On an estimate by Sergej Kuksin concerning a partial differential equation on a torus with variable coefficients

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1 First Version Jan 31, 1996 Updated Oct 8, 1996 On an estimate by Sergej Kuksin concerning a partial differential equation on a torus with variable coefficients THOMAS KAPPELER Universität Zürich JÜRGEN PÖSCHEL Universität Stuttgart We consider the first order partial differential equation Statement of the Result. i ω u + λu + b(x)u = f, x T n, (1) for functions on the torus T n = R n /2πZ n, where ω = n ω ν xν, ω = (ω 1,...,ω n ) R n. ν=1 We make the following assumptions. (A) ω is diophantine: there are constants α >, τ > n and l > such that for all = k Z n. Also, λ αl. k ω + λ αl k τ, k ω α k τ, estimate :3

2 2 Estimate by Sergej Kuksin (B) b is analytic on some complex strip U(s) = { x : Im x < s } C n around T n with mean value zero: [b] = n b(x) dx =, and b s,τ def = k ˆb k k τ e k s γα (C) with some γ>. We also assume s 1 for simplicity. f is analytic on U(s) and bounded: f s def = k ˆ f k e k s <. Proposition (Kuksin [1]) Under assumptions (A), (B), (C) equation (1) has a unique solution u = L f that is analytic on U(s) and satisfies Moreover, if then also u s σ c αl f s e2γ, <σ s. (2) σ τ γ λ σ 5 ω, u s σ c αl f s 1, <σ s. (3) σ τ+n+3 The constant c depends only on τ in the first case, and only on τ and n in the second case. The second estimate which is the difficult one is not needed if there is a uniform bound on γ. But if a whole family of equations with no such bound is considered, then it gives a uniform bound for u s σ provided λ is substantially larger than γ. Corollary If λ ω γ 1+β

3 Estimate by Sergej Kuksin 3 with some β>, then u = L f satisfies the estimate u s σ c αl f s e2(5/σ )1/β, <σ s, σ τ+n+3 where c is a constant which depends only on n and τ. Indeed, with this assumption we have Hence, the second estimate (3) applies, if λ γ ω γ β. γ β 5 σ. Otherwise, the first estimate (2) applies, with γ (5/σ ) 1/β. Combinig these two estimates we obtain the estimate of the Corollary. The proof of the Proposition has two parts. First, the solution u = Lf is constructed by converting (1) via an integrating factor into an equation with constant coefficients. This also provides the first estimate (2). Second, for γ/λ sufficiently small, one considers a sequence of equations (1) where the frequency vector ω is approximated by rational ones and the variable coefficient b by trigonometric polynomials. Each of these equations has a unique periodic solution, which can be represented by an oscillatory integral with a complex valued phase function. By a contour deformation using Cauchy s theorem, these oscillatory integrals are estimated uniformly. Taking limits one obtains the second estimate (3). This intricate and ingenious scheme was first described by S. Kuksin [1]. We reproduce it here in our style, first to make the result more accessible, and second to apply in [2]. Before giving the details we observe that we may divide equation (1) by α and thus replace ω,λ,b and f by α 1 ω,α 1 λ, α 1 b and α 1 f, respectively. In the assumptions (A), (B) and (C), α is then replaced by 1. Henceforth, we can assume the normalization α = 1 for the rest of our considerations. The proof of the Proposition now takes N steps, N = 7.

4 4 Estimate by Sergej Kuksin Then 1. Existence and Uniqueness. To obtain an integrating factor B in (1), solve ω B = b, [B] =. B s = ˆB k e k s = ˆb k k ω e k s k = k = k = ˆb k k τ e k s b s,τ γ. Hence, B is analytic in U(s). Now set u = e ib v, f = e ib g to obtain i ω v + λv = g. This equation has a unique solution v by comparing Fourier coefficients: (k ω + λ) ˆv k =ĝ k, k Z n. We get v s σ = k = k k ˆv k e k (s σ) ĝ k k ω + λ e k (s σ) 1 l ĝ k (1 + k τ )e k (s σ) 1 l g s sup (1 + t τ )e σ t t 1 l g s c σ τ with some constant c depending only on τ. Going back we obtain the unique analytic solution u = e ib v of (1) with the estimates u s σ e B s v s σ e γ v s σ, g s e B s f s e γ f s,

