Dispersion Relation

Module 4: Space and Pattern Formation

The Turing conditions tell you whether some mode grows. The dispersion relation tells you which modes grow and how fast (murray2003mathematical? and 2.6).

Growth Rate Versus Wavenumber

For each wavenumber $k$, the growth rates are the roots of

\[ \sigma^2 - \operatorname{tr}A(k^2)\,\sigma + h(k^2) = 0, \qquad A(k^2) = \gamma J - k^2\operatorname{diag}(1, d), \tag{1}\]

with $h(k^2) = \det A(k^2) = d\,k^4 - \gamma (d f_u + g_v) k^2 + \gamma^2 \det J$, as on the Turing Conditions page. The function $k \mapsto \operatorname{Re}\sigma(k^2)$, taking the larger root, is the dispersion relation.

The Unstable Band

$\sigma > 0$ exactly when $h(k^2) < 0$. The roots of $h$ give the band

\[ k_1^2 < k^2 < k_2^2, \qquad k_{1,2}^2 = \frac{\gamma}{2d}\left[(d f_u + g_v) \mp \sqrt{(d f_u + g_v)^2 - 4 d \det J}\right]. \tag{2}\]

The band has width proportional to $\gamma$ in $k^2$. At $d = d_c$ it shrinks to the single value $k_c^2 = \gamma (d_c f_u + g_v)/(2 d_c)$ (murray2003mathematical?).

The Fastest-Growing Mode

The fastest-growing wavenumber $k_{\max}$ maximizes $\operatorname{Re}\sigma(k^2)$. Near onset it is close to $k_m$, the minimizer of $h$, but the two are not equal in general, because $\sigma$ also depends on the trace. You find $k_{\max}$ numerically. The corresponding wavelength $2\pi/k_{\max}$ predicts the spacing of the first visible pattern. Nonlinear effects can change the spacing later (murray2003mathematical?).

On a finite domain only discrete wavenumbers exist. With Neumann conditions on $(0, L)$, $k_n = n\pi/L$. The mode that grows fastest is the admissible $k_n$ with the largest $\sigma$. If no $k_n$ falls into the band, the uniform state stays stable even when the Turing conditions hold.

Figure 1: Dispersion relation of Gierer-Meinhardt with $a = 0.4$, $b = 1$, $\gamma = 1$. Dots mark the Neumann modes $k_n = n\pi/L$ for $L = 40$.

For $d = 20$ every mode decays. At $d = d_c \approx 26.24$ the curve touches zero. For $d = 30$ a band of modes grows.

import numpy as np

from amlab.turing import dispersion_relation, gierer_meinhardt_jacobian, unstable_band

jac = gierer_meinhardt_jacobian(0.40, 1.00)
d = 30
k1_sq, k2_sq = unstable_band(*jac, d)
k = np.linspace(np.sqrt(k1_sq), np.sqrt(k2_sq), 2001)
k_max = k[np.argmax(dispersion_relation(k**2, *jac, d))]

print(f"unstable band: {np.sqrt(k1_sq):.3f} < k < {np.sqrt(k2_sq):.3f}")
print(f"fastest-growing k = {k_max:.3f}, wavelength = {2 * np.pi / k_max:.1f}")
print(f"closest Neumann mode on L = 40: n = {k_max * 40 / np.pi:.2f}")

unstable band: 0.349 < k < 0.523
fastest-growing k = 0.429, wavelength = 14.6
closest Neumann mode on L = 40: n = 5.46

The fastest wavenumber $k_{\max} \approx 0.43$ corresponds to a wavelength of about $14.6$, which lies between the Neumann modes $n = 5$ and $n = 6$ on $L = 40$. Both modes grow, at almost the same rate, and mode $5$ is slightly faster. The laboratory checks which one the simulation selects.

Exercise

Keep $d = 30$ and $\gamma = 1$ with Neumann conditions. Mode $n$ is unstable when $k_1 < n\pi/L < k_2$. For which domain lengths $L < 20$ is no mode $n \ge 1$ unstable?

Solution

The code above gives $k_1 \approx 0.349$ and $k_2 \approx 0.523$. Mode $n$ is unstable for $n\pi/k_2 < L < n\pi/k_1$:

$n = 1$: $6.01 < L < 8.99$,
$n = 2$: $12.02 < L < 17.99$,
$n = 3$: $18.03 < L < 26.98$.

No mode is unstable for $L \le 6.01$, for $8.99 \le L \le 12.02$, and in the narrow gap $17.99 \le L \le 18.03$. Beyond that the intervals overlap. Small domains and some intermediate sizes stay uniform even though the Turing conditions hold (murray2003mathematical?).

