# PDE: Laplace (?) Problem? Sturm Liouville?

1. Mar 10, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
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}
2. Relevant equations

3. 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.

2. Mar 10, 2016

### LCKurtz

You also have the homogeneous boundary condition $u(x,0) = X(x)Y(0) = 0$ so you have $Y(0)=0$
$Y(0)=0$ implies $D_1=0$.

You should have $u(x,y) = \sum_{n-1}^\infty c_n\sin(\frac{n\pi x}{L})\sinh(\frac {n\pi} L y)$ and now let's see the calculation of the Fourier coefficients to make $u(x,L) = x(L-x)$. Also, it really isn't that much more work to type Latex.

3. Mar 10, 2016

### RJLiberator

Er, fourier coefficients
b_n = a_n*cosh(n*pi)+β_n*sinh(pi*n)

Is that what you mean?

4. Mar 10, 2016

### LCKurtz

No. Just read the last two lines of post #2 and finish it from there.

5. Mar 10, 2016

### RJLiberator

I understand what you are saying --- by calculating the fourier coefficients we can find the form that I am looking for.

But I dont understand two things.
1: You want to calculate the fourier coefficients (implying there is more than 1) I only see 1 being Cn.
2: Where does x(L-x) come in to play?

6. Mar 10, 2016

### LCKurtz

$c_n,~n=1..\infty$ is infinitely many constants to use in your sum. You need the formula for $c_n$ to finish the problem.

Look at the first 3 lines in your original post. You have a square region with 4 boundary conditions. 3 of them are zero, which you have satisfied. You have to make the last one work. Also, I have been using $L$ instead of $l$ just from habit.

7. Mar 12, 2016

### RJLiberator

@LCKurtz I had some time to look at the material and study.

I finally understand a lot of the steps that I've taking to get to this point, and what you are saying. I see where I made somewhat of a mistake in overlooking Y(0) = 0 point.

I see now we need to calculate that Cn coefficients.

So here is where I am. Calculating the Cn coefficients.
The general form to find Cn = 1/(2*pi) integral from -pi to pi of f(x)*e^(-inx)dx.
I think we use f(x) = x(L-x)
So, C_0 is thus, pi^2/3

But that doesn't seem to help me too much to get it into form. At least from just observation.

I have:
C_0 = pi^2/3
C_1 = (2-i*L)
C_2 = i(L+i)/2
C_3 = (2-3i*L)

I don't see a whole lot going on here to help, no?

8. Mar 12, 2016

### LCKurtz

You have the series $u(x,y) = \sum_{n-1}^\infty c_n\sin(\frac{n\pi x}{L})\sinh(\frac {n\pi} L y)$ as your proposed solution, but you need to know what the $c_n$'s are. You need to have $u(x,L)$ which is $u(x,L) = \sum_{n-1}^\infty c_n\sin(\frac{n\pi x}{L})\sinh(n\pi)= x(L-x)$. If you call $b_n=c_n\sinh(n\pi)$ this looks like $\sum_{n-1}^\infty b_n\sin(\frac{n\pi x}{L})= x(L-x)$. This needs to work on $(0,L)$, not $(0,\pi)$. You should recognize this as a half range sine expansion. Look in your book for the formulas for half range expansions. You don't want to use the exponential version of the FS for half range expansions. You should be able to figure out $b_n$, hence $c_n$, hence $u(x,y)$. Once you get that far, come back and I will help you get the final form for the $u$ series if you don't see it.

9. Mar 12, 2016

### RJLiberator

Based on this video:

near the end of the video they talk about the half range sine series.

Seems like b_n = (2/L) integral from 0 to L of f(x)*sin(n*pi*x/L)*dx

So if that is b_n, then we set it in the equation b_n = c_n*sinh(n*pi)

So c_n = b_n/sinh(n*pi)

Am I on the right track here?

10. Mar 12, 2016

### RJLiberator

I think I am on the right track. I got this when I plug my work in:

This looks very much like the form they have in the back of the book now.

The only difference is WHY did the back of the book change from a simple n to (2n-1) I understand that this is for n to be all odd numbers, but that doesn't make entire sense to me? Is it because all even n values produce a value of 0?

And then why do I get a 2 in front instead of an 8.

11. Mar 13, 2016

### LCKurtz

If you will simplify the expression $(\pi n \sin(\pi n)+ 2\cos(n\pi) -2))$ you might answer your own questions.

12. Mar 13, 2016

### RJLiberator

Ah, it is -4 for n=odd and 0 for n = even. I can see how this helps out.

Thank you for providing me help here. I am pretty damn close to really understanding the problem. I will put this at rest now.