( k∑i,j=1

gij γiγj

)1/2

we observe that, while the first one depends on the manifold, the second one depends on g.The equality suggests that the representation does not depend on the charts.

Lemma 18. The representation

L(c) =

∫ T

0

( k∑i,j=1

gij γiγj

)1/2

dt

where

gij(x) =n∑

m=1

∂fm∂xi

∂fm∂xj

does not depend of the choice of f, U,W .

Proof:

Consider two different charts, f : U → W and f : U → W such that c([0, T ]) ⊂ f(U) andc([0, T ]) ⊂ f(U). By definition, we can find two continuous maps, γ, γ, such that

c(t) = f(γ(t)) = f(γ(t)) , ∀t ∈ [0, T ]

Define ϕ : f−1(f(U) ∩ f(U))→ f−1(f(U) ∩ f(U)) by

ϕ = f−1 f

93

We want to show thatk∑

i,j=1

gij(x)γiγj =k∑

i,j=1

gij(x) ˙γi ˙γj

By definition, for x ∈ U

gij(x) =n∑

m=1

∂fm∂xi

∂fm∂xj

Sincef(γ(t)) = c(t) = f(γ(t))

thenγ(t) = ϕ(γ(t))

Therefore, we can compute the derivatives, using the chain rule

˙γi =d

dt(ϕ(γ(t))i =

k∑l=1

∂ϕi∂xl

(γ)γl (38)

On the other hand

gij(x) =n∑

m=1

∂fm∂xi

∂fm∂xj

==n∑

m=1

∂(f(ϕ))m∂xi

∂(f(ϕ))m∂xj

since f = f ϕ. Carrying out the derivatives

∂fm∂xi

=k∑l=1

∂fm∂xl

∂ϕl∂xi

and as a consequence, we obtain a relation between the two metric tensors

gij(x) =n∑

m=1

( k∑l=1

∂fm∂xi

∂ϕl∂xi

)( k∑t=1

∂fm∂xt

∂ϕt∂xj

)Finally, using (38)

k∑i,j=1

gij(x)γiγj =k∑

i,j=1

k∑l,t=1

glt(x)∂ϕl∂xi

∂ϕt∂xj

γiγj

=k∑

l,t=1

glt(x) ˙γl ˙γt

Hence we conclude that the expressions for L and E are independent of the chart.

♦

94

Next, we will extend the notions for the case when the curve doesn’t lie in a single elementof the chart. Let c : [0, 1] → M , and take any partition P = (t0, ..., tk) of the interval [0, 1].Choose a parametrization of the curve, such that

c([ti, ti+1]) ⊂ fi(Ui)

For each fi : Ui → Wi, 1 ≤ i ≤ n, is a single chart and therefore has an associated metric tensor(gνij(x)). By the additivity property of the length

L(c) =m∑1

L(c|tνtν−1) =

∑∫ tν

tν−1

(∑gνij(γν(t))γ

iν γ

jν

)1/2

where c(t) = fν(γν(t)) for t ∈ [tν−1, tν ]. Nevertheless, to avoid extra difficulties in the notation,we will abreviate

L(c) =

∫ 1

0

( k∑i,j=1

gij(γ(t))γiγj

)1/2

dt

and likewise for E.

We want to find the curve of shortest length joining two arbitrary points of a k-dimensionaldifferentiable manifold M of Rn. The expressions we derived for L and E, using the localcoordinates of M , are undoubtly messier, but we no longer have to deal with the constraintc(t) ∈ M for all t ∈ [0, T ]: from now on, it is enough to consider curves, γ, defined on a openset of Rk. The purpose of this is, if we want to derive the differential equation satisfied by aminimizer, v, of L, to argue as before, we need to be sure that v+δϕ ∈M for some appropriateϕ. This may not be an easy task to do.

To solve the problem, for x, y ∈M , consider

Γxy = q ∈ W 1,2([0, T ],M) : q(0) = x , q(T ) = y

As remarked, we want to find the curve of Γxy of shortest length. Lemma 17 suggests that itis better to minimize the energy, E(c), instead, since its minimizers may be parametrized withrespect to arclength. Nevertheless, we need to be sure that the curves that minimize E, alsominimize L.

Proposition 13. Let q ∈ Γxy be a minimizer of E over Γxy. Then

1. q is parametrized proportionally to arclength.

2. q minimizes L(q) over the class Γxy.

Proof:

1. If this is was not the case, then

L(q) =

∫ T

0

| ˙q| dt ≤(∫ T

0

| ˙q|2dt)1/2

=√

2TE(q)

95

If q∗ is the parametrization of q over [0, 1] proportional to arclength, as a consequence of Lemma17

L(q) = L(q∗) =√

2TE(q∗) <√

2TE(q)

implying E(q∗) < E(q) which is a contradiction.

2. If not, there exists q ∈ Γ with L(q) < L(q). Since parametrization doesn’t change length,we can assume that q is parametrized proportional to arclength. But then, Lemma 17 implies√

2TE(q) <√

2TE(q)

which is a contradiction.♦

Assume for the moment, that the minimizers of E are sufficiently regular. Since

E(c) =

∫ T

0

k∑i,j=1

gij(γ)γiγjdt

we know how to derive the associated Euler-Lagrange equations

d

dt

∂

∂γl

( k∑i,j=1

gij(γ)γiγj

)− ∂

∂γl

( k∑i,j=1

gij(γ)γiγj

)= 0

for 1 ≤ l ≤ k. Carrying out the partial derivatives

2d

dt

k∑j=1

glj(γ)γj −∂

∂γl

( k∑i,j=1

gij(γ)γiγj

)= 0

which is equivalent to

k∑j=1

glj γj +k∑

i,j=1

∂glj∂γi

γiγj −1

2

k∑i,j=1

∂gij∂γl

(γiγj) = 0 (39)

Since G is a positive definite, symmetric matrix, it is nonsingular, and therefore it admits aninverse matrix, G−1. Let G−1 = (gij)ki,j=1, where

k∑j=1

gijgjl = δil =

1 if i = l0 if i 6= l

Multiplying all the terms in (39) by the inverse matrix, we obtain

γl +1

2

k∑i,j,p=1

gpl(2∂glj∂γi

γiγj −∂gij∂γl

γiγj) = 0

Recall that the metric tensor depends of the partial derivatives of f , which is at least C3(Rk,Rn),and as a consequence, the coefficients of the differential equation are at least C1 functions. Thus,we can conclude that the associated initial value problem, has a unique local solution.

96

Definition 20. The solutions of the above system of ODE’s, are called geodesics.

The usual form of the Geodesic equation is

γl +k∑

i,j=1

Γlij(γ(t))γiγj = 0

for 1 ≤ l ≤ k, where

Γlij =1

2

k∑p=1

glp(gpj,i + gip,j − gij,p)

and

gij,l =∂gij∂xl

The terms Γpk,l are denoted the Christoffel Symbols.

Example 35. Compute the geodesics in sphere, i.e., for

M = S2 = x ∈ R3 : |x| = 1

The solutions of the Euler equation are denoted Great Circles. (It takes 2 coordinate patches:Ω1 = S2 \ N and Ω2 = S2 \ S.)

