Math 273 (51) - Final

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Name: Id #: Math 273 (5) - Final Autumn Quarter 26 Thursday, December 8, 26-6: to 8: Instructions: Prob. Points Score possible 25 2 25 3 25 TOTAL 75 Read each problem carefully. Write legibly.

1 Name: Id #: Math 273 (5) - Final Autumn Quarter 26 Thursday, December 8, 26-6: to 8: Instructions: Prob. Points Score possible TOTAL 75 Read each problem carefully. Write legibly. Show all your work on these sheets. Feel free to use the opposite side. This exam has 2 pages, and 3 problems. Please make sure that all pages are included. You may not use books, notes, calculators, etc. Cite theorems from class or from the texts as appropriate. Proofs should be presented clearly (in the style used in lectures) and explained using complete English sentences. Good luck!

2 Math 273 (5) - Final Autumn Quarter 26 Page 2 of 2 Question. (Total of 25 points) a) Consider the IVP { x = Ax where x() = x 2 2 /7 A := 2 and x := /2. 8 i) ( points) Find the real solution to this problem. ii) (4 points) By plugging your solution into the system, check that it solves the ODE. b) (3 points) Sketch the trace-determinant diagram for classifying the dynamics of planar linear systems. c) Consider the first order ODE x = f k (x) where f k : R R is given by f k (x) := x(x ) kx. i) (4 points) Sketch the graphs of x(x ) and the graphs of kx for various values of k. Use this picture, together with an explanation, to conclude a bifurcation occurs at some value k. What is the value of k? ii) (4 points) Draw a bifurcation diagram for the ODE. Solution. a) i) Step : Find the eigenvalues of A. Observe that 2 λ 2 det(a λi) = det λ 2 8 λ ( ) λ 2 = (λ 2) det = (λ 2)(λ 2 2λ + 7) 8 λ so from the quadratic formula the eigenvalues are λ := 2, λ 2 = + 4i, λ 3 = 4i. Hence we have a real eigenvalue and a pair of conjugate nonreal eigenvalues. Step 2: Find the corresponding eigenvectors. It is clear that the eigenspace for 2 is spanned by v := (,, ). To find

3 Math 273 (5) - Final Autumn Quarter 26 Page 3 of 2 the eigenspaces for the complex eigenvalues note that 4i 2 4i 2 A ( + 4i)I = 4i 2 4 2i R 2 i R 2 8 4i 8 4i 4i 2 4 2i R 3 R 3 2R 2 where we have applied a sequence of elementary row operations to the matrix. Hence, any (complex) eigenvector v 2 = (z, z 2, z 3 ) C 3 satisfies z 2 = i 2 z 3 and z = 2 4i z 3. Letting z 3 = we have 2/7 8i/7 2/7 + 8i/7 v 2 := i/2 and v 3 := v 2 = i/2 is are eigenvectors. Step 3: Transform the matrix into canonical form and express the inital data in terms of the new basis. Consider the real and imaginary parts w 2 and w 3 of v 2 given by 2/7 8/7 w 2 := and w 3 := /2 From lectures we know that if T = (v w 2 w 3 ) is the 3 3 matrix with columns v, w 2 and w 3, then T AT = C is the canonical form of A, which is given by 2 C = 4. 4 Recall that the initial data is given by /7 2/7 8/7 x = /2 = + /2 and so x = T b where b = (,, ).

4 Math 273 (5) - Final Autumn Quarter 26 Page 4 of 2 Step 4: Find the general solution to the system corresponding to C. We know that the solutions are given by y(t) := exp(tc)y for some choice of y R 3. From lectures we know that the matrix exponential is given by e 2t exp(tc) = e t cos 4t e t sin 4t. e t sin 4t e t cos 4t Step 5: Find the solution to the IVP by transforming the above solution to y = Cy. We know that the solution to the IVP is given by x(t) = T exp(tc)b, which is 2/7 8/7 e 2t x(t) = /2 e t cos 4t e t sin 4t e t sin 4t e t cos 4t 6/7 /7 = e t sin 4t /2 + e t cos 4t /2 ii) If x(t) is as defined in the previous part, then by the product and chain rules x (t) = e t sin 4t[ 6/7 /7 /2 4 /2 ] e t cos 4t[ /7 6/7 /2 + 4 /2 ] 46/7 4/7 = e t sin 4t 5/2 + e t cos 4t 3/2 3 5 On the other hand, 2 2 6/7 46/7 2 /2 = 5/2 8 3 and 2 2 /7 4/7 2 /2 = 3/2 8 5 so that x = Ax, whilst clearly x() = x. b) The trace-determinant diagram can be found on p. 64 of the course textbook.

