Lecture 3: Lie symmetry methods for PDEs and PEs - Symmetry Methods for Differential ... ·...

Lecture 3: Lie symmetry methods for PDEs and P∆Es

Lecture 3: Lie symmetry methods for PDEs andP∆Es

Symmetry Methods for Differential and Difference Equations

Peter Hydon

University of Kent


Outline

1 Lie symmetries of scalar PDEs

2 Lie symmetries of scalar P∆Es

3 Invariant solutions

4 Linearization using Lie point symmetries

5 Summary: the main results in Lecture 3


Lie symmetries of scalar PDEs


Point symmetries of PDEs are defined in much the same way asthose of ODEs. For simplicity, let us start by considering PDEswith one dependent variable, u, and two independent variables, xand t. A point transformation is a locally-defined diffeomorphism

Γ : (x , t, u) 7→(x(x , t, u), t(x , t, u), u(x , t, u)

).

This transformation maps any surface u = f (x , t) to the followingsurface (which is parameterized by x and t):

x = x(x , t, f (x , t)

),

t = t(x , t, f (x , t)

),

u = u(x , t, f (x , t)

).



Lie point transformations can be written in the form:

x = x + εξ(x , t, u) + O(ε2),

t = t + ετ(x , t, u) + O(ε2),

u = u + εη(x , t, u) + O(ε2).

For such transformations, the surface u = f (x , t) is mapped to

u = f (x , t) + εη(x , t, f (x , t)

)+ O(ε2),

x = x − εξ(x , t, f (x , t)

)+ O(ε2),

t = t − ετ(x , t, f (x , t)

)+ O(ε2),

which amounts to

u = f (x , t) + εQ(x , t, f (x , t)

)+ O(ε2),

whereQ = η(x , t, u)− u,x ξ(x , t, u)− u,t τ(x , t, u)

is the characteristic. Any surface on which Q = 0 is invariant.



Now consider the effect of a point transformation on a given scalarPDE,

A(x , t, [u]) = 0; (1)

for differential equations, square brackets around an expressiondenote the expression and its derviatives.

The transformation is prolonged to derivatives of u via the chainrule, by differentiating x , t and u with respect the parameters xand t. To do this, use the total derivative operators

Dx = ∂x + ux∂u + uxx∂ux + uxt∂ut + · · · ,Dt = ∂t + ut∂u + uxt∂ux + utt∂ut + · · · .

(Total derivatives treat the dependent variable and its derivativesas functions of the independent variables.)



The transformation is a symmetry if it maps the set of solutions toitself. This gives the symmetry condition,

A(x , t, [u]) = 0 when (1) holds. (2)

For Lie point symmetries, the prolonged infinitesimal generator is

X = ξ ∂x + τ ∂t +η ∂u +ηx ∂u,x +ηt ∂u,t +ηxx ∂u,xx +ηxt ∂u,xt + · · · ,

where ξ, τ and η stand for ξ(x , t, u), τ(x , t, u) and η(x , t, u)respectively, and where the prolongation terms are

ηx = DxQ + ξuxx + τuxt , ηt = DtQ + ξuxt + τutt ,

ηxx = D2xQ + ξuxxx + τuxxt , ηxt = DxDtQ + ξuxxt + τuxtt ,

and so on.



Explicitly,

ηx = ηx + (ηu − ξx)ux − τxut − ξuu2x − τuuxut ,

ηt = ηt − ξtux + (ηu − τt)ut − ξuuxut − τuu2t ,

ηxx = ηxx + (2ηxu − ξxx)ux − τxxut + (ηuu − 2ξxu)u2x

− 2τxuuxut − ξuuu3x − τuuu2

xut + (ηu − 2ξx)uxx

− 2τxuxt − 3ξuuxuxx − τuutuxx − 2τuuxuxt ,

ηxt = ηxt + (ηtu − ξxt)ux + (ηxu − τxt)ut − ξtuu2x

+ (ηuu − ξxu − τtu)uxut − τxuu2t − ξuuu2

xut − τuuuxu2t

− ξtuxx − ξuutuxx + (ηu − ξx − τt)uxt − 2ξuuxuxt

− 2τuutuxt − τxutt − τuuxutt .