Let’s now summarize a litle bit:

(1) If M is a metric space, for c ∈ C([0, T ],M), we defined

L(c) = supP∈P

L(Pf)

and showed that, if M is complete, connected and compact, there is a curve of shortestlength, joining two arbitrary points of M.

(2) If M is a differentiable, k-dimensional manifold of Rn, for c ∈ AC([0, T ],M), we defined

E(c) =1

2

∫ T

0

|c(t)|2dt

and

L(c) =

∫ T

0

|c(t)| dt

=

p∑ν=1

∫ tν

tν−1

( k∑i,j=1

gνij(γ)γiν γjν

)1/2

dt

where the intervals (tν−1, tν) were chosen so that the image of c|tνtν−1lies entirely in one

coordinate chart, i.e., c(t) = fν(γν(t)) for t ∈ [tν−1, tν ], where fν : Uν → M and Uν ⊂Rk. We showed that the minimizers of E are minimizers of L which are parametrizedproportionally to arclength. Moreover, the C2 minimizers of E are solutions of the Eulerequations, which are denoted by geodesics.

97

Next, we will bring these two notions together. By introducing a metric in M , our manifoldbecomes a metric space, and by our earlier theory, there is a curve of shortest length. We willshow, that the curve is a solution of the geodesic equation.

Definition 21. For x, y ∈M , consider

Γxy = q ∈ AC([0, T ],M) : q(0) = x , q(T ) = y

and define the distanced(x, y) = infL(c) : c ∈ Γxy

Proposition 14. As defined, d(·, ·) is a metric on M , i.e., satisfies

(i) For all x, y ∈M , d(x, y) ≥ 0 and d(x, y) = 0 iff x = y;

(ii) d(x, y) = d(y, x) for all x, y ∈M ;

(iii) d(x, z) ≤ d(x, y) + d(y, z), for all x, y, z ∈M .

Proof:

(i) Since L(c) ≥ 0, for all c ∈ Γxy, it is obvious that d(x, y) ≥ 0, and also, if x = y,d(x, y) = 0. Suppose now that d(x, y) = 0. Then there exists a sequence (cm) ⊂ γxy such that

L(cm) =

∫ T

0

|cm(t)| dt→ 0 as m→∞

Since, for all m ∈ N, cm ∈ AC([0, t],M),

cm(t) = cm(0) +

∫ t

0

|cm(s)| ds , ∀t ∈ [0, T ]

In particular, for t = T

cm(T ) = cm(0) +

∫ T

0

|cm(s)| ds

Also, by definition of Γxy, for all m ∈ N, cm(T ) = y and cm(0) = x. Then

y = x+

∫ T

0

|cm(s)| ds

Letting m→∞, we obtain x = y.

(ii) This is obvious by reparametrization of the curves.

(iii) By definitiond(x, z) = infL(c) : c ∈ Γxz

Consider, for y ∈M

Γxz = γ ∈ Γxz : ∃t0 ∈ [0, T ] such that γ(t0) = y

98

i.e., Γxz is the subclass of Γxz of the curves passing through y. It is obvious that

infL(c) c ∈ Γxz = infL(c1) : c1 ∈ Γxy+ infL(c2) : c2 ∈ Γyz = d(x, y) + d(y, z)

But, Γxz ⊂ Γxz, and therefore

infL(c) c ∈ Γxz ≥ infL(c) : c ∈ Γxz

Hence, we conclude that the triangular inequality is verified by d.

♦

If M is a compact manifold, M ⊂ Rn, gets the induced topology from Rn (the open sets in Mare U ∩M where U is open in Rn. On the other hand once we introduce a metric in M , wedefine another topology in M . It can be proved that these two topologies are equivalent.

For x ∈M let dx = d(x, ∂f(U)) and define rx = 13dx. We can cover M by Brx(x), and since

M is compact we can find x1, ..., xj ∈M such that

M ⊂ Br1(x1) ∪ ...Brj(xj)

Define r0 = miniri.

Theorem 17. Let M be a compact, connected, differentiable k-dimensional manifold of Rn. If,for x, y ∈M , d(x, y) ≤ r0 then there exists a shortest geodesic joining x and y.

Proof:

For x ∈M , there exists xi ∈M such that x ∈ Bri(xi). Then

d(y, xi) ≤ d(y, x) + d(x, xi) ≤ r0 + ri ≤ 2ri ≤2

3dx

and therefore x, y lie in the same coordinate patch - the one associated to xi.

Let us minimize E(c) over Γxy, where

E(c) =1

2

∫ T

0

|c(t)|2dt =1

2

∫ T

0

k∑i,j=1

gij(γ)γiγjdt

and γ : [0, T ] → Uxi is defined by c(t) = f(γ(t)), with (f, U) the local coordinates associatedto xi. Thus, we are minimizing E over the class

Γxy = γ ∈ W 1,2([0, T ],Rk) : γ(0) = f−1(x) , γ(T ) = f−1(y)

Pick a minimizing sequence (γm) ⊂ Γxy. Then

E(γm) =1

2

∫ T

0

k∑i,j=1

gij(γm)γimγjmdt

99

Since the matrix G is positive definite, there is β > 0, such that

E(γm) ≥ 1

2β

∫ T

0

|γm|2dt

and therefore (γm) is a bounded sequence in L2([0, T ],Rk). On the other hand, γm(0) = f−1(x),for all m ∈ N, and therefore, for t ∈ [0, T ]

γm(t) = γm(0) +

∫ t

0

|γm| dt

implying|γm(t)| ≤ |f−1(x)|+ T 1/2‖γm‖L2([0,T ],Rk)

Then, (γm) is bounded in L∞([0, T ],Rk), and as a consequence (γm) is bounded in L2([0, T ],Rk).Finally, (γm) is bounded in W 1,2([0, T ],Rk), and there is γ ∈ W 1,2([0, T ],Rk) such that γm →γ weakly in W 1,2([0, T ],Rk) and strongly in L∞([0, T ],Rk). The strong convergence impliesthat γ(t) ⊂ U and γ(0) = f−1(x), γ(T ) = f−1(y). Let us prove that E is weakly lowersemicontinuous. Suppose γm γ in W 1,2([0, T ],Rk) (which as seen before, implies γm → γ inL∞([0, T ],Rk)). Then

E(γm) =1

2

∫ T

0

k∑i,j=1

gij(γm)γimγjmdt

=1

2

∫ T

0

( k∑i,j=1

gij(γ)γimγjm +

1

2

k∑i,j=1

(gij(γm)− gij(γ))γimγjm

)dt

The strong convergence implies

gij(γm)− gij(γ)→ 0 ; m→∞

Thus ∫ T

0

k∑i,j=1

(gij(γm)− gij(γ))γimγjmdt→ 0 as m→∞

since∫ T

0

∑i, j = 1kγimγ

jmdt is uniformly bounded. Finally

lim inf E(γm) = lim inf1

2

∫ T

0

k∑i,j=1

gij(γ)γimγjmdt

= E(γ) +1

2lim inf

∫ T

0

k∑i,j=1

gij(γ)(γimγjm − γiγj) dt

≥ E(γ) + β lim inf

∫ T

0

|γm − γ|2dt

100

as a consequence of the positive definiteness of G. Finally, the weak convergence implies theresult. If this is the case,

