Stability, the Maslov Index, and Spatial Dynamics

Transcription

1 Stability, the Maslov Index, and Spatial Dynamics Margaret Beck Boston University joint work with Graham Cox (MUN), Chris Jones (UNC), Yuri Latushkin (Missouri), Kelly McQuighan (Google), Alim Sukhtayev (Miami) MSRI, October 9, 2018

2 Stability for PDEs General framework: w t = F(w), ϕ = stationary solution of interest, F(ϕ) = 0 Analyze behavior of perturbations: w(x, t) = ϕ(x) + u(x, t) with u(x, 0) small u t = Lu + N (u), L = DF(ϕ) Stability of ϕ: does the perturbation u(t) 0 as t (or stay small t)? Types of stability: Spectral stability: λ σ(l) Re(λ) < 0? Linear stability: u t = Lu u(t) 0 as t? Nonlinear stability: u t = Lu + N (u) u(t) 0 as t? Focus on spectral stability for this talk.

3 Example to keep in mind Reaction diffusion equation with gradient nonlinearity: w t = w + G(w), x Ω R d, w R n, G : R n R Solution of interest: localized stationary solution 0 = ϕ + G(ϕ), lim x Ω ϕ(x) = 0 Perturbation Ansatz: w(x, t) = ϕ(x) + u(x, t) u t = Lu + N (u) Lu = u + 2 G(ϕ(x))u N (u) = G(ϕ + u) G(ϕ) 2 G(ϕ(x))u = O(u 2 ). Spectral stability: σ(l) = σ ess(l) σ pt(l) Essential spectrum relatively easy to compute; assume it is stable. Are there unstable eigenvalues?

4 Case 1: one spatial dimension, scalar equation Sturm-Liouville eigenvalue problem: λu = u xx + g (ϕ(x))u = Lu, x (a, b) Prüfer coordinates: define (r, θ) via To obtain u(a) = u(b) = 0 u(x; λ) = r(x; λ) sin θ(x; λ), u (x; λ) = r(x; λ) cos θ(x; λ) r = r(1 + λ g (ϕ(x))) cos θ sin θ θ = cos 2 θ + (g (ϕ(x)) λ) sin 2 θ Observe: {r = 0} is invariant, so for a nontrivial solution, u(x; λ) = 0 if and only if θ = jπ, j Z For λ 1, θ > 0, so solutions will be forced to oscillate Let θ(a; λ) = 0 be the initial condition, evolve in x; is θ(b; λ) {jπ}? If so, this corresponds to an eigenfunction with eigenvalue λ.

5 Case 1: one spatial dimension, scalar equation Looking for eigenfunctions and eigenvalues via θ = cos 2 θ + (g (ϕ(x)) λ) sin 2 θ, x (a, b) Initial condition: θ(a; λ) = 0; flow forward and see if θ(b; λ) {jπ} For some λ 1 there must be an eigenvalue. Fix such a λ k : θ(b; λ k ) = (k + 1)π. Increase λ until you again land in {jπ}, which is the eigenvalue λ k 1. :... (b; ): (k + 1) k k k 1 0 R Process stops at largest λ 0; θ no longer can complete one half-rotation

6 Case 1: one spatial dimension, scalar equation Using these ideas one can show: λu = u xx + g (ϕ(x))u = Lu, x R, u L 2 (R) There exists a decreasing sequence of simple eigenvalues g 00 ('(1))... k k 1 0 R Corresponding eigenfunctions u k (x) have k simple zeros Quickly conclude the pulse is unstable: u t = u xx + g (u) Observe that x[0 = ϕ xx + g (ϕ)] 0 = (ϕ ) xx + g (ϕ)ϕ = Lϕ Qualitatively, ϕ and ϕ look like '(x) ' 0 (x) x x Therefore, ϕ = u 1 and λ 1 = 0, and so λ 0 > 0 and pulse is unstable

7 Case 1: one spatial dimension, scalar equation Related concept of conjugate points: θ = cos 2 θ + (g (ϕ(x)) λ) sin 2 θ, x (a, s) Initial condition: θ(a; λ) = 0; flow forward and see if θ(s; λ) {jπ} Fix λ = λ k to be an eigenvalue, so if s = b we know θ(b; λ) = (k + 1)π Decrease s until you again land in {jπ}, which is the conjugate point s k 1. s : (s; k) : a s 0... s k 1 k b R s k = b (k + 1) Process stops at largest s 0; θ no longer can complete one half-rotation

8 Case 1: one spatial dimension, scalar equation Square : Relationship between eigenvalues and conjugate points: b s eigenvalues conjugate points s 0 s 1 no conjugate points s 2 a no eigenvalues One can prove: No eigenvalues for s = a; no time to oscillate. No conjugate points for λ = λ large; ODE or spectral analysis. Number of conjugate points for λ = λ equals the number of eigenvalues λ > λ.

