Existence and stability of solitary-wave solutions to nonlocal equations

Transcription

1 Existence and stability of solitary-wave solutions to nonlocal equations Mathias Nikolai Arnesen Norwegian University of Science and Technology September 22nd, Trondheim

2 The equations u t + f (u) x (Lu) x = 0, (1) u t + f (u) x + (Lu) t = 0, (2) where u and f are real valued functions, and L is Fourier multiplier operator with symbol m: Lu(ξ) = m(ξ)û(ξ).

3 The equations u t + f (u) x (Lu) x = 0, (1) u t + f (u) x + (Lu) t = 0, (2) Backround: Equations of the forms (1) and (2) arise in the modelling of propagation of waves. Typically "long waves" on the surface of an incompressible, irrotational uid of nite depth with a regular, impermeable bottom.

4 Examples KdV: u t + ( 1 2 u2 ) x ( 2 x u ) x = 0 (m(ξ) = ξ2 ) BBM: u t + (u u2 ) x + ( 2 x u ) t = 0 (m(ξ) = ξ2 )

5 Examples KdV: u t + ( 1 2 u2 ) x ( x 2 u ) x = 0 (m(ξ) = ξ2 ) BBM: u t + (u u2 ) x + ( x 2 u ) t = 0 (m(ξ) = ξ2 ) fkdv: u t + ( 1 2 u2 ) x (D s u) x (m(ξ) = ξ s ) fbbm: u t + (u u2 ) x + (D s u) t = 0 (m(ξ) = ξ s )

6 Motivational example The Whitham equation with surface tension, u t + ( 1 2 u2 ) x (Lu) x = 0, where the operator L has symbol (1 + β ξ m(ξ) = 2 ) tanh(ξ). ξ Here β 0 is the strength of the surface tension.

7 Solitary waves We are only interested in solitary wave solutions of (1) and (2). Therefore assume u(x, t) = u(x ct) and u(x ct) 0 as x ct ±. Assuming u is a solitary wave solution of (1) or (2), we get, respectively: Lu + cu f (u) = 0, (3) c(lu + u) f (u) = 0. (4)

8 Background: related results Albert '99: (1) for f (u) = u p, p (1, 2s + 1), and s = 1. Zeng '03: (2) for f (u) = u + u p, p 2 integer, and s 1. Albert, Bona & Saut '97 (Weinstein (87)): remark that their method can be extended to f (u) = u p, p 2 integer, and s 1 given some conditions. Ehrnström, Groves & Wahlén '12: (1) for smoothing operators L (s < 0), in particular for the Whitham equation. These results leave the gap 0 s < 1.

9 Background: related results Albert 1999: (1) for f (u) = u p, p (1, 2s + 1), and s = 1. Zeng 2003: (2) for f (u) = u + u p, p 2 integer, and s 1. Albert, Bona & Saut 1997 (Weinstein 1987): remark that their method can be extended to f (u) = u p, p 2 integer, and s 1 given some conditions. Ehrnström, Groves & Wahlén 2012: (1) for smoothing operators L (s < 0), in particular for the Whitham equation. These results leave the gap 0 s < 1. Remark: Existence has since been established for 0 < s < 1, but only for m(ξ) = ξ s, i.e. the fractional laplace operator. The proofs rely on commutator estimates specic to this operator. (Frank & Lenzmann 2010 and Linares, Pilod & Saut 2015)

10 Assumptions (A) L is a Fourier multiplier operator with symbol m, Lu(ξ) = m(ξ)û(ξ), the function m is piecewise continuous with a nite number of discontinuities, s > 0, and A 1 ξ s m(ξ) A 2 ξ s for ξ 1, 0 m(ξ) A 2 for ξ 1 for some constants A 1, A 2 > 0. Assumption (A) implies that L : H s (R) L 2 (R), and Lu L 2 (R) u H s (R).

11 Assumptions (B) The nonlinearity f is of the form (B1) f (u) = c p u u p 1, c p > 0, or (B2) f (u) = c p u p, c p 0, where p (1, 2s + 1) or p (1, 1+s 1 s ) ( 1+s 1 s One can also consider nonlinearities g(u) = u + f (u). = when s 1).

