Solving 1-d wave equation

• Mar 29th 2009, 01:57 AM
Ali Sura
Solving 1-d wave equation
Hey guys,
i have a small question. 1D-Wave equation is given as:

$\displaystyle \frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}$

i need to find $\displaystyle u_s(x,t)=X(x)sinwt$

From the equation form above , U_xx=$\displaystyle X''(x)sinwt$
U_tt=$\displaystyle -w^2X(x)sinwt$ are obtained as given as a solution.

My question is how $\displaystyle -w^2$ is obtained from $\displaystyle u_s(x,t)=X(x)sinwt$

thank you,
• Mar 29th 2009, 11:42 AM
HallsofIvy
What are your boundary conditions? That is what determines k.
• Mar 29th 2009, 04:27 PM
Ali Sura
Two Boundary conditions are:

1)
$\displaystyle u(0,t)=0$

2)
$\displaystyle u(1,t)=sinwt$

thanks,
• Mar 30th 2009, 05:01 AM
HallsofIvy
If $\displaystyle u_s(x,t)= X(x)sin(\omega t)$

Then $\displaystyle u_s(0,t)= X(0)sin(\omega t)= 0$ so X(0)= 0.

Also $\displaystyle u_s(1,t)= X(1)sin(\omega t)= sin(\omega t)$ so X(1)= 1.

Now, $\displaystyle u_xx= X_xx sin(\omega t)$ and $\displaystyle u_tt= -\omega^2 X sin(\omega t)$ so the partial differential equation becomes $\displaystyle X_xx sin(\omega t)= -\omega^2 X sin(\omega t)$ for all x and t and so we must have $\displaystyle X_xx= -\omega^2 X$ for all x.\

Can you solve the ordinary differential equation $\displaystyle X_xx+ \omega^2 X$ with the boundary conditions X(0)= 0, X(1)= 1?