9 Case 1: one spatial dimension, scalar equation To analyze stability, choose λ = 0: b s eigenvalues conjugate points s 0 s 1 s 2 a no eigenvalues no conjugate points = Number of conjugate points = number of unstable eigenvalues = Morse(L) This is a simple case of what is often called the Morse Index Theorem, and it goes back to the work of Morse, Bott, etc.

10 Case 1: one spatial dimension, scalar equation To summarize, when we have λu = u xx + g (ϕ(x))u = Lu, u R, x R Spectral stability can be determined almost immediately using qualitative properties of the underlying solution ϕ, with little knowledge of g. A pulse is necessarily unstable. Similarly a front in necessarily stable. '(x) '(x) x x pulse front Positive eigenvalues can be counted by instead counting the number of conjugate points when λ = 0. Conjugate points and eigenvalues can be analyzed via the winding of a phase in R 2. Monotonicity in λ and s is key. Can this be generalized to u R n or x R d?

11 Case 2: one spatial dimension, system of equations Eigenvalue equation: λu = u xx + 2 G(ϕ(x))u = Lu, u(x) R n, x R Assume: X = L 2 (R), dom(l) = H 2 (R). ϕ is a pulse, lim x ϕ(x) = ϕ 0. σ( 2 G(ϕ 0)) < 0. Which implies L is self-adjoint, λ R. σ ess(l) is stable. Write as a first-order system: ( ) ( ) ( ) d u 0 I u = dx v (λ 2 G(ϕ(x))) 0 v ( ) ( ) ( ) 0 I (λ 2 G(ϕ(x))) 0 u = I 0 0 I v [Arnol d 85]: generalized notion of phase via the Maslov index and proved oscillation theorems

12 Case 2: one spatial dimension, system of equations First-order eigenvalue problem: ( ) d u = JB(x; λ) dx v ( ) 0 I J =, B(x; λ) = I 0 ( ) u, v ( (λ 2 G(ϕ(x))) 0 0 I ), B(x; λ) = B(x; λ) Assumption σ( 2 G(ϕ 0)) < 0 implies JB (λ) is hyperbolic: ( ) ( ) ( ) 0 I u u (λ 2 = ν (λ 2 G(ϕ G(ϕ 0)) 0 v v 0))u = ν 2 u. If µ j σ( 2 G(ϕ 0)), µ j < 0, j = 1,... n, then for λ > 0 ν ± j = ± λ µ j, ν + j > 0 > ν j, j = 1,..., n. Dimension of asymptotic stable/unstable subspaces is n: dim(e s,u (λ)) = n

13 Case 2: one spatial dimension, system of equations First-order eigenvalue problem: ( ) d u = JB(x; λ) dx v ( ) 0 I J =, B(x; λ) = I 0 ( ) u, v ( (λ 2 G(ϕ(x))) 0 0 I ), B(x; λ) = B(x; λ) For an eigenfunction (u, v)(x; λ) we need lim x (u, v)(x; λ) = (0, 0). {( ) ( ) ( ) } E u u u 0 (x; λ) = (x; λ) solution : (x; λ) as x. v v 0 Space of solutions asymptotic to the unstable subspace of E u (λ). This is a Lagrangian subspace with respect to the symplectic form ω(u, V ) = U, JV R 2n. Lagrangian-Grassmanian: Λ(n) = {l R 2n : dim(l) = n, ω(u, V ) = 0 U, V l}.

14 Case 2: one spatial dimension, system of equations d dx ( ) u = JB(x; λ) v ( ) u, B(x; λ) = B(x; λ), J = J = J 1 v If U, V E u (x; λ), then Moreover, d dx ω(u(x), V (x)) = U (x), JV (x) + U(x), JV (x) = JBU(x), JV (x) + U(x), J 2 BV (x) = BU(x), V (x) BU(x), V (x) = 0. lim U(x), V (x) = 0 lim ω(u(x), V (x)) = 0 x x and so ω(u(x), V (x)) = 0 x R.

15 Case 2: one spatial dimension, system of equations [B., Cox, Jones, Latushkin, McQuighan, Suhktayev 18]: Proved square relating eigenvalues to conjugate points. Proved a pulse solution is necessarily unstable. Key ideas in paper: λu = u xx + 2 G(ϕ(x))u = Lu, u R n, dom(l) = H 2 (R) L 2 (R). Compactify domain: σ(s) = tanh(s), s(σ) = 1 ( ) 1 + σ 2 ln, s [, ], σ [ 1, 1] 1 σ Will not comment on this further.