12 The general method of proving existence Formulate (3) and (4) as (constrained) variational problems. Use concentration-compactness principle of Lions to redeem the lack of compactness on R. This also yields (conditional energetic) stability if the involved functionals are time invariant w.r.t. the ivp of the relevant equation ((1) or (2))

13 Variational setting I Let F (u) = f (u). We dene functionals E : H s/2 (R) R and Q : L 2 (R) R by E(u) = 1 ulu dx F (u) dx, Q(u) = 1 u 2 dx. 2 R R 2 R For q > 0 we dene the quantity I q := inf{e(u) : u H s/2 (R), Q(u) = q}. Denote by D q the set of minimizers of I q. Elements of D q are solutions to (3), c being the Lagrange multiplier.

14 Variational setting I Problem: if p 2s + 1 then I q = for any q > 0. For the capillary Whitham equation, s = 1/2 and p = 2. Thus this variational formulation cannot be used to nd solitary waves for this equation.

15 Variational setting II For κ > 0, dene functionals J κ : H s/2 (R) R and U : L p+1 (R) R by J κ (u) = 1 ulu + κu 2 dx, U(u) = F (u) dx. 2 R For λ > 0 we consider Γ λ (κ) = Γ λ = inf{j κ (u) : u H s/2 (R), U(u) = λ}. (5) Denote by G λ the set of minimizers of Γ λ. Clearly, Γ λ 0 for any λ > 0, and if p (1, 1+s 1 s ) then Γ λ > 0. R

16 If u G λ, then Lu + κu γf (u) = 0, where γ is the Lagrange multiplier. Using the homogeneity of f, letting β 1 v = u β p 1 = γ, we get that if u is a minimizer of Γ λ, then: For κ = 1, u solves (4) with c = 1/γ. Considering L in J κ, where κ 1 L = L, v solves (4) with c = κ. For any κ > 0, v solves (3) with c = κ.

17 Concentration-compactness Lemma 1 (P.L. Lions '84) Let {ρ n } n L 1 (R) be a sequence that satises ρ n 0 a.e. in R and R ρ n = µ for a xed µ > 0 and all n N. Then there exists a subsequence {ρ nk } k satisfying one of three properties: (1) (Compactness). There exists a sequence {y k } k R such that for every ε > 0, there exists r < satisfying for all k N: or (next page...) yk +r y k r ρ nk dx µ ε.

18 Concentration-compactness cont. (2) (Vanishing). For all r <, y+r lim sup ρ nk dx = 0 k y R y r (3) (Dichotomy). There exists µ (0, µ) such that for every ε > 0 there exists a k 0 1 and two sequences of positive functions {ρ (1) k } k, {ρ (2) k } k L 1 (R) satisfying for k k 0 ρ nk R R ( ρ (1) k + ρ (2) k ρ (1) k dx µ ε ) L 1 ε ρ (2) k dx (µ µ) ε dist(supp(ρ (1) k ), supp(ρ(2) )). k

19 Rough outline of the proof step 1 Take a minimizing sequence (of I q or Γ λ ) and nd something reasonable to apply Lemma 1 to. step 2 Preclude vanishing and dichotomy to conclude that compactness occurs. step 3 Prove that any minimizing sequence has a subsequence that converges in H s/2 (R) to a minimizer using standard convergence and lower semi-continuity arguments.

20 Remarks Much of the argumentation is "standard" and relies only on R ulu + u2 dx u 2, allowing one to use Sobolev H s/2 (R) embeddings. This is guaranteed by the growth condition in (A). The main challange is the preclude dichotomy, where the non-local nature of L comes into play.

21 Dichotomy. If {u n } n is a minimizing sequence for I q, then ρ n = 1 2 u2 n obvious candidate for Lemma 1, with µ = q. is an

22 Dichotomy. If {u n } n is a minimizing sequence for I q, then ρ n = 1 2 u2 n obvious candidate for Lemma 1, with µ = q. Choose ϕ C satisfying is an ϕ(x) = { 1, if x < 1, 0, if x > 2, and ψ C such that ϕ 2 + ψ 2 = 1 and 0 ϕ, ψ 1.

