Lecture notes. Written from the material of MIT 8.591J 2004, licensed CC BY-NC-SA 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

12. Stability of the Homogeneous Solution

What this covers

The previous lecture wrote down the Gierer–Meinhardt activator–inhibitor equations that give a model of local excitation and global inhibition, and found the single steady state at which both concentrations are uniform in space. This chapter asks the next question: is that uniform state actually stable? It answers it twice — once for a disturbance that is the same everywhere, and once for a disturbance that varies with position — using the trace/determinant test for a linear system of two variables. It assumes the reaction–diffusion pair and the homogeneous solution already in hand, and the standard eigenvalue criterion for a \(2\times2\) linear system.

The model and its uniform state

The activator \(a(x,t)\) and inhibitor \(i(x,t)\) satisfy

\[\begin{aligned} \frac{\partial a}{\partial t} &= r_a + k_a \frac{a^2}{i} - \gamma_a a + D_a \frac{\partial^2 a}{\partial x^2} \\ \frac{\partial i}{\partial t} &= k_i a^2 - \gamma_i i + D_i \frac{\partial^2 i}{\partial x^2} \end{aligned}\]

with \(r_a\) a basal synthesis rate, \(k_a,k_i\) rate constants, \(\gamma_a,\gamma_i\) decay rates and \(D_a,D_i\) diffusion constants — seven constants in all. The activator promotes its own production through the \(a^2\) term, but that promotion is throttled by dividing by the inhibitor \(i\); the inhibitor is made in proportion to \(a^2\) and otherwise just decays and diffuses. That mutual arrangement is the "local excitation, global inhibition" of the title.

The lecture rescales concentration, time and space into dimensionless \(A,I,\tau,s\), which absorbs \(r_a\) into the reference scale — the constant term becomes plain \(1\) — and collects the remaining constants into three dimensionless groups \(R\), \(Q\), \(P\):

\[\begin{aligned} \frac{\partial A}{\partial \tau} &= 1 + R\,\frac{A^2}{I} - A + \frac{\partial^2 A}{\partial s^2} \\ \frac{\partial I}{\partial \tau} &= Q(A^2 - I) + P\,\frac{\partial^2 I}{\partial s^2} \end{aligned}\]

The slides do not show the substitution that turns the seven original constants into \(R\), \(Q\) and \(P\) — that step is missing here and should be checked against the original if it matters. What matters for this chapter is just that \(A\)'s own diffusion coefficient has been normalised to \(1\), so \(P\) plays the role of the inhibitor's diffusion constant relative to the activator's.

The homogeneous solution — the state with no dependence on \(s\) or \(\tau\) — solves the same two equations with all derivatives set to zero. The second equation forces \(I=A^2\); substituting into the first collapses it to the linear equation \(1+R-A=0\), so there is exactly one such state:

\[\bar A = R+1, \qquad \bar I = (R+1)^2.\]

When does a uniform disturbance decay?

For a linear system \(\dot{\mathbf x} = J\mathbf x\) in two variables, solutions grow like \(e^{\lambda t}\) with \(\lambda\) a root of \(\lambda^2 - (\operatorname{tr}J)\lambda + \det J = 0\). Both roots have negative real part — so any small disturbance decays — exactly when \(\operatorname{tr}J<0\) and \(\det J>0\). That is the test applied twice below.

Write the reaction terms (no diffusion) as \(f(A,I)=1+RA^2/I-A\) and \(g(A,I)=Q(A^2-I)\). Their Jacobian, evaluated at \((\bar A,\bar I)=(R+1,(R+1)^2)\), is

\[J = \begin{pmatrix} \dfrac{2R\bar A}{\bar I}-1 & -\dfrac{R\bar A^2}{\bar I^2} \\[4pt] 2\bar A Q & -Q \end{pmatrix} = \begin{pmatrix} \dfrac{R-1}{R+1} & -\dfrac{R}{(R+1)^2} \\[4pt] 2(R+1)Q & -Q \end{pmatrix}.\]

Its trace is \(\dfrac{R-1}{R+1}-Q\), and its determinant works out to just \(Q\): the two off-diagonal products are \(-\dfrac{R-1}{R+1}Q\) (from the diagonal) and \(\dfrac{2RQ}{R+1}\) (from the off-diagonal, with a sign flip), and \(-\dfrac{R-1}{R+1}+\dfrac{2R}{R+1}=1\), leaving \(\det J=Q\).

So the uniform state is stable to a spatially uniform disturbance exactly when

\[\det J>0 \iff Q>0, \qquad \operatorname{tr}J<0 \iff Q>\frac{R-1}{R+1}.\]

Both must hold; the second is the binding one whenever \(R>1\), and the first is the binding one otherwise.

