Bifurcation Theory. Solutions to Problem Sheet 2. u(0) = u(1) = 0

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1 Karlsruhe Institute of Technology (KIT) Institute for Analysis Dr. Rainer Mandel M.Sc. Janina Gärtner, M.Sc. Dominic Scheider Summer Term Bifurcation Theory Solutions to Problem Sheet Problem 4 (On the Energy Method) Find such continuous functions g : R R R with g(, µ) = for all µ R that { u = g(u, µ) in (, 1), u() = u(1) = (1) has the asserted properties. (a) There is such µ R that (1) does not admit a nontrivial solution for µ µ. (b) There exist µ R and sequences (u n, µ n ) n N, (ũ n, µ n ) n N C ([, 1]) R of solutions of (1) which bifurcate from the trivial family at (, µ ) and satisfy u n, ũ n, µ n < µ < µ n for all n N. (c) We have g C (R R, R) and there exist α >, j N, λ R with g( z, µ ) = g(z, µ ), g(z, µ )z > for < z < α ; (G(α, µ ) G(z, µ )) 1 < for < α < α ; g z (, µ ) = π (j + 1). According to Theorem II.3 from the lecture, the transversality condition g zµ (, µ ) implies that j-nodal solutions of (1) bifurcate from (, µ ) in C ([, 1]). Demonstrate that bifurcation need not occur if the transversality condition is violated. Solution (a) For arbitrary µ R, consider g(z, µ) := (µ µ)z. Then a direct calculation shows that, for µ µ, the differential equation in (1) has a solution of the form u(t) = α cosh( µ µ t) + β sinh( µ µ t) with coefficients α, β R. The homogeneous boundary conditions now give = u() = α, = u(1) = α cosh( µ µ ) + β sinh( µ µ ) = β sinh( µ µ ), hence α = β = and u. Bifurcation Theory.5.17

2 (b) Consider, as a slight modification of the pendulum equation discussed in the lecture, g(z, µ) = (λ j + (µ λ j ) ) sin(z) where λ j = ((j + 1)π) for some j N, and consider µ := λ j. We know from the lecture that, for u = λ sin(u), u() = u(1) =, j-nodal solutions bifurcate at the point (, λ j ), parametrized as (α, λ j (α)) <α<α with λ j (α) λ j as α. With g chosen as above, we find two parameters λ corresponding to each value of α via λ j (α) = λ j + (µ λ j ) µ = λ j ± λ j (α) λ j, which yields to distinct families of bifurcating j-nodal solutions of (1) parametrized as ( ) ( ) α, λ j + λ j (α) λ j, α, λ j λ j (α) λ j. <α<α <α<α In the context of this problem, we choose α n, < α n < α and let µ n := λ j λ j (α n ) λ j, µ n := λ j + λ j (α n ) λ j, u n = ũ n with the suitable j-nodal solution of u = λ j (α n ) sin(u), u() = u(1) = with u n = α n. (c) Again, we modify the pendulum equation. We introduce g(z, µ) = (λ j (µ λ j ) ) sin(z) where λ j = ((j + 1)π) for some j N, and consider µ := λ j. However, using the notation from part (b), the equation λ j (α) = λ j (µ λ j ) does not have solutions, and the bifurcation diagram for u = λ sin(u), u() = u(1) = from the lecture reveals that there is no bifurcation of (1) at (, λ j ). As in the lecture, we see that all conditions of Theorem II.3 are satisfied, in particular but for the transversality condition g z (, λ j ) = λ j cos() = π (j + 1), g zµ (, λ j ) = (λ j λ j ) cos(1) =. Bifurcation Theory.5.17

3 Problem 5 (An application of the Energy Method) Consider the boundary value problem { u + λ(u u 3 ) = u() = u(π) =. in (, π), () Prove that, for k N, nontrivial (k 1)-nodal solutions of () bifurcate from the trivial branch at the point λ k = k. Solution We intend to apply Theorem II.3 and define g : R R R, g(z, λ) := λ(z z 3 ). Then g C (R R, R), g(, λ) = for every λ R. Fixing λ >, we check the assumptions of Theorem II.3: (i) For all z R, we have g( z, λ) = g(z, λ). (ii) g(z, λ) z = λ(z z 4 ) = λz (1 z ) > holds for z R with < z < 1. (iii) We let G(z, λ) := z g(s, λ)ds = 1 4 λ(z z 4 ) be the primitive of g(, λ). For α (, 1), we calculate (G(α, λ) G(z, λ)) 1 α = λ (α α 4 z + z 4 ) 1 = ((1 z ) (1 α ) ) 1. λ In order to prove finiteness of this integral, we use the mean value theorem to find ξ z (z, α) with (1 z ) (1 α ) d = (z α) dξ (1 ξ ) = (α z) 4ξ z (1 ξz) ξ=ξz and then estimate as follows: (α z) 4z(1 α ), ((1 z ) (1 α ) ) 1 ((α z) 4z(1 α )) 1 = = α + (α z) z 1 α α + α z 1 α α 1 α z = α α α z z 1 α α z α 1 α α 1 α α <. Bifurcation Theory.5.17

