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!
 A: 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 $\quad f(x)=x+c\:\tau(x)$
