I'm currently reading J.D. Murray's Mathematical Biology I, Chapter 13.2 discussing travelling wave solutions to the Fisher-KPP equation, and in particular their dependence on initial conditions $u(x, 0)$ as $x \to \infty$.
We say that $u(x,t)$ is a travelling wave solution with speed $c$ to a PDE where for some $U(z)$, $$ u(x,t)=U(x-ct). $$ Our non-dimensionalised PDE is given by $$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + u(1-u), $$ where $u(x,t) \to 1$ as $x \to -\infty$, $u(x,t) \to 0$ as $x \to \infty$.
On the basis that we can always find a travelling wave solution, we want to analyse the wavespeed $c$ of solutions where for $A, a > 0$, $u(x,0) \sim Ae^{-ax}$ as $x \to \infty$.
To do this, we consider $u(x,t)$ such that $u(x,t) \sim Ae^{-a(x-ct)}$ as $x \to \infty$. For $x$ large enough, as $u^2 \ll u$ we see that $$ ca=a^2+1, $$ and so solutions of this form are possible with $c = a + a^{-1}$. Indeed, this indicates that $c \ge 2$, with this minimum occurring for $a = 1$.
In my mind, this seems like the end of the problem, indicating that $c = a + a^{-1}$ is the true behaviour of the wave speed where the travelling wave has asymptotic behaviour as described. Yet Murray indicates the following (paraphrasing):
For $a < 1$, $e^{-ax} > e^{-x}$, and so the velocity of propagation depends on the leading edge of the wave, giving $c = a + a^{-1}$.
For $a > 1$, $e^{-ax} < e^{-x}$, and there is a front with wavespeed $c = 2$ which is a solution.
I'm confused by this last part, as it seems to imply that $u(x,t) \sim Ae^{-(x-2t)}$ as $x \to \infty$ satisfies $u(x,0) \sim Ae^{-ax}$, yet this seems obviously false as $e^{-x}/e^{-ax} = e^{x(a-1)} \to \infty \neq 1$ as $x \to \infty$.
Any help in explaining why we should have wavespeed $c = 2$ for $a \ge 1$, or why we wouldn't expect $c = a + a^{-1}$ to be a solution for $a > 1$, would be much appreciated.
Another attempt
I've had another attempt at this problem, but it's led me back to the same point of confusion.
With boundary condition $U(z) \sim Ae^{-az}$, taking $z$ sufficiently large we see it should tend to the solution of the linearised ODE $$ U'' + cU' + U = 0. $$ Assuming that $c \neq 2$, this gives the general solution $$ U_1(z) = C_1 e^{\frac{-c + \sqrt{c^2 - 4}}{2} z} + C_2 e^{\frac{-c - \sqrt{c^2 - 4}}{2}z}, $$
At the same time, we see we can also assume that $c = 2$, which gives the general solution $$ U_2(z) = (D_1 + D_2 z)e^{-z}. $$ and thus by superposition we have a more general solution $$ U(z) = U_1(z) + U_2(z), $$ and so we satisfy the boundary condition provided as $z \to \infty$, $$ A^{-1}e^{az}\left(C_1 e^{\frac{-c + \sqrt{c^2 - 4}}{2}z} + C_2 e^{\frac{-c - \sqrt{c^2 - 4}}{2}z} + D_1 e^{-z} + D_2 z e^{-z}\right) \to 1. $$ We can distinguish between cases now:
For $a < 1$, the $D_1$ and $D_2$ terms die out, so as the $C_1$ term dominates the $C_2$ term, we require that $C_1 = A$ and $a = \frac{c - \sqrt{c^2 - 4}}{2}$, so $c = a + a^{-1}$.
For $a = 1$, $D_2 = 0$ as otherwise the term blows up, and $c = 2$ to prevent the $C_1$ term blowing up. From then $C_1$ and $D_1$ are both attached to $e^{-z}$ so we just require $C_1 + D_1 = A$.
For $a > 1$, $D_1 = D_2 = 0$ to prevent terms blowing up. Again the $C_1$ term dominates the $C_2$ term so we want $a = \frac{c - \sqrt{c^2 - 4}}{2}$, yet this is impossible as this would give $a \le 1$. Thus we need to set $C_1 = 0$, $C_2 = A$, $c = a + a^{-1}$.
This brings me back to my initial problem, in that it seems impossible to get a travelling wave to satisfy initial conditions decaying at rate $a > 1$ and also with speed $c > 2$.
