Soliton Stability in 2D NLS: Linear Stability Analysis and High-Frequency Limit

Soliton Stability in 2D NLS

Natalie Sheils
sheilsn@seattleu.edu
April 10, 2010

Natalie Sheils Soliton Stability in 2D NLS

Outline

1. Introduction to NLS


Outline

Trivial-Phase Solutions of NLS


Outline

Soliton Solution of NLS


Outline

2. Linear Stability


Outline

2. Linear Stability
3. High-Frequency Limit


Outline

2. Linear Stability
3. High-Frequency Limit
4. Future Work


Introduction to NLS

The two-dimensional cubic nonlinear Schr¨dinger equation (NLS),
o

iψt + ψxx − ψyy + 2|ψ|2 ψ = 0.


Introduction to NLS

o

iψt + ψxx − ψyy + 2|ψ|2 ψ = 0.

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


Introduction to NLS

o

iψt + ψxx − ψyy + 2|ψ|2 ψ = 0.

pulse propagation along optical ﬁbers.


Introduction to NLS

o

iψt + ψxx − ψyy + 2|ψ|2 ψ = 0.

pulse propagation along optical ﬁbers.
surface waves on deep water.


Introduction to NLS

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



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


A speciﬁc NLS solution ψ is called a soliton solution.

ψ(x, t) = sech(x)e it
Ψ x, 0
1.0

0.8

0.6

0.4

0.2

x
20 10 10 20


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 + u + i v + O( 2 ))

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


Linear Stability

We substitute into NLS and simplify. We know is small, so the
terms with the lowest order of are dominant. The O( 0 ) terms
cancel out so O( 1 ) is the leading order.

−ut = vxx − vyy + (2φ2 (x) − 1)v
(1)
vt = uxx − uyy + (6φ2 (x) − 1)u


Linear Stability

Since the coeﬃcients of the perturbed system (1) do not explicitly
depend 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.


Linear Stability


v (x, y , t) = V (x)e iρy +Ωt + c.c.

Ω gives us the following conditions for spectral stability:


Linear Stability


v (x, y , t) = V (x)e iρy +Ωt + c.c.

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


Linear Stability


v (x, y , t) = V (x)e iρy +Ωt + c.c.

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


Linear Stability


v (x, y , t) = V (x)e iρy +Ωt + c.c.

If all Ω have negative real parts, the solution is stable.
If all Ω are purely imaginary, the solution is stable.


Linear Stability

Then, U and V satisfy the following diﬀerential equations.

ΩU = V − ρ2 V − 2V φ2 − V
(2)
−ΩV = U − ρ2 U − 6Uφ2 − U


Linear Stability


High-Frequency Limit



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

U(x) ∼ u0 (µx) + ρ−1 u1 (µx) + ρ−2 u2 (µx) + . . .
V (x) ∼ v0 (µx) + ρ−1 v1 (µx) + ρ−2 v2 (µx) + . . .
µ ∼ ρ + µ0 + µ1 ρ−1 + µ2 ρ−2 + . . .
Ω ∼ ω−2 ρ2 + ω3 ρ−3 .
Pick z = µx.



At leading order in ρ, equation (2) becomes:
(4) 2
v0 + 2v0 + (1 + ω−2 )v0 = 0.



At leading order in ρ, equation (2) becomes:
(4) 2
v0 + 2v0 + (1 + ω−2 )v0 = 0.

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



Now we have
√ √ √ √
−w −1
v0 = c1 e z + c2 e −z −w −1
+ c3 e z w −1
+ c4 e −z w −1

where ci ’s are complex constants.



Now we have
√ √ √ √
−w −1
v0 = c1 e z + c2 e −z −w −1
+ c3 e z w −1
+ c4 e −z w −1

where ci ’s are complex constants.

If v0 is bounded, u0 is bounded and we ﬁnd u0 to be
√ √ √ √
−w −1
u0 = −ic1 e z − ic2 e −z −w −1
+ ic3 e z w −1
+ ic4 e −z w −1
.



The next order of ρ is O(ρ):
(4) (4)
v1 + 2v1 + (1 − w 2 )v1 = −2iw µ0 u0 − 2µ0 v0 − 2µ0 v0 .



(4) (4)
v1 + 2v1 + (1 − w 2 )v1 = −2iw µ0 u0 − 2µ0 v0 − 2µ0 v0 .

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



(4) (4)
v1 + 2v1 + (1 − w 2 )v1 = −2iw µ0 u0 − 2µ0 v0 − 2µ0 v0 .

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

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

µ0 = 0.



Now we have
√ √ √ √
−w −1
v1 = c5 e z + c6 e −z −w −1
+ c7 e z w −1
+ c8 e −z w −1

and
√ √ √ √
−w −1
u1 = −ic5 e z − ic6 e −z −w −1
+ ic7 e z w −1
+ ic8 e −z w −1
.



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



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



We need to make sure the particular solution of the
nonhomogeneous equation is bounded.
µ1 ∈ R
µ2 =0.


Future Work


Future Work

Continue looking at orders of ρ.


Future Work

Continue looking at orders of ρ.
We hope to ﬁnd that ω3 is the ﬁrst ωi with nonzero real part.


Acknowledgments

Dr. John Carter of Seattle University


Acknowledgments

Seattle University College of Science and Engineering


Acknowledgments

Seattle University College of Science and Engineering
Paciﬁc Northwest Section of the Mathematical Association of
America


Questions

Questions?


Soliton Stability in 2D NLS: Linear Stability Analysis and High-Frequency Limit

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