16 Case 2: one spatial dimension, system of equations First prove square on half-line with Dirichlet BCs: λu = u xx + 2 G(ϕ(x))u = L L u, x (, L) L dom(l L ) = {u H 2 (, L) : u(l) = 0}. s eigenvalues conjugate points no conjugate points (L L ) ( 1, kr 2 G(')k 1 ] 1 0 no eigenvalues 1 E u ( 1; ) \ D = {0} E u (s; λ) is a path of Lagrangian planes, homotopic to the trivial loop, for (s, λ) [, L] [0, λ ] Reference Lagrangian plane - Dirichlet plane: ( ) {( ) } 0 u D = = R 2n : u = 0. I v

17 Case 2: one spatial dimension, system of equations L s eigenvalues no conjugate points conjugate points (L L ) ( 1, kr 2 G(')k 1 ] 1 0 no eigenvalues 1 E u ( 1; ) \ D = {0} Maslov index: counts crossings of the path E u (s; λ) with the reference plane D. Homotopy argument Mas(E u (s; λ)) = 0. No crossings on bottom or right side of square; eigenvalues contribute negatively, conjugate points contribute positively: Morse(L L ) = number of conjugate points on (, L)

18 Case 2: one spatial dimension, system of equations Φ : [a, b] Λ(n) path of Lagrangian planes, D reference plane. A crossing is a t 0 [a, b] such that Φ(t 0) D = {0}. Generically Φ(t) is transversal to D for all t [t 0 ɛ, t 0 + ɛ], and φ(t) : Φ(t 0) D so that Φ(t) = graph φ(t) = {V + φ(t)v : V Φ(t 0)} D? (t) (t)v (V, (t)v ) V (t 0 ) Crossing form [Robbin, Salamon 93]: Q(U, V ) = d dt ω(u, φ(t)v ) t=t 0, U, V Φ(t 0) D.

19 Case 2: one spatial dimension, system of equations D? (t) (t)v (V, (t)v ) V (t 0 ) Crossing form [Robbin, Salamon 93]: Q(U, V ) = d dt ω(u, φ(t)v ) t=t 0, U, V Φ(t 0) D. Q R k k symmetric, where k = dim(φ(t 0) D). t 0 is regular if detq 0; generic crossings are regular and isolated. Signature of Q: signq = n +(Q) n (Q), n ±(Q) = number of positive/negative eigenvalues

20 Case 2: one spatial dimension, system of equations Maslov index for single crossing: if t 0 [a 0, b 0] is the only crossing of Φ with D, n (Q) if t 0 = a 0 Mas(Φ [a0,b 0 ], D) = signq = n +(Q) n (Q) if t 0 (a 0, b 0) n +(Q) if t 0 = b 0 Endpoint convention is somewhat arbitrary; affects intermediate results but not our end result. Define Maslov index of a regular smooth path by defining it on segments around each crossing and summing. If all crossings of a path Φ : [a, b] Λ(n) with D are positive, ie Q > 0, then Mas(Φ [a,b], D) = dim(φ(t) D) a<t b Similarly, if all crossings are negative, then Mas(Φ [a,b], D) = a t<b dim(φ(t) D)

21 Case 2: one spatial dimension, system of equations L s eigenvalues no conjugate points conjugate points (L L ) ( 1, kr 2 G(')k 1 ] 1 0 no eigenvalues 1 E u ( 1; ) \ D = {0} Key aspect of proof is showing monotonicity, ie crossings at conjugate points are positive, and crossings at eigenvalues are negative. Path of Lagrangian planes: E u (s; λ). Parameter is s or λ depending on side. Negative crossings in λ: need to show for s = L fixed Q(U, V ) = d dλ ω(u, φ(λ)v ) λ=λ 0 < 0, U, V E u (L; λ 0) D.

22 Case 2: one spatial dimension, system of equations Suffices to check that Q(V, V ) = d dλ ω(v, φ(λ)v ) λ=λ 0 < 0, V E u (L; λ 0) D. Let W (L; λ) E u (L; λ) so that We have W (L; λ 0) = V, W (L; λ) = V + φ(λ)v. Q(V, V ) = d dλ ω(v, φ(λ)v ) λ=λ 0 = d dλ ω(v, V + φ(λ)v ) λ=λ 0 = d dλ ω(w (L; λ0), W (L; λ)) λ=λ 0 = ω(w (L; λ 0), W λ (L; λ 0)). Recall: d dx W = JB(x; λ)w d dx W λ = JB(x; λ)w λ +MW, M = ( ) 0 0. I 0

