Thread: Pde

1. Pde

The displacement for the transverse vibrations of a beam satisfies the differential equation
utt + (b^2)uxxxx = 0
Find u(x,t) if u(0,t) = 0, u(L,t) = 0, uxx(0,t) = 0, uxx(L,t) = 0, u(x,0) = f(x), ut(x,0) = g(x)

I've never encountered a differential equation like this before. How do you solve it?

2. Originally Posted by JimmyT
The displacement for the transverse vibrations of a beam satisfies the differential equation
utt + (b^2)uxxxx = 0
Find u(x,t) if u(0,t) = 0, u(L,t) = 0, uxx(0,t) = 0, uxx(L,t) = 0, u(x,0) = f(x), ut(x,0) = g(x)

I've never encountered a differential equation like this before. How do you solve it?
Same way as with $u_t = u_{xx}$ and $u_{xx} + u_{yy} = 0$ - separation of variables.

3. So I get d^2T/dt^2 = (b^2)T*lambda and d^4X/dx^4 = X*lambda

How do I solve it from here?

4. Originally Posted by JimmyT
So I get d^2T/dt^2 = (b^2)T*lambda and d^4X/dx^4 = X*lambda

How do I solve it from here?
Pick $X^{(4)} = \lambda X$ first. Because of your boundary conditions $\lambda = \omega^4$ so $X^{(4)} - \omega^4 X = 0$. The characteristic equation $m^4 - \omega^4 = 0\;\Rightarrow\; m = \pm i, \pm 1$ so the solutions are

$X = c_1 \sin \omega x + c_2 \cos \omega x + c_3 \sinh \omega x + c_4 \cosh \omega x$.

Your bc's will determine the constants and the $\omega$. Then go the the T equation.