E(γ) = infΓE

We still have to show that the minimizer of the energy (and of the length) satisfies the following:

γ ⊂ Uxi ⇒ c ⊂ f(Uxi)

If that is not the case, there exists t0 such that c(t0) ∈ ∂(f(Uxi)). On one hand

r0 ≥ d(x, y) = L(c) = L(c|t00 ) + L(c|1t0)

Notice, that c|t00 and c|1t0 are also minimizers of the length, over the class of curves joining x toc(t0) and c(t0) to y respectively. Then

r0 ≥ d(x, c(t0)) + d(c(t0), y) ≥ d(x, c(t0) ≥ dx

On the other hand, as a consequence of c(t0) ∈ ∂(f(Uxi)), we can choose z ∈ ∂f(Uxi), suchthat d(x, z) = dx and d(xi, z) ≥ dxi . Then

d(xi, z) ≤ d(x, xi) + d(x, z) = d(x, xi) + dx

implyingdx ≥ d(xi, z)− d(x, xi) ≥ dxi − rxi = 2rxi ≥ 2r0

Then r0 ≥ dx and dx ≥ 2r0 which is obviously contradictory.

♦

We still need to prove the regularity of the minimizer: once we do that it will be a solution ofthe geodesic equation, i.e., c is really a shortest geodesic. Let us assume that for the moment.The previous Theorem was preparatory to the next one.

Theorem 18. Let M be as in Theorem 17. Then, for all x, y ∈ M , there exist a shortestgeodesic joining x to y.

Proof:

As remarked before, we can assume that all the curves are parametrized proportionally toarclength. Consider (cm) ⊂ Γxy a minimizing sequence for L, i.e., L(cm) d(x, y) as m→∞.We can assume that, for all m ∈ N, L(cm) ≤ d(x, y) + 1 ≡ K. For x0, ...xj ∈ M such that∪iBri(xi) ⊃ M , for fixed m, consider Pm = (tm0 , ...., t

mj ) a partition of the interval [0, T ], such

thatcm|

tmitmi−1∈ fi(ui)

where (fi, Ui) are the local coordinates associated to xi, 1 ≤ i ≤ j. Moreover, we can assume

tmi − tmi−1 ≤r0

K

101

Since cm is parametrized proportional to arclength,

d(cm(tmi−1), cm(tmi )) ≤ Kr0

K= r0

We are in the hypothesis of Theorem 17, and conclude that cm(tmi−1) and cm(tmi ) can be joinedby a shortest geodesic. Define cm : [0, T ]→M by the curve obtained by gluing all the shortestgeodesics, i.e., for t ∈ [tmi−1, t

mi ], cm(t) is the shortest geodesic joining cm(tmi−1) to cm(tmi ). By

constructionL(cm) ≤ L(cm)

and thus (cm) ⊂ Γxy is a new minimizing sequence, which we can assume to be reparametrizedproportional to arclength. Notice that for each 1 ≤ i ≤ j, (cm(tmi )) is a bounded sequence, andtherefore, as m → ∞, tmi → ti ∈ [0, T ], possibly along a subsequence. On the other hand, fors, t ∈ [0, T ],

d(cm(t), cm(t)) ≤ L(cm)|s− t| ≤ K|s− t|implying that (cm) is an equicontinuous sequence. On the other hand, for t ∈ [0, T ]

d(x, cm(t)) ≤ k|t| ≤ KT

and as a consequence (cm) is uniformly bounded. By Ascoli-Arzela Theorem, cm → c inC([0, T ],M), as m→∞. The continuity implies

cm(tmi )→ c(ti) , m→∞

We claim that c is the shortest geodesic joining x to y. By construction, c is the shortest geodesicconnecting c(ti−1 to c(ti), for all 1 ≤ i ≤ j. But we could find s ∈ (ti−1, ti), σ ∈ (ti, ti+1) and acurve c|ti+1

ti−1through c(s) and c(σ), such that

L(c|ti+1

ti−1) < L(c|titi−1

) + L(c|ti+1

ti )

If this was the case, we could define a new partition (t0, ..., ti−1, s, σ, ti+1, ..., tj) and a newminimizing sequence (cm) ⊂ Γxy, with

L(cm) < L(cm)

which is a contradiction to the definition of cm.

♦

To finish these proofs, we have to show the regularity of the minimizers. We will do it by aseries of Lemmas.

Lemma 19. (DuBois-Raymond) Let f ∈ L1([a, b],R) such that∫ b

a

f(x)ϕ′(x) dx = 0

for all ϕ ∈ C0([a, b],R), such that ϕ(x) exists a.e. for x ∈ [a, b]. Then, there exist c ∈ R, suchthat f(x) = c almost everywhere for x ∈ [a, b].

102

Proof:

Recall, x is a Lebesgue point for f , if for all family Ωx of measurable sets containing x

1

|Ωx|

∫Ωx

f(y) dy → f(x) as |Ωx| → 0

Moreover, if f ∈ L1([a, b],R), almost all points in [a, b] are Lebesgue points for f . Considery < z Lebesgue points in (a, b), and choose

ϕ(x) =

1 if x ∈ (y, z)0 if x ∈ (a, y − ε) ∪ (z + ε, b)1ε(x− y + ε) if x ∈ (y − ε, y)

1ε(z − x+ ε) if x ∈ (z, z + ε)

Then, ϕ is continuous in [a, b], ϕ(a) = ϕ(b) = 0 and ϕ ∈ C([a, b] \ y, z,R). By hypothesis

0 =

∫ b

a

f(x)ϕ′(x) dx =1

ε

∫ y

y−εf(x) dx− 1

ε

∫ z+ε

z

f(x) dx

Letting ε → 0 and since y, z are Lebesgue points we obtain f(y) = f(z). If we let z changein (y, b], we proved that f(x) = c for all Lebesgue points, and since they are dense in [a, b],f(x) = c a.e. on [a, b].

♦

Lemma 20. Suppose that v ∈ C1([a, b],Rn), F = F (t, u, p) ∈ C1([a, b]× Rn × Rn,R) and∫ b

a

(Fu(t, v, v

′) · ϕ+ Fp(t, v, v′) · ϕ′

)dt = 0

for all ϕ ∈ C10([a, b],Rn) (notice that we are requiring that the Frechet derivative of I is 0).

Then, there exists c ∈ Rn such that

Fp(t, v, v′) = c+

∫ t

a

Fu(s, v(s), v′(s)) ds

for all t ∈ [a, b]

Let us study first the meaning of the Lemma. Since Fu(s, v(s), v′(s)) is continuous, theintegral

∫ taFu(s, v(s), v′(s)) ds is a differentiable function and as a consequence of the equality

Fp(t, v, v′) is a differentiable function (as a composition). Therefore we can differentiate both

sides of the equation and obtain another version of the Euler equation

d

dtF = Fu

Nevertheless we can’t use the chain rule in the left hand side, since we know that the compositeis in C1, but we know nothing about the pieces.

