Stability of travelling wavesusing the Evans function

Jitse Niesenin collaboration with Simon Malham

Industrial and Applied Mathematics SeminarHeriot–Watt University, 8 February 2006

Outline

1. Travelling waves and their stability.

2. The Evans function.

3. Computing the Evans function.

4. The limit λ →∞.

Travelling wave solutions

Consider a PDE like the reaction–diffusion equation

yt = Kyxx + F (y)

where y(t, x) ∈ Rd with x ∈ R.

Travelling waves are solutions to the equation that move withconstant speed c while maintaining their shape.

In the moving frame ξ = x − ct, the PDE is

yt = Kyξξ + cyξ + F (y).

Travelling wave solutions are of the form y(t, ξ) = y(ξ).Hence, they satisfy Kyξξ + cyξ + F (y) = 0.

Stability of travelling waves

Given a travelling wave y , what is the fate of solutions whoseinitial conditions are small perturbations of y? If any such solutionstay close to y , we say that the travelling wave is stable.

Technical detail: We must specify function spaces to define “small”and “close”, e.g., C 0

unif (bounded uniformly continuous) or L2.

Orbital stability: If y(ξ) is a travelling wave solution, then so is itstranslate y(ξ − ξ0). So, we require for stability that smallperturbations of y(ξ) stay close to the family y(ξ − ξ0) : ξ0 ∈ R.

Examples: Signals in a nerve (John W. Evans)Neuronal networks (Gabriel Lord et al.)Autocatalytic fronts (Simon Malham et al.)Deformations of an elastic filament (Jocelyn Lega)Vegetation stripes (Jonathan Sherratt)

Linear stability analysis

A natural approach to stability analysis is to linearize the equationabout the travelling wave.

Suppose that y(t, ξ) = y(ξ) + y(t, ξ) is a perturbation of thetravelling wave at t = 0.

y solves the full PDE: Kyξξ + cyξ + f (y) = yt

Travelling wave y solves: Kyξξ + cyξ + f (y) = 0Subtracting: Kyξξ + cyξ + f (y + y)− f (y) = yt

Linearized stability analysis: Kyξξ + cyξ + Df (y) y = yt

Set y(ξ, t) = eλt y(ξ) to get: Kyξξ + cyξ + Df (y) y = λy

So, we need to study the spectrum of the differential operator

L = K∂ξξ + c∂ξ + DF (y).

If L has an eigenvalue λ with Re λ ≥ 0, then the wave is unstable.

The spectrum of LThe spectrum of L can be decomposed in two parts.

The point spectrum σpt consists of eigenvalues λ, for whichLy = λy has a solution. We have 0 ∈ σpt.

The essential spectrum σess typically contains open sets, and isasymptotically (as λ →∞) inside the parabolic regionλ : (Im λ)2 ≤ −C Re λ.

σess

Im λ

Re λ

σpt

Nonlinear stability vs linear stability

If the spectrum of L is contained in the left half of the complexplane and bounded away from the imaginary axis, except for asimple eigenvalue at the origin, then the travelling wave is(nonlinearly) stable. (Henri ’81)

If the essential spectrum touches the imaginaryaxis at the origin, then one typically has tointroduce weighted norms to prove stability.

For many equations, the essential spectrum is easily computed.Hence, we now assume that σess is in the left half-plane, and weconcentrate on the point spectrum.

Solving the eigenvalue problem

The usual approach for solving the eigenvalue problem

Kyξξ + cyξ + Df (y) y = λy , y → 0 as ξ → ±∞

is to discretize the differential operator L by finite differences,finite elements, or some spectral method, and solve the resulting(huge) matrix eigenvalue problem.

The discretization may create spurious eigenvalues, but sometimesone can prove that the spectrum of the matrix converges to thespectrum of L as the grid size ∆ξ → 0.

Secondly, the infinite domain needs to be truncated to ξ ∈ [−L, L].The exact asymptotic boundary conditions are nonlinear; replacingthem by linear BCs leads again to spurious eigenvalues.

(Bridges, Derks & Gottwald ’02)

Outline

1. Travelling waves and their stability.

2. The Evans function.

3. Computing the Evans function.

4. The limit λ →∞.

The miss-distance function

Consider the Sturm–Liouville problem

−u′′ + q(x) u = λu for x ∈ [a, b]

with boundary conditions u(a) = u(b) = 0.

We can rewrite this in first-order form:

y ′ = A(x , λ)y where A(x , λ) =

[0 1

q(x)− λ 0

](∗)

with boundary conditions y1(a) = y1(b) = 0.

