Turing Conditions

Module 4: Space and Pattern Formation

Reading

Compulsory: Turing (1952).
Recommended: (murray2003mathematical?), Chapter 2, Sections 2.1 to 2.6.

Turing (1952) showed that diffusion can destabilize a uniform state that is stable without diffusion. This is called diffusion-driven or Turing instability. This page derives the four conditions for it in the notation of (murray2003mathematical?).

The Question

Take the linearized system from Laplacian Eigenmodes. Each mode with wavenumber $k$ grows like $e^{\sigma t}$, where $\sigma$ solves

\[ \det\left(\gamma J - k^2 D - \sigma I\right) = 0, \qquad D = \operatorname{diag}(1, d). \]

You want the uniform state to be:

stable to the flat mode $k = 0$ (stable without diffusion), and
unstable to at least one mode with $k \neq 0$.

Trace and Determinant of the Mode Matrix

For $A(k^2) = \gamma J - k^2 D$,

\[ \operatorname{tr} A = \gamma (f_u + g_v) - (1 + d)k^2, \]

\[ h(k^2) \equiv \det A = d\,k^4 - \gamma\,(d f_u + g_v)\,k^2 + \gamma^2 (f_u g_v - f_v g_u). \tag{1}\]

Stability at $k = 0$ requires $\operatorname{tr} J < 0$ and $\det J > 0$. Then $\operatorname{tr} A < 0$ for every $k$, so the only way to get $\operatorname{Re}\sigma > 0$ is $h(k^2) < 0$ for some $k^2 > 0$, which gives one real positive $\sigma$.

$h$ is a quadratic in $k^2$ with positive constant term. It becomes negative only if its minimum lies at $k^2 > 0$ and is negative. The minimum is at

\[ k_m^2 = \frac{\gamma\,(d f_u + g_v)}{2d}, \qquad h_{\min} = \gamma^2\left[\det J - \frac{(d f_u + g_v)^2}{4d}\right]. \tag{2}\]

The Four Conditions

Collecting the requirements gives the Turing conditions (murray2003mathematical?):

\[ \begin{aligned} &f_u + g_v < 0, \\ &f_u g_v - f_v g_u > 0, \\ &d f_u + g_v > 0, \\ &(d f_u + g_v)^2 - 4 d\,(f_u g_v - f_v g_u) > 0. \end{aligned} \tag{3}\]

All derivatives are evaluated at the steady state. The conditions do not contain $\gamma$ or the domain size. They are necessary for patterns. On a finite domain you also need an admissible wavenumber inside the unstable band (see Dispersion Relation).

What the Signs Tell You

The first and third conditions force $d \neq 1$. If both species diffuse equally, no Turing instability exists.
They also force $f_u$ and $g_v$ to have opposite signs. Call the species with $f_u > 0$ the activator. Then $g_v < 0$ and $d > -g_v / f_u > 1$: the inhibitor must diffuse faster than the activator.
With $f_u g_v < 0$, the second condition needs $f_v g_u < 0$. Either the activator promotes the inhibitor and the inhibitor suppresses the activator ($g_u > 0$, $f_v < 0$), or the opposite cross signs hold (murray2003mathematical?).

The usual summary is local activation with long-range inhibition (Gierer and Meinhardt 1972; kondo2010reaction?).

The Critical Diffusion Ratio

At the onset $h_{\min} = 0$, so the critical ratio $d_c$ solves

\[ f_u^2\,d_c^2 + 2\,(2 f_v g_u - f_u g_v)\,d_c + g_v^2 = 0, \tag{4}\]

taking the root with $d_c f_u + g_v > 0$ (murray2003mathematical?). For $d > d_c$ a band of wavenumbers is unstable.

