MATH 1280 HOMEWORK 2 SOLUTIONS

Section 2.5

Problem 5: Consider the initial value problem x = |x|p/q, x(0) = 0, where p and q are positive integraerswith no common factors.

(a) Show that there are an infinite number of solutions if p < q.(b) Show that there is a unique solution if p > q.

Solution: First note that since x ≥ 0 and x(0) = 0, it is only necessary to consider the case x ≥ 0 in ourcalculations. The given system is separable, so we can integrate it directly to find

t + C =

∫|x|−p/qdx =

x−p/q+1

1− pq

⇒(

1− p

q

)(t− t0)q/(q−p)

(a) If p < q, then a piecewise x defined as

x =

{0 if t < t0(

1− pq

)(t− t0)q/(q−p) if t ≥ t0

is a solution for any nonnegative value of t0. Thus, there are an infinite number of solutions. (This is unsurpris-ing, since the vector field is not Lipschitz for p < q, so we know that we couldn’t guarantee uniqueness throughthe regular means.)

(b) If q < p, the solution we found above by integrating is negative, and thus not a valid solution. Alterna-tively, we can note that the vector field is Lipshitz in this case, and we can apply the Picard-Lindelof theoremto get uniqueness.

Date: February 23, 2014.

1

2 MATH 1280 HW2

Section 2.7

Problem 6: For x = r +x−x3, plot the potential function V (x) and identify all the equilibrium points andtheir stability for various values of r.

Solution: Since x = f(x) = −dV

dx, we find V (x) =

x4

4− x2

2− rx. There are 6 qualitatively different

situations: |r| < 2

3√

3, |r| = 2

3√

3, |r| > 2

3√

3.

A cartoon of the first three situations is sketched below, with the opposite inequalities corresponding to themirror image of the cartoons. The minima of V (x) correspond to stable fixed points of the system, and maximacorrespond to unstable fixed points.

|r| < 2

3√

3r =

2

3√

3r >

2

3√

3

x

V

x

x

x

V

x

x

x

V

x

x

MATH 1280 HW2 3

Section 3.1

Problem 1: Collect the bifurcation information for x = 1 + rx + x2.

Solution: Fixed points occur at the roots of 1 + rx + x2, giving x∗ =−r ±

√r2 − 4

2. We see immediately

that there are three cases of interest.

Case (i): |r| < 2. In this case there are no real values for x∗.Case (ii): |r| = 2. In this case there is a single fixed point at x∗ = − r

2 . This is where the saddle nodebifurcation occurs.

Case (iii): |r| > 2. In this case both possibilities for x∗ are valid. Checking the derivative of 1 + rx + x2 willshow that one is stable and one is unstable, so I leave that to you.

These cases are shown below for r > 0.

|r| < 2 |r| = 2 |r| > 2

The bifurcation diagram is shown below. Stable branches are shown as solid lines, unstable branches areshown as dashed lines.

x∗

r

2−2

4 MATH 1280 HW2

Problem 4: Collect the bifurcation information for x = r +x

2− x

1 + x.

Solution: In terms of r, the fixed points are x∗ =1− 2r ±

√4r2 − 12r + 1

2. We see that the number of solu-

tions will be dependent on the discriminant, 4r2−12r+1. The three qualitatively different cases are shown below.

x

xr chosen so the discriminant isnegative – no solutions

x

x r chosen so the discriminant iszero – a single solution, saddlenode bifurcation

x

xr chosen so the discriminant ispositive – two real solutions

The saddle node bifurcations occur precisely when 4r2−12r+1 = 0⇒ rc =3±√

2

2. The bifurcation diagram

is given below.

x∗

rrcrc

MATH 1280 HW2 5

Section 3.2

Problem 2: Collect the bifurcation information for x = rx− ln(1 + x).Solution: It is simplest to view the fixed points as intersections of rx and ln(1 + x). There are three

qualitatively different cases of interest.Case (i): r < 0. There is only a single intersection, at the origin. It is stable.Case (ii): 0 < r < 1. There are two intersections, one at the origin and one for a positive value of x. The

origin is stable, the positive point is unstable.Case (iii): r > 1. There are again two intersections, one at the origin and one for a negative value of x. The

origin is unstable, the negative point is stable. We see that the origin has undergone a transcritical bifurcationat r = 1.

Vector fields for the three cases are shown below, as is a bifurcation diagram.

x

r < 0

x

0 < r < 1

x

r > 1

1 r

x∗

6 MATH 1280 HW2

Problem 4: Collect the bifurction information for x = x(r − ex).Solution: We notice that x∗ = 0 is always a fixed point, and the other fixed point will occur at x∗ = ln r.

Immediately we see that x∗ = 0 is the only fixed point for r ≤ 0.This is one of the few cases where I think its easiest to see stability directly from the calculation rather than

from a sketch of the vector field, so let’s calculate directly.If we let f(x) = x(r − ex), then f ′(x) = r − ex(1 + x).For x∗ = 0, f ′(0) = r − 1, so this will be stable for r < 1 and unstable for r < 1.For x∗ = ln r, f ′(ln r) = −r ln r. This will be unstable (f ′(x∗) > 0) when r < 1, and stable (f ′(x∗) < 0) when

r > 1.The qualitatively different vector fields are below, as well as the bifurcation diagram.

r < 0

x = 0 x = 0

0 < r < 1

x = 0

r > 1

1 r

x∗

Problem 6: The problem statement is fairly involved, so I won’t reproduce it here.Solution:(a) Given x = X + bX3 +O(X4), we have x3 = X3 +O(X4). Then x + cx3 = X + bX3 + cX3 +O(X4), so

choosing c = −b gives the appropriate inversion.

(b) Using the result from part (a) and collecting to O(x4) we have

X = x− bx3 +O(x4)

X2 = x2 +O(x4)

X3 = x3 +O(x4)

Then from the problem statement we have

X = RX −X2 + aX3 +O(X4)

= Rx−Rbx3 − x2 + ax3

Applying the chain rule to x and writing everything in terms of x, we have

MATH 1280 HW2 7

x = X + 3bX2X +O(X4)

= Rx−Rbx3 − x2 + ax3 + 3bx2(Rx−Rbx3 − x2 + ax3) +O(x4)

= Rx− x2 + (2bR + a)x3 +O(x4)

(c) Choosing b = − a

2Rgives k = 0.

(d) The assumption that R 6= 0 was definitely needed for this procedure to work. As we can see in part (c),our choice of b has an R in the denominator, so R = 0 causes a problem. Put another way, if R = 0 the cubicterm in our transformation is ax3, and we no longer have a free parameter to try and eliminate the a term.

Jeff Dunworth, Thackeray 517

E-mail address: jbd20@pitt.edu