103

Proof:

Integrating by parts∫ b

a

Fu(t, v, v′) · ϕ(t) dt =

(∫ t

a

Fu(s, v, v′) ds

)· ϕ(t)|ba −

∫ b

a

∫ t

a

(Fu(t, v, v

′) ds)· ϕ′(t) dt

Since, by hypothesis, ϕ(a) = ϕ(b) = 0∫ b

a

Fu(t, v, v′) · ϕ(t) dt = −

∫ b

a

(∫ t

a

Fu(s, v, v′) ds

)· ϕ′(t) dt

Then ∫ b

a

(Fp(t, v, v

′)−∫ t

a

Fu(s, v, v′) ds

)· ϕ′(t) dt = 0

for all ϕ ∈ C10([a, b],Rn). By Lemma 19, there exists c ∈ Rn such that

Fp(t, v, v′)−

∫ t

a

Fu(s, v, v′) ds = c

♦

Homework 24. Prove that Lemma 20 is verified a.e. in t, if we replace v ∈ C1([a, b],Rn) byv lipschitz continuous form [a, b] to Rn. (note that |v′| ≤ k).

Consider F = F (t, u, p) ∈ C2([a, b]× Rn × Rn,R) verifying the conditions

(F1) there are positive constants, α and β, such that

α|p|2 ≤ F (t, u, p) ≤ β|p|2

for all (t, u, p) ∈ Ω ≡ [a, b]× Rn × Rn;

(F2) there exists M(R) > 0, such that for (t, u, p) ∈ Ω and |u| < R

|Fu(t, u, p)|+ |Fp(t, u, p)| ≤M(R)(1 + |p|2)

i.e., we assume a growth condition on the derivatives;

(F3) for all (t, u, p) ∈ Ω and ξ ∈ Rn \ 0∑Fpipj(t, u, p)ξiξj > 0

which is a convexity assumption.

104

You can easily check, that

F (t, u, p) =k∑

i,j=1

gi,j pipj

verifies conditions (F1)-(F3). Suppose further, that v is a local minimum of

I(u) =

∫ b

a

F (t, u, u′) dt

in W 1,2([a, b],Rn) satisfying some prescribed boundary conditions.

Proposition 15. If F verifies (F1)-(F3) and v ∈ C1([a, b],Rn), then v ∈ C2([a, b],Rn).

Proof:

Since v is a local minimum of I, I ′(v) = 0, and Lemma 20 implies

Fp(t, v(t), v′(t)) = c+

∫ t

a

Fu(s, v(s), v′(s)) ds ≡ f(t) (40)

By hypothesis, Fu(s, v(s), v′(s)) is continuous, and the Fundamental Theorem of calculus allowus to conclude that Fp(t, v(t), v′(t)) is continuously differentiable for t ∈ (a, b). Define

A(t, p) = Fp(t, v, p)− f(t)

As remarked, A ∈ C1([a, b]× Rn,Rn) and by (40)

A(t, v′) = 0 , ∀t ∈ [a, b]

On the other handAp(t, v

′) = Fpp(t, v, v′)

and by hypothesis (F3), the matrix Fpp(t, v, v′) is nonsingular. The Implicit Function Theorem

implies that for each t0 ∈ [a, b], there exists ξ0 ∈ C1(Nt0 ,R) such that, the solutions of theequation A(t, p) = 0 near (t0, v

′(t0)) are given by (t, ξ0(t)) for t near t0. Since the zeros of Aare precisely (t, v′(t)), we conclude ξ0(t) = v′(t) for t near t0. The differentiability of ξ0 impliesv′ ∈ C2([a, b],Rn) as wanted.

♦

Theorem 19. If F ∈ C2([a, b]× Rn × Rn,R) verifies (F1)-(F3), and v is a local minimum ofI, then v ∈ C2([a, b],Rn) and satisfies the associated Euler equation. Moreover F ∈ Ck impliesv ∈ Ck.

Proof:

If v is a local minimum of I, there is a real constant M such that I(v) ≤M . Using (F1)

M ≥ α

∫ b

a

|v′(t)|2dt

105

implying that v′ ∈ L2([a, b],Rn). As we have seen before in previous examples, if v verifies someboundary conditions and v′ ∈ L2([a, b],Rn), then v ∈ W 1,2([a, b],Rn). By the Sobolev Theo-rem, we conclude v ∈ L∞([a, b],Rn). Using hypothesis (F2) and the Dominated ConvergenceTheorem, we conclude that both Fu(t, v, v

′) and Fp(t, v, v′) are bounded in L1([a, b],Rn). Since

I ′(v) = 0, v is a solution of the weak Euler equation∫ b

a

(Fu(t, v, v′)ϕ+ Fp(t, v, v

′)ϕ′) dt = 0 , ∀ϕ ∈ C10([a, b],Rn)

and by Lemma 20, there exists a real number, c, for which

Fp(t, v, v′) = c+

∫ t

a

Fu(s, v, v′) ds (41)

We claim that the map Fp(·, ·, p) : Rn → Rn, is a global diffeomorphism. For fixed p0, (F3)implies that Fpp(·, ·, p0) is a nonsingular matrix, and as a consequence Fp(·, ·, p) is a localdiffeomorphism. All we have to show is that it is globally 1-1 and onto. If it is not globally 1-1,there are p, p1 ∈ Rn such that

Fp(·, ·, p) = Fp(·, ·, p1)

Then

0 = (Fp(·, ·, p)− Fp(·, ·, p1)) · (p− p1)

=

∫ 1

0

d

dsFp(·, ·, sp+ (1− s)p1) · (p− p1) ds

=

∫ 1

0

Fpp(·, ·, sp+ (1− s)p1)(p− p1) · (p− p1) ds

Using hypothesis (F3), we obtain the absurd 0 > 0. Thus Fp(·, ·, p) is globally 1-1 (notice thatthis is a direct consequence of the convexity assumption). To prove that is onto, for w ∈ Rn, wewant to show that there exists z ∈ Rn (which will be unique as a consequence of the injectivity),such that

Fp(·, ·, z) = w

We will prove this fact, minimizing F (·, ·, x) − w · x over Rn. Hypothesis (F1) implies thatF (t, u, p) grows at least quadratically in p, and therefore F (·, ·, x) − wx is bounded by below.As a consequence, it admits a minimizer which we will denote by z. On the other handF (·, ·, x)− wx is a continuously differentiable function, implying

(F (·, ·, x)− wx)′|x=z = 0

ThenFp(·, ·, z) = w

as wanted, and we proved that Fp(·, ·, p) is a global diffeomorphism. If this is the case, equation(41) becomes

w(t) = c+

∫ t

a

Fu(t, v, v′) dt

106

for a unique z such thatFp(t, v(t), z(t)) = w(t)

As we defined it, w ∈ C([a, b],Rn). We will show that also z ∈ C([a, b],Rn). Let (tm) ⊂ [a, b]such that tm → t, as m→∞. By (F1)

α|z(tm)|2 − wz(tm) ≤ F (tm, v(tm), z(tm))− wz(tm)

and since z(tm) is a minimizer of F (·, ·, p)− wp

F (tm, v(tm), z(tm))− wz(tm) ≤ F (tm, v(tm), 0) ≤ c

Thenα|z(tm)|2 − wz(tm) ≤ c

implying that (z(tm)) is bounded in Rn, and therefore it will converge (possibly along a subse-quence), for some ζ ∈ Rn. On the other hand,

Fp(tm, v(tm), z(tm)) = w(tm)

and letting m→∞Fp(t, v(t), ζ) = w(t)

and, by the definition of w, ζ is in fact the minimizer of F (t, v(t), x) − w(t)x, implying thatζ = z(t), and therefore z is continuous. Moreover, v′(t) = z(t) a.e. for t ∈ [a, b]. Finally, sincev ∈ W 1,2([a, b],Rn), for all t ∈ [a, b]

v(t) = v(a) +

∫ t

a

|v′(s)|ds = v(a) +

∫ t

a

z(s) ds

The Fundamental theorem of calculus and the continuity of z, imply v ∈ C1([a, b],Rn), andProposition 15, implies v ∈ C2([a, b],Rn).

♦

6.2 Geodesics in the Torus

Consider the n torus Πn = Rn\Zn. We are intersted in finding closed geodesics in the torus,i.e., curves γ ∈ C([a, b],Πn), such that, for some a, b ∈ Rn, γ(a) is identified with γ(b). InRn, they correspond to curves γ ∈ C(R,Rn), verifying γj = γ + j for some j ∈ Zn. For eachk ∈ Zn \ 0 define a homotopy class of the curves winding ki times in the xi direction, andsuppose that the metric tensor gij(x) is smooth, and satisfies gij(x+ el) = gij(x) for all l ∈ Zn(is periodic in the coordinate directions). We can use g to define a metric tensor

L(γ) =

∫ b

a

√∑gij(γ)γiγjdt

107

for γ ∈ AC([a, b],Πn) or γ ∈ W 1,2([a, b],Πn). As a consequence

d(x, y) = infL(γ) : γ ∈ W 1,2([a, b],Rn) , γ(a) = x , γ(b) = y

induces a metric in Πn or in Rn, and we can extend this notion to the length of a continuousfunction. If γ ∈ C([0, 1],Rn) then

L(γ) = supP

∑d(γ(ti−1), γ(ti))

for any partition P of the interval [0, 1]. The next Theorem states the existence of a curve ofhomotopy type k of shortest length.

Theorem 20. For each k ∈ Zn \ 0, define Γk as the set of absolutely continuous curves ofhomotopy type k, i.e.,

Γk = γ ∈ W 1,2(R,Rn) : ∃T = T (γ) for which γ(t+ T ) = γ(t) + k

(after time T they differ by k). Then there exists γ∗k ∈ Γk such that

ck = infγ∈Γk

L(γ) = L(γ∗k)

Proof:

We start by noticing that if γ ∈ Γk, then τθγ(t) = γ(t − θ) ∈ Γk for all θ ∈ Zn. As aconsequence, we have the freedom to choose γ(0). We will assume γ(0) ∈ [0, 1]n, and considerthat all the curves are parametrized with respect to arclength. Let (γm) ⊂ Γk be a minimizingsequence. Then, for some positive M , L(γm) ≤ M , for all m ∈ N, and by Lemma 12, for alls, t ∈ R

d(γm(t), γm(s)) ≤ L(γm)|s− t| ≤M |t− s|We can conclude that (γm) is an equicontinuous sequence with L(γm) uniformly bounded. ByAscoli-Arzela Theorem, there exists γ∗k ∈ W

1,2(R,Rn) such that γm → γ∗k uniformly, possiblyalong a subsequence. Since L is lower semicontinuous

L(γ∗k) ≤ lim inf L(γm) = ck (42)

On the other hand, by definition of Γk,

γm(t+ T (γm)) = γm(t) + k

Taking the limit as m→∞γ∗k(t+ T (γ)) = γ∗k(t) + k

and we conclude that γ∗k ∈ Γk. (42) implies L(γ∗k) = ck.

♦

Since we proved that any minimizer of L is C2([a, b],Rn), for all a, b ∈ R, γ∗k ∈ C2(R,Rn), andconsequently is a solution of the geodesic equation.

108

Definition 22. Two geodesics on Πn, parametrized by arclength, will be called geometricallydistinct if they do not differ by a phase shift.

If γ∗k ∈ Γk is a minimizer of L,

ck = L(γ∗k) = L(γ∗k + j)

for any j ∈ Zn, which means, that in Rn, there is an infinite number of representatives,geometrically distinct, for the minimizer of L over Γk. Consider the case n = 2, and define

Mk = γ ∈ Γk : L(γ) = ck

Definition 23. A geodesic, γ, is said to be minimal if for all a < b ∈ R, γ|ba minimizes L(·)over the class of curves in Rn joining a to b, that is

L(γ|ba) = d(γ(a), γ(b))

Theorem 21. (Morse’s Theorem)

If γ∗k ∈Mk, then γ∗k is a minimal geodesic.

To prove Theorem 21, we need a preliminary result.

Proposition 16. For l ∈ Nclk = lck

Proof:

Choose some homotopy type k = (k1, k2), and γ∗k a geodesic for k. It is obvious, that γ∗krunned l times, is an element of Γlk, and as a consequence

clk = infγ∈Γlk

L(γ) ≤ lck = L(γ∗k runned l times) (43)

We will prove, by induction on l, that clk = lck. The case l = 1 is trivially true. For l = 2, (43)implies

2ck =

∫ 2T (γ∗k)

0

√g(γ∗k, γ

∗k)dt ≥ c2k

We will prove, that alsoc2k ≥ 2ck

Let γ∗2k be the minimizer of L over Γ2k, and assume that we can find a pair of numbers τ < s,τ, s ∈ [0, T (γ∗2k)], such that

γ∗2k(τ) + k = γ∗2k(s) (44)

By definition, γ∗2k|sτ ∈ Γk, and as a consequence

L(γ∗2k|sτ ) ≥ ck

109

If we consider the curve γ obtained by gluing γ∗2k|sτ to γ∗2k|T (γ2k+σs , it is easy to verify that

γ(T (γ∗2k) + τ) = γ(τ) + 2k = γ(s) + k

and thereforeγ∗2k|

T (γ∗2k+σs ∈ Γk

implying

L(γ∗2k|T (γ∗2k)+σs ) ≥ ck

Thenc2k = L(γ2k = L(γ∗2k|sτ ) + L(γ∗2k|

T (γ∗2k)+σs ) ≥ 2ck

as wanted. This proves the case l = 2, assuming we can find τ , s verifying property 44. Denoteγ by γ∗2k for simplicity in notation. γ lies between two lines of slope k2

k1, that we will denote

by L and L, respectively. Shift the lines until tangent to γ. Without loss of generality assumeγ(0) ∈ L. By construction there exists

σ = mint ∈ (0, T (γ)) : γ(t) ∈ L

If γ(0) + k ∈ γ then we are done. Otherwise γ(0) + k 6∈ γ, but notice that γ(t) + k lies in theregion bounded by γ(t) and L for t near 0, and is in the region bounded by γ(t) and L for tnear τ . By continuity it must intersect γ for some θ ∈ [0.τ ], which proves the result for l = 2.

