Laplacian Eigenmodes

Module 4: Space and Pattern Formation

The linearized system $\partial_t \eta = D\,\Delta\eta + \gamma J\eta$ couples every point to its neighbours. You decouple it by expanding $\eta$ in eigenfunctions of the Laplacian, which depend on the domain and the boundary conditions (murray2003mathematical?; britton2003essential?).

The Eigenvalue Problem

Look for functions $\phi$ and numbers $k^2 \ge 0$ with

\[ \Delta \phi + k^2 \phi = 0 \quad \text{in } \Omega, \tag{1}\]

plus the boundary condition of the model. The eigenvalue of $\Delta$ is $\lambda = -k^2$, and $k$ is the wavenumber. The wavelength of the mode is $2\pi / k$.

On an Interval $(0, L)$

Boundary condition	Condition	Eigenfunctions	Wavenumbers
Neumann (zero flux)	$\phi'(0) = \phi'(L) = 0$	$\cos(n\pi x / L)$	$k_n = n\pi/L$, $n = 0, 1, 2, \dots$
Dirichlet (fixed value)	$\phi(0) = \phi(L) = 0$	$\sin(n\pi x / L)$	$k_n = n\pi/L$, $n = 1, 2, \dots$
Periodic	$\phi(0) = \phi(L)$, $\phi'(0) = \phi'(L)$	$\cos(2n\pi x/L)$, $\sin(2n\pi x/L)$	$k_n = 2n\pi/L$, $n = 0, 1, 2, \dots$

Mode $n$ with Neumann conditions fits $n$ half-wavelengths into the interval. Mode $0$ is the flat perturbation. For periodic conditions each $n \ge 1$ has two independent eigenfunctions with the same $k_n$.

Figure 1: First four Neumann eigenfunctions $\cos(n\pi x/L)$ on $(0, L)$.

On a Rectangle $(0, L_x) \times (0, L_y)$

Separation of variables gives products of the 1D eigenfunctions. With Neumann conditions on all four edges,

\[ \phi_{n,m}(x, y) = \cos\left(\frac{n\pi x}{L_x}\right)\cos\left(\frac{m\pi y}{L_y}\right), \qquad k_{n,m}^2 = \left(\frac{n\pi}{L_x}\right)^2 + \left(\frac{m\pi}{L_y}\right)^2. \tag{2}\]

For Dirichlet conditions replace cosines by sines and start at $n, m \ge 1$. For periodic conditions replace $\pi/L$ by $2\pi/L$. Different pairs $(n, m)$ can share almost the same $k^2$. That is why 2D domains can select stripes, spots, or mixtures (murray2003mathematical?).

Mode Decomposition

Expand the perturbation as

\[ \eta(x, t) = \sum_n c_n(t)\,\phi_n(x), \qquad c_n(t) \in \mathbb{R}^2. \]

Because $\Delta \phi_n = -k_n^2 \phi_n$, the modes separate and each coefficient solves a $2 \times 2$ linear ODE:

\[ \dot c_n = A_n c_n, \qquad A_n = \gamma J - k_n^2 D. \tag{3}\]

So $c_n(t) \propto e^{\sigma t}$, where $\sigma$ is an eigenvalue of $A_n$. The uniform state is linearly stable when every admissible mode has $\operatorname{Re}\sigma < 0$. The next session turns this into the Turing conditions and the dispersion relation.

Exercise

On $(0, 40)$ with Neumann conditions, which mode $n$ has wavenumber closest to $k = 0.4$? How many half-wavelengths does it fit into the domain?

Solution

$k_n = n\pi/40$, so $n = 40k/\pi \approx 5.09$. The closest mode is $n = 5$, with five half-wavelengths, which is two and a half full wavelengths.

Next session: Turing Conditions.

