# direct differentiation in PDE

How can we check by direct differentiation that the formula $u(x, t) = \varphi(z)$, where $z$ is given implicitly by $x − z = ta(\varphi(z))$, does indeed provide a solution of the PDE $u_t + a(u)u_x = 0$?

So here's my intuition:

find value of $z$ from $x − z = ta(\varphi(z))$, by isolating it and then substitute; I only can think of characteristics, but really can't see about the a(u)

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I added some TeX for you here ;) – Nikita Evseev Oct 15 '12 at 7:44
Thanks for your help – Buddy Holly Oct 15 '12 at 7:50
Could you calculate $z_x$ and $z_t$? – Nikita Evseev Oct 15 '12 at 7:53

I'll just continue with nikita2's solution. To find $u_x$, different the implicit equation for $z$ with respect to $x$:

\begin{align*} 1 - z_x & = ta'(\varphi(z))\varphi'(z)z_x \\ z_x & = \frac{1}{1 + ta'(\varphi(z))\varphi'(z)} \end{align*}

Substitute into $u_x = \varphi'(z)z_x$ and $u_t = \varphi'(z)z_t$ to get $u_t + a(u)u_x = 0$.

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Ok, first of all $u_t = \varphi'(z)z_t$,

then differentiate the equation $x − z = ta(\varphi(z))$ and have

$-z_t=a(\varphi(z)) + ta'(\varphi(z))\varphi'(z)z_t$

Thus $z_t = -a(\varphi(z))/(ta'(\varphi(z))\varphi'(z)+1)$.

Doing the same to find $u_x = \varphi'(z)z_x$ you will obtain the aim.

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So does that mean that I can change everywhere with "t" put an x and the same entire proof argument followed through? – Buddy Holly Oct 15 '12 at 19:31
And when I tried it, I just got a mess with the a expression. I still don't see what to do – Buddy Holly Oct 16 '12 at 4:08