23 Case 2: one spatial dimension, system of equations Q = ω(w (L; λ 0), W λ (L; λ 0)), d dx W λ = JB(x; λ)w λ + MW. L Q = JW (L; λ 0), W λ (L; λ 0)) = L = = = = L L L [ J 2 BW, W λ ) + JW, JBW λ + MW ]dx JW, MW dx ( ) 0 I W, I 0 (W 1(x; λ 0)) 2 dx < 0. ( ) 0 0 W dx I 0 d dx JW (x; λ0), W λ(x; λ 0)) dx

24 Case 2: one spatial dimension, system of equations L s eigenvalues no conjugate points conjugate points (L L ) ( 1, kr 2 G(')k 1 ] 1 0 no eigenvalues 1 E u ( 1; ) \ D = {0} This monotonicity implies: 0 = Mas(E u (x; λ) square, D) = Mas(E u (x; λ) left, D) + Mas(E u (x; λ) top, D) = {number of conjugate points} {number of eigenvalues}. Hence, {number of conjugate points} = {number of eigenvalues} = Morse(L L ).

25 Case 2: one spatial dimension, system of equations Remaining steps: λu = u xx + 2 G(ϕ(x))u = Lu, u R n, x R. Extend result to full line R: show for L > L large, Morse(L) = Morse(L L ) Follows because you can approximate the point spectrum of an operator on R using a large subdomain. Prove any pulse solution is necessarily unstable: Show there is at least one conjugate point. Uses reversibility arguments applied to the original PDE u t = u xx + G(u), ie x x symmetry. Remark: very few actual applications of the Maslov index in stability analysis!

26 Case 3: multiple spatial dimensions, one equation Eigenvalue problem: u + V (x)u = λu, x Ω R d, u R n, λ R Family of domains [Smale 65]: Hilbert space u Ω = 0 {Ω s : 0 s 1}, Ω 1 = Ω, Ω 0 = {x 0}. H = H 1/2 ( Ω) H 1/2 ( Ω), ω((f 1, g 1), (f 2, g 2)) = g 2, f 1 g 1, f 2 Path of subspaces in the Fredholm-Lagrangian Grassmannian of H: {( Φ(s) = u, u ) } Ωs : u H 1 (Ω s), u + V (x)u = λu, x Ω s n Reference subspace: {( D = u, u ) ( Ω = 0, u ) } Ω : u H 1 (Ω s) n n [Deng, Jones 11], [Cox, Jones, Latushkin, Suhktayev 16],... show one can compute Morse(L) by counting conjugate points in this context; also results for more general boundary conditions.

27 Case 3: multiple spatial dimensions, one equation Future work: does this suggest a spatial dynamics for R d? 0 = u + F (u), x Ω R d Family of domains parameterized by family of diffeomorphisms: ψ s : Ω Ω s, s [0, 1], Ω 1 = Ω, Ω 0 = {x 0}. Define boundary data via f (t; y) = u(ψ t(y)), and trace map g(t; y) = u (ψt(y)), t [0, 1], y Ω n Tr tu = (f (t), g(t)). Can (formally) obtain a first-order system ( ) ( ) d f F(f, g) = dt g G(f, g) Can we make sense of the above equation and put it to good use?

28 Summary We ve seen how the Maslov index can be used in stability analysis to obtain results of the form number of conjugate points = Morse(L). Due to recent work, many abstract results exist for systems of equations with x R d, d 1. Current/future work: Find more examples! In some sense, these results are only useful if they can be used to actually determine stability in situations of interest. I know of two examples for x R (one mentioned today - pulse instability; other is [Chen, Hu 14]) and none for x R d. Further understand relationship between these results and the Evans function. Develop a spatial dynamics for R d.

29 Lagrangian subspace calculation Φ(s) = {( u, u ) } Ωs : u H 1 (Ω s), u + V (x)u = λu, x Ω s n If u, v Φ then ω(u, v) = = = = v n, u Ω Ω Ω u n, v ( v n u u n v ) ds (( u v + u v) ( u v + v u)) dx (u(λv Vv) v(λu Vu)) dx = 0.

30 Pulse - existence of conjugate point 0 = ϕ xx + G(ϕ(x)), Generically, ϕ(x) will be unique as a solution (up to translation) asymptotic to the fixed point ϕ 0 = lim x ± ϕ(x) Equation invariant under x x, so ϕ( x) is also a solution. By uniqueness, we therefore have This implies But then ϕ(x) = ϕ( x + δ) ϕ(δ/2 + x) = ϕ(δ/2 x) x R. d ϕ(δ/2 + x) x=0 = ϕ(δ/2 x) x=0 ϕx(δ/2) = 0. dx Since ϕ x is a eigenfunction with eigenvalue λ = 0, we have ( ) ϕx(x) E u ϕ (x; 0) E u (δ/2, 0) D = {0} xx(x) which is our conjugate point.