fourier-like method for solution of linearized korteweg-de vries...

Fourier-like method for solution of linearizedKorteweg-de Vries Equation

Tyler Bolles

Math 651

April 17, 2018

Tyler Bolles (Math 651) Fourier-like method for solution of linearized Korteweg-de Vries EquationApril 17, 2018 1 / 24

Outline

1 Introduction

2 Review of Spectral Theory

3 Review Of Scattering

4 Sachs’ Result


Introduction

Consider the KdV equation

qt + qxxx − 6qqx = 0 (1)

Let q(x, t) be a solution and consider a perturbation q + u for u small:

(q + u)t + (q + u)xxx − 6(q + u)(q + u)x ≈ ut + uxxx − 6(qu)x (2)

How to find exact solution to the initial value problem

ut + uxxx − 6(qu)x = 0

u(x, 0) = u0(x)(3)

Analysis of solution concerns stablity of q as a solution to KdV.

Sachs [4] found a Fourier-like transform involving Jost solutions valid forwide class of initial conditions!


Spectral Theory in finite dimensions (See [2])

The Cauchy integral formula

f (z) =1

2πi

∫C

f (w)

w− zdw (4)

for complex valued functions. What about operator valued functions?

f (T) =1

2πi

∫C

f (λ)

λ− Tdλ (5)

where T is an operator on a complex Banach space X. For the moment, let’ssuppose X is finite dimensional.

DefinitionThe spectrum σ(T) of the operator T ∈ L(X) is the set of complex numbers λfor which the operator λI − T is not invertible. On σ(T)c, denoteR(λ; T) = (λI − T)−1 the resolvent. The index ν(λ) is the smallestnon-negative integer for which (λI − T)νx = 0 for all x ∈ ker(λI − T)ν+1.


TheoremIf P and Q are polynomials, P,Q : L(X)→ L(X), thenP(T) = Q(T)⇔ (P− Q)(λ) : C→ C has a zero of order at least ν(λ) ateach point λ ∈ σ(T).

Proof.The minimum polynomial of T is mT(λ) = (λ− λ1)ν(λ1) · · · (λ− λn)ν(λn) forσ(T) = {λ1, . . . , λn}.

Next an important concept:

DefinitionDenote F(T) the family of functions which are analytic on some opencovering of σ(T). The covering may be dependent on the function.


TheoremIn finite dimensions, {f (T)|f ∈ F(T)} is the finite dimensional vector spaceof polynomials in T

Proof.Suppose for each λi ∈ σ(T), f has a Taylor seriesf (λ) =

∑∞n=0

f (n)(λi)n! (λ− λi)

n. Then there exists a polynomial P(λ) of degree≤∑

λi∈σ(T)(ν(λi)− 1) such that Pm(λi) = f m(λi) form = 0, 1, . . . , ν(λi)− 1. In this case, the Weierstrass factorization theoremimplies f (λ)−P(λ)

mT(λ) ∈ F(T)⇒ f (T) = P(T) wherever f exists even thoughP(λ) 6= f (λ) for λ 6∈ σ(T)

Images of analytic functions are defined only up to so many terms in theirTaylor series.


Example: Let T = I and f be analytic on a neighborhood of one and definep(λ) = f (1) = 1. Then f (I)− p(I) =

∑∞n=1

f (n)(λi)n! (I − I)n = 0.

Define eλ0(λ) being one on a neighborhood of λ0 and zero in a neighborhoodof σ(T) \ {λ0}.

Projections: E(λ0) ≡ eλ0(T)

E(λ0) 6= 0⇔ λ0 ∈ σ(T),

E(λ0)E(λ1) = 0 if λ0 6= λ1 and E(λ0)2 = E(λ0), and∑λi∈σ(T) E(λi) = 1.

Jordan normal form decomposition:

X = E(λ1)X ⊕ · · · ⊕ E(λn)X

E(λi)X = {x ∈ X|(T − λiI)ν(λi)x = 0}

f (T) =∑

λi∈σ(T)

∑ν(λi)−1n=0

f (n)(λi)n! (T − λiI)nE(λi)


Back to Cauchy integral formula.

Suppose f ∈ F(T) is defined on an open set U ⊃ σ(T) and for λ 6∈ σ(T),define r(ξ) = (λ− ξ)−1 on U.

⇒ r(T) =∑

λi∈σ(T)

ν(λi)−1∑n=0

(T − λiI)n

n!(λ− λi)n+1 E(λi)

so that

12πi

∫∂U

f (λ)r(T)dλ =1

2πi

∫∂U

f (λ)∑

λi∈σ(T)

ν(λi)−1∑n=0

(T − λiI)n

n!(λ− λi)n+1 E(λi)dλ

=1

2πi

∑λi∈σ(T)

ν(λi)−1∑n=0

(T − λiI)n

n!