The linearized symmetry condition (LSC) for Lie point symmetriesis

X(A)

= 0 when A = 0,

where A is shorthand for A(x , t, [u]).

Just as for ODEs, the LSC is solved by first using the given PDEto reduce the LSC to an identity on (x , t, u) and some derviativesof u, then splitting this identity according to each term’sdependence on the remaining derivatives of u.

This process works when the PDE can be put into the following(generalized) Kovalevskaya form: it can be written (possibly after achange of independent variables) as

A = Dpx u − ω(x , t, [u]),

where ω has no derivatives of u of order ≥ p with respect to x .



Example The LSC for Burgers’ equation,

uxx − ut − uux = 0, (3)

isηxx − ηt − uηx − ηux = 0 when (3) holds.

Once uxx has been replaced by the left hand side of (3), thehighest-order derivative terms in the LSC have a factor uxt . Westart by writing down those terms alone:

−2(τx + τuux)uxt = 0 −→ τx = τu = 0.

This early result removes many terms from the LSC, leaving only

0 = ηxx − ηt − uηx + (2ηxu + ξt − ξxx − uξx − η)ux

+ (ηuu − 2ξxu − 2uξu)u2x − ξuuu3

x +(τt − 2ξx − 2ξuux

)ut .



In particular, the terms multiplied by ut split into

ξu = 0, ξx = 12τ′(t).

Henceξ = 1

2τ′(t)x + α(t),

for some locally smooth function α.The remaining terms in theLSC are split by equating powers of ux .

A routine calculation then shows that the Lie algebra of pointsymmetry generators is spanned by

X1 = ∂x , X2 = ∂t , X3 = t∂x + ∂u,

X4 = x∂x + 2t∂t − u∂u, X5 = xt∂x + t2∂t + (x − ut)∂u.



Generalized symmetries of PDEs

For Lie point symmetries, the infinitely prolonged infinitesimalgenerator decomposes into X = ξDx + τDt + X , where the redterms generate trivial symmetries and the nontrivial component is

X = Q∂u+(DxQ)∂u,x +(DtQ)∂u,t +(D2xQ)∂u,xx +(DxDtQ)∂u,xt +· · · .

A useful notation, which generalizes immediately to an arbitrarynumber of independent variables, is

X =∑

J

(DJQ)∂

∂u,J, J = (j , k), DJ = D j

xDkt .

Generalized symmetries are generators X whose characteristic Qhas any given dependence on x , t and [u]. The LSC in all cases is

X(A)

= 0 when [A = 0].



Example: The potential Burgers equation,

A = u,xx − u,t + u2,x = 0,

has a generalized symmetry characteristic,

Q = u,xxx + 3u,xu,xx + u3,x . (Olver)

Definition Two generalized symmetry characteristics for a givenPDE, A = 0, are equivalent if their difference vanishes on allsolutions of A = 0.

The characteristic in the example above is not equivalent to acharacteristic of Lie point symmetries; it is genuinely higher-order.


Lie symmetries of scalar P∆Es


Notation (scalar P∆Es with two independent variables)

mn

u

Independent variables: m, n; dependent variable: u.

Shift operators:

Sm : (m, n)→ (m + 1, n); Sn : (m, n)→ (m, n + 1).

Shorthand: ui ,j = u(m + i , n + j).



The method for finding Lie symmetries of a given P∆E is muchthe same as the method for O∆Es. For example, to find the Liepoint symmetries of a P∆E of the form

u1,1 = ω(m, n, u, u1,0, u0,1), (4)

substituteu = u + εQ(m, n, u) + O(ε2)

into (4) and expand to first order in ε. This yields the LSC,

Q(m + 1, n + 1, ω) = ω,1Q(m, n, u) + ω,2Q(m + 1, n, u1,0)

+ ω,3Q(m, n + 1, u0,1).

Just as for O∆Es, the LSC is a functional equation, which can besolved by the method of differential elimination.



