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Let $u(t,x)$ be a function in the coordinates $t$ and $x$. Now $u$ is to be expressed in coordinates $t$ and $\xi$ instead where $\xi=x-ct$.

Do we then have $$ u_{xx}=u_{\xi\xi}+cu_{\xi}? $$

I tried to answer this but did not come along. Maybe you can help me?

M. Meyer
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1 Answers1

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It helps to have different names for different things. If you write $$ U(t, \xi) = u(t, \xi + ct) $$ then you can compute \begin{align} U_{\xi}(t, \xi) &= u_x(t, \xi + ct) \cdot \frac{d(\xi + ct)}{d\xi} \\ &= u_x(t, \xi + ct) \end{align} and similarly, \begin{align} U_{\xi\xi}(t, \xi) &= u_{xx}(t, \xi + ct). \end{align}

Post-comment addition:

On the other hand, if you compute $U_t$, you get

\begin{align} U_{t}(t, \xi) &= u_t(t, \xi + ct) \cdot \frac{d(t)}{dt} + u_x(t, \xi + ct) \cdot \frac{d(\xi + ct)}{dt} \\ &= u_t(t, \xi + ct) + u_x(t, \xi + ct) c \\ \end{align} so that \begin{align} u_t(t, \xi + ct) &= U_{t}(t, \xi) - u_x(t, \xi + ct) c \end{align} And since $u_x = U_{\xi}$ (modulo shuffling arguments a little), this can be rewritten \begin{align} u_t(t, \xi + ct) &= U_{t}(t, \xi) - c ~U_{\xi}(t, \xi) \end{align}

John Hughes
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  • I am just wondering about the following. The original PDE, i.e. in coordinates $t$ and $x$ is given by $u_t=u_{xx}+f(u)-w$, Now, after going over to coordinates $t$ and $\xi=x-ct$, the ODE is stated to be $u_t=u_{\xi\xi}+ cu_{\xi}+f(u)-w$. And I am wondering where the $cu_{\xi}$ comes from. – M. Meyer Dec 01 '15 at 14:35
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    See "post-comment additions" above. Once you substitute in the expression for $u_{xx}$ and $u_{t}$ into the original PDE, you'll get a new PDE for the function $U$ that'll look like $U_t = U_{\xi\xi} + cU_{\xi} + F(U) - w$, where $F(U(t, \xi))$ is defined to be $f(u(t, x))$, using the coordinate transformation above. Calling "f" and "F" by the same name, and calling $u$ and $U$ by the same name...that's typical (alas), and has confused generations of students. – John Hughes Dec 01 '15 at 15:19
  • Ah, thats the multi-dimensional chain rule that you use to compute $U_t(t,\xi)$. – M. Meyer Dec 01 '15 at 15:26
  • Exactly right. $~$ – John Hughes Dec 01 '15 at 16:41