∫∂U

f (λ)

(λ− λi)n+1 E(λi)dλ

=∑

λi∈σ(T)

ν(λi)−1∑n=0

(T − λiI)n

n!f (n)(λi)

= f (T)

(6)Tyler Bolles (Math 651) Fourier-like method for solution of linearized Korteweg-de Vries EquationApril 17, 2018 8 / 24

Infinite dimensions, T a bounded operator

DefinitionThe resolvent set ρ(T) is the set of complex numbers λ for which theresolvent R(λ; T) = (λI − T)−1 exists as a bounded operator with domain X.The spectrum σ(T) = ρ(T)c.

Lemmaρ(T) is open, R(λ; T) is analytic on ρ(T) and supσ(T) ≤ ‖T‖.

Proof.

Let λ ∈ ρ(T) and µ < ‖R(λ; T)‖−1. Then ‖µR‖ < 1 and

∞∑k=0

(−µ)kR(λ; T)−(k+1) ≡ S(µ) = ((µ+ λ)I − T)−1 (7)

showing λ+ µ ∈ ρ(T) and R(λ+ µ; T) = S(µ) is analytic at µ = 0.


Let U be a neighborhood of σ(T) and suppose f (λ) is analytic on U, i.e.,f ∈ F(T). It will now be a matter of definition that

f (T) =1

2πi

∫∂U

f (λ)R(λ; T)dλ (8)

DefinitionA point λ0 ∈ σ(T) is isolated if there exists a neighborhood U of λ0 such thatU ∩ σ(T) = {λ0}. An isolated point λ0 is said to be a pole of T of orderν(λ0) if it is a pole of order ν(λ0) of R(λ; T). The set σp(T) of λ ∈ σ(T) suchthat λI − T is not injective is called the point spectrum. The set σc(T) ofλ ∈ σ(T) such that λI − T is injective, (λI − T)X is dense in X but(λI − T)−1 is not bounded is called the continuous spectrum.

Notice λI − T is not injective implies there is an x ∈ X such that Tx = λx.Thus λ ∈ σp(T)⇔ ker(λI − T) is nontrivial.


Theorem (Spectral Mapping Theorem)If f ∈ F(T), then f (σ(T)) = σ(f (T)).

Notice the above theorem says that if f (T) = 0, then f (σ(T)) = 0 but it doesnot imply that f (λ) ≡ 0 on every neighborhood of σ(T).

Theorem (Minimal Equation)Let f , g ∈ F . Then f (T) = g(T) if and only if f (λ) = g(λ) on aneighborhood of σ(T) or, if σ ≡ {λ1, . . . , λk} is a collection poles of orderν(λi), a neighborhood of σ(T) \ σ. In this case, f − g must have a zero oforder at least ν(λi).

Indicator functions play same role:

eλ0(T) = 12πi

∫Cλ0

(λI − T)−1dλ = 0⇔ λ 6= λ0

eλ0(λ)eλ1(λ) ≡ 0⇒ eλ0(T)eT(λ) = 0 by above theorem, and

eλ0(λ)2 = eλ0(λ)⇒ eλ0(λ)(eλ0(λ)− 1) ≡ 0⇒ eλ0(T)(eλ0(T)− 1) = 0⇒ eλ0(T)2 = eλ0(T)


Proof of Minimal Equation.(Sketch) WLOG, set g(T) = 0.(⇒) Let f (z) = 0 on an open set containing σ(T) \ σ;

f (T) =1

2πi

k∑i=1

∫Ci

f (λ)R(λ; T)dλ (9)

Apply Cauchy’s theorem: f has a zero at λi of at least order ν(λi)⇒ f (T) = 0(⇐) Suppose f (λ) is analytic on a neighborhood U ⊃ σ(T) s.t. f (T) = 0. UseCompactness of σ(T) (boundedness of T) to show f (λ) ≡ 0 on neighborhoodsof accumulation points of σ(T) since σ(f (T)) = σ(0) = 0 = f (σ(T)). If λ1isolated and f ≡ 0 on every nbhd of λ1, done. Suppose f (λ) 6≡ 0 around λ1.Spectral mapping shows ∃n > 0 s.t. λ1 is a zero of order n of f . Therefore,(λ− ξ)n/f (ξ) is analytic⇒ (λ1I − T)keλ1(T) = (λ−T)k

f (T) f (T) = 0 for all k ≤ nimplying λ1 is a pole of order at most n of T .