Example For the autonomous dpKdV equation,

u1,1 = u +1

u1,0 − u0,1, (5)

the LSC can be written as an identity in u, u1,0 and u0,1:

Q(m+1, n+1, ω) = Q(m, n, u)+Q(m, n + 1, u0,1)− Q(m + 1, n, u1,0)

(u1,0 − u0,1)2,

where ω is the right-hand side of (5). Applying the operator∂/∂u1,0 + ∂/∂u0,1 to the LSC, then simplifying, we get

Q ′(m, n + 1, u0,1)− Q ′(m + 1, n, u1,0) = 0. (6)

Thus Q ′′(m + 1, n, u1,0) = 0, so Q is linear in its third argument:

Q(m, n, u) = A(m, n) u + B(m, n)

for some functions A and B.



The intermediate equation (6) gives

A(m, n + 1)− A(m + 1, n) = 0 −→ A(m, n) = α(m + n).

Finally, the LSC gives the following (simplified) conditions:

α(m + n + 1) = −α(m + n),

B(m + 1, n + 1) = B(m, n), B(m + 1, n) = B(m, n + 1).

Therefore the set of all characteristics of Lie point symmetries isspanned by

Q1 = 1, Q2 = (−1)m+n, Q3 = (−1)m+nu.

Note This is a relatively simple example: differential eliminationusually removes only one term at a time. Even so, it illustrates thegeneral method.



The method is easily extended to generalized symmetries, for whichQ depends on m, n and [u]. (For difference equations, squarebrackets around an expression denote the expression and its shifts.)

Exercise (Fill in the gaps!) The P∆E

u1,1 =u0,1(u1,0 − 1)

u+ 1 (7)

has no Lie point symmetries, so let us look for characteristics ofthe form

Q = Q(m, n, u, u1,0).

The P∆E (7) prolongs to the following, whose right-hand side iswritten in terms of a set of ‘initial variables’, I = {ui ,j : ij = 0}:

u2,1 ={u0,1(u1,0 − 1) + u}(u2,0 − 1)

uu1,0+ 1. (8)



Now write the LSC as an identity in terms of I, differentiate theresult with respect to u2,0 and simplify to obtain

u1,1

u1,1 − 1Q,2(m+1, n+1, u1,1, u2,1) =

u1,0

u1,0 − 1Q,2(m+1, n, u1,0, u2,0).

(9)Here u1,1 and u2,1 are defined by (7) and (8). To eliminate theleft-hand side, differentiate with respect to u1,0, keeping u1,1 andu2,1 fixed. This yields the PDE{

∂

∂u1,0+

u2,0 − 1

u1,0

∂

∂u2,0

}(u1,0

u1,0 − 1Q,2(m + 1, n, u1,0, u2,0)

)= 0,

whose solution, taking (9) into account, is

Q(m+1, n, u1,0, u2,0) = (u1,0−1)A

(m + 1,

u2,0 − 1

u1,0

)+B(m+1, n, u1,0).



Having split Q, it is now fairly easy to continue the differentialelimination and obtain the general solution of the LSC:

Q(m, n, u, u1,0) = f (m)u +f (m + 1)u(1− u)

u1,0 − 1. (10)

Here f is an arbitrary function, so there is an infinite family ofcharacteristics. This may seem surprising, given that the P∆E (7)has no Lie point symmetries. (The reason is that the P∆E isDarboux integrable.)

Note An essential step in finding symmetry characteristics forP∆Es (resp. PDEs) is that we can eliminate some shifts (resp.derivatives) of u to write the LSC as an identity in some initialvariables.

For a PDE that can be put in Kovalevskaya form, the existence ofsuch a set is guaranteed. There is an analogue for P∆Es.



An important technicality: Kovalevskaya form

i

jstencil points are black

vertices are ringed

The stencil is the set of points (i , j) such that ui ,j appears in theP∆E. Each of the following P∆Es has the above stencil:

u2,0 = (u1,1 − 2u + u0,2)2 , (11)

u1,1 = (u2,0 − 2u + u0,2)2 . (12)

A scalar P∆E can be put into Kovalevskaya form if u at a vertexis uniquely determined by the values of u elsewhere on the stencil.

