# biharmonic equation

• May 16th 2011, 03:19 AM
ulysses123
biharmonic equation
verify that the following is a solution to the biharmonic equation

F(x,y)=(Acosbx+Bsinbc+Cxcosbx+Dxsinbx)(Ecoshby+Fsi nhby)

i have tried taking the derivates and have gotten up to the laplacian but this is a very long and tedious way,and the result up to second order is really long, like a page long,there must be an easier way. Can someone suggest an alternative method to prove this
• May 16th 2011, 04:19 AM
Jester
The biharmonic eqiation is

$u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$

Your solution is of the form

$u = A(x)B(y)$

Note that when you substitute into the biharmonic you get

$A^{(4)}(x)B (y)+ 2 A''(x)B''(y) + A(x) B^{(4)}(y) = 0$

With your $B(y)$ you have $B^{(4)}(y) = b^2B''(y) = b^4B(y)$ so what you really want to show is

$A^{(4)}(x)+ 2b^2 A''(x) + b^4A(x) = 0$.

Next, notice that this is

$\left(A'' + b^2A\right)'' + b^2\left(A'' + b^2A \right) = 0$

so compute $C = A'' + b^2A$ first and then show this satisfies $C'' + b^2C = 0$. It might make it a little more manageable.
• May 16th 2011, 03:18 PM
ulysses123
thanks thats way easier