Stability of travelling waves using the Evans function

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Stability of travelling waves using the Evans function Jitse Niesen in collaboration with Simon Malham Industrial and Applied Mathematics Seminar Heriot–Watt University, 8 February 2006

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Stability of travelling waves using the Evans functionJitse Niesen in collaboration with Simon Malham
Industrial and Applied Mathematics Seminar Heriot–Watt University, 8 February 2006
Outline
2. The Evans function.
4. The limit λ →∞.
yt = Kyxx + F (y)
where y(t, x) ∈ Rd with x ∈ R.
Travelling waves are solutions to the equation that move with constant 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 whose initial conditions are small perturbations of y? If any such solution stay 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 its translate y(ξ − ξ0). So, we require for stability that small perturbations 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 equation about the travelling wave.
Suppose that y(t, ξ) = y(ξ) + y(t, ξ) is a perturbation of the travelling wave at t = 0.
y solves the full PDE: Kyξξ + cyξ + f (y) = yt
Travelling wave y solves: Kyξξ + cyξ + f (y) = 0 Subtracting: 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 L The spectrum of L can be decomposed in two parts.
The point spectrum σpt consists of eigenvalues λ, for which Ly = λy has a solution. We have 0 ∈ σpt.
The essential spectrum σess typically contains open sets, and is asymptotically (as λ →∞) inside the parabolic region {λ : (Im λ)2 ≤ −C Re λ}.
σess
Nonlinear stability vs linear stability
If the spectrum of L is contained in the left half of the complex plane and bounded away from the imaginary axis, except for a simple eigenvalue at the origin, then the travelling wave is (nonlinearly) stable. (Henri ’81)
If the essential spectrum touches the imaginary axis at the origin, then one typically has to introduce 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 we concentrate on the point spectrum.
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 sometimes one can prove that the spectrum of the matrix converges to the spectrum 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; replacing them by linear BCs leads again to spurious eigenvalues.
(Bridges, Derks & Gottwald ’02)
2. The Evans function.
4. The limit λ →∞.
−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
Denote by y−(x) the solution of (∗) with y−(a) = [
0 1
The matching point ξ
y ′ = A(x , λ)y where A(x , λ) =
[ 0 1
Denote by y−(x) the solution of (∗) with y−(a) = [
0 1
Denote by y+(x) the solution of (∗) with y+(b) = [
0 1
] .
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
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
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
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.
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 ⊂ C such that for all λ ∈
I the matrices A±(λ) = lim x→±∞
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 → +∞.
D(λ) = det [ y−1 (ξ) · · · y−k (ξ) y+
1 (ξ) · · · y+ n−k(ξ)
] .
It is analytic in and its zeros corresponds to eigenvalues. (Evans ’75; Alexander, Gardner & Jones ’90; Sandstede ’02)
Outline
2. 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 condition y(−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 can use the argument principle to count the number 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 and locate 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 are analytic 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 the first part, and below for the second part.
I If Re µ−1 > Re µ−2 , then y−1 grows faster than y−2 , so any errors in the y−1 direction will dominate the y−2 solution.
Solution: Do not look at the y−i individually, but look at the subspace S = span{y−1 , . . . , y−k } and lift the equation y ′ = 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) = (n k
) .
Example: The autocatalysis system
ut = δuxx − uvm with δ = 0.1 and m = 8 vt = vxx + uvm (Aparicio, Malham & Olver ’05)
D−→−→−→
Example: The autocatalysis system
ut = δuxx − uvm with δ = 0.1 and m = 9 vt = vxx + uvm (Aparicio, Malham & Olver ’05)
D−→−→−→
Outline
2. The Evans function.
4. The limit λ →∞.
This equation has a travelling front solution:
u(x , t) = u(ξ) = 1(
√ 6 t.
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 =
] ,
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, and the global error is
Gk =
] ,
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 coming from 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 = −∞ is the 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
{ λ−1/2h4 (analysis)
λ−1h4 (numerics)
Something special happens when evaluating the Evans function with Gauss–Legendre, akin to numerical quadrature of periodic functions with the trapezium rule.
Conclusions
I Close to the essential spectrum, Magnus outperforms Gauss–Legendre.
I Away from the essential spectrum, GL is better (but not completely 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 perform well, with Magnus better close to σess.
I The big question is, whether the Evans function is overall a good‘idea. My impression is that it is more robust than the matrix approach.
I Problem: dim Λk(Cn) = (n k
) .