# Math Help - wave equation

1. ## wave equation

The question is Solve the one dimensional wave equation

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

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

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

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

$
u_tt(x,t) = X''(x)T(t)$
and $u_tt(x,t) = X(x)T''(t)
$

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

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

$
X''(x) = w^2 X(x) = 0$

Using Auxilliary Equations:
$
m^2 - w^2 = 0$

So $m = +- w$
And $X(x) = C_1 e^{wx} + C_2 e^{-wx}$

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

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

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

$X''(x) + w^2X(x) = 0$

Auxilliary Equation: $m^2 + w^2 = 0, m = +-iw$\
So $C_5coswx +C_6sinwx = 0$
$C_5 = 0,$ So $
C_5cos3w + C_6sin3w = 0$

$sin3w = 0$
Therefore, $3w = \frac{n\Pi}{3}$ and $V = \frac{n^2\Pi^2}{9}$
Thus, $X(x) = B sin$ $\frac{xn\Pi}{3}$

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:

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

so integrating twice gives

$X = A x +B$

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

Case 3: you have correctly found that:

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

Consequently

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

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

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

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

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

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

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

where $c_n = a_n b_n$ and

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

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

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

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 $a_n$. At the end substitute back into your general solution the results for $a_n$ and $c_n = a_n b_n$ 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