soliton stability of the 2d nonlinear schrödinger equation

Soliton Stability in 2D NLS

Natalie Sheils

[email protected]

April 10, 2010

Natalie Sheils Soliton Stability in 2D NLS

Outline

1. Introduction to NLS

I Trivial-Phase Solutions of NLSI Soliton Solution of NLS

2. Linear Stability

3. High-Frequency Limit

4. Future Work

Outline

1. Introduction to NLSI Trivial-Phase Solutions of NLS

I Soliton Solution of NLS

2. Linear Stability

4. Future Work

Outline

1. Introduction to NLSI Trivial-Phase Solutions of NLSI Soliton Solution of NLS

2. Linear Stability

4. Future Work

Outline

2. Linear Stability

4. Future Work

Outline

2. Linear Stability

4. Future Work

Outline

2. Linear Stability

4. Future Work

Introduction to NLS

The two-dimensional cubic nonlinear Schrodinger equation (NLS),

iψt + ψxx − ψyy + 2|ψ|2ψ = 0.

Among many other physical phenomena, NLS arises as a model of

I pulse propagation along optical fibers.

I surface waves on deep water.

Introduction to NLS

iψt + ψxx − ψyy + 2|ψ|2ψ = 0.

Introduction to NLS

iψt + ψxx − ψyy + 2|ψ|2ψ = 0.

Introduction to NLS

iψt + ψxx − ψyy + 2|ψ|2ψ = 0.

Introduction to NLS

Figure: Wave tank in the Pritchard Fluid Mechanics Labratory in theMathematics Department at Penn State University.

Trivial-Phase Solutions of NLS

NLS admits a class of 1-D trivial-phase solutions of the form

ψ(x , t) = φ(x)e iλt

where φ is a real-valued function and λ is a real constant.

A specific NLS solution ψ is called a soliton solution.

ψ(x , t) = sech(x)e it

-20 -10 10 20x

1.0ΨHx, 0L

A specific NLS solution ψ is called a soliton solution.

ψ(x , t) = sech(x)e it

-20 -10 10 20x

1.0ΨHx, 0L

Linear Stability

In order to examine the stability of trivial-phase solutions to NLS,

ψ(x , y , t) = φ(x)e it

we add two-dimensional perturbations

ψ(x , y , t) = e it(φ+ εu + iεv +O(ε2))

where ε is a small real constant and u = u(x , y , t) andv = v(x , y , t) are real-valued functions.

Linear Stability

We substitute into NLS and simplify. We know ε is small, so theterms with the lowest order of ε are dominant. The O(ε0) termscancel out so O(ε1) is the leading order.

−ut = vxx − vyy + (2φ2(x)− 1)v

vt = uxx − uyy + (6φ2(x)− 1)u(1)

Linear Stability

Since the coefficients of the perturbed system (1) do not explicitlydepend on y or t, we may separate variables, and let

u(x , y , t) = U(x)e iρy+Ωt + c .c .

v(x , y , t) = V (x)e iρy+Ωt + c .c .

Ω gives us the following conditions for spectral stability:

I If any Ω has positive real part, the solution is unstable.

I If all Ω have negative real parts, the solution is stable.

I If all Ω are purely imaginary, the solution is stable.

Linear Stability

u(x , y , t) = U(x)e iρy+Ωt + c .c .

v(x , y , t) = V (x)e iρy+Ωt + c .c .

Linear Stability

u(x , y , t) = U(x)e iρy+Ωt + c .c .

v(x , y , t) = V (x)e iρy+Ωt + c .c .

Linear Stability

u(x , y , t) = U(x)e iρy+Ωt + c .c .

v(x , y , t) = V (x)e iρy+Ωt + c .c .

Linear Stability

u(x , y , t) = U(x)e iρy+Ωt + c .c .

v(x , y , t) = V (x)e iρy+Ωt + c .c .

Linear Stability

Then, U and V satisfy the following differential equations.

ΩU = V − ρ2V − 2Vφ2 − V ′′

−ΩV = U − ρ2U − 6Uφ2 − U ′′(2)

Linear Stability

High-Frequency Limit

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01234567051849:!;70<51:5!;1!

=8052<8706<<996!;7059684:!;1!

In our linear stability problem (2) we assume ρ is large and

U(x) ∼ u0(µx) + ρ−1u1(µx) + ρ−2u2(µx) + . . .

V (x) ∼ v0(µx) + ρ−1v1(µx) + ρ−2v2(µx) + . . .

µ ∼ ρ+ µ0 + µ1ρ−1 + µ2ρ

−2 + . . .

Ω ∼ ω−2ρ2 + ω3ρ

Pick z = µx .

At leading order in ρ, equation (2) becomes:

v(4)0 + 2v ′′0 + (1 + ω2

−2)v0 = 0.

In solving this equation, we want v0 to be bounded. This impliesthat ω−2 is purely imaginary and −1 < iω−2 < 1. Pick w = iω−2 .

At leading order in ρ, equation (2) becomes:

v(4)0 + 2v ′′0 + (1 + ω2

−2)v0 = 0.

In solving this equation, we want v0 to be bounded. This impliesthat ω−2 is purely imaginary and −1 < iω−2 < 1. Pick w = iω−2 .

Now we have

v0 = c1ez√−w−1 + c2e

−z√−w−1 + c3e

w−1 + c4e−z√

where ci ’s are complex constants.

If v0 is bounded, u0 is bounded and we find u0 to be

u0 = −ic1ez√−w−1 − ic2e

−z√−w−1 + ic3e

w−1 + ic4e−z√

w−1.

Now we have

v0 = c1ez√−w−1 + c2e

−z√−w−1 + c3e

w−1 + c4e−z√

where ci ’s are complex constants.

If v0 is bounded, u0 is bounded and we find u0 to be

u0 = −ic1ez√−w−1 − ic2e

−z√−w−1 + ic3e

w−1 + ic4e−z√

w−1.

The next order of ρ is O(ρ):

v(4)1 + 2v ′′1 + (1− w2)v1 = −2iwµ0u

′′0 − 2µ0v

(4)0 .

We want v1 to be bounded so we require the right-hand side of theequation to be orthogonal to the solution of the homogeneousequation.

In this case, we require our right-hand side to be zero. Then wehave the following restriction:

µ0 = 0.

v(4)1 + 2v ′′1 + (1− w2)v1 = −2iwµ0u

′′0 − 2µ0v

(4)0 .

µ0 = 0.

v(4)1 + 2v ′′1 + (1− w2)v1 = −2iwµ0u

′′0 − 2µ0v

(4)0 .

µ0 = 0.

Now we have

v1 = c5ez√−w−1 + c6e

−z√−w−1 + c7e

w−1 + c8e−z√

u1 = −ic5ez√−w−1 − ic6e

−z√−w−1 + ic7e

w−1 + ic8e−z√

w−1.

I For the next few orders of ρ the general solution of thehomogeneous problem is the same as the previous orders.

I We need to make sure the particular solution of thenonhomogeneous equation is bounded.

I µ1 ∈ RI µ2=0.

Future Work

I Continue looking at orders of ρ.

I We hope to find that ω3 is the first ωi with nonzero real part.

Future Work

Acknowledgments

I Dr. John Carter of Seattle University

I Seattle University College of Science and Engineering

I Pacific Northwest Section of the Mathematical Association ofAmerica

Acknowledgments

Questions

Questions?

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