= Solution
For a nonconstant invasion <travelling wave>, the relevant end states are $w(-\infty)=k$ and $w(+\infty)=0$. These fronts do not exist at every nonnegative speed: the <phase plane> calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv0$ and $w\equiv k$ are, of course, nonnegative <travelling waves> for every speed, including zero.
Put $z=x-ct$ and $p=w'$. The <travelling-wave reduction of a reaction-diffusion system> gives
$$
w''+cw'+w(k-w)=0,\qquad
\begin{pmatrix}w'\\p'\end{pmatrix}=\begin{pmatrix}p\\-cp-w(k-w)\end{pmatrix}.
$$
The <equilibrium points> in the <phase plane> are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative <eigenvalue>. Thus it is a <saddle equilibrium>. Its descending <unstable manifold> has $w<k$, $p<0$ and locally
$$
p=r_+(w-k)+o(|w-k|),\qquad r_+=\frac{-c+\sqrt{c^2+4k}}2>0.
$$
At $(0,0)$ the <eigenvalues> are
$$
r=\frac{-c\pm\sqrt{c^2-4k}}2.
$$
For $c>2\sqrt{k}$ it is a <stable node>; for $0<c<2\sqrt{k}$ it is a <stable spiral>. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.
Suppose $c\geq2\sqrt{k}$. Choose $a>0$ with $a(c-a)\geq k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular region
$$
\mathcal T=\{(w,p):0\leq w\leq k,\ -aw\leq p\leq0\}.
$$
On the horizontal side $p=0$, the vector field has $p'=-w(k-w)\leq0$, pointing into $\mathcal T$. On the vertical side $w=k$, it has $w'=p\leq0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,
$$
h'=p'+aw'=(a-c)p-w(k-w)=w[a(c-a)-k+w]\geq w^2\geq0.
$$
Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an <equilibrium point> and cannot be crossed in finite time by a nonconstant trajectory, by <uniqueness theorem for ordinary differential equations>. This proves forward invariance of $\mathcal T$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.
The descending <unstable manifold> of $(k,0)$ initially enters $\mathcal T$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p'<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell\in[0,k)$. To identify the limit, its <omega-limit set> lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w'=0$; its second equation then forces $\ell(k-\ell)=0$. Since $\ell<k$, it follows that $\ell=0$, and the only possible <omega-limit set> is $(0,0)$. Backward in $z$, the <unstable manifold> tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing <heteroclinic orbit> and hence a nonnegative <travelling wave> for every $c\geq2\sqrt{k}$.
For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu=\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero <equilibrium point> the system becomes
$$
X'=-\frac c2X+\nu Y,\qquad Y'=-\nu X-\frac c2Y+\frac{X^2}{\nu}.
$$
For a nonzero trajectory its polar angle satisfies
$$
\vartheta'=\frac{XY'-YX'}{X^2+Y^2}
=-\nu+\frac{X^3}{\nu(X^2+Y^2)}=-\nu+O(\sqrt{X^2+Y^2}).
$$
A trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integral
$$
E=\frac12p^2+\frac k2w^2-\frac13w^3
$$
is conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no <heteroclinic orbit> connects them. Consequently
$$
\boxed{\text{nonnegative invasion fronts exist exactly for }c\geq c_*=2\sqrt{k}.}
$$
This is the <phase-plane construction of a Fisher travelling front>, including the critical speed. The statement for all $c\geq0$ must distinguish these nonconstant fronts from the two constant <travelling waves>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-64-phase-plane.png]
{title=Invariant triangle and descending Fisher-front trajectory at a supercritical speed, compared with sign-changing motion at a subcritical speed}
Back to article page