5 Estimate by Sergej Kuksin 5 where we made use of the fact that the norm is multiplicative: uv s u s v s. Hence, u s σ c l f s e 2γ σ τ as stated in (2). 2. Approximation. To obtain estimate (3) we now assume that γ λ σ 5 ω, (4) and approximate ω by rational frequency vectors ω ν. The following is proven at the end; recall that α = 1. Approximation Lemma There exists a sequence of frequency vectors ω ν = 2π T ν m ν with T ν,m ν Z n, and a sequence K ν such that for all ν, (a) ω ω ν 2π T ν, (b) k ω ν 1 2 k τ, < k < K ν, (c) k ω ν + λ l, k =, 2 k τ+n+2 Consequently, also (d) γ λ σ 4 ω ν for large ν. 3. The periodic problem. For each large ν we now consider the approximate, periodic problem i ων u + λu + b ν (x)u = f,

6 6 Estimate by Sergej Kuksin where ω ν = 2π T ν m ν, b ν = < k <K ν ˆb k e ik x. Fixing ν we drop it from the notation in the following, which hopefully does not lead to confusions. We may solve again ω B = b with [B] =. Using (b) of the Approximation Lemma we get, as before, B s 2 b s,τ 2γ, and with u = e ib v, f = e ib g, we obtain the equation Its solution has the integral representation i ω v + λv = g. (5) v(x) = η T e iλt g(x + tω) dt, η T = i e iλt 1. (6) To verify this write w(t) = v(x + tω) and h(t) = g(x + tω). Both functions are periodic in t with period T. Substituting into (5) we find d w(t) + iλw(t) = ih(t). dt With an integrating factor e iλt and a subsequent integration over [, T ] this leads to (6). From this and the substitutions results the representation of u as an oscillatory integral: u(x) = η T = η T e iλt+ib(x+tω) ib(x) f (x + tω) dt e iλ(t+δ[b(x+tω) B(x)]) f (x + tω) dt (7) with δ = 1 λ.

7 Estimate by Sergej Kuksin 7 Our aim is to obtain a uniform estimate for x U(s 2σ). In the following we therefore fix x with Im x < s 2σ and seek such an estimate for u(x). 4. Stationary Phase Transformation. We want to find a change of the integration variable t = φ(r) with φ() = and φ(t ) = T such that We then obtain φ(r) + δ ( B(x + φ(r)ω) B(x) ) = r. (8) u(x) = η T e iλr f (x + φ(r)ω)φ (r) dr, which we will use to estimate u. To find φ we lift the problem to the torus T n and write φ(r) = r + ψ(rω), ψ : T n C. Also, let rω = ξ T n. From (8) we obtain the equation ψ(ξ) + δ ( B(x + ξ + ψ(ξ)ω) B(x) ) = (9) which we solve by a contraction argument. Consider the space A of functions ψ : U(σ ) C with Define a map T on A by (i) ψ analytic and periodic on U(σ ), (ii) ψ() =, (iii) ψ σ = sup ψ(ξ) 4γδ. ξ U(σ ) T (ψ)(ξ) = δ ( B(x + ξ + ψ(ξ)ω) B(x) ) To show that T is well defined we use that γδ = γ λ by (d) in the Approximation Lemma. Then σ 4 ω Im(ψ(ξ)ω) ψ σ ω 4γδ ω σ,

8 8 Estimate by Sergej Kuksin hence Im(x + ξ + ψ(ξ)ω) < s 2σ +σ +σ = s. It follows that B(x +ξ +ψ(ξ)ω) and thus T (ψ) are well defined. T maps A into A: (i) and (ii) clearly hold, and, as B σ B s 2γ, Further, T is a contraction: T (ψ) σ 2δ B σ 4γδ. T (ψ)(ξ) T (χ)(ξ) = δ ( B(x + ξ + χω) B(x + ξ + ψω) ) = δ = δ 1 1 db(x + ξ(s)),(ψ χ)ω ds b(x + ξ(s))(ψ χ)ds with ξ(s) = ξ + ((1 s)χ + sψ)ω, hence, using assumption (B), T (ψ) T (χ) σ δ b s,τ ψ χ σ γδ ψ χ σ 1 2 ψ χ σ, since γδ σ/4 ω 1/4 ω, and ω 1 2 for all approximating frequencies. Thus there exists a unique analytic solution ψ A of equation (9). Consequently, φ with φ(r) = r + ψ(rω) solves equation (8). In particular, φ id has period T. Moreover, from the fixed point equation (9) one gets and, again using assumption (B), ω ψ = δ db,ω + ( ω ψ)ω = δb(1 + ω ψ) ω ψ σ γδ(1 + ω ψ σ ) 1 2 (1 + ωψ σ ), hence ω ψ σ Transformation and Estimate. We observe that φ maps the interval [, T ] into a curve Ɣ in U(σ ) with the same endpoints as [, T ]. By Cauchy s theorem, the integral in (7) over [, T ] is the same as over Ɣ. Thus, substituting t = φ(r) = r + ψ(rω) in the integral (7) we obtain u(x) = η T e iλr f (x + φ(r)ω)φ (r) dr = η T e iλr h(x; ξ) ξ = rωdr