--- title: "Dispersion Relation" subtitle: "Module 4: Space and Pattern Formation" format: html --- The Turing conditions tell you whether some mode grows. The dispersion relation tells you which modes grow and how fast [@murray2003mathematical, Sections 2.5 and 2.6]. ## Growth Rate Versus Wavenumber For each wavenumber $k$, the growth rates are the roots of $$ \sigma^2 - \operatorname{tr}A(k^2)\,\sigma + h(k^2) = 0, \qquad A(k^2) = \gamma J - k^2\operatorname{diag}(1, d), $$ {#eq-dr-quadratic} with $h(k^2) = \det A(k^2) = d\,k^4 - \gamma (d f_u + g_v) k^2 + \gamma^2 \det J$, as on the [Turing Conditions](turing-conditions.qmd) page. The function $k \mapsto \operatorname{Re}\sigma(k^2)$, taking the larger root, is the dispersion relation. ## The Unstable Band $\sigma > 0$ exactly when $h(k^2) < 0$. The roots of $h$ give the band $$ k_1^2 < k^2 < k_2^2, \qquad k_{1,2}^2 = \frac{\gamma}{2d}\left[(d f_u + g_v) \mp \sqrt{(d f_u + g_v)^2 - 4 d \det J}\right]. $$ {#eq-dr-band} The band has width proportional to $\gamma$ in $k^2$. At $d = d_c$ it shrinks to the single value $k_c^2 = \gamma (d_c f_u + g_v)/(2 d_c)$ [@murray2003mathematical, Section 2.3]. ## The Fastest-Growing Mode The fastest-growing wavenumber $k_{\max}$ maximizes $\operatorname{Re}\sigma(k^2)$. Near onset it is close to $k_m$, the minimizer of $h$, but the two are not equal in general, because $\sigma$ also depends on the trace. You find $k_{\max}$ numerically. The corresponding wavelength $2\pi/k_{\max}$ predicts the spacing of the first visible pattern. Nonlinear effects can change the spacing later [@murray2003mathematical, Section 2.6]. On a finite domain only discrete wavenumbers exist. With Neumann conditions on $(0, L)$, $k_n = n\pi/L$. The mode that grows fastest is the admissible $k_n$ with the largest $\sigma$. If no $k_n$ falls into the band, the uniform state stays stable even when the Turing conditions hold. ```{python} #| label: fig-dr-gm #| fig-cap: 'Dispersion relation of Gierer-Meinhardt with $a = 0.4$, $b = 1$, $\gamma = 1$. Dots mark the Neumann modes $k_n = n\pi/L$ for $L = 40$.' #| fig-width: 7 #| fig-height: 4 #| echo: false import matplotlib.pyplot as plt import numpy as np from amlab.turing import ( critical_diffusion_ratio, dispersion_relation, gierer_meinhardt_jacobian, ) jac = gierer_meinhardt_jacobian(0.40, 1.00) d_c = critical_diffusion_ratio(*jac) k = np.linspace(0, 1.0, 500) k_n = np.arange(13) * np.pi / 40 fig, ax = plt.subplots(figsize=(7, 4)) for d in (20, d_c, 30): (line,) = ax.plot(k, dispersion_relation(k**2, *jac, d), label=f"d = {d:.2f}") ax.plot(k_n, dispersion_relation(k_n**2, *jac, d), "o", color=line.get_color(), ms=4) ax.axhline(0, color="black", lw=0.8) ax.set_ylim(-0.15, 0.04) ax.set_xlabel("wavenumber k") ax.set_ylabel(r"Re $\sigma(k^2)$") ax.legend() plt.show() plt.close() ``` For $d = 20$ every mode decays. At $d = d_c \approx 26.24$ the curve touches zero. For $d = 30$ a band of modes grows. ```{python} import numpy as np from amlab.turing import dispersion_relation, gierer_meinhardt_jacobian, unstable_band jac = gierer_meinhardt_jacobian(0.40, 1.00) d = 30 k1_sq, k2_sq = unstable_band(*jac, d) k = np.linspace(np.sqrt(k1_sq), np.sqrt(k2_sq), 2001) k_max = k[np.argmax(dispersion_relation(k**2, *jac, d))] print(f"unstable band: {np.sqrt(k1_sq):.3f} < k < {np.sqrt(k2_sq):.3f}") print(f"fastest-growing k = {k_max:.3f}, wavelength = {2 * np.pi / k_max:.1f}") print(f"closest Neumann mode on L = 40: n = {k_max * 40 / np.pi:.2f}") ``` The fastest wavenumber $k_{\max} \approx 0.43$ corresponds to a wavelength of about $14.6$, which lies between the Neumann modes $n = 5$ and $n = 6$ on $L = 40$. Both modes grow, at almost the same rate, and mode $5$ is slightly faster. The laboratory checks which one the simulation selects. ::: {.callout-note} ## Exercise Keep $d = 30$ and $\gamma = 1$ with Neumann conditions. Mode $n$ is unstable when $k_1 < n\pi/L < k_2$. For which domain lengths $L < 20$ is no mode $n \ge 1$ unstable? ::: ::: {.callout-tip collapse="true"} ## Solution The code above gives $k_1 \approx 0.349$ and $k_2 \approx 0.523$. Mode $n$ is unstable for $n\pi/k_2 < L < n\pi/k_1$: - $n = 1$: $6.01 < L < 8.99$, - $n = 2$: $12.02 < L < 17.99$, - $n = 3$: $18.03 < L < 26.98$. No mode is unstable for $L \le 6.01$, for $8.99 \le L \le 12.02$, and in the narrow gap $17.99 \le L \le 18.03$. Beyond that the intervals overlap. Small domains and some intermediate sizes stay uniform even though the Turing conditions hold [@murray2003mathematical, Section 2.5]. :::