Denote by y−(x) the solution of (∗) with y−(a) =[

01

].

The miss-distance function is

D(λ) = y+1 (b).

Eigenvalues correspond to zeros of the miss-distance function.

The matching point ξ

We are looking at the Sturm–Liouville problem

y ′ = A(x , λ)y where A(x , λ) =

[0 1

q(x)− λ 0

](∗)

with boundary conditions y1(a) = y1(b) = 0.

Denote by y−(x) the solution of (∗) with y−(a) =[

01

].

Denote by y+(x) the solution of (∗) with y+(b) =[

01

].

The SLP (∗) has a solution if y+ is a multiple of y−.

The miss-distance function, evaluated at ξ ∈ [a, b], is

D(λ) = det

[y−1 (ξ) y+

1 (ξ)y−2 (ξ) y+

2 (ξ)

].

Eigenvalues correspond to zeros of the miss-distance function.

For ξ = b, we get D(λ) = y−1 (b), as before.

Problems on an infinite interval

Consider the following Sturm–Liouville problem with x ∈ R:

y ′ = A(x , λ)y where A(x , λ) =

[0 1

q(x)− λ 0

](∗)

with boundary conditions y(x) → 0 as x → ±∞.

Assume that q(x) → 0, then eigenvalues of A(±∞, λ) are ±√−λ.

Assume that λ ∈ C \ [0,∞), then v−u =[ 1√

−λ

]is the eigenvector

corresponding to the unstable eigenvalue µ−u =√−λ at x = −∞.

Let y− be the sol-n of (∗) with eµ−u xy−(x) → v−u as x → −∞.Let y+ be the sol-n of (∗) with eµ+

s xy+(x) → v+s as x → +∞.

Define D(λ) = det[y−(ξ) y+(ξ)

], with ξ ∈ R arbitrary.

Eigenvalues of (∗) correspond to zeros of D.

Increasing the number of dimensions

Consider y ′ = A(x , λ) y with y(x) ∈ Cn for x ∈ R.

We assume that there is a region Ω ⊂ C such that for all λ ∈ Ω

I the matrices A±(λ) = limx→±∞

A(x , λ) exist and are hyperbolic;

I A−(λ) has k unstable eigenvalues µ−1 , . . . , µ−k ;

I A+(λ) has n − k stable eigenvalues µ+1 , . . . , µ+

n−k .

Let y−i be a solution with eµ−i xy−i (x) → v−i as x → −∞.

Let y+i be a solution with eµ+

i xy+i (x) → v+

i as x → +∞.

The Evans function is defined by

D(λ) = det[y−1 (ξ) · · · y−k (ξ) y+

1 (ξ) · · · y+n−k(ξ)

].

It is analytic in Ω and its zeros corresponds to eigenvalues.(Evans ’75; Alexander, Gardner & Jones ’90; Sandstede ’02)

Outline

1. Travelling waves and their stability.

2. The Evans function.

3. Computing the Evans function.

4. The limit λ →∞.

Evaluating the Evans function numerically

The definition can be used as a basis for a numerical algorithm:

I Compute the unstable eigenvectors v−1 , . . . , v−k of A−.

I For i = 1, . . . , k, solve y ′ = A(ξ, λ) y with initial conditiony(−L) = v−i (where L is large) to get y−i (ξ).

I Compute y+i (ξ) similarly, and calculate the determinant.

The Evans function is analytic, so we canuse the argument principle to count thenumber of eigenvalues in a given region.

(Evans & Feroe ’77)

Im λ

Re λ

Alternatively, we can use Newton’s method to solve D(λ) = 0 andlocate the eigenvalues. (Pego, Smereka & Weinstein ’93)

Problems

I Unstable space =⇒ solutions grow exponentially;for example, y−i grows with rate ≈ µ−i .

Solution: Find y−i by solving y ′ =(A(ξ, λ)− µ−i I

)y .

I Evans function is only analytic if the eigenvectors v±i areanalytic functions in λ. This won’t happen automatically,and is impossible if eigenvalues µ±i coalesce.

Solution: See (Brin ’02) and (Bridges et al. ’02) for thefirst part, and below for the second part.

I If Re µ−1 > Re µ−2 , then y−1 grows faster than y−2 , so anyerrors in the y−1 direction will dominate the y−2 solution.

Solution: Do not look at the y−i individually, but look atthe subspace S = spany−1 , . . . , y−k and lift the equationy ′ = A(ξ, λ) y to S ′ = `(A(ξ, λ))S .

Lifting to the exterior product space

Let V be a vector space with basis e1, . . . , en.The exterior product space Λk(V ) is a vector space with basis

ei1 ∧ . . . ∧ eik : 1 ≤ i1 < · · · < ik ≤ n.