23 Dichotomy. Dichotomy q (0, q) and a subsequence {u n } n of {u n } n such that for any ε > 0, N N, sequences {y n } n R and R n, such that setting we have for all n N. ϕ n (x) = ϕ((x y n )/R n ), ψ n (x) = ψ((x y n )/R n ), u (1) n = ϕ n u n, u (2) n = ψ n u n, Q(u (1) n ) q < ε, Q(u (2) n ) (q q) < ε, R n x y n 2R n u 2 n dx ε

24 In general one can show that for q 1, q 2 > 0: I q1 +q 2 < I q1 + I q2. We have: lim inf n [ ] E(u n (1) ) + E(u n (2) ) I q + I q q + ε, and lim inf E(u n) = I q. n [ ] What is the relation between E(u n ) and E(u n (1) ) + E(u n (2) )?

25 E(u n (1) ) + E(u n (2) ) =E(u n ) + ϕ n u n (L(ϕ n u n ) ϕ n Lu n ) dx (6) R + ψ n u n (L(ψ n u n ) ψ n Lu n ) dx (7) R [ + (ϕ 2 n ϕ p+1 n ) + (ψn 2 ψn p+1 ) ] F (u n ) dx. R Need (6) and (7) to be smaller than ε for large n.

26 E(u n (1) ) + E(u n (2) ) =E(u n ) + ϕ n u n (L(ϕ n u n ) ϕ n Lu n ) dx (6) R + ψ n u n (L(ψ n u n ) ψ n Lu n ) dx (7) R [ + (ϕ 2 n ϕ p+1 n ) + (ψn 2 ψn p+1 ) ] F (u n ) dx. R Need (6) and (7) to be smaller than ε for large n. Note: This is not true without a continuity assumption on m.

27 Main results Theorem 2 (Existence) Assume L satises (A) and f satises (B). (i) Let p (1, 2s + 1) and q > 0. Then the set D q is non-empty. Moreover, if {u n } n H s/2 (R) is a minimizing sequence for I q, then there exists a sequence {y n } n R such that the sequence {ũ n } n dened by ũ n (x) = u n (x + y n ) has a subsequence that converges in H s/2 (R) to an element of D q. (ii) If p (1, 1+s 1 s ), then for any κ > 0 and λ > 0 the set G λ is non-empty and the statement above holds for minimizing sequences of Γ λ. If f satises (B2), then the result holds also for λ < 0. Remark: By the structure of the equations, any H s/2 (R) solution will, in fact, lie in H s (R).

28 Main results Corollary 3 (conditional energetic stability) For any q > 0, the set D q is a stable set for the initial value problem of (1) in the following sense: for every ε > 0 there exists δ > 0 such that if inf u 0 w H w D s/2 (R) < δ, q where u(x, t) solves (1) with u(x, 0) = u 0 (x), then inf u(, t) w H w D s/2 (R) < ε, q for all t R. For any λ > 0, all positive scalings of the set G λ are stable sets for the initial value problem of (2) in the sense above.

29 Comments on stability Corollary 3 is a simple consequence of Theorem 2 if the functionals are time invariant for solutions with H s/2 (R) initial data. If u solves (1) with u 0 H s/2 (R), then E(u) and Q(u) are independent of t. J (u) and U(u) are time invariant for solutions of (2), but not for solutions of (1). (the proof of) stability of solitary-wave solutions to (1) fails at the critical exponent p = 2s + 1 and beyond. The solutions found lie in H s (R), but we have stability w.r.t. the H s/2 (R) norm.

30 Main results: viewer friendly version Theorem 4 (Summary) (i) If p (1, 1+s 1 s ), there exists sets of Hs/2 (R) solution to (3) for any given wave speed c > 0. (ii) If p (1, 2s + 1), there exists stable sets of H s/2 (R) solutions to (3), where the wave speed c is a Lagrange multiplier. (iii) If p (1, 1+s 1 2 ), there exists stable sets of Hs/2 (R) solutions to (4) for any given wave speed c > 0.

31 Solitary-waves of the capillary Whitham equation Theorem 2 (ii) gives existence of solitary-waves solutions to the capillary Whitham equation for any positive wave-speed c. However, we are in the critical case where our proof of stability fails. It is therefore still an open question whether the capillary Whitham equation admits stable (sets of) solitary-wave solutions.

32 Thank you for your attention!