Worked Example: Gierer-Meinhardt

From Linearization, $f_u = b(1-a)/(1+a)$, $f_v = -b^2/(1+a)^2$, $g_u = 2(1+a)/b$, $g_v = -1$, and $\det J = b$. So $f_u > 0$ needs $a < 1$, and $u$ is the activator. With $a = 0.4$ and $b = 1$:

from amlab.turing import critical_diffusion_ratio, gierer_meinhardt_jacobian, turing_conditions

fu, fv, gu, gv = gierer_meinhardt_jacobian(a=0.40, b=1.00)
print(f"f_u = {fu:.4f}, f_v = {fv:.4f}, g_u = {gu:.4f}, g_v = {gv:.4f}")
print(f"d_c = {critical_diffusion_ratio(fu, fv, gu, gv):.2f}")
for d in (20, 30):
    print(f"d = {d}: Turing conditions hold? {bool(turing_conditions(fu, fv, gu, gv, d))}")

f_u = 0.4286, f_v = -0.5102, g_u = 2.8000, g_v = -1.0000
d_c = 26.24
d = 20: Turing conditions hold? False
d = 30: Turing conditions hold? True

So $d = 20$ gives no Turing instability and $d = 30$ does. You will test this prediction in the laboratory.

Exercise

For Gierer-Meinhardt with $b = 1$, show that the fourth condition reads $d f_u - 1 > 2\sqrt{d}$, and solve it for $\sqrt{d}$.

Solution

With $g_v = -1$ and $\det J = b = 1$, the third and fourth conditions together say $d f_u - 1 > 2\sqrt{d}$. This is a quadratic inequality in $s = \sqrt d$: $f_u s^2 - 2s - 1 > 0$. For $f_u > 0$ it holds for $s > \left(1 + \sqrt{1 + f_u}\right)/f_u$. With $a = 0.4$, $f_u = 3/7$ and $s > 5.122$, so $d > 26.24$.

References

Gierer, Alfred, and Hans Meinhardt. 1972. “A Theory of Biological Pattern Formation.” Kybernetik 12 (1): 30–39. https://doi.org/10.1007/BF00289234.

Turing, Alan M. 1952. “The Chemical Basis of Morphogenesis.” Philosophical Transactions of the Royal Society B 237 (641): 37–72. https://doi.org/10.1098/rstb.1952.0012.

