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Consider $$ u_t=u_{xx}+f(u)-w,~~w_t=\varepsilon (u-\gamma w).~~(*) $$

Consider the ansatz $$ (u(x,t),w(x,t))=(u(\xi),w(\xi)),~~\xi=x+ct, c\in\mathbb{R}. $$

Putting this ansatz into $(*)$, it is said that we get $$ u'=v\\ v'=cv-f(u)+w\\ w'=\frac{\varepsilon}{c}(u-\gamma w), $$ where $'=d/d\xi$.

I do not see that.

Consider $\frac{d}{dt}u(\xi)$, then the inner derivative is $c$, but what is the outer derivative?

mathfemi
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1 Answers1

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The point of this is to convert the partial differential equations with functions of two variables into a system of ordinary differential equations with functions of one variable. It's a bit misleading to use $u(x, t) = u(\xi)$ because the $u$'s are actually different functions.

Suppose we define $u(x,t) = U(x + c t)$ and $w(x, y) = W(x + c t)$. Now the capital letters are functions of a single variable being applied to $\xi = x + c t$.

Looking at the first two equations, we have $u_{xx} = U''(x + c t)$, $u_t = c U'(x + c t)$, and $w_t = c W'(x + c t)$. Now you can rearrange the equations into:

\begin{align*} U'(\xi) &= U''(\xi) + f(U(\xi)) - W(\xi)\\ c W'(\xi) &= \varepsilon(U(\xi) - \gamma W(\xi)) \end{align*}

Using the substitution $V(\xi) = U'(\xi)$ you can get the system you stated. This is a first order nonlinear system or ODEs. Whether or not you can get a closed form solution will depend on the nature of $f$.

Paul Castle
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  • My problem is to see that $u_{xx}=U''(x+ct), u_t=cU'(x+ct)$, where $'=d/d\xi$. – mathfemi Jul 10 '16 at 11:34
  • Using the chain rule, you have $\frac{\partial}{\partial t} u(x, t) = \frac{\partial}{\partial t} U(\xi(x, t)) = \frac{\partial U}{\partial \xi} \frac{\partial \xi}{\partial t} = U'(\xi) c$. – Paul Castle Jul 10 '16 at 11:43
  • What I do not understand is: We want to have the derivative with respect to $t$. So the inner function surely is given by $x+ct$ and if we take the derivative w.r.t. to t, we surely have $c$. But I do not understand why for the outer function, we take the derivative with respect to $\xi$. – mathfemi Jul 10 '16 at 11:48
  • @mathfemi Because this is how chain rule applies in this case, because you have series of mappings $(x, t) \mapsto \xi \mapsto U$. – Evgeny Jul 11 '16 at 05:28