4 As g C 1 (R R, R), Theorem II.3 (with T = π and α = 1) gives the following necessary condition for bifurcation of (k 1)-nodal solutions from the trivial family at a point (, λ k ): z g(, λ k ) = ( ) πk, ( ) T and equivalently λ k = k. Further, since the transversality condition (ii) of Theorem II.3, λ z g(, λ k ) = 1 holds, the condition ( ) is sufficient for bifurcation, and hence the proof is complete. Bifurcation Theory.5.17

5 Problem 6 (A problem without periodic Solutions) Let g C(R R, R) with g(, λ) = (λ R). Consider b C(R R, R) with b(x, λ) for all x, λ R. Prove that the differential equation u + b(x, λ) u = g(u, λ) in R, (3) does not admit a periodic solution in C (R) unless it is constant. Solution First, we note that by assumption, b is continuous and does not have any zero, so b is either negative or positive on all of R R. We assume that u C (R) is a periodic solution of (3). We introduce E : R R, E(t) := 1 [u (t)] + G(u(t), λ) where G(z, λ) := z g(s, λ) ds for λ, z R. Then, E C1 (R), and for t R E (t) = u (t) (u (t) + g(u(t), λ)) = b(t, λ) [u (t)] ( ) where we have inserted the differential equation in the last step. Since b does not change sign, this implies that E is a monotone function. As u is periodic, so is E, and we conclude that E is constant. Hence, E, and by ( ), u. So u is constant, and the assertion is proved. Bifurcation Theory.5.17

6 Problem 7 (Variational Methods) Let n N. We consider the n-dimensional bifurcation problem x I(x, λ) = (4) where I : R n R R, I(x, λ) := 1 x Ax λ sin( x p ) for some p (1, ) and a symmetric, positive definite matrix A R n n. (a) Show that, for λ R, I(, λ) is differentiable with respect to x and calculate x I(, λ). Verify that x I(, λ) = for all λ R. (b) Prove that, for λ R, I(, λ) admits a minimizer x λ R n. (c) Prove that (, ) is a bifurcation point for problem (4). Solution (a) We fix λ R and i {1,..., n}. For x R n, we have I(x, λ) = 1 n a jk x j x k λ sin( x p ). j,k=1 For x, the chain rule gives partial differentiability and, using the symmetry of A, I xi (x, λ) = 1 = n a jk (δ ij x k + x j δ ik ) λ cos( x p ) px i x p j,k=1 For x =, we compute directly n a ij x j λ cos( x p ) px i x p = j I(he i, λ) I(, λ) I xi (, λ) = lim h h lim I x i (h, λ) = h [ ] Ax λ cos( x p ) p x p x. i 1 = lim a iih λ sin( h p ) h h since p > 1. We conclude that I is indeed continuously differentiable with { Ax λ cos( x p ) p x p x, λ R, x, x I(x, λ) =, λ R, x = =, and and in particular, we have the trivial family of solutions x I(, λ) =, λ R. Bifurcation Theory.5.17

7 (b) For x R n, λ > we estimate I(x, λ) α x λ sin( x p ) λ, ( ) hence j λ := inf x R n I(x, λ) λ and we find a minimizing sequence (x (n) λ ) n N, i.e. lim n I(x(n) λ, λ) = j λ. Taking another look at the first estimate in ( ), we see that (x (n) λ ) n N is necessarily bounded and hence, working in finite dimension, the Bolzano-Weierstrass theorem asserts that we can assume without loss of generality (possibly after passing to a subsequence) that (x (n) λ ) n N converges to some element x λ R n. It then follows from the continuity of I that j λ = lim n I(x (n) λ, λ) = I(x λ, λ), and thus x λ is a minimizer for I(, λ). (c) To find non-trivial solutions of (4), we consider the minimizers x λ of I(, λ), which are stationary points of I(, λ) and hence provide solutions of (4). We will prove the following: Assertion 1 : For every λ >, x λ. Assertion : We have x λ as λ. This finally yields that (, ) is a bifurcation point of problem (4) since it is a point where the non-trivial family {(x λ, λ) : λ > } meets the trivial branch of solutions. We now prove the assertions 1 and. First of all, since A is positive definite, we find β α > with α x x Ax β x for x R n. ( ) Proof of Assertion 1 : If x λ = held true, then we would find j λ = and therefore I(, λ). We aim to show, however, that I(, λ) attains negative values. To this end, we recall that sin(ξ) [ π ξ for ξ, π ]. For x R n with x p π, we estimate I(x, λ) β x λ π x p = β ( x p x p 4λ ). βπ { ( Hence, choosing x R n with < x < min π ) 1 p, ( 4λ βπ ) } 1 p, we have I(x, λ) <, and the assertion is proved. Note that such a choice is possible because p <. Bifurcation Theory.5.17

8 Proof of Assertion : Since x λ, we have for all λ > x I(x λ, λ) Ax λ = λ cos( x λ p ) p x λ p x λ. Multiplying from the left with x λ, we find in view of ( ): ( <) α λ cos( x λ p ) p x λ p β. Letting λ, the estimate from below can only hold if cos( x λ p ) x λ p. As the cosine is bounded and p <, we infer x λ as λ. Bifurcation Theory