Extension of the method of characteristics

Solve the pde (advection equation): $$u_t + c u_x = 0, \hspace{0.5cm} t>0, \hspace{0.5cm} x\in \mathbb{R}$$ with the condition given on an arbitrary curve $t=\tau (x)$, that is: $$u(x,\tau (x))= \phi (x), \hspace{0.5cm} \hspace{0.5cm} x\in \mathbb{R}.$$ How to generally approach this problem? I know the method of characteristics (in this example $x=ct + \xi$ and $u$ is constant on characteristics) but I have only solved problems with initial condition: $$u(x,0)=\phi(x)$$ or the boundary condition: $$u(0,t)=\psi (t)$$ Can someone explain to me (on this example) how to extend the method of characteristics to conditions of the mentioned form? Thanks in advance!

In fact, thanks to the method of characteristics (or other method), the general solution of the PDE is found on the form : $$u(x,t)=F(x-ct)\quad\text{ any derivable function } F$$ The question now is to determine the particular function $F$ according to the particular boundary condition which is: $$u\left(x\:,\: \tau(x) \right)=\phi(x) \quad \text{with known (given) functions } \tau(x) \text{ and } \phi(x)$$ So, the condition is : $$\phi(x)=F\left(x-c\:\tau(x)\right)$$ With this equation and known $\tau(x)$ and known $\phi(x)$ we have to determine the unknown function $F$.
Let $\quad \theta=x-c\:\tau(x)=f(x)\quad$ Hense $f(x)$ is a known function.
The key point is here : Whe have to consider the inverse function of $\quad \theta=f(x)\quad$ that is : $\quad x=f^{-1}(\theta)\quad$. Or in other words $x=$the root(s) of the equation $\quad x-c\:\tau(x)-\theta=0\quad$ , that is $x$ as a function of $\theta$.
$$\phi\left(f^{-1}(\theta) \right)=F(\theta)$$ So, the function $F$ is determined. The solution, according to the boundary condition is : $$u(x,t)=\phi\left(f^{-1}(x-ct) \right)$$ where $f^{-1}$ is the inverse function of $\quad f(x)=x+c\:\tau(x)$
• Thank you for your answer. I have one question though: what exactly is $X = x-c \tau (x)$ - is it somehow connected to the characteristics equation? My typical approach for this kind of problems is to write system of diff. eq, namely: $$\frac{dX}{dt}, \hspace{0.5cm} X(0)=\xi \\ \frac{dU}{dt} = 0, \hspace{0.5cm} U(0)=u(\xi,0)=\phi(\xi) \\ \text{(for initial condition u(x,0)=\phi(x))}$$ where $U(t)$ is equal to $u(x,t)$ on characteristic, namely: $U(t)=u(X(t),t)$. I'm sorry if I made a mistake somewhere - I'm taking now first course in pde and just trying to figure this out. – Mat Dyl Mar 4 '16 at 15:58
• Do not confuse my variable $X$ with your variable $X$. I will change of symbol in me first answer in order to avoid the confusion. – JJacquelin Mar 4 '16 at 18:05
• Ok thank you, but could you explain to me what is $\theta$ exactly? – Mat Dyl Mar 4 '16 at 18:10
• $\theta=x-ct$ is a solution of the PDE $u_t+c\:u_x=0$ which takes the particular form $x-c\:\tau(x)=\phi(x)$ on the boundary $t=\tau(x)$ in the system $(x,t)$. – JJacquelin Mar 4 '16 at 18:23