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For the equation

$\frac{\partial^2}{{\partial}x^2}(IE\frac{{\partial^2}u}{{\partial}x^2}) = \mu(\frac{{\partial^2}u}{{\partial}t^2})$

with $E$ a function of x, derive two ODE's by separation of variables.


I greatly appreciate your help/advice. I'm pretty unsure with how to handle derivatives like this. This is my best attempt, I've been working on this one for quite a while.

http://mathbin.net/51398

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Can you tell what $I$ is? Is $\mu$ a function? I guess not, but better ask. –  Jonas Teuwen Sep 1 '10 at 12:27
    
I and mu are constants. E is a function of x –  gurk Sep 1 '10 at 12:58
    
I would just fill in $u(x,t) = f(x)g(t)$ and compute it directly. It will give you a fourth derivative in $f$, then you can just separate them and solve the ODEs. I tried it and just got a fourth order ODE of $f$ and a second order of $g$. Where are you stuck in this process? –  Jonas Teuwen Sep 1 '10 at 13:06
    
Your Ansatz is perfectly correct. –  Tobias Kienzler Sep 1 '10 at 14:57
    
I am struggling with this problem... So far this is what I have. Is this correct? I really appreciate the help provided so far. Thank you. Putting $u(x,t) = X(x)T(t)$ as mentioned: $\frac{d^{2}}{dx^{2}}\left( I\mbox{E}\left( x \right)\frac{d^{2}X}{dx^{2}}T \right)\; =\; \mu \frac{d^{2}T}{dt^{2}}X$ Now: $\frac{d^{4}X}{dx^{4}}\left( I\mbox{E}\left( x \right)T \right)\; =\; \mu \frac{d^{2}T}{dt^{2}}X$ Simplifying again: $\frac{d^{4}X}{dx^{4}}\frac{\mbox{E}\left( x \right)}{\mu X}\; =\; \frac{d^{2}T}{dt^{2}}\frac{1}{TI}$ Is this correct so far? I know the final step put it = to a constant –  gurk Sep 1 '10 at 15:31

1 Answer 1

Let $u(x,t)=X(x)T(t)$ ,

Then $\dfrac{\partial^2}{\partial x^2}\biggl(IE(x)\dfrac{\partial^2(X(x)T(t))}{\partial x^2}\biggr)=\mu\dfrac{\partial^2(X(x)T(t))}{\partial t^2}$

$T(t)\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)=\mu X(x)\dfrac{d^2T(t)}{dt^2}$

$\dfrac{1}{\mu X(x)}\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)=\dfrac{1}{T(t)}\dfrac{d^2T(t)}{dt^2}=-(f(s))^2$

$\begin{cases}\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)+\mu(f(s))^2X(x)=0\\\dfrac{d^2T(t)}{dt^2}+(f(s))^2T(t)=0\end{cases}$

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