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I have $$\begin{align} u_x + u_t &= 0\\(\rho u)_x + \rho_t &= 0 \\ \left(Eu\right)_{x}+E_{t}+pu_{x}&=0\end{align} $$ satisfying these boundary conditions: $u\left(x,0\right)=x,\ \rho\left(x,0\right)=\left(x-1\right)^{2},\ E\left(x,0\right)=E_{0}\left(x\right)$ where $p$ is a constant and $E_0$ is arbitrary.

My attempt at a solution so far is as follows:

The first PDE I can solve easily enough by inspection to get $u=f(x-t)$. Substituting this into the second equation $u\rho_{x}+u_{x}\rho+\rho_{t}=0$, I get $$(x-t)\rho_{x}+\rho+\rho_{t}=0.$$

Apparently the integral curves of $$(x-t)\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial t}$$ are $x=x_0e^t+t$ but I'm lost and don't know what I'm doing. I feel like I need someone to walk me through how to solve it properly because I couldn't do it again on my own.

Sorry about this formatting, I tried to use the FAQ and learn some latex but yeah :p

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You're not starting out quite right. The general solution of the first equation is not $u=x-t$ but $u=f(x-t)$, where $f$ is an arbitrary (differentiable) function of one variable. – Hans Lundmark Sep 21 '11 at 14:52
My mistake, should have been more careful before submitting, thanks :) – user16453 Sep 21 '11 at 15:01
Welcome to the site! It is generally a good idea to be more specific in your question title. Plain ol' "Partial Differential Equations" won't do to pique readers' interest, and repeats the information already available in the tag (pde). It is better to use a more detailed/informative question title. I've tried to give it one, please see if you can come up with one that is even better. Cheers! – Willie Wong Sep 21 '11 at 15:58
try method of characteristics – Alice Sep 21 '11 at 16:06
If you know that $u(x,t) = f(x-t)$ and $u(x,0) = 0$, what does that tell you about $f$? – Craig Sep 21 '11 at 18:11

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