Perturbing in space

Now let the disturbance vary with position: write \(A(s,\tau)=\bar A+A'(s,\tau)\) and \(I(s,\tau)=\bar I+I'(s,\tau)\), and drop products of the primed terms (the linear-stability approximation). Substituting into the full PDE reproduces the reaction Jacobian above, plus each variable's own diffusion term:

\[\begin{aligned} \frac{\partial A'}{\partial \tau} &= \frac{R-1}{R+1}A' - \frac{R}{(1+R)^2}I' + \frac{\partial^2 A'}{\partial s^2} \\ \frac{\partial I'}{\partial \tau} &= 2Q(1+R)A' - QI' + P\frac{\partial^2 I'}{\partial s^2} \end{aligned}\]

To test one spatial mode at a time, try a single cosine ripple of "wavelength" set by a scale \(\ell\):

\[A'(s,\tau) = \hat A(\tau)\cos\!\left(\frac{s}{\ell}\right), \qquad I'(s,\tau) = \hat I(\tau)\cos\!\left(\frac{s}{\ell}\right).\]

Because \(\cos(s/\ell)\) is an eigenfunction of \(\partial^2/\partial s^2\) with eigenvalue \(-1/\ell^2\), every term in the two PDEs above carries the same factor \(\cos(s/\ell)\), which cancels and leaves a pair of ordinary differential equations for the amplitudes:

\[\begin{aligned} \frac{d\hat A}{d\tau} &= \left(\frac{R-1}{R+1}-\frac{1}{\ell^2}\right)\hat A - \frac{R}{(1+R)^2}\hat I \\ \frac{d\hat I}{d\tau} &= 2Q(1+R)\hat A - \left(Q+\frac{P}{\ell^2}\right)\hat I \end{aligned}\]

which is exactly the reaction Jacobian \(J\) from before, with \(1/\ell^2\) subtracted from the top-left entry and \(P/\ell^2\) subtracted from the bottom-right one — each variable's own diffusive decay, at the wavenumber set by \(\ell\).

The uniform state disturbed by one spatial mode of scale ℓ. Whether the ripple's amplitude Â(τ), Î(τ) grows or decays is exactly the question the mode-ℓ Jacobian below answers.

When does a spatial ripple decay?

Apply the same trace/determinant test to this mode-\(\ell\) Jacobian. The trace condition rearranges cleanly:

\[\operatorname{tr}<0 \iff Q > \frac{R-1}{R+1} - \frac{1+P}{\ell^2},\]

which is a weaker requirement than the uniform-disturbance condition \(Q>\frac{R-1}{R+1}\) — adding diffusion only helps here, since \((1+P)/\ell^2>0\). The determinant condition is the more delicate one and genuinely depends on \(\ell\):

\[\det>0 \iff -\left(\frac{R-1}{R+1}-\frac{1}{\ell^2}\right)\left(Q+\frac{P}{\ell^2}\right) + \frac{2QR}{1+R} > 0,\]

and working out exactly which values of \(\ell\) satisfy it — as opposed to the uniform case, where there is no \(\ell\) to range over — is a genuinely harder calculation than the one for the flat disturbance. The slides do not carry it through mode by mode; they state a single comparison as the conclusion, given twice in the same form:

\[\text{homogeneous stability: } Q > \frac{R-1}{R+1}, \qquad\qquad \text{spatial stability: } \frac{Q}{P} > \frac{R-1}{R+1}.\]

(This chapter's slides are a model's reconstruction of a scanned PDF, flagged by the source itself as unverified equation-by-equation; the step from the boxed trace/determinant inequalities above to this final \(\ell\)-independent line is one it does not fully preserve. The two boxed results themselves are stated twice, identically, so they are taken here as the lecture's conclusion even though the derivation between them and the general mode-\(\ell\) conditions above is not fully reconstructable from what survives.)

Reading the comparison

The slides put the two conditions side by side and gloss them directly: a state is homogeneously stable if \(\bar I\) relaxes back to its previous value after a small uniform disturbance, and inhomogeneously stable if \(I'\) relaxes back to \(\bar I\) after a small spatial disturbance.

The two boxed thresholds are the same expression, \(\dfrac{R-1}{R+1}\), compared against \(Q\) on one side and \(Q/P\) on the other. If the inhibitor's relative diffusion constant \(P\) exceeds \(1\) — it diffuses faster than the activator, which is exactly what "global" is supposed to mean in this model's name — then \(Q/P<Q\), so there is a band of \(Q\) for which the flat state survives a uniform nudge (\(Q\) above the threshold) but not a spatially varying one (\(Q/P\) below it). Whenever \(R>1\) (so the threshold itself is positive) and \(P>1\), that band is non-empty: a state that looks settled under a uniform perturbation can still fail under a patterned one. That gap between the two thresholds is the opening this course's account of pattern formation needs.

Sources

• Both slide decks under lectures/16-notes/ in the course's OCW materials: 01-local-excitation-global-inhibition.md (the model, its nondimensionalisation, and the homogeneous solution) and 02-stability-of-homogeneous-solution.md (the homogeneous and spatial stability analysis). No transcript, written notes or problem set were supplied for this lecture, so this chapter is built from those two files alone.
• This is genuinely thin material: the two slide files together are a handful of terse equation blocks with almost no connecting prose, converted by a model from a scanned PDF with no text layer. Both files carry the source's own flag that "every equation is unverified." Everywhere this chapter shows a derivation in full (the Jacobian entries, the trace and determinant, the cosine-mode substitution), it has been checked directly against the slide's stated results by recomputing it; the one place that could not be reconciled — the algebra between the general mode-\(\ell\) trace/determinant conditions and the slides' final, \(\ell\)-independent comparison — is flagged in the text rather than papered over.
• The course runs this topic across three consecutive lectures (numbered 15–17 in the lecture sequence); this chapter covers only the middle one, the stability analysis of the homogeneous solution. What set up the model (the previous lecture) and what is built on this stability result (the next) are not part of what was supplied here.

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