1. ## wave equation

The question is Solve the one dimensional wave equation

$\displaystyle \frac{1}{c^2}$

on $\displaystyle 0<=x<=3$, subject to the conditions

$\displaystyle u(x,0) = \frac{1}{2}x$, $\displaystyle u_t(x,0)=x(3-x), u(0,t)=u(3,t)=0$

well i started it, but im not sure if this is right coz i don't really understand wave or heat equations SIGH...

$\displaystyle u_tt(x,t) = X''(x)T(t)$ and $\displaystyle u_tt(x,t) = X(x)T''(t)$
$\displaystyle \frac{1}{c^2}$

Case 1 let $\displaystyle v = w^2 > 0$
$\displaystyle \frac{X''(x)}{X(x)} = w^2$
$\displaystyle X''(x) = w^2 X(x) = 0$

Using Auxilliary Equations:
$\displaystyle m^2 - w^2 = 0$
So $\displaystyle m = +- w$
And $\displaystyle X(x) = C_1 e^{wx} + C_2 e^{-wx}$

As $\displaystyle u(0,t) = u(3,t) = 0, X(0) = V(3) = 0$
Therefore, $\displaystyle C_1 + C_2 = 0$ and $\displaystyle C_1 = - C_2$
Therefore, no solutions...

Case 2 $\displaystyle v = w^2=0$
$\displaystyle X''(x) - 0X(x) = 0$
Auxilliary Equations $\displaystyle m^2 - 0 = 0$, $\displaystyle m = +- 0$
Therefore, $\displaystyle X(x) = C_3 e^0 + C_4 e^{-0}$
As $\displaystyle X(0) = X(2) = 0, X(0) = C_3, X(3) = 2C_3 + C_4$
So $\displaystyle 2C_3 + C_4 = 0$
Therefore, no solution as $\displaystyle 2C_3 = -C_4$

Case 3 $\displaystyle v = -w^2$ as $\displaystyle v<0$
$\displaystyle \frac{X''(x)}{X(x)} = -w^2$
$\displaystyle X''(x) + w^2X(x) = 0$

Auxilliary Equation: $\displaystyle m^2 + w^2 = 0, m = +-iw$\
So $\displaystyle C_5coswx +C_6sinwx = 0$
$\displaystyle C_5 = 0,$ So $\displaystyle C_5cos3w + C_6sin3w = 0$
$\displaystyle sin3w = 0$
Therefore, $\displaystyle 3w = \frac{n\Pi}{3}$ and $\displaystyle V = \frac{n^2\Pi^2}{9}$
Thus, $\displaystyle X(x) = B sin$ $\displaystyle \frac{xn\Pi}{3}$

Now Im lost, am i doing it right so far, if so, then how do i go about completing this, coz i have no idea what to do now.. Thank you

2. Case 1: correct.

Case 2: you have come to the right result but via the wrong method. For case 2 you have:

$\displaystyle \frac{d^2 X}{dx^2} = 0$

so integrating twice gives

$\displaystyle X = A x +B$

and to satisfy BCs $\displaystyle A = B = 0$ hence no solution as you concluded (for the wrong reasons).

Case 3: you have correctly found that:

$\displaystyle X_n (x) = B_n \sin \left( \frac{n \pi}{3} x \right)$

Consequently

$\displaystyle \frac{d^2 T}{dt^2} + c^2 \left(\frac{n \pi}{3}\right)^2 T = 0$

so then, after finding the auxiliary equation, we find the solution for $\displaystyle T$ is

$\displaystyle T_n = C \sin \left(\omega_n c t \right) + D \cos \left(\omega_n c t \right)$

where $\displaystyle \omega_n = \frac{n \pi}{3}$ so the general solution is given by:

$\displaystyle u(x,t) = \sum_{n=0}^{n=\infty} a_n(\sin \left(\omega_n c t \right) + b_n \cos \left(\omega_n c t \right)) \sin \left( \omega_n x \right)$

Applying the initial conditions on u(x,0) and u'(x,0) we have

$\displaystyle \sum_{n=1}^{n=\infty} c_n \sin \left( \omega_n x \right) = \frac{1}{2} x$

where $\displaystyle c_n = a_n b_n$ and

$\displaystyle \sum_{n=1}^{n=\infty} a_n \omega_n c \sin \left( \omega_n x \right) = x(3-x)$.

In the first of these summations you can find $\displaystyle c_k$ by multiplying by $\displaystyle \sin( \omega_k x)$ on both sides and then integrating wrt x from 0 to 3. Doing this should give you that:

$\displaystyle c_n = (-1)^{n+1} \frac{3}{n \pi}$ .

I may have made arithmetic errors here so double check the result in case.

For the second summation it's exactly the same procedure to give you $\displaystyle a_n$. At the end substitute back into your general solution the results for $\displaystyle a_n$ and $\displaystyle c_n = a_n b_n$ and you're done!

An alternative, and easier way would have been to transform the problem to Fourier space and then solve.

Hope that helped!

3. that's awesome!! Thank you SO much