= Solution
The <equal-diffusivity Fisher reduction> uses equal <diffusion coefficients> and the invariant relation $v=1-u$. The second equation then requires $u_\tau=\delta u(1-u)+u_{\xi\xi}$, while the first becomes $u_\tau=(1-\beta)u(1-u)+u_{\xi\xi}$. They agree when
$$
\boxed{\beta=1-\delta,\qquad v=1-u.}
$$
This reduction requires compatible initial and boundary data on that invariant relation. Tuning a parameter alone does not force arbitrary solutions of the two-field system onto it. For positive physical $\beta$, the displayed normalization also requires $0<\delta<1$. More generally the line $v=C(1-u)$ works when $\beta C=1-\delta$; the same positivity restriction remains for $C>0$.
Let $z=\xi-c\tau$ and $u=U(z)$ with $U(-\infty)=0$, $U(+\infty)=1$. The profile equation is $U''+cU'+\delta U(1-U)=0$. Multiplication by $U'$ and integration gives
$$
c\int_{-\infty}^\infty(U')^2\,dz=-\delta\int_0^1U(1-U)\,dU=-\frac\delta6.
$$
Consequently \b[$c<0$: the increasing front moves left], with the $u=1$ state invading $u=0$. <Linearization> at its leading edge gives real monotone exponents only when $|c|\geq2\sqrt\delta$, yielding the usual Fisher front speed range. A convenient exact example is
$$
U(z)=\left(1+e^{-\sqrt{\delta/6}\,z}\right)^{-2},\qquad c=-5\sqrt{\delta/6}.
$$
Substitution verifies it and the figure shows its leftward displacement. The source uses $x$ for the end limits after defining $\xi$; they denote the same spatial coordinate here.
\Image[/past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2010-fisher-front.png]
{title=An increasing Fisher front translates left as the occupied state invades}
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