For the general case, if the result is true for l ∈ 1, ...,m− 1 we want to show that

cmk = mck

Once again it is obvious that mck ≥ cmk, since γ∗k runned m times belongs to Γmk. To obtainthe reverse inequality, it suffices to show the existence of 0 < τ < s ≤ T (γ∗mk) such that

γ∗mk(τ) + k = γ∗mk(s)

If this is the case, γ∗mk|sτ ∈ Γk (since the end points are k), and also

γ∗mk|T (γ∗mk)+τs ∈ Γ(m−1)k

By induction

cmk = L(γ∗mk) = L(γ∗|sτ ) + L(γ∗mk|T (γ∗mk)+τs ) ≥ ck + (m− 1)ck = mck

To find τ and s we argue as for the case l = 2.

♦

We can now prove Morse’s Theorem (Theorem 16)

Proof:

110

Take any pair of points a, b ∈ R, a < b. We want to show that

d(γ∗k(a), γ∗k(b)) = L(γ∗k|ba)

where γ∗k minimizes L over Γk. If not, by definition, there exists γ : [a, b]→ R2, such that

d(γ∗k(a), γ∗k(b)) = L(γ|ba) < L(γ∗k|ba)

and γ(a) = γ∗k(a), γ(b) = γ∗k(b). We will consider two distinct cases.

(i) Assume that, for some l ∈ N

γ∗k(b) = γ∗k(a) + lk

ThenL(γ∗k|ba) = lck

and by Proposition 16,L(γ∗k|ba) = clk

which implies that γ∗k|ba minimizes L over Γlk. Since by construction γ(a) = γ∗k(a) and γ(b) =γ∗k(b), also γ ∈ Γlk. Thus

clk ≤ L(γ|ba) < L(γ∗k|ba) = clk

is an absurd.

(ii) Assume that, for all l ∈ N

γ∗k(b) 6= γ∗k(a) + lk

If this is the case, extend γ∗k to the interval [b, t], such that

γ∗k(t) = γ∗k(a) + lk

for some l ∈ N. Consider the curve γ : [a, t]→ R2, defined by gluing γ|ba to γ∗k|tb. By constructionγ ∈ Γlk, and

L(γ) = L(γ|ba) + L(γ∗k|tb) ≥ clk

On the other handL(γ) < L(γ∗k|ta) = lck = clk

These two inequalities imply that clk < clk, which is obviously a contradiction.

♦

Definition 24. A geodesic γ : R → Rn, which is asymptotic to geodesics γ± as t → ±∞ iscalled a heteroclinic geodesic from γ− to γ+.

Theorem 22. (Morse-Hedlund)

For any k ∈ Z2 \ 0, let v+, v− be adjacent minimizers of L in Mk. Then there exists aminimal geodesic heteroclinic from v+ to v− and a minimal geodesic heteroclinic from v− to v+.

111

We will prove Theorem 22, using a variational argument. Without loss of generlity, assumeT (v±) = 1. For k fixed, pick another homotopy type l 6= k. Then a curve of homotopytype l, which we will denote by z0 (and whose existence is guaranteed by Theorem 20, will betransversal to v±. We will assume, z0(0) ∈ v+(R) and z0(T (z0)) = v−(0). Moreover, for i ∈ Z,consider

zi = z0 + ik

i.e., the translates of z0. Define Ri as the region of R2, with boundaries v+, zi+1, v− and zi andconsider

Πi = γi ∈ W 1,2([0, 1],R2) : γi(0) ∈ zi , γi(1) ∈ zi+1 and γi([0, 1]) ⊂ Ri

Furthermore, assume that, for all i ∈ Z, γi is parametrized proportionally to arclength. Let

Π = γ = (γi)i∈Z : γ verifies (Π1)− (Π3)

where

(Π1) γi ∈ Πi for all i ∈ Z;

(Π2) γi+1(0) = γi(1) for all i ∈ Z ;

(Π3) For zi(ti(γi)) = γi(0) and zi+1(ti+1(γi)) = γi(1) we require that

ti+1(γ) ≤ ti(γ) , ∀i ∈ Z

Notice that Πi is endowed with the maximum topology and Π is endowed with the producttopology. Define the length functional over this class (which as before is a renormalized func-tional), as

J(γ) =∑i∈Z

(L(γi)− ck)

where ck = infΓkL = L(v±).

Lemma 21. 1) There exists γ ∈ Π such that J(γ) <∞.

2) J is bounded by below on Π.

3) If γ ∈ Π is such that J(γ) ≤M then∑i∈Z

|L(γi)− ck| ≤M + 2cl

where cl = infΓlL = L(z0).

4) J is weakly lower semicontinuous.

112

Proof:

1) Consider γ as the curve obtained by gluing v−|0−∞ to z0|T0 to v+|∞1 . It is obvious thatγ ∈ Π, and

J(γ) =

∫ T

0

(L(z0)− ck) dt = cl − ck

2) Let αi = γi(0), αi+1 = γi+1(1) and consider βi+1 = α1 + k (the translate by k of αi). Then

γi + zi+1|βi+1αi+1 is an element of Γk, and as a consequence

L(γi) + L(zi+1|βi+1αi+1

) ≥ ck

implyingL(γi)− ck ≥ −L(zi+1|αi+1

βi+1)

Summing on i ∈ Z, we obtain

J(γ) ≥ −∑i

L(zi+1|αi+1

βi+1) ≥ −L(z0) = −cl (45)

3) ConsiderN− = i ∈ Z : L(γi)− ck ≤ 0

For i ∈ N−, using (45)

−∑i∈N−

(L(γi)− ck) ≤ cl

For n ∈ N fixed

n∑i=−n

(L(γi)− ck) =∑

i∈N−,|i|≤n

(L(γi)− ck) +∑

i∈N+,|i|≤n

(L(γi)− ck)

where N+ = Z \N−. Hence

∑i∈N+,|i|≤n

(L(γi)− ck) =n∑

i=−n

(L(γi)− ck)−∑

i∈N−,|i|≤n

(L(γi)− ck)

≤ J(γ)−∑|i|>n

(L(γi)− ck) + cl

Since, by hypothesis, J(γ) < M , for ε > 0 arbitrary, we can choose n sufficiently large, so that

−∑|i|>n

(L(γi)− ck) ≤ ε

and as a consequence ∑i∈N+,|i|≤n

(L(γi)− ck) ≤M + ε+ cl

113

Finally

n∑i=−n

|L(γi)− ck| =∑

i∈N+,|i|≤n

(L(γi)− ck)−∑

i∈N−,|i|≤n

(L(γi)− ck) ≤M + ε+ 2cl

and letting ε→ 0n∑

i=−n

|L(γi)− ck| ≤M + 2cl

We conclude ∑i∈Z

|L(γi)− ck| = limn→∞

n∑i=−n

|L(γi)− ck| ≤M + 2cl

as wanted.

