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RJLiberator
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Homework Statement
Solve ∇^2u=0 in D subject to the boundary conditions
u(x,0) = u(0,y) = u(l,y) = 0,
u(x,l) = x(l-x)
where D = {(x,y): 0≤x≤l, 0≤y≤l}
Homework Equations
The Attempt at a Solution
So, I've looked at the notes and the book and have a gameplan to attack this problem. However, the back of the book answer does not match with mine.
First, the back of the book answer is.
I start by considering the the square with the necessary boundary conditions.
Length, l, is the same on all sides.
I first consider g_1=g_2 = 0 which is the boundary conditions for the left and right sides of the square. I say f_1 is the bottom side and f_2 is the top side of the square.
U_xx+U_yy 0
u(0,y) = u(l,y) = 0
u(x,0) = f_1(x)
u(x,l) = f_2(x)
So now,
u(x,y) = X(x)Y(y)
X''Y+Y''X = 0
Y''/Y = -X''/X = v^2
We have Sturm-Liouville Problem
X''+v^2X = 0, X(0) = X(l) = 0
Y''-v^2Y = 0
eigenvalue: (pi*n/l)^2 = v^2
sin(n*pi*x/l) is the eigenfunction
General solution: D_1*cosh(n*pi*y/l) + D_2*sinh(n*pi*y/l)
u(x,y) = sum from n = 1 to infinity of (sin(n*pi*x/l)(α_n*cosh(n*pi*y/l)+β_b*sinh(n*pi*y/l))
f_1(x) = sum from 1 to infinity of a_n*sin(n*pi*x/l)
f_2(x) = sum from 1 to infinity of b_n*sin(n*pi*x/l)
At y=0, we see u(x,0) = f_1(x)
At y=l, u(x,l) = f_2(x)
b_n = ancosh(n*pi) + β_b*sinh(n*pi)
We divide by sinh(pi*n)
u(x,y) = sum from 1 to n of (sin(n*pi*x/l)(a_n*cosh(n*pi*y/l)+[b_n*csch(n*pi)-a_n*coth(n*pi)]sin(n*pi*y/l))
But this is not the same answer as the answer from the back of the book (or is it with some manipulation)?
I know my writing is horrible, it's like due to my ill understanding of the problem. It's sloppy, all over the place, but hopefully someone can either confirm my answer or tell me where I went wrong.