9 Estimate by Sergej Kuksin 9 with h(x; ξ) = f (x + ξ + ψ(ξ)ω)(1 + ω ψ(ξ)) analytic on U(σ ), and h(x; ) σ 2 f s. The above integral for u(x) is now easily evaluated. Since h is periodic in ξ, h(x; ξ) = k ĥ k (x)e ik ξ, ĥ k h σ e k σ. Moreover e iλr e ik ξ ξ = rωdr = e i(λ+k ω)r dr = i eiλt 1 k ω + λ = η 1 T k ω + λ. Hence we obtain u(x) = k ĥ k (x) k ω + λ. With (c) of the Approximation Lemma, one obtains for every x U(s 2σ) the estimate u(x) 2 l k τ+n+2 e k σ h σ + 1 λ h σ 1 l f s k = c σ τ+n+3, where c depends only on n and τ. We used that λ l by hypotheses (A) for α = Taking Limits. Summarizing the preceding arguments we obtain the following. For every frequency vector ω ν = 2π T ν m ν, m ν Z n, satisfying (b), (c) and (d) the equation i ων u +λu +b ν u = f has a unique analytic solution u ν satisfying u ν s 2σ 1 l f c s, <σ s. σ τ+n+3 Letting ν and taking (a) of the Approximation Lemma into account, we obtain approximating sequences ω ν ω, b ν b. The corresponding u ν are uniformly

10 1 Estimate by Sergej Kuksin bounded. Hence we can choose a convergent subsequence converging to some solution u of i ω u + λu + bu = f with u s 2σ 1 l f s c σ τ+n+3. From this the second estimate (3) of the proposition follows. Incidentally, since the solution u is unique, actually the whole sequence u ν converges to u. 7. Proof of the Approximation Lemma. Suppose ω satisfies assumption (A). For each t -interval I =[T, T + ], there is an integer vector m Z n such that Hence, T = 2π ω tω 2πm 2π, t I. ω ω t 2π T, ω t = 2π t m, for each t I. This gives (a). To obtain (b) we estimate, for t I arbitrary, k ω t k ω k (ω ω t ) 1 k τ k 2π T 1 2 k τ for 4π k τ+1 T, that is, for k K T def = This gives (b). For (c) we estimate similarly ( ) T 1/(τ+1). 4π k ω t + λ k ω + λ k (ω ω t ) l k τ k 2π T l 2 k τ for t I and 4π k τ+1 lt. So it remains to consider the case 4π k τ+1 lt. (1)

11 Estimate by Sergej Kuksin 11 Assume that for some t I, k ω t + λ l 2, otherwise there is nothing to do. In particular, we have k ω t l/2, since λ l. As long as this holds, we have for ϕ(t) def = k ω t + λ the estimate ϕ (t) = 1 t k ω t l 2t l 4T. Hence the measure of the subset of the t -interval I where can be estimated by k ω t + λ l 2 k τ+n+2 (11) 4T l l 2T = 2 k τ+n+2 k τ+n+2. Summing over all k Z n with (1) the total measure of the subset of t -values in I satisfying (11) can be estimated by c(n) T K τ+2, where c(n) depends only on n, and K is the lower bound of k in (1). That is, Hence the measure is bounded by K = c(l)t 1/(τ+1). c(n, l)t 1/(τ+1) < 2π ω = I, if T is sufficiently large. Note that the dependence of the constant on l is irrelevant here, since we want to make T large anyhow, after l is given. Hence for each large T we can find t [T, T + ] so that (11) does not hold. That is, for such t,(c) holds. This proves the lemma.

12 12 References References [1] S.B. KUKSIN, A KAM theorem for equations of the Korteweg-de Vries type. Preprint (1995). [2] T. KAPPELER &J.PÖSCHEL, Perturbations of KdV equations. In preparation. Universität Zürich, Institut für Reine Mathematik, Winterthurerstrasse 19, CH-857 Zürich tk@math.unizh.ch Mathematisches Institut A, Universität Stuttgart, Pfaffenwaldring 57, D 7569 Stuttgart poschel@mathematik.uni-stuttgart.de