4) Let γm γ in Π, as m→∞. For ε > 0 arbitrary, there exists j0 = j0(ε) ∈ N, such that

j∑i=−j

(L(γi)− ck) ≥ J(γ)− ε

for all j ≥ j0. By the product topology, for all i ∈ Z, γmi → γi in L∞([0, 1],R2), as m → ∞.Then

lim infm→∞

j∑−j

(L(γmi )− ck) ≥j∑−j

(L(γi)− ck) ≥ J(γ)− ε

since L is weakly lower semicontinuous. Let j →∞, we obtain

lim infm→∞

J(γm) ≥ J(γ)− ε , ∀ε > 0

implyinglim infm→∞

J(γm) ≥ J(γ)

as wanted.♦

Note that, by property (3), if J(γ) is convergent, then the series is absolutely convergent.

Proposition 17. If γ ∈ Π and J(γ) <∞, then

‖γj − v+‖L∞[j,j+1] → 0 as j →∞;

‖γj − v−‖L∞[j,j+1] → 0 as j → −∞.

Proof:

114

For j > 0, we can consider γj ∈ C([0, 1]) (just translate them). Since J(γ) <∞, L(γj)→ cas j → ∞. By construction, for all j ∈ Z, γj ∈ Rj, all the functions are uniformly bounded,and they are all parametrized proportionally to arclength. Hence

L(γj|ts) ≤ L(γj)|t− s|

and as a consequence (γj) are uniformly bounde and equicontinuous. Ascoli-Arzella Theoremimply the existence of γ such that γj → γ as j →∞. Since L is weakly lower semicontinuous

L(γ) ≤ lim inf L(γj) = c

Moreover γ ∈Mk (since also γ(1)− γ(0) = k). Then

L(γ) = c

which implies that γ is a minimizer of L in R0. By construction of R0, γ can only be one of v+

or v− (they are adjacent curves in Mk). By construction of the sequence sj it must necessarilybe v+. By the uniqueness of limit all the sequence must converge.

♦

Proof:

Let (γm) be a minimizing sequence. Then there exist M such that

m ≥ J(γm) c+

By (3) of Lemma 0 ∑|L(γmj )− c| ≤M + 2L(z)

implying that|L(γmj )− c| ≤M + 2L(z)

for all m, j. For fixed jL(γmj ) ≤ c+M + 2L(z) ∀m

This is a uniform bound for L(γmj ) in m, which gives us an upper bound in the Lipschitzconstant. Furthermore the fact that, by construction γmj ∈ Rj gives us an L∞ bound. Thenwe can use Ascolli-Arzella, for each j there exists γ∗j ∈ AC([0, 1]) such that γmj → γ∗j . We wantto show that γ∗j ∈ Πj. By continuity, it is obvious that γ∗j (0) ∈ Zj and γ∗j (1) ∈ zj+1. If also γ∗jis parametrized proportional to arclength then γ∗j ∈ Πj. For that purpose, it suffices to showthat

L((γ∗j |t0) = tL(γ∗j )

for all t ∈ [0, 1]. Since γmj → γ∗j in L∞, we will have L(γ∗j ) ≤ lim inf L(γmj ). We claim thatequality is verified. If not, then

L(γ∗j ) < lim inf L(γmj )

115

Since we already know that

γ∗j (0) = lim γmj (0) and γ∗j (1) = lim γmj (1)

we can choose m large enough so that

L(γ∗j ) + ε < lim inf L(γmj )

Then we can replace γmj by γmj (0) glued to γ∗j (0) to γmj (1) to γ∗j (1). This is a new minimizingsequence, denoted by γmj , and by construction

L(γmj ) < L(γmj )

which is a contradiction. The same argument holds for [0, 1] replaced by [0, t]. Hence

L(γ∗j ) = lim inf L(γmj )

ButL(γmj |t0) = tL(γmj )

Letting m → ∞, we obtain L(γ∗j |t0) = tL(γ∗j ) which implies, as remarked earlier that γ∗j ∈ Πj.The monotonocity properties are consequence of the L∞ convergence. Then, by definition ofthe product topology

γm → γ∗

in Π. Hence γ∗ ∈ Π andJ(γ∗) ≤ lim inf J(γm) = c∗

As a consequence J(γ∗) = c∗. Next we have to show that γ∗ is a solution of (HS). We will startby showing that γj is a solution of the geodesic equation, for all j.

Remark:

Recall the geodesic equation

γi +∑i

(...)quadratic term inγi = 0

If γ(t) is a solution of the equation, by the form of the equation γ(αt) is also a solution for allnon zero α. This is the reason that we can always parametrize proportional to arclength.

To proceed the curve, look at all curves joining the endpoints. we proved earlier that thereexists a minimizer for this problem, which is C2 and must lie in Ri. We claim that γ∗i is such aminimizer. It is obvious that it must be, otherwise we could replace γ∗ by a curve with smallerlength which contradicts our previous arguments. Then γ∗i is C2 and therefore a solution ofthe geodesic equation. It remains to prove that there is no problem in the boundary of theRi. Let’s reparametrize γ∗ by arclength, taking γ∗(0) ∈ z0 (we do change of variables so that[0, 1]→ [0, L(γ∗)), [1, 2]→ [L(γ∗), 2L(γ∗)] and so on). We will show that γ∗ is C2 through 0.

picture

116

Use the same arguments as before, but we have to worry that γ∗(0) intersects z0 in the rightway. If not then we can find points k apart and use this to shorten the minimizer.

Our next claim is that γ∗ is a minimal geodesic.

Considerb = L(z) = inf

Ml

L

and choose a point p lying between v+ and v− such that

c = infMk,p∈γ

L(γ) > c+ 2b (46)

When does this condition hold?

picture

Take p0 = p and consider the translates pi by k. We claim

p0 6∈ z0 ∪ z1

Suppose this was not the case. Then p0 ∈ z0 implying that p1 ∈ z1. Then, by definitiond(p0, p1) = c. Consider the curve z0|0p0 glued to v+|T0 glued to z1|p10 . This also a curve joiningp0 to p1 and therefore

c ≤ 2α + c (47)

On the other hand, consider the curve z0|0p0 glued to v−|T0 glued to z1|p10 which is such curve,which yields

c ≤ 2β + c (48)

Summing (47) and (48), we obtainc < b+ c

contradicting assumptiom (46) In a similar way, we can prove the existence of functions h+, h−

homoclinic to v+ and v− respectively.

