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.

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- May 10th 2011, 09:27 AMalexandrabel90laplace
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, 09:35 AMTheEmptySet
I think you have a typo. Did you mean

Quote:

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. - May 10th 2011, 09:46 AMalexandrabel90
ya i meant u(o,y) = 0.

- May 10th 2011, 09:59 AMalexandrabel90
Attachment 21533

this is my working with the help that you gave. im still confused with how to find the solution :( - May 10th 2011, 10:35 AMTheEmptySet
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

$\displaystyle u(x,y)=\frac{3}{4\sinh(\pi)}\sin(\pi x)\sinh(y)-\frac{1}{4\sinh(3\pi)}\sin(3 \pi x)\sinh(3y)$ - May 10th 2011, 11:14 AMalexandrabel90
i think c1 and c2 should be both multiplied by 2 right?

- May 10th 2011, 01:34 PMTheEmptySet