I'm having trouble solving these coupled partial differential equations:
$$\frac{\partial}{\partial t}f(x,t)-c\frac{\partial}{\partial x}f(x,t)-Ap(x,t)=0,$$
$$\frac{\partial}{\partial t}p(x,t)+c\frac{\partial}{\partial x}p(x,t)+Af(x,t)=0,$$
with $A,c$ real constants.
What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.
Best Answer
$$\begin{cases} \frac{\partial}{\partial t}f(x,t)-c\frac{\partial}{\partial x}f(x,t)-Ap(x,t)=0 \\ \frac{\partial}{\partial t}p(x,t)+c\frac{\partial}{\partial x}p(x,t)+Af(x,t)=0 \end{cases}$$ Regularised form with $\begin{cases} T=At \\ X=\frac{A}{c}x \end{cases} \quad\to\quad \begin{cases} \frac{\partial}{\partial T}f(X,T)-\frac{\partial}{\partial X}f(X,T)-p(X,T)=0 \\ \frac{\partial}{\partial T}p(X,T)+\frac{\partial}{\partial X}p(X,T)+f(X,T)=0 \end{cases}$
$$\begin{cases} f_T-f_X-p=0 \\ p_T+p_X+f=0 \end{cases}$$ $p=f_T-f_X \quad\to\quad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$\frac{\partial^2 f}{\partial T^2}-\frac{\partial^2f}{\partial X^2}+f(X,T)=0$$ Solving this hyperbolic PDE leads to $f(X,T)=f\left(At\:,\:\frac{A}{c}x \right)$
Then $\quad p(X,T)=\frac{\partial f}{\partial T}-\frac{\partial f}{\partial X}=p\left(At\:,\:\frac{A}{c}x \right)$
For example of solving see : Finding the general solution of a second order PDE This method leads to the integral form of solution : $$f(X,T)=\int c(s)e^{\sqrt{\alpha(s)-\frac{1}{2}}\:X +\sqrt{\alpha(s)+\frac{1}{2}} \:T } ds$$ $c(s)$ and $\alpha(s)$ are arbitrary real or complex functions.