Picture

To prove the existence of h−, for example, we will minimize the renormalized functional Jover an appropriate class of functions. Define

Π−i = γi ∈ AC([0, 1]) : γi(0) ∈ zi , γi(1) ∈ zi+1

where we consider γi parametrized with respect to arclength. Then

Π− = (γi)i∈Z : (Π1)− (Π4) are verified

for

Π1 γi ∈ Π−i for all i ∈ Z;

Π2 γi+1(0) = γi(1);

117

Π3 γi lies below Pi for all i 6= 0 and γ0 lies above P0;

Π4 γi gets closer to v− as i→ ±∞.

Picture

Except from the region P−1 to P1 the γ′is lie below ui and in this region lies above u−1, u0.We consider the renormalized functional

J(γ) =∑l∈Z

(L(γi)− c)

As in our previous example, we can obtain a lower bound for J in Π−, i.e., for all γ ∈ Π−

J(γ) ≥ −2L(z)

Since the curve obtained by gluing v− up to z0, to z0 up to v+, to v+ up to z1 to v+ is anelement of Π−, and the value of J at this curve is 2b, we can consider

infγ∈Π−

J(γ) ≤ 2b

Similarly to Proposition —, we can prove the result which will give us the right asymptotiocbehavior of the curves in Π−.

Proposition 18. If γ ∈ Π− is such that J(γ) <∞, then

‖γi − v−‖L∞([i,i+1]) → 0 as i→ ±∞.

Then we can prove the result stating the existence of such curves, as being the minimizersof J over Π−.

Theorem 23. There exists h−

Pi− such that J(h−) = infΠ−

J , and by construction h− is a homoclinic geodesic.

The proof of this theorem follows closely the proof of Theorem —. What can go wrong?

Picture

It may happen that the limit of a minimizing sequence hits P0. We will prove that is thisis the case we have a contradiction.

Suppose that this is the case. Since J(h−) <∞, Proposition 18 gives the right asymptoticbehavior. Consider εn as the distance between the intersections of h− and v−.

picture

As remarked above εn → 0 as n→ ±∞. Look at the length of the curve h−|tnt−n , where t±ndenote the time of intersection with z±n respectively. we can find two points in the curve whichare k apart. Cut out that piece, and shift by k. Then

L(h−|tnt−n) ≥ L(new curve) + c

118

Repeat this procedure 2n times,

L(h−|tnt−n) ≥ L(last curve) + 2nc ≥ c+ 2nc

ThenJ(h−|tnt−n) ≥ c− c

and thereforeJ(h−) ≥ c− c+ small error

By assumption ...J(h−) ≥ 2b+ small error

But we proved thatinfΠ−

J ≤ 2b

and therefore we got the contradiction, i.e., h− can’t run through P0.

We can generalize this type of solutions. We will associate to each type, a symbolic sequencewhich gives us the homotopy type of the geodesic.

LetΣ = σ = (σi)i∈Z : σi ∈ +,−

Corresponding to each σ ∈ Σ there is a function hσi + ik (translates of h±. From the strip, wewill get the following information

to a + corresponds a piece of the curve lying between h+ and v−,

to a − corresponds a piece of the curve lying between h− and v−.

Finally we define Dσ as the remaining region (note that there are no P − i′s in this region).

Theorem 24. If c > c+ 2b, then, for all σ ∈ Σ there exists a geodesic of homotopy type σ onDσ.

Proof:

Let ρ, r ∈ +,−, and define the subclass of Σ

Σρ,r = σ ∈ Σ : σi = ρ ∀i ≤ −L and σi = r ∀i ≥ L

where L is a large positive integer. Then to each σ ∈ Σρ,r it corresponds a geodesic in Dσ,which is asymptotic to vρ as t→ −∞ and is asyptotic to vr as t→∞. We will start to sketchthe proof for this special case and mimics the previous proof. We define

Πσi = γi ∈ AC([0, 1]) : γi(0) ∈ zi , γi(1) ∈ zi+1 and γi lies in Dσ ∩Ri

119

and Πσ as the class of functions obtained by gluing together the γi ∈ Πσi so that the curves

match at the end points, the monotone intersections (in the regions where ρ, r start to repeat).We minimize

J(γ) =∑i∈Z

(L(γi)− c)

over Πσ and obtain the minimizers using the same arguments. Notice, however that the proofis a litle simpler: there is no danger in running through pi since the minimizer can’t intersectthe boundary of Dσ. It remains to prove the case σ ∈ Σ \ Σρ,r. We can’t use the previousarguments since J(γ) may not be bounded. We will use an approximation argument instead.For any σ ∈ Σ \ Σρ,r and for j ∈ N consider σj = (σji )i∈Z, defined by

σji =

σi if |i| < jσj if i ≥ jσ−j if i ≤ −j

It is easy to check that σj ∈ Σρ,r and therefore, for each j ∈ N, there is a minimal geodesic(Gj)j∈N of homotopy type σj. We claim that we can find bounds for Gj in C2

loc. If so, byAscolli-Arzela we can find a convergent subsequence in C1, which will be a solution of thegeodesic equation. This will be a minimal geodesic of homotopy type σ. It remains to provethe claim:

- Gj is bounded in L∞loc for all j, since by construction they all are in the region Dσ;

- suppose we can find bound for Gj in L∞loc,

- By the geodesic equation

Gj = stuff depending onGj + quadratic onGj

Then we will get L∞loc on Gj, which proves that (Gj) is bounded in C2loc. It remains to show

that Gj is bounded in L∞loc. Let xj−n and xjn the intersections of Gj with z−n and zn respectively.Then

L(Gj|xjn

xj−n) =

∫ ∑aik(Gj)GijGjk dt

and since the matrix (aij) is positive definite

L(Gj|xjn

xj−n) ≥ β

∫|Gij|2dt

To get an upper bound for L(Gj|xjn

xj−n), compute the length of some curve with the same end

points, which will be greater since Gj is a geodesic. Then we have upper bounds for ‖Gj‖L2 .The L∞ bounds on Gj plus the L2 bounds on Gj gives us (via the differential equation) L1

bounds on Gj. Considering τj, θj defined by Gj(τj) = xj−n and Gj(θj) = xjn

Gj(θj)− Gj(τj) =

∫ θ

τ

|Gj dt

120

implying

|Gj(θj)| ≤ |Gj(τj)|+∫ θ

τ

|Gj dt

Integrate with respect to τj and divide by the length interval

|Gj(θj)| ≤1

θj − τj‖Gj‖L1 +

∫ θ

τ

|Gj dt

and via the Schwarz inequality we obtain the desired bound, if the difference in the times issufficiently large. To finish up, let’s obtain the lower bound on the time difference.

|γ(t−n)− γ(tn)| ≥ (2n+ 1) min dist(neighboring( zi)

since in this time interval it winds (2n+ 1) times around the torus. On the other hand∫|γ|dt ≤ |tn − t−n|1/2(

∫|γ|2dt)1/2

Putting these inequalities together

1

|t−n − tn|≤ k

∫|γ|2dt ≤ c(n)

We proved the existence of some sort of chaotic dynamics - highly sensitivity to initial condi-tions.

Finally we remark that in a similar form we can prove the existence of periodic solutions,by requiring that σi + l = σi for all i and some l.

121