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- #1

- Apr 13, 2013

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I want to check by direct differentation that the formula $u(x,t)=\phi(z)$, where $z$ is given implicitly by $x-z=t a(\phi(z))$, does indeed provide a solution of the pde $u_t+a(u) u_x=0$.

I have tried the following, but we do not get the desired result. Have I done something wrong?

$x-z=ta\phi (z)\Rightarrow \phi (z)=\frac{x-z}{ta}$

$u(x,t)=\phi (z)=\frac{x-z}{ta}$

$u_t=-\frac{x-z}{at^2}$

$u_x=\frac{1}{ta}$

$u_t+a(u)u_x=-\frac{x-z}{at^2}+a\frac{1}{ta}=-\frac{x-z}{at^2}+\frac{at}{t^2a}=\frac{-x+z+at}{at^2} $.