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Given two first order PDE's

$$\frac{\partial v(x,t)}{\partial x} = R~i(x,t) + L~\frac{\partial i(x,t)}{\partial t}\tag{1}$$

$$\frac{\partial i(x,t)}{\partial x} =G~v(x,t) + C\frac{\partial v(x,t)}{\partial t}\tag{2}$$

How to combine these two PDE as a single PDE of the form:

$$\frac{\partial^2f(x,t)}{\partial x^2} = RG f(x,t) + (RC +LG)\frac{\partial f(x,t)}{\partial t} + LC\frac{\partial^2 f(x,t)}{\partial t^2}$$

my attempt is this:

substitute (2) into (1).

$$\frac{\partial v(x,t)}{\partial x} = R~i(x,t) + L~(G~v(x,t) + C\frac{\partial v(x,t)}{\partial t})$$

$$\frac{\partial v(x,t)}{\partial x} = R~i(x,t) + LG~v(x,t) + LC\frac{\partial v(x,t)}{\partial t}$$

?? doesn't really look the same...

the textbook claims that f(x,t) can be either v(x,t) or i(x,t)....

pico
  • 941
  • In your attempt, you replaced $\large \frac{\partial i}{\partial t}$ with $\large \frac{\partial i}{\partial x}$, which is not valid. You cannot replace one thing with a different, inequivalent thing. – anon Jan 03 '20 at 21:03
  • whoops... didn't notice that... – pico Jan 03 '20 at 21:08

1 Answers1

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You may write your system as

$$ \begin{cases} \displaystyle \frac{\partial v}{\partial x} = \left(R+L\frac{\partial}{\partial t}\right)i \\ \displaystyle \frac{\partial i}{\partial x} = \left(G+C\frac{\partial}{\partial t}\right)v \end{cases} $$

So evaluate $\displaystyle \frac{\partial^2 v}{\partial x^2}$ by writing $\displaystyle\frac{\partial}{\partial x}\left(\frac{\partial v}{\partial x}\right)$, substituting for $\displaystyle\frac{\partial v}{\partial x}$, use the fact that $\displaystyle\frac{\partial}{\partial x}$ and $\displaystyle\frac{\partial}{\partial t}$ commute, then substitute again to get a $2$nd order DE with derivatives with respect to $t$.

anon
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