3
$\begingroup$

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!

$\endgroup$

1 Answer 1

3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Mat Dyl
    Commented Mar 4, 2016 at 15:58
  • $\begingroup$ 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. $\endgroup$
    – JJacquelin
    Commented Mar 4, 2016 at 18:05
  • $\begingroup$ Ok thank you, but could you explain to me what is $\theta$ exactly? $\endgroup$
    – Mat Dyl
    Commented Mar 4, 2016 at 18:10
  • $\begingroup$ $\theta=x-ct$ 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 $(x,t)$. $\endgroup$
    – JJacquelin
    Commented Mar 4, 2016 at 18:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .