Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / B / 5 / c

For $z=ct-x$ the chain rule gives ${\partial }_t=cd/dz$, ${\partial }_x=-d/dz$, ${\partial }_{xx}=d^2/d{z}^2$. ThusAs $z$ increases from the material ahead of the wave towards that behind it, the singular pulse construction is $A\to B\to C\to D\to A$. The front $AB$ raises $v$ at approximately fixed $w=0$ from the low root to the high root. On the plateau $BC$, $f(v)\simeq w$ on the right attracting branch while $w$ rises slowly. The back $CD$ decreases $v$ at approximately fixed $w=w_0$ to the left attracting branch. The recovery tail then lowers $w$ back to zero on that branch, returning to $A=(0,0)$. Both sharp transitions have thickness $O(ϵ)$ in $z$; the other segments have the slower recovery scale.

The figure labels all four points and the front/back orientations. This frozen-recovery construction needs $ϵ\gamma \ll 1$ as well as $ϵ\ll 1$ if $\gamma$ is large, so that the recovery variable remains nearly fixed in an inner transition. It also needs an admissible positive front speed for the chosen orientation, addressed below.

Put $X=z/ϵ$. The travelling equations become $c{v}_X=f(v)-w+v_{XX}$ and $c{w}_X=ϵ(\gamma v-w)$. Hence, for nonzero $O(1)$ speed and $ϵ\gamma \ll 1$, $w$ is constant to leading order across the front.

Writing $p=v_X$ gives the planar system $v_X=p$, $p_X=cp-f(v)+w$. At the outer cubic roots $A=(V_{-},0)$ and $B=(V_{+},0)$, $f^{\prime }(V_{\pm })<0$. The Jacobian determinant is $f^{\prime }(V_{\pm })<0$, so both are saddle points. The increasing front is a heteroclinic orbit from the unstable manifold of $A$ to the stable manifold of $B$. Varying $c$ supplies the shooting parameter required to match these manifolds. For this cubic one can also construct it explicitly, as in the next solution.

Multiply the second-order equation by $v_X$ and integrate across the front. Because $v_X\to 0$ at both ends, the term $\int v_{XX}{v}_{X}dX=[v_X^2/2{]}_{-\infty }^{\infty }$ vanishes. The remaining identity isThe denominator is positive for a nonconstant front, so its direction is determined by the signed area between the cubic and the horizontal recovery line: Maxwell balance for a scalar reaction-diffusion front gives zero speed when that area vanishes.

The derivative in the denominator must use the inner coordinate $X$. If one instead uses the source's original $z$ and denotes $J_z=\int v_z^{2}dz$, then $J_z={ϵ}^{-1}\int v_X^{2}dX$ and the identity is $c=\int (f-w)dv/(ϵJ_z)$. Thus the printed $v^{\prime }(z)$ is valid only if $z$ has silently been reused for the rescaled inner coordinate; with the original coordinate it misses a factor $ϵ$.

Let $c_{\uparrow }(w)$ denote the increasing-front speed from the preceding solution. Reversing that heteroclinic reverses its speed, so a decreasing back has speed $c_{\downarrow }(w)=-c_{\uparrow }(w)$. A rigidly travelling pulse has one speed throughout, requiringThis is the correct use of the integral formula: compare the oppositely oriented transitions, including their signs, rather than equating two increasing-front speeds.

For completeness, write $f(v)-w=-(v-r_1)(v-r_2)(v-r_3)$, $r_1<r_2<r_3$. The explicit increasing frozen-recovery cubic front isSubstitution, using $v_X=(v-r_1)(r_3-v)/\sqrt{2}$, verifies the equation. Since $r_1+r_2+r_3=1+a$, speed matching givesFor $0<a<1/2$ this selects the appropriate positive back level within the cubic's three-root range. The high branch connects $B$ to $C$ slowly, and the low branch supplies recovery after $D$.

The stated positive-speed, rising-front orientation imposes $a<1/2$ at leading order. The printed range $0<a<1$ by itself does not guarantee that pulse construction: at $a=1/2$ the leading front is stationary, while for $a>1/2$ the calculated increasing-front speed is negative. Existence and stability of a full finite-$ϵ$ pulse also require the slow connections; speed matching is its leading necessary construction, rather than a proof for every parameter in the printed range.