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If we have $u_t + c(x,t) u_x = 0 \; \; $ describes uni-directional wave propagation in a medium with variable wave speed.

a) Explain how to solve it by the method of charichtaristics for general $c(x,t)$ and Cauchy data $u(x,0)=f(x)$.

b) If $c=1+\epsilon \sin x$ with small parameter $\epsilon$ goes to $0$, find the explicit form of the solution including terms up to $O(\epsilon)$.

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@Gortaur: I wouldn't bother linking to an unrelated question with the same meta-ish discussion about imperative mode. It's a definitely a recurring theme; just point out how the tone appears to others and let it go from there. lio: That doesn't look like a wave equation. Are you sure you don't mean $u_{tt}+c(x,t)u_{xx}$? – anon Sep 29 '11 at 10:05
@anon: thank you. – Ilya Sep 29 '11 at 10:10
@lio, you're using first derivatives of $u$ which is not a wave equation. Did you mean $u_{tt} + c(x,t) u_{xx} = 0\;$? – rcollyer Sep 29 '11 at 11:19
At least can't find the general method of solving $u_t+c(x,t)u_x=0$ . – doraemonpaul Oct 24 '12 at 2:43
$u_{tt}+c(x,t)u_{xx}=0$ is difficult to solve it generally unless when $c(x,t)=f(x)g(t)$ so that can solve it by separation of variables. – doraemonpaul Oct 24 '12 at 3:10

If you really want to solve $u_t+c(x,t)u_x=0$ :

By method of characteristics,


With reference to,

For $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$


For $\dfrac{du}{ds}=0$ , letting $u(0)=f(x(0))$ , we have $u(x(t),t)=f(x(0))$ , where $x(t)$ satisflies $\dfrac{dx}{dt}=c(x,t)$

But $\dfrac{dx}{dt}=c(x,t)$ is a general first-order ODE, as no general method of solving first-order ODE are known, we can only express like this.

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