5 Math 273 (5) - Final Autumn Quarter 26 Page 5 of 2 c) i) For a fixed value of k the equilibrium points correspond to values of x at which the line y = kx intersects with the graph of x(x ). We therefore see that there exists a unique value of k where there is a unique equilibrium point and for all other values of k there are precisely two. The value k corresponds to the tangent to the curve y = x(x ) at x = and is therefore given by k = 2x x= =. ii) It is easy to see the equilibrium points for the system occur at x = and x = k +. Plotting these lines on the (k, x)-plane and analysing the sign of the vector field we obtain the following bifurcation diagram:

6 Math 273 (5) - Final Autumn Quarter 26 Page 6 of 2 Question 2. (Total of 25 points) Consider the planar system where a R. x = ax y (x + y)(x 2 + y 2 ) y = x + ay + (x y)(x 2 + y 2 ) a) (4 points) Write down the linearised system at the equilibrium point (, ). Compute the eigenvalues of the linear system and determine the nature of the phase portrait. b) (3 points) State the Hartman-Grobman theorem. c) (3 points) Determine the behavior of the original system near the equilibrium point (, ) for values of a. Justify your answer. d) (4 points) Show that the system is given by in polar coordinates. r = r(a r 2 ) θ = + r 2 e) (2 points) Using the polar form, determine the behavior of the original system near the equilibrium point (, ) for a =. f) (6 points) Sketch phase portraits of the system for a < and a >. g) (3 points) Describe in detail the bifurcation which occurs when a =. Solution. a) The linearised system is ( ) x = y ( a a ) ( x y The characteristic polynomial of the matrix is λ 2 2aλ + a 2 + and so the eigenvalues are a ± i. To determine the nature of the phase portrait we consider three regimes. If a <, then the system is a spiral sink. If a =, then the system is a center. If a >, then the system is a spiral source. b) The Hartman-Grobman theorem states: Suppose f : R n R n is a C vector field and x is a hyperbolic equilibrium point. The flow is locally conjugate to the flow of the linearised system y = Df x y around x. ).

7 Math 273 (5) - Final Autumn Quarter 26 Page 7 of 2 c) The equilibrium point (, ) is hyperbolic for a and so by the Hartman- Grobman theorem the system shares the same qualitative behavior as the linearised system near the equilibrium point. Hence for a < we have a sink and for a > we have a source. d) Letting (x, y) = (r cos θ, r sin θ) it follows by the chain and product rules that x = r cos θ rθ sin θ, y = r sin θ + rθ cos θ. On the other hand, if (x, y) solves the system, then ( ) ( ) cos θ sin θ r + rθ = sin θ cos θ ( ( ) ) r a cos θ sin θ (cos θ + sin θ)r 2 r ( cos θ + a sin θ + (cos θ sin θ)r 2) ( ) ( ) cos θ sin θ = r(a r 2 ) + r( + r 2 ) sin θ cos θ Comparing the coefficients of these vectors we see that r = r(a r 2 ) θ = + r 2. e) If a =, then the system becomes r = r 3, θ = + r 2. Since r < for r > and θ, it follows that the solutions spiral away from the origin in the anticlockwise direction. Thus (, ) is a source in this case. f) Note that θ so solutions always wind around the origin. If a <, then a r 2 < for all r R and so r < away from the origin. Thus is a global sink, as pictured. If a >, then r = whenver r = a and so the circle centred at of radius a forms a periodic solution. If r > a, then r < and if r < a, then r >. Consequently, all non-equilibrium solutions tend towards this periodic solution, as pictured.

8 Math 273 (5) - Final Autumn Quarter 26 Page 8 of 2 g) The solution undergoes a Hopf bifurcation: for a < we have a stable equilibrium at (, ). When a >, the equilibrium is now unstable and a periodic solution arises in a neighbourhood of the equilibrium point.

