Wave Equation using seperation of variables

**flaming** · Mar 26th 2009, 01:54 PM

Hello i was wondering if someone could help me out and show me how to solve the following heat wave problem.We dont have a book for this class and i couldnt make it to my last class so i dont knoow how to go about this question

Solve the wave equation using seperation of variables and show thte the solution reduces to D'Alembert's solution

$\mu_{tt} = c^2\mu_{xx}$
$\mu(0,t) = 0$
$\mu(L,t) = 0$
$\mu(x,0) = f(x)$
$\mu_t(x,0) = 0$

**HallsofIvy** · Mar 28th 2009, 07:27 AM

Originally Posted by flaming

Hello i was wondering if someone could help me out and show me how to solve the following heat wave problem.We dont have a book for this class and i couldnt make it to my last class so i dont knoow how to go about this question

Solve the wave equation using seperation of variables and show thte the solution reduces to D'Alembert's solution

$\mu_{tt} = c^2\mu_{xx}$
$\mu(0,t) = 0$
$\mu(L,t) = 0$
$\mu(x,0) = f(x)$
$\mu_t(x,0) = 0$

I would feel better if you would at least show that you know what "separation of variables" is!

Assume $\mu(x,t)= X(x)T(t)$ . Then $\mu_{tt}= XT^{''}$ and $\mu_{xx}= X^{''} T$ so the equation becomes $XT^{''} = c^2X^{''} T$ . Dividing on both sides of the equation by XT gives $\frac{T^{''} }{T}= c^2\frac{X^{''} }{X}$ . Since the left side depends only on t and the right side only on x, to be equal they must each equal a constant.

That gives two ordinary equations:
$c^2\frac{X^{''} }{X}= \lambda$
or
$c^2X^{''} = \lambda X$
and
$\frac{T^{''} }{T}= \lambda$
or
$T^{''} = \lambda T$ .
You should be able to find the general solution to each of those, depending on $\lambda$ of course.
What must $\lambda$ be in order that X(0)= 0 and X(L)= 0?

I have a feeling I have repeated what your text book says!

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