# laplace

• May 10th 2011, 10:27 AM
alexandrabel90
laplace
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.
• May 10th 2011, 10:35 AM
TheEmptySet
Quote:

Originally Posted by alexandrabel90
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
Quote:

u(x,y)=0
$u(0,y)=0$

Your answer should not have trig functions rasied to any power as they are not part of the basis!

Hint 1:
$\sin^3(\pi x)=\frac{3}{4}\sin(\pi x)-\frac{1}{4}\sin(3 \pi x)$

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.
• May 10th 2011, 10:46 AM
alexandrabel90
ya i meant u(o,y) = 0.
• May 10th 2011, 10:59 AM
alexandrabel90
Attachment 21533

this is my working with the help that you gave. im still confused with how to find the solution :(
• May 10th 2011, 11:35 AM
TheEmptySet
So you have

$u_n(x,y)=c_n\sin(n \pi x)\sinh(n \pi y)$

Now using the boundary condition we have

$u_n(x,1)=\frac{3}{4}\sin(x)-\frac{1}{4}\sin(3 \pi x)=c_n \sin(n \pi x)\sinh(\pi n)$

Now using the inner product we can solve for the constants. Multiply both sides by
$\sin(n \pi x)$
and integrate you will need a 2 in front of your integral to make the set orthonormal.

$2\int_{0}^{1}\sin(n \pi x)\left( \frac{3}{4}\sin(\pi x)-\frac{1}{4}\sin(3 \pi x)\right)dx=c_n\sinh(\pi n)$

The left hand side is only non zero when n=1 or n=3

This gives

$c_1=\frac{3}{4\sinh(\pi)} \quad c_3=-\frac{1}{4\sinh(3\pi)}$

all of the other c's are zero so the sum only contains two terms and gives

$u(x,y)=\frac{3}{4\sinh(\pi)}\sin(\pi x)\sinh(y)-\frac{1}{4\sinh(3\pi)}\sin(3 \pi x)\sinh(3y)$
• May 10th 2011, 12:14 PM
alexandrabel90
i think c1 and c2 should be both multiplied by 2 right?
• May 10th 2011, 02:34 PM
TheEmptySet
Quote:

Originally Posted by alexandrabel90
i think c1 and c2 should be both multiplied by 2 right?

No becuse

$2\int_{0}^{1}\sin^2(n \pi x)dx=1$