So (11) can be put into Kovalevskaya form, but (12) cannot.



shear rotate

Lemma A transformation preserves the product structure of Z2 ifand only if it is affine. A stencil point can be made rightmost byan affine transformation if and only if it is a vertex.

Consequently, a given P∆E that can be put into Kovalevskayaform may be written (possibly after an affine transformation of theindependent variables) as

A = up,0 − ω(m, n, [u]),

where ω contains only those ui ,j for which 0 ≤ i < p. Then theLSC can be written as an identity in I0 = {ui ,j : 0 ≤ i < p}.



Note Kovalevskaya form guarantees the existence of a set ofinitial values. However, for some classes of P∆E, it is convenient(for computational reasons) to choose a set of initial values thatdiffers from I0.

If a given P∆E of the form

u1,1 = ω(m, n, u, u1,0, u0,1), (m, n) ∈ Z2,

can also be written as equations for each of u, u1,0 and u0,1, it isusually easiest to use the initial values I = {ui ,j : ij = 0}.


Invariant solutions

Invariant solutions of a given PDE

Main idea: Given a PDE and its Lie symmetries, seek aninvariant solution that satisfies Q = 0.

For Lie point symmetries, the condition Q = 0 amounts to

η(x , t, u)− u,x ξ(x , t, u)− u,t τ(x , t, u) = 0,

which is solvable by the method of characteristics. The solution isan arbitrary (locally smooth) function of the first integrals of thePfaffian system

dx

ξ(x , t, u)=

dt

τ(x , t, u)=

dx

η(x , t, u).


Invariant solutions

Example The Lie symmetries of the Thomas equation,

u,xt = u,xu,t − 1, (13)

include ξ(x , t, u) = x , τ(x , t, u) = −t, η(x , t, u) = −1/2.

Therefore, the Pfaffian system governing invariant solutions is

dx

x=

dt

−t=

du

−1/2−→ r =

√xt, v = u − 1

2 ln |t|.

As r = c cannot be a solution, every invariant solution must be ofthe form v = F (r). Substitute u = 1

2 ln |t|+ F (r(x , t)) into (13):

F ′′(r)−(F ′(r)

)2+ 4 = 0 −→ F (r) = − ln

∣∣c1e2r + c2e

−2r∣∣.


Invariant solutions

Generalized symmetries can also be used to give invariant solutionsof a given PDE, A = 0. In general, the method of characteristicsno longer works, but one may be able to use differential algebra toreduce the overdetermined system

A = 0, Q = 0

to something simpler.


Invariant solutions

Example: The potential Burgers equation,

A = u,xx + u2,x − u,t = 0,

has a generalized symmetry characteristic,

Q = u,xxx + 3u,xu,xx + u3,x . (Olver)

Although Q = 0 is a parametrized ODE, it is not easy to solveexplicitly! (This is even though it is a second-order ODE forv = ux with two independent Lie point symmetry generators.)

However, we seek only those solutions that also satisfy A = 0. Usethis PDE to eliminate u,xxx and u,xx from Q = 0, giving a PDEthat linearizes to the wave equation (using u = ln |v |):

u,xt + u,xu,t = 0 −→ v,xt = 0 −→ u = ln |f (x) + g(t)|.

The constraint A = 0 yields u = ln |c1(x2 + 2t) + c2x + c3|.


Invariant solutions

Invariant solutions of a given P∆E

Exactly the same idea works for P∆Es. The main distinction isthat, for Lie point symmetries, Q = 0 is an algebraic equation.

Example Earlier, we found all Lie point symmetries of theautonomous dpKdV equation,

u1,1 = u +1

u1,0 − u0,1. (14)

Every characteristic that involves u is a multiple of

Q = (−1)m+nu + c1 + c2(−1)m+n,

so any invariant solutions must be of the form

u = −c1(−1)m+n − c2.

No such solution exists, as it would give u1,0 − u0,1 = 0.


