how do you solve a laplace equation given that u(x,0) =0, u(x,1) = sin ^3 (pi x)
u(x,y)=0 and u(1,y) =0 for 0<x<1, 0<y<1?
i got the final answer of sum {sin ^3(pi x) sinh(n pi y)} over (sinh(pi x))
but this seems to look wrong.
I think you have a typo. Did you mean
$\displaystyle u(0,y)=0$u(x,y)=0
Your answer should not have trig functions rasied to any power as they are not part of the basis!
Hint 1:
$\displaystyle \sin^3(\pi x)=\frac{3}{4}\sin(\pi x)-\frac{1}{4}\sin(3 \pi x)$
Using the above the only terms that will survive when you expand the x's with will the two given above.
Please post what you have done.
Untitled.pdf
this is my working with the help that you gave. im still confused with how to find the solution
So you have
$\displaystyle u_n(x,y)=c_n\sin(n \pi x)\sinh(n \pi y)$
Now using the boundary condition we have
$\displaystyle u_n(x,1)=\frac{3}{4}\sin(x)-\frac{1}{4}\sin(3 \pi x)=c_n \sin(n \pi x)\sinh(\pi n)$
Now using the inner product we can solve for the constants. Multiply both sides by
$\displaystyle \sin(n \pi x)$
and integrate you will need a 2 in front of your integral to make the set orthonormal.
$\displaystyle 2\int_{0}^{1}\sin(n \pi x)\left( \frac{3}{4}\sin(\pi x)-\frac{1}{4}\sin(3 \pi x)\right)dx=c_n\sinh(\pi n)$
The left hand side is only non zero when n=1 or n=3
This gives
$\displaystyle c_1=\frac{3}{4\sinh(\pi)} \quad c_3=-\frac{1}{4\sinh(3\pi)}$
all of the other c's are zero so the sum only contains two terms and gives
$\displaystyle u(x,y)=\frac{3}{4\sinh(\pi)}\sin(\pi x)\sinh(y)-\frac{1}{4\sinh(3\pi)}\sin(3 \pi x)\sinh(3y)$