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Given PDE $u_t +u^2u_x=\epsilon u_{xx}$. Consider solutions of the form $f(x-st)$ with $f(+\infty)=f_R$ and $f(-\infty)=f_L$. I want to know for what values of $\epsilon$, $f_L$, $f_R$, and $s$ will the ODE have a solution?

My work: The ODE is $$\epsilon f'' + f'[s-f^2]=0$$ Integrating, $$\epsilon f' + f[s-\frac{f^2}{3}]=C$$ Applying boundary conditions yields necessary condition, $$s = \frac{f_R^3-f_L^3}{3(f_R-f_L)}$$ But I do not see how to get a sufficient condition for when the ODE will have a solution?

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Are you sure of the "Integrating" passage? I don't understand what you have done. Beware that the integral of a product is not the product of the integrals. –  Giuseppe Negro Mar 3 '13 at 23:10
    
@GiuseppeNegro That part looks okay, $f'f^2$ integrates to $f^3/3$. –  user53153 Mar 3 '13 at 23:19
    
Oh, ok. I thought it was a mistake, sorry about that. Well, then you are finished it seems. You have found a relation that speed $s$ and amplitudes $f_L,\ f_R$ need satisfy if the traveling wave is to be a solution of the PDE. –  Giuseppe Negro Mar 3 '13 at 23:25

1 Answer 1

up vote 1 down vote accepted

We are looking for a solution of the boundary value problem $$\epsilon y' + sy - y^3/{3}=C , \ \ \ y(-\infty)=f_L, \ y(+\infty)=f_R\tag1$$ Since (1) is autonomous and of 1st order, we consider the inverse function, which (should it exist) satisfies $$\epsilon^{-1} \frac{dx}{dy} = (C- sy + y^3/{3})^{-1} , \ \ \ x(f_L)=-\infty, \ x(f_R)=+\infty \tag2 $$ Clearly, $C$ must be chosen so that $C- sy + y^3/3=0$ at both $y=f_L$ and $y=f_R$. This is a system of two equations for two unknowns $C,s$, from which we get $$ C=\frac{f_Lf_R(f_L+f_R)}{3},\quad s=\frac{f_L^2+f_R^2+f_Lf_R}{3} \tag3$$ Of course, this is the same value of $s$ that you got. Now (2) simplifies to $$ \frac{dx}{dy} = \frac{3\epsilon}{(y-f_L)(y-f_R)(y+f_L+f_R)} \tag4 $$ which we can easily integrate to find the required function (which will be monotone, hence invertible), as long as the denominator does not vanish between $f_L$ and $f_R$. The latter is assured when $f_L$ and $f_R$ are of the same sign.

So, you have a sufficient condition: $f_L,f_R$ of the same sign, ans $s$ is as in your post. This may also be necessary, if you can reverse-engineer the above to show that functions that try to satisfy (1) with boundary values of opposite sign will blow up somewhere. The cubic term will be responsible for the blow up.

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Wow, you're a genius. Thank-you! –  Stuart Mar 4 '13 at 1:50

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