Invariant solutions

However, the autonomous dpKdV equation (14) also has (amongmany others) the generalized symmetry characteristic

Q = 1− 2(u1,0 − u−1,0)−1,

which is zero if and only if

u = m + f (n) + g(n)(−1)m+n,

for some functions f and g . Then (14) holds if and only if

0 = (u1,1−u)(u0,1−u1,0)+1 = (f1−f )2−(g1−g)2−2(−1)m+n(g1−g).

Comparing terms with equal dependence on m splits this condition:

g1 − g = 0, f1 − f = 0, −→ u = m + c1 + c2(−1)m+n.

This is among the simplest of a rich family of invariant solutionsthat stem from generalized symmetry characteristics.


Linearization using Lie point symmetries


As well as yielding invariant solutions, Lie point symmetries may beused to detect whether a given nonlinear PDE or P∆E islinearizable by a point transformation.

By the superposition principle, every linear homogeneous scalarPDE (with the variables x , t and u) is invariant under theone-parameter Lie group

(x , t, u) =(x , t, u + εU(x , t)

),

where u = U(x , t) is an arbitrary solution of the PDE.

Hence, the Lie algebra includes the family of generatorsXU = U(x , t)∂u that involves all solutions of the linear PDE.



The set of point symmetries is preserved under any pointtransformation (which can be regarded as a change of coordinates).Consequently, if a nonlinear PDE A = 0 can be mapped to a linearhomogeneous PDE by a point transformation, the Lie algebras ofthe two equations must be related by this change of variables.

In particular, the Lie algebra for A will contain generators thatdepend on solutions of the linear PDE, enabling one to identify thelinearizing transformation.

The same idea works equally for PDEs with any number ofvariables and for P∆Es.



Example For the Thomas equation, u,xt = u,xu,t − 1, the Liealgebra L of point symmetry generators is spanned by

X1 = ∂x , X2 = ∂t , X3 = x∂x − t∂t , X4 = ∂u,

XV = V (x , t)eu∂u, where V,xt = V .

Having found that there is an underlying linear equation, v,xt = v ,calculate its Lie algebra, L′:

X ′1 = ∂x , X ′2 = ∂t , X ′3 = x∂x − t∂t , X ′4 = v∂v ,

X ′V

= V (x , t)∂v , where V,xt = V .

Use the change of variables formula to determine a transformationthat maps L to L′; for simplicity, choose v = exp(−u). This is alinearizing transformation.



A slight modification is to use the characteristics, as follows.

Example Every characteristic of Lie point symmetries for theP∆E

u1,1 =uu1,0 − 2u1,0 + 1

u − u1,0(15)

is of the form

Q(m, n, u) = c1(u − 1) + V (m, n) (u − 1)2, (16)

where v = V (m, n) is an arbitrary solution of

v1,1 = v1,0 − v . (17)

The Lie point symmetries of (17) have characteristics

Q(m, n, v) = c1v + V (m, n), (18)

where v = V (m, n) is an arbitrary solution of (17).



If the point transformation v = ψ(m, n, u) linearizes (15) to (17),every pair (c1,V ) must correspond to a pair (c1, V ). The changeof variables formula

Q(m, n, v) = Xψ(m, n, u) = Q(m, n, u)ψ′(m, n, u),

amounts to

c1ψ(m, n, u)+V (m, n) ={c1(u − 1) + V (m, n) (u − 1)2

}ψ′(m, n, u).

This holds for all pairs (c1,V ) if and only if it splits as follows:

c1ψ(m, n, u) = c1(u − 1)ψ′, V (m, n) = V (m, n) (u − 1)2ψ′.

It is convenient to choose V (m, n) = V (m, n), which yields

ψ(m, n, u) = 1/(1− u), c1 = −c1.

This point transformation maps (15) to (17).


Summary: the main results in Lecture 3

Summary: the main results in Lecture 3

Provided that the LSC can be written as an identity, one canfind its Lie symmetries by the same methods as for ODEs orO∆Es.

Besides Lie point symmetries, PDEs and P∆Es may also havegeneralized symmetries.

Invariant solutions, which satisfy the given equation andQ = 0, may be constructed from Lie point or generalizedsymmetries.

Lie point symmetries can be used to construct linearizingpoint transformations for a given PDE or P∆E, provided suchtransformations exist.

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