9 Math 273 (5) - Final Autumn Quarter 26 Page 9 of 2 Question 3. (Total of 25 points) a) Let f : R n R n be a C vector field, x R n and consider the IVP { x = f(x) x() = x. i) (4 points) Define the sequence of Picard iterates (u k ) k=. Show, by induction, that these functions are all continuous. ii) (4 points) Suppose u k x uniformly as k for some function x: R n R. Show that x is differentiable and solves the IVP. b) Let v : R 2 R be the function v(x, y) := sin(πx) sin(πy) i) (3 points) Sketch the contours of v on the region [, ] 2. ii) (5 points) Sketch the phase portrait for the gradient system x = v(x). (Hint: consider the direction of the vector field along the horizontal lines y = ±/2 and the vertical lines x = ±/2). c) Let f : R 2 R 2 be a C vector field and consider the planar system x = f(x). i) (3 points) For x R 2 define the set of ω-limits ω(x). ii) (6 points) Suppose γ is a periodic solution curve for the system bounding the open region U and let z γ. Show that z cannot simultaneously belong to α(x) and ω(x) for any x U.(Hint: consider a local section through z). Solution. a) i) The functions (u k ) k= are defined recursively as follows. We let u (t) := x for all t R. Supposing u,..., u k have already been defined and are continuous we let u k (t) := x + t f u k (s) ds. Since u k and f are both continuous, f u k is continuous and therefore, by the Fundamental Theorem of Calculus, u k is differentiable (and hence continuous). ii) Since the convergence is uniform, x is a continuous function and x(t) = lim k u k (t) = lim k x + t Since f is continuous we therefore have x(t) = x + f u k (s) ds = x + t f x(s) ds, t lim f u k (s) ds. k

Math 273 (5) - Final Autumn Quarter 26 Page of 2 which will be differentiable by the Fundamental Theorem of Calculus with x (t) = f x(t) and x() = x. Hence x solves the IVP, as required.

10 Math 273 (5) - Final Autumn Quarter 26 Page of 2 which will be differentiable by the Fundamental Theorem of Calculus with x (t) = f x(t) and x() = x. Hence x solves the IVP, as required. b) i) It is an easy exercise to see that the contour diagram is given by: ii) It is easy to see that there is an equilibrium point at each lattice point of the form (k, l)+(/2./2) where k, l Z. This can be seen by direct computation, or noting that the contours through these points are singular (in particular, they are just points) on the contour diagram. The vector field is given by f(x, y) := π ( ) cos(πx) sin(πy) cos(πy) sin(πx) and so for x = ±/2 we have cos(πx) = and so f(±/2, y) = (, π cos(πy)). This means the vector field is tangential to the lines x = ±/2 and it is easy to determine the direction of the vector field along these lines by considering the sign of the cosine function for various values. One can apply a similar analysis along the y = ±/2 lines to obtain the following preliminary phase portrait.

Math 273 (5) - Final Autumn Quarter 26 Page of 2 The full phase portrait can then be sketched by considering the diagram from the previous problem and recalling that the solutions must pass through

11 Math 273 (5) - Final Autumn Quarter 26 Page of 2 The full phase portrait can then be sketched by considering the diagram from the previous problem and recalling that the solutions must pass through the level curves of v in their normal direction. c) i) We say y R 2 is an ω-limit point for x if there exists a sequence of times (t n ) n= such that t n and φ tn (x) y as n. We then define ω(x) to be the set of all ω-limit points for x. ii) Let S z be a local section through z and V z be a flow box round z. Suppose z ω(x) so that there exists a sequence of times (t n ) n= such that t n as n and φ tn (x) z as n. Thus, by passing to the tail of the sequence we can assume φ tn (x) V z for all n N. By the property of flow boxes, there exists a sequence of times (t n) n= such that t n t n < C and φ t n (x) S z for all n N and φ t n (x) z as n. If we also have z α(x), then by a similar argument we have a sequence of

12 Math 273 (5) - Final Autumn Quarter 26 Page 2 of 2 times (s n) n= such that s n as n, φ s n (x) S z for all n N and φ s n (x) z as n. But clearly this contradicts the monotonicity property of local sections.

Math 163 (23) - Midterm Test 1

Math 163 (23) - Midterm Test 1 Name: Id #: Math 63 (23) - Midterm Test Spring Quarter 208 Friday April 20, 09:30am - 0:20am Instructions: Prob. Points Score possible 26 2 4 3 0 TOTAL 50 Read each problem carefully. Write legibly. Show