# wave equation

• May 28th 2009, 03:37 AM
gconfused
wave equation
The question is Solve the one dimensional wave equation

$\frac{1}{c^2}$ http://www.mathhelpforum.com/math-he...363cc0f4-1.gif

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)
$

$\frac{1}{c^2}$ http://www.mathhelpforum.com/math-he...363cc0f4-1.gif

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
• May 28th 2009, 08:54 AM
the_doc
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!
• May 28th 2009, 08:45 PM
gconfused
that's awesome!! Thank you SO much :)