Unbounded, closed operators

DefinitionLet the domain of T D(T) ⊂ X a Banach space. An operator is said to beclosed if its graph (x,Tx) is closed, i.e., if xn → x and Txn → y, thenx ∈ D(T) and Tx = y.

Now F(T) will denote the set of functions analytic on a neighborhood ofσ(T) and at complex infinity.

How to get f (T) using the Cauchy integral for unbounded σ(T)?

Map to a bounded set: let α ∈ ρ(T) and define A = (T − αI)−1 andµ = Φ(λ) = (λ− α)−1.

LemmaFor α ∈ ρ(T), Φ(σ(T) ∪∞) = σ(A) and each φ ∈ F(A) is in one-to-onecorrespondence with an f ∈ F(T) via φ(µ) = f (Φ−1(µ)). It follows byboundedness of A that σ(T) is closed.


DefinitionFor f ∈ F(T), define f (T) = φ(A) where φ(µ) = f (Φ−1(µ))

The meaning for the above definition is that since α ∈ ρ(T) and ρ(T) is open,Φ(σ(T) ∪∞) = σ(A) is bounded. Hence φ(µ) can be defined as a contourintegral enclosing σ(T), from which f can be recovered.

TheoremIf f ∈ F(T), then f (T) is independent of α ∈ ρ(T). Let V ⊃ σ(A) be openwith boundary Γ being the finite union of Jordan arcs such that f is analyticon V ∪ Γ. Let Γ have positive orientation w.r.t. the possibly unbounded set V.Then

f (T) = f (∞)I +1

2πi

∫Γ

f (λ)R(λ; T)dλ (10)

Proof.Notice the integral is independent of α. In fact, since f (λ)R(λ; T) is analyticon ρ, we may assume α 6∈ V ∪ Γ. Integral then follows from change ofvariables.


Direct Scattering

The KdV equation has a Lax pair as q is a solution of (1) is equivalent to

dLdt

= [B,L] (11)

whereL = −∂2

x + q, B = −4∂3x + 3q∂x + 3∂xq (12)

In light of (11) the spectrum of L is time-invariant. This in turn motivates theauxiliary problem at time t=0:

Lψ = −∂2xψ + q(x, t)ψ = λψ (13)


If ‖(1 + |x|)q(x)‖1 <∞, σp(L) = {λ ∈ R : Lψ = λψ and λ < 0} is finite

Setting λ = k2, if Im k = 0:

Jost solutions satisfy (L− k2I)f± = 0 and form a basis of solutions withboundary data f±(x, k) ∼ e±ikx as x→ ±∞‖f±‖L2(R) =

∥∥(L− λI)−10∥∥

L2(R)=∞

f± correspond to σc(L)

If Im k > 0:

Lψ = k2ψ ⇒ ψ(x) ∼ e−√−λ|x| as |x| → ∞ a possibility

Jost solution are a single L2 eigenfunction iff their Wronskian=0

T(k) has simple poles in this case

If ψ ∈ L2 so assume ‖ψ‖2 = 1. Then there are unique ck s.t.limx→∞ ψ(x)eikx = ck


f+(x, k) a solution for⇒ Im k = 0⇒ f ∗+ (complex conjugate) a solution

Wronskian [f+, f ∗+] = 2ik implies f+ and f ∗+ form a basis.

There exists a(k) and b(k) s.t. for f−(x, k) ∼ e−ikx as x→ −∞,

f− = af ∗+ + bf+ (14)

By Cramer’s rule,

a(λ) =[f+, f−]

[f+, f ∗+]=

[f+, f−]

2ik(15)

b(λ) =[f−, f ∗+]

[f+, f ∗+]=

[f−, f ∗+]

2ik. (16)

We can re-write (14) as

T(k)f+ = f ∗+ + R(k)f+ (17)

where T = 1/a and R = b/a are the transmission and reflection coefficients.


The scattering data for KdV is the collection R(k), σ(L) and the ck.

Nice properties:

σ(L) is conserved in time

ck(t) = ck(0)e4k3t

R(k, t) = R(k, 0)e8ik3t, analytic on Im k > 0

T(k) conserved in time, meromorphic for Im k > 0, simple poles onσp(L).

f±(x, k, t) ∼ exp{±i(kx + 4k3t)

}as x→ ±∞ analytic on Im k > 0

|f±/e±ikx − 1| ≤ exp{

C1|k|−1}C2|k|

|∂x(f±/e±ikx)−∫∞

x e±2ik(y−x)Q(y)dy| ≤ C3(1 + |k|)−1

T(k) = 1 +O(k−1) as |k| → ∞.

