Benjamin-Ono equation: Lax pair and simplicity of eigenvalues

Transcription

1 Benjamin-Ono equation: Lax pair and simplicity of eigenvalues

2 The Benjamin-Ono equation is u t + 2uu x + Hu x x = where the Hilbert transform H is defined as Hϕx) = P.V. π ϕy) x y dy. Unfortunately the definition with the opposite sign also appears in the literature, and the whole thing becomes a bit confusing. Let s stick with this definition for now. The best way to think about the Lax pair is to regard it as being decomposed with respect to H ±, the Hardy spaces of L 2 functions whose Fourier transforms are supported on either the positive or the negative half line. On H +, we have On H, we have where Lv i v x C + uc + v) = λv Bv 2λv x + iv xx + 2C + u x )C + v) Lv i v x + C uc v) = λv Bv 2λv x iv xx + 2C u x )C v) C ± ϕ = ϕ ± ihϕ 2 are the Cauchy projections onto H ±. Notice that when u is real, the equations on H is essentially just the complex conjugate of the equations on H +. Here is the calculation with the + sign, the case with the sign is similar. One first has and L t v = C + u t C + v) = 2C + uu x C + v) + ic + u xx C + v) [L, B]v = ic + u xx C + v) + 2λC + u x C + v) + 2iC + u x C + v x ) + 2C + u x )C + uc + v)) 2C + uc + u x )C + v) eplacing λv by Lv, one eventually gets L t v + [L, B]v =2C + u x uc + v) + 2C + u x )C + uc + v)) 2C + C + u x )uc + v) 2C + u x C + uc + v)) which is seen to be zero since C + fg) + C + f)c + g) C + fc + g) C + gc + f) =.

3 One strange thing is that B does not seem to be skew-adjoint. Consider the H + case. Assuming u to be real, the skew-adjointness of B implies that û x η ξ)ŵξ) dξ = for all η >, which is certainly not true. In the following, we could think of the potential u as belonging to the Schwartz class. Following the previous calculation of the Lax pair, let us consider the operator L = i x C + uc + on H +. Here are a few useful things to know. Lemma. i x C + uc + is a relatively compact perturbation of i x, hence is self-adjoint on H + H. By Weyl s theorem, the essential spectrum of L is [, ). L is also bounded from below. Proof. Standard. Lemma 2. Equivalence of differential and integral equations.. If ϕ H + H, λ <, then if and only if i xϕ C + uϕ) = λϕ ϕ = G λ uϕ). 2. If ϕ H + H, λ <, then if and only if i xϕ ) C + uϕ )) = λϕ ) + C + u ϕ = + G λ uϕ). where G λ x) = e ixξ 2π ξ λ dξ. By the Sobolev embedding theorem, any ϕ satisfying the above equations is bounded and continuous. Proof. Fourier transform. Lemma 3. If λ <, the Hilbert-Schmidt norm of G λ u ) on H + is Proof. Plancherel. λ u 2. 2

4 Lemma 4. G λ x) = where B, B 2 are bounded functions Proof. Direct calculation. See []. Lemma 5. If ϕ L and λ <, as x ±. G λ uϕ)x) = 2πiλx Proof. Direct calculations and estimates. { B λx) if λx > B 2 λx) + log λx if λx ) uy)ϕy) dy + O x 2 Since L is bounded from below and has essential spectrum [, ), it has a discrete set of negative eigenvalues bounded from below, which could possibly accumulate at. We now prove a key formula. I haven t seen this formula in the literature. This might be new. Lemma 6. If λ < is an eigenvalue of L, and ϕ is an eigenvector, then Proof. We have Fourier transform to get In other words uϕ dx 2 = 2π λ ϕ 2 dx. i xϕ C + uϕ) = λϕ. ûϕχ +ξ) = ξ λ) ˆϕ. ûϕ = ξ λ) ˆϕ ) when ξ >. Since uϕ decays, from ) we see that ˆϕ is C on [, ) and ˆϕ as ξ because uϕ L. Consider ˆϕ 2 λ 2 d dξ ˆϕ 2 = 2 [ ˆϕ λ ˆϕ ) ˆϕ + ˆϕ λ ˆϕ ) ˆϕ] = 2 [ûϕ ξ ˆϕ ) ˆϕ + ûϕ ξ ˆϕ ) ˆϕ] 3

5 The last equality follows from the derivative of ). We now integrate, remembering that ˆϕ is supported on + : = 2 = 2 ˆϕ 2 dξ + λ 2 ˆϕ) 2 [ûϕ ξ ˆϕ ) ˆϕ + ûϕ ξ ˆϕ ) ˆϕ] dξ ) ξ ˆϕ 2 ) dξ + ûϕ ˆϕ + ûϕ ˆϕ) dξ Here the second term vanishes by Plancherel s identity. The first term evaluates to ˆϕ 2 dξ. 2 Hence we have By ), ûϕ) = λ ˆϕ). Therefore λ ˆϕ 2 dξ = λ ˆϕ) 2. ˆϕ 2 dξ = ûϕ) 2, which by the Plancherel identity is nothing but the claim. Theorem. All the eigenvalues in the discrete spectrum of L are simple. Proof. Suppose λ < is an eigenvalue. It certainly has finite multiplicity. Assume that the multiplicity is N. Pick an orthogonal basis {φ j } of the eigenspace, and let each φ j be normalized so that uφ j dx = 2πiλ. Note that the left hand side is non-zero by Lemma 6. Let ζ) be the resolvent of L at ζ, we have in a neigborhood of λ see [2]): ζ) = P + hζ). Here P is projection onto the eigenspace, and h is holomorphic at λ. Let W ζ) solve W = + G ζ uw ). 2) 4

6 We have by Lemma 2 W ζ) =ζ)c + u = = = i uφ j dx φ j ) φj φ j + Hζ) ) 2πλφj uφ + Hζ) j dx φ j + Hζ). Here we have used Lemma 6 and the normalization of φ j. We now plug this into 2) to get Hζ) i Hζ) = G ζ uhζ) + )) φ j = G ζ u Hζ) i φ j + )) i G ζ G λ ) u This is because φ j = G λ uφ j ). Take the limit as ζ λ to get Hλ) = G λ u Hλ)+ )) +x By the normalization of φ j we have Hλ) x This equation can be rewritten as where φ j G λ u x φ j ). ) ) φ j + i 2πλ φ j G λ u Hλ) + x u ) ) φ j = N. V G λ uv ) = N)G λ u V = Hλ) x φ j + N. Notice that V L 2 since H L 2 and x φ j dx. φ j + N L 2 by Lemma 5 and normalization of φ j. Denote I G λ u ) by I T. We have N)G λ u) ani T ) ani T ) = KerI T ). 5

7 Apparently T = ug λ. It is not hard to see that KerI T ) = Span N {uφ j }. If N, we must have G λ u) uφ j for every j, a simple calculation shows this implies uφ j dx =, which is impossible by the normalization of φ j. Hence N must be. 6

8 Bibliography [] Coifman,.., and Wickerhauser, M. V. The scattering transform for the Benjamin-Ono equation. Inverse Problems 6, 5 99), 825. [2] Kato, T. Perturbation theory for linear operators, vol. 32. springer, 995. [3] eed, M., and Simon, B. Methods of Modern Mathematical Physics: Vol.: 4.: Analysis of Operators. Academic press New York,