For example, Λ2(C4) is six-dimensional with basis

e1 ∧ e2, e1 ∧ e3, e1 ∧ e4, e2 ∧ e3, e2 ∧ e4, e3 ∧ e4.

The wedge product is associative, bilinear and alternating on V .

Identify subspace span y1, . . . , yk ⊂ Cn

with the decomposable k-form y1 ∧ · · · ∧ yk ∈ Λk(Cn).

Lift the equation y ′ = A(ξ, λ) y in Cn to exterior product space

z ′ = A(k)(ξ, λ) z z(ξ) ∈ Λk(Cn).

(Brin ’00, Allen & Bridges ’02)

Problem: dim Λk(Cn) =(nk

).

Example: The autocatalysis system

ut = δuxx − uvm with δ = 0.1 and m = 8vt = vxx + uvm (Aparicio, Malham & Olver ’05)

D−→−→−→

Example: The autocatalysis system

ut = δuxx − uvm with δ = 0.1 and m = 9vt = vxx + uvm (Aparicio, Malham & Olver ’05)

D−→−→−→

Outline

1. Travelling waves and their stability.

2. The Evans function.

3. Computing the Evans function.

4. The limit λ →∞.

The Fisher equation

The Fisher equation is: ut = uxx + u − u2.

This equation has a travelling front solution:

u(x , t) = u(ξ) =1(

1 + eξ/√

6)2

, where ξ = x − 56

√6 t.

Determine its stability by solving the corresponding EVP.

Methods for solving the ODE are:

I an explicit Runge–Kutta method (classical RK4),

I a stable but implicit Runge–Kutta method (Gauss–Legendre),

I a method based on the Magnus series (standard M4).

The essential spectrum is σess = (−∞,−1].The Magnus method performs very well for λ near σess.

(Moan ’98; Aparicio, Oliver & Malham ’05)

Error in Evans function along imaginary axis

Blue: Gauss–Legendre; red: Magnus; green: classical RK.All methods are fourth order and have step size h = 0.2.

Analysis of the Magnus method

For large λ, the ODE is stiff.

After a coord. transformation which puts the stiffness in the 2nd

component, the local error in the limit h → 0, |λh2| → ∞ is

Lk =

[|λ|−1/2h5(. . .) +O(λ−1/2h7, λ−1h5)

|λ|−1/2h2(. . .) +O(λ−1/2h4, λ−1h2)

],

So, we have order reduction in the stiff component:The local error is not h5 but h2.

However, the error in the stiff component does not propagate, andthe global error is

Gk =

[|λ|−1/2h4(. . .) +O(λ−1/2h6, λ−1h4)

|λ|−1/2h2(. . .) +O(λ−1/2h4, λ−1h2)

],

Hence, the method is second order when λ h−2.

Analysis of the Magnus method — II

The global error is Gk =

[|λ|−1/2h4(. . .) +O(λ−1/2h6, λ−1h4)

|λ|−1/2h2(. . .) +O(λ−1/2h4, λ−1h2)

].

To evaluate the Evans function, we match the solution comingfrom ξ = −∞ with the one from ξ = +∞:

D(λ) = y− ∧ y+.

Hence, the error is ED = G− ∧ y+ + y− ∧ G+.

However, the stiff component when integrating from ξ = −∞ isthe nonstiff component when integrating from ξ = +∞.

This results in ED = h4(. . .) +O(h6, λ−1/2h4).The Evans function does not see the order reduction.

Magnus (left) vs Gauss–Legendre (right)

Local error:

[λ−1/2h5

λ−1/2h2

] [λ−1/2h5

λ−1h2

]

Global error:

[λ−1/2h4

λ−1/2h2

] [λ−1/2h4

λ−1h2

]

Error in Evans function: h4

λ−1/2h4 (analysis)

λ−1h4 (numerics)

Something special happens when evaluating the Evans functionwith Gauss–Legendre, akin to numerical quadrature of periodicfunctions with the trapezium rule.

Conclusions

I Explicit methods run into stability problems.

I Close to the essential spectrum, Magnus outperformsGauss–Legendre.

I Away from the essential spectrum, GL is better (but notcompletely understood).

I The analysis is only valid for scalar equations. However,experiments show similar behaviour for the vector case:explicit method run into trouble and Magnus and GL performwell, with Magnus better close to σess.

I The big question is, whether the Evans function is overall agood‘idea. My impression is that it is more robust than thematrix approach.

I Problem: dim Λk(Cn) =(nk

).