See [3] for scattering, [1] for proofs of inequalities.


The solution at a later time q(x, t) is

q(x, t) = −2∂xK(x, x) (18)

where K satisfies the Gel’fand-Levitan integral equation

K(x, y) + B(x + y) +

∫ ∞x

B(y + z)K(y, z)dz = 0 (19)

where

B(ξ) =

N∑n=1

c2neiknξ

︸︷︷︸point spectrum

+1

2π

∫R

b(k)eikξdk︸︷︷︸continuous spectrum

(20)


Sach’s Result

The formal adjoint of the linearized KdV equation is

vt + vxxx − 6qvx = 0. (21)

So if v is a solution to (21), we have

vxt + vxxxx − 6∂x(qvx) = 0 (22)

or by writing u = vx,ut + uxxx − 6(qu)x = 0, (23)

the linearized KdV equation. Suffices to solve (21).

If ψ a solution to the Schrodinger equation, then for each t,q(x, t) = ψxx(x,t)+k2ψ(x,t)

ψ(x,t) so that

0 = qt + qxxx − 6qqx

= ∂t

(ψxx + k2ψ

ψ

)+ ∂xxx

(ψxx + k2ψ

ψ

)− 6

(ψxx + k2ψ

ψ

)∂x

(ψxx + k2ψ

ψ

)(24)


This gives R ≡ ψt + ψxxx − 3(q + k2)ψx satisfies LR = k2R

If ψ ∼ exp{±i(kx + 4k3t)

}as x→ ±∞, R→ 0 as x→ ±∞ implying R is

the zero solution.

If ψ1 and ψ2 are (not necessarily linearly independent) solutions ofSchrodinger with correct asymptotics, then direct calculation using R = 0shows ψ = ψ1ψ2 is a solution to the adjoint operator

⇒ ∂x(ψ1ψ2) solves linearized KdV.

ψ = ψ1ψ2 also satisfies

Tψx ≡ −14ψxxx + qψx +

12

qxψ = k2ψx. (25)

T has spectrum of L but with eigenfunctions being derivatives of quadraticterms. The (formal) resolvent of T r(k) = (k2I − T)−1 could then be

integrated around a neighborhood of σ(T) to obtain the identity in terms ofsquared eigenfunctions.

Sachs uses this informal analogy to motivate the coming integral relation:Tyler Bolles (Math 651) Fourier-like method for solution of linearized Korteweg-de Vries EquationApril 17, 2018 21 / 24

Theorem

If φ ∈ L1(R) and continuous and if∫R(1 + x2)|q(x, t)|dx <∞, then

φ(x) =

∫R

∫R

T2(k)

4πik∂x(f 2+(x, k)f 2

−(y, k)− f 2−(x, k)f 2

−(y, k))φ(y)dydk

+ residue terms (26)

where the residue occur from k integration crossing the poles of T(k)2 fork ∈ σp(L).

Proof.(Sketch) Carry out y integration. f± → exp(±ikx) and T → 1 as |k| → ∞ sodeform path of k integration on to a semicircle in the upper half plane. Correctfor residue terms at poles of T . Resulting integral reads

1π

∫CR

(∫R

e2ik|x−y|φ(y)dy)

dk +O(R−1) as R→∞ (27)


In light of last theorem, the solution to the initial value problem

ut + uxxx − 6(qu)x = 0

u(x, 0) = u0(x)(28)

is

u(x, t) =

∫R

∫R

T2(k)

4πik∂x(f 2

+(x, t, k)f 2−(y, 0, k)

− f 2−(x, t, k)f 2

−(y, 0, k))u0(y)dydk + residue terms (29)


Bibliography

Percy Deift and Eugene Trubowitz.Inverse scattering on the line.Communications on Pure and Applied Mathematics, 32(2):121–251,1979.

Nelson Dunford and Jacob T Schwartz.Linear operators part I: general theory, volume 7.Interscience publishers New York, 1958.

Clifford S Gardner, John M Greene, Martin D Kruskal, and Robert MMiura.Method for solving the korteweg-devries equation.Physical Review Letters, 19(19):1095, 1967.

Robert L Sachs.Completeness of derivatives of squared schrodinger eigenfunctions andexplicit solutions of the linearized kdv equation.SIAM Journal on Mathematical Analysis, 14(4):674–683, 1983.Tyler Bolles (Math 651) Fourier-like method for solution of linearized Korteweg-de Vries EquationApril 17, 2018 24 / 24

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