--- title: "Turing Conditions" subtitle: "Module 4: Space and Pattern Formation" format: html --- ::: {.callout-note} ## Reading - **Compulsory:** @turing1952chemical. - **Recommended:** @murray2003mathematical, Chapter 2, Sections 2.1 to 2.6. ::: @turing1952chemical showed that diffusion can destabilize a uniform state that is stable without diffusion. This is called diffusion-driven or Turing instability. This page derives the four conditions for it in the notation of @murray2003mathematical [Section 2.3]. ## The Question Take the linearized system from [Laplacian Eigenmodes](laplacian-eigenmodes.qmd). Each mode with wavenumber $k$ grows like $e^{\sigma t}$, where $\sigma$ solves $$ \det\left(\gamma J - k^2 D - \sigma I\right) = 0, \qquad D = \operatorname{diag}(1, d). $$ You want the uniform state to be: - stable to the flat mode $k = 0$ (stable without diffusion), and - unstable to at least one mode with $k \neq 0$. ## Trace and Determinant of the Mode Matrix For $A(k^2) = \gamma J - k^2 D$, $$ \operatorname{tr} A = \gamma (f_u + g_v) - (1 + d)k^2, $$ $$ h(k^2) \equiv \det A = d\,k^4 - \gamma\,(d f_u + g_v)\,k^2 + \gamma^2 (f_u g_v - f_v g_u). $$ {#eq-tc-h} Stability at $k = 0$ requires $\operatorname{tr} J < 0$ and $\det J > 0$. Then $\operatorname{tr} A < 0$ for every $k$, so the only way to get $\operatorname{Re}\sigma > 0$ is $h(k^2) < 0$ for some $k^2 > 0$, which gives one real positive $\sigma$. $h$ is a quadratic in $k^2$ with positive constant term. It becomes negative only if its minimum lies at $k^2 > 0$ and is negative. The minimum is at $$ k_m^2 = \frac{\gamma\,(d f_u + g_v)}{2d}, \qquad h_{\min} = \gamma^2\left[\det J - \frac{(d f_u + g_v)^2}{4d}\right]. $$ {#eq-tc-hmin} ## The Four Conditions Collecting the requirements gives the Turing conditions [@murray2003mathematical, Section 2.3]: $$ \begin{aligned} &f_u + g_v < 0, \\ &f_u g_v - f_v g_u > 0, \\ &d f_u + g_v > 0, \\ &(d f_u + g_v)^2 - 4 d\,(f_u g_v - f_v g_u) > 0. \end{aligned} $$ {#eq-tc-conditions} All derivatives are evaluated at the steady state. The conditions do not contain $\gamma$ or the domain size. They are necessary for patterns. On a finite domain you also need an admissible wavenumber inside the unstable band (see [Dispersion Relation](dispersion-relation.qmd)). ## What the Signs Tell You - The first and third conditions force $d \neq 1$. If both species diffuse equally, no Turing instability exists. - They also force $f_u$ and $g_v$ to have opposite signs. Call the species with $f_u > 0$ the activator. Then $g_v < 0$ and $d > -g_v / f_u > 1$: the inhibitor must diffuse faster than the activator. - With $f_u g_v < 0$, the second condition needs $f_v g_u < 0$. Either the activator promotes the inhibitor and the inhibitor suppresses the activator ($g_u > 0$, $f_v < 0$), or the opposite cross signs hold [@murray2003mathematical, Section 2.3]. The usual summary is local activation with long-range inhibition [@gierer1972theory; @kondo2010reaction]. ## The Critical Diffusion Ratio At the onset $h_{\min} = 0$, so the critical ratio $d_c$ solves $$ f_u^2\,d_c^2 + 2\,(2 f_v g_u - f_u g_v)\,d_c + g_v^2 = 0, $$ {#eq-tc-dc} taking the root with $d_c f_u + g_v > 0$ [@murray2003mathematical, Section 2.3]. For $d > d_c$ a band of wavenumbers is unstable. ## Worked Example: Gierer-Meinhardt From [Linearization](linearization.qmd), $f_u = b(1-a)/(1+a)$, $f_v = -b^2/(1+a)^2$, $g_u = 2(1+a)/b$, $g_v = -1$, and $\det J = b$. So $f_u > 0$ needs $a < 1$, and $u$ is the activator. With $a = 0.4$ and $b = 1$: ```{python} from amlab.turing import critical_diffusion_ratio, gierer_meinhardt_jacobian, turing_conditions fu, fv, gu, gv = gierer_meinhardt_jacobian(a=0.40, b=1.00) print(f"f_u = {fu:.4f}, f_v = {fv:.4f}, g_u = {gu:.4f}, g_v = {gv:.4f}") print(f"d_c = {critical_diffusion_ratio(fu, fv, gu, gv):.2f}") for d in (20, 30): print(f"d = {d}: Turing conditions hold? {bool(turing_conditions(fu, fv, gu, gv, d))}") ``` So $d = 20$ gives no Turing instability and $d = 30$ does. You will test this prediction in the laboratory. ::: {.callout-note} ## Exercise For Gierer-Meinhardt with $b = 1$, show that the fourth condition reads $d f_u - 1 > 2\sqrt{d}$, and solve it for $\sqrt{d}$. ::: ::: {.callout-tip collapse="true"} ## Solution With $g_v = -1$ and $\det J = b = 1$, the third and fourth conditions together say $d f_u - 1 > 2\sqrt{d}$. This is a quadratic inequality in $s = \sqrt d$: $f_u s^2 - 2s - 1 > 0$. For $f_u > 0$ it holds for $s > \left(1 + \sqrt{1 + f_u}\right)/f_u$. With $a = 0.4$, $f_u = 3/7$ and $s > 5.122$, so $d > 26.24$. :::