Past exam of the topics in applied mathematics course of the University of Oxford / 2011 / A / 2

Let $J$ be population flux. For any fixed control volume $V\subseteq \Omega$, mass conservation with local production givesAssume local, isotropic random dispersal, no imposed advection, and constant diffusion coefficient $D$. Fick's first law is $J=-D\nabla u$. The divergence theorem and arbitrariness of $V$ yieldThe coefficient has units length squared per time; $f$ has units population density per time. Smoothness sufficient to pass from the integral balance to the local equation is understood. This is a reaction-diffusion equation.

The travelling-wave reduction of a reaction-diffusion system of Fisher's equation isMultiplying by $U^{\prime }$ and integrating gives $c{\int }_{\mathbb{R}}(U^{\prime }{)}^{2}dz=1/6$, so the decreasing invasion front has $c>0$. The linearization at $U=0$ has spatial eigenvaluesFor $0<c<2$ they are complex: approaching zero spirals in the $(U,U^{\prime })$ plane and changes the sign of $U$, which is inadmissible for a density. Thus $c\ge 2$ is necessary.

It is also sufficient. Write $V=U^{\prime }$ and choose $\ell =(c-\sqrt{c^2-4})/2>0$. The triangleis forward invariant for $U^{\prime }=V$, $V^{\prime }=-cV-U(1-U)$. On $V=0$, $V^{\prime }\le 0$. On $V=-\ell U$,The boundary $U=1$ points left, and the boundary at $U=0$ consists only of the origin. The unstable branch from the saddle $(1,0)$ enters this triangle, where $U$ decreases and remains nonnegative. Its limiting state must have $V=0$ and $U(1-U)=0$; since it has left $U=1$, it tends to $(0,0)$. Therefore a monotone positive front exists for every $c\ge 2$, including the double-root endpoint:This is the Fisher-KPP minimum wave speed, obtained with an explicit invariant-region existence argument.

Set $Y=U^{q/2}$ and $k=2b/q$. The proposed profile with $a,b>0$ givesSubstituting into $U^{\prime \prime }+c{U}^{\prime }+U(1-U^q)=0$ and dividing in the interior by $U(1-Y)$ yieldsEquating the constant and $Y$ coefficients givesHence the exact generalized Fisher front isThe free positive $a$ translates the front. It connects one on the left to zero on the right and is strictly decreasing. At $q=1$, its speed is $5/\sqrt{6}$.

The explicit travelling wave speed increases with $q$:Its two tails change differently. Near the leading empty state, $U\sim a^{-2/q}{e}^{-kz}$ with $k=\sqrt{2/(q+2)}$ decreasing as $q$ grows: that tail lengthens. Near the filled state, $1-U\sim (2a/q)e^{bz}$, with $b=q/\sqrt{2(q+2)}$ increasing: that tail becomes steeper. Thus the change is not a uniform sharpening of the whole profile. For example the half-height slope is $-\frac{k}{2}(1-2^{-q/2})$.

For meaningful comparison, the figure aligns all fronts at $U(0)=1/2$ by choosing $a=2^{q/2}-1$. It displays the speed increase and the unequal changes of the two tails. Since $f^{\prime }(0)=1$ for every $q$, the minimum admissible speed remains two; the explicit family has $c(q)>2$ for $q>0$. The invariant-triangle proof above extends with boundary derivative $U^{q+1}\ge 0$.