I am working with the Fisher equation which I have non-dimensionalised as \begin{equation} \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} +u(1-u) \end{equation}
I am looking for travelling wave solutions for the previous equation. I let $u(x,t)=U(z)$ with $z=x-ct$ so that $c$ is the speed of the travelling wave with $x$ from -100 to 100. and sudying for $t \geq 0$. I found the second order ODE for $U(z) as from the last eq, this being \begin{align} U'=V \end{align} \begin{align} V' = -cV - U(1-U) \end{align}
This is the part that I am having issues with. I am given the following $$ u_0(x)=u(x,0)= \frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^2} $$ For some positive parameter $\beta$. I must show that there is some unique $c$, and state it's value, such that $U(z)=u_0(z)$ for some arbitrary $\beta$.
I would appreciate any help with this problem or an approach that will lead me in the right direction. Thank you all in advance.
I agree with what you found : $\quad\begin{cases} U'=V\\ V' = -cV - U(1-U) \end{cases}$ $$U''=-cU'-U(1-U)\tag 1$$ With initial condition : $$ u_0(x)=u(x,0)= \frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^2}=U(x) \tag 2 $$ because this condition must satisfy the ODE $(1)$.
$$U'(x)=\frac{-2}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\frac{\beta}{\sqrt{6}}e^{\frac{x}{\sqrt{6}}}\tag 3$$ $$U''(x)=\frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^4}\beta^2e^{2\frac{x}{\sqrt{6}}}) -\frac{1}{3(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\beta e^{\frac{x}{\sqrt{6}}}\tag 4$$ Putting $(2),(3),(4)$ into equation $(1)$ after simplification leads to : $$\frac{c\sqrt{3}-5}{3(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\beta e^{\frac{x}{\sqrt{6}}}=0$$ This can be satisfied any $\beta$ only with $$c=\frac{5}{\sqrt{3}}$$