--- title: "Laplacian Eigenmodes" subtitle: "Module 4: Space and Pattern Formation" format: html --- The linearized system $\partial_t \eta = D\,\Delta\eta + \gamma J\eta$ couples every point to its neighbours. You decouple it by expanding $\eta$ in eigenfunctions of the Laplacian, which depend on the domain and the boundary conditions [@murray2003mathematical, Section 2.3; @britton2003essential, Chapter 7]. ## The Eigenvalue Problem Look for functions $\phi$ and numbers $k^2 \ge 0$ with $$ \Delta \phi + k^2 \phi = 0 \quad \text{in } \Omega, $$ {#eq-eig-problem} plus the boundary condition of the model. The eigenvalue of $\Delta$ is $\lambda = -k^2$, and $k$ is the wavenumber. The wavelength of the mode is $2\pi / k$. ## On an Interval $(0, L)$ | Boundary condition | Condition | Eigenfunctions | Wavenumbers | |---|---|---|---| | Neumann (zero flux) | $\phi'(0) = \phi'(L) = 0$ | $\cos(n\pi x / L)$ | $k_n = n\pi/L$, $n = 0, 1, 2, \dots$ | | Dirichlet (fixed value) | $\phi(0) = \phi(L) = 0$ | $\sin(n\pi x / L)$ | $k_n = n\pi/L$, $n = 1, 2, \dots$ | | Periodic | $\phi(0) = \phi(L)$, $\phi'(0) = \phi'(L)$ | $\cos(2n\pi x/L)$, $\sin(2n\pi x/L)$ | $k_n = 2n\pi/L$, $n = 0, 1, 2, \dots$ | Mode $n$ with Neumann conditions fits $n$ half-wavelengths into the interval. Mode $0$ is the flat perturbation. For periodic conditions each $n \ge 1$ has two independent eigenfunctions with the same $k_n$. ```{python} #| label: fig-eig-neumann #| fig-cap: 'First four Neumann eigenfunctions $\cos(n\pi x/L)$ on $(0, L)$.' #| fig-width: 7 #| fig-height: 3.5 #| echo: false import matplotlib.pyplot as plt import numpy as np length = 1.0 x = np.linspace(0, length, 400) fig, ax = plt.subplots(figsize=(7, 3.5)) for n in range(4): ax.plot(x, np.cos(n * np.pi * x / length), label=f"n = {n}") ax.set_xlabel("x / L") ax.set_ylabel(r"$\phi_n(x)$") ax.legend(loc="lower left", ncols=4) plt.show() plt.close() ``` ## On a Rectangle $(0, L_x) \times (0, L_y)$ Separation of variables gives products of the 1D eigenfunctions. With Neumann conditions on all four edges, $$ \phi_{n,m}(x, y) = \cos\left(\frac{n\pi x}{L_x}\right)\cos\left(\frac{m\pi y}{L_y}\right), \qquad k_{n,m}^2 = \left(\frac{n\pi}{L_x}\right)^2 + \left(\frac{m\pi}{L_y}\right)^2. $$ {#eq-eig-rectangle} For Dirichlet conditions replace cosines by sines and start at $n, m \ge 1$. For periodic conditions replace $\pi/L$ by $2\pi/L$. Different pairs $(n, m)$ can share almost the same $k^2$. That is why 2D domains can select stripes, spots, or mixtures [@murray2003mathematical, Section 2.5]. ## Mode Decomposition Expand the perturbation as $$ \eta(x, t) = \sum_n c_n(t)\,\phi_n(x), \qquad c_n(t) \in \mathbb{R}^2. $$ Because $\Delta \phi_n = -k_n^2 \phi_n$, the modes separate and each coefficient solves a $2 \times 2$ linear ODE: $$ \dot c_n = A_n c_n, \qquad A_n = \gamma J - k_n^2 D. $$ {#eq-eig-mode-matrix} So $c_n(t) \propto e^{\sigma t}$, where $\sigma$ is an eigenvalue of $A_n$. The uniform state is linearly stable when every admissible mode has $\operatorname{Re}\sigma < 0$. The next session turns this into the Turing conditions and the dispersion relation. ::: {.callout-note} ## Exercise On $(0, 40)$ with Neumann conditions, which mode $n$ has wavenumber closest to $k = 0.4$? How many half-wavelengths does it fit into the domain? ::: ::: {.callout-tip collapse="true"} ## Solution $k_n = n\pi/40$, so $n = 40k/\pi \approx 5.09$. The closest mode is $n = 5$, with five half-wavelengths, which is two and a half full wavelengths. ::: Next session: [Turing Conditions](turing-conditions.qmd).

Boundary condition	Condition	Eigenfunctions	Wavenumbers
Neumann (zero flux)	\(\phi'(0) = \phi'(L) = 0\)	\(\cos(n\pi x / L)\)	\(k_n = n\pi/L\), \(n = 0, 1, 2, \dots\)
Dirichlet (fixed value)	\(\phi(0) = \phi(L) = 0\)	\(\sin(n\pi x / L)\)	\(k_n = n\pi/L\), \(n = 1, 2, \dots\)
Periodic	\(\phi(0) = \phi(L)\), \(\phi'(0) = \phi'(L)\)	\(\cos(2n\pi x/L)\), \(\sin(2n\pi x/L)\)	\(k_n = 2n\pi/L\), \(n = 0, 1, 2, \dots\)

Laplacian Eigenmodes

The Eigenvalue Problem

On an Interval \((0, L)\)

On a Rectangle \((0, L_x) \times (0, L_y)\)

Mode Decomposition