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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.

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  • $\begingroup$ I suppose you could take $\partial_{x}$ (and $\partial_{t}$) of the first equation, rearrange for $\partial_{x} p$ and $\partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$. $\endgroup$ – Mattos Mar 11 '16 at 16:03
  • $\begingroup$ @Mattos What do you mean by rearranging $\partial_x p$? $\endgroup$ – new guy Mar 11 '16 at 16:05
  • $\begingroup$ From the first equation $$\partial_{x} p = \frac{1}{A} \partial_{x} \bigg( \partial_{t} f - c \partial_{x} f \bigg)$$ You can get something similar for $\partial_{t} p$. $\endgroup$ – Mattos Mar 11 '16 at 16:06
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$$\begin{cases} \frac{\partial}{\partial t}f(x,t)-c\frac{\partial}{\partial x}f(x,t)-Ap(x,t)=0 \\ \frac{\partial}{\partial t}p(x,t)+c\frac{\partial}{\partial x}p(x,t)+Af(x,t)=0 \end{cases}$$ 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.

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Hint.

We have

$$ \mathcal{D}_1 f = A p\\ \mathcal{D}_2 p = -A f $$

then

$$ \mathcal{D}_2\mathcal{D}_1 f = A\mathcal{D}_2 p = -A^2 f\\ \mathcal{D}_1\mathcal{D}_2 p = -A \mathcal{D}_1f = -A^2p $$

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