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I need to find an ordinary differential equation for the travelling-wave solution $$\hat{u}(\xi)=\hat{u}(x-b\cdot t)=u(x,t)$$ of $$\frac{\partial u}{\partial t}=\frac{\partial ^2 u}{\partial x^2} - u(1-u)(a-u).$$

Draw a phase plane for $\hat{u}$ and investigate when a solution exists subject to the conditions $$\lim_{\xi \to \infty} \hat{u}(\xi) \to 1$$ and $$\lim_{\xi \to -\infty} \hat{u}(\xi) \to a.$$

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I think this is discussed in Joel Smoller's book. – Hans Engler Jun 2 '13 at 15:22

You have $$ \xi=x-bt, $$

$$ \frac{d}{dx}\xi=1,\;\frac{d}{dt}\xi=-b, $$

$$ u_x=u_\xi \xi_x=u_\xi,\;\; u_t=u_\xi \xi_t=-u_\xi b. $$ In the same manner you obtain the second derivative of $u$. Thus you obtain a second order ODE, right? In particular, you have $$ -bu'=u''+u(1-u)(a-u). $$ This can be written as $$ u'=v $$ $$ v'=-u(1-u)(a-u)-bv\;\;(=u'') $$ Notice that $(u,v)=(0,0),(1,0)$ and $(a,0)$ are equilibrium points for the ODE. You can plot the phase portrait. Now you need to obtain $b$ such that there exists a trajectory leaving $a$ and entering into 1. I hope this can help you.

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