# Laplacian in [0,L] x [0, ∞]

1. Jul 9, 2017

### SqueeSpleen

1. The problem statement, all variables and given/known data
I have to solve the following problem
$$\left\{ \begin{array}{ll} \dfrac{ \partial^{2} u }{ \partial x^{2} } + \dfrac{ \partial^{2} u }{ \partial y^{2} } =0 & \qquad \forall x \in (0, L), y > 0 \\ & \\ \dfrac{ \partial u }{ \partial x } (0,y) =0, & \qquad \forall y > 0 \\ & \\ u(L,y)=0, & \qquad \forall y > 0, \\ & \\ u(x,0) = V, & \qquad \forall x \in (0,L) \\ & \\ u(x,y) \text{ is bounded as } & y \rightarrow + \infty \end{array} \right.$$
Where $V$ is a real constant.

My attemp:

I proposed a solution of the kind
$$u(x,y) = \sum_{n=1}^{\infty} K_{n} X_{n} (x) Y_{n} (y)$$
Arrived to
$$\dfrac{X''(x)}{X(x)} = - \dfrac{Y''(y)}{Y(y)} = - \lambda$$
So now I have
$$X''(x) + \lambda X(x) = 0$$
With $X'(0)=0$ and $X(L)=0$.
And
$$Y''(x) - \lambda Y(x) = 0$$
Negative eigenvalues and zero one didn't work for the problem in $x$, so I proposed a trigonometric series of the kind
$$A \cos ( \sqrt{\lambda} x ) + B \sin ( \sqrt{\lambda} x )$$
The thing is, my boundary conditions forced a cosine series, and the condition on $y$ forced this series to be equal to a constant. But a cosine series without the constant term (as zero eigenvalue was discarded) can only be constant if it's zero. So I'm with empty hands.

I thought that if I were to swap boundary conditions on $X(x)$ I should be able to solve it with a sine series. What's the way to do this? Well, to invert the order of the sine.
Something like
$$\sin ( \lambda (L-x) )$$

So I proposed a sighly different kind of series as solution, and arrived to:

$$u(x,y) = 4 V \sum_{n=1}^{\infty} \dfrac{ sen(\sqrt{ \lambda_{n} } (L-x) )}{2n+1} e^{ - \sqrt{ \lambda_{n} } y }$$
Where
$$\lambda_{n} = \dfrac{ \pi^{2} }{L^{2}} \left( n+\dfrac{1}{2} \right)^{2}$$

Where I used to bounding condition to discard exponentials with positive power.

Is this right? I have never seen done in a book and I don't see why it might fail, after all if I didn't make any major mistakes I'm sure that I should be able to solve the problem where $\dfrac{ \partial u }{ \partial x } (L,y) =0$ and $u(0,y)=0$ are the conditions on $x$.

Sorry if this is silly, I'm self studying this. Oh, also in the title I should have written $[0, \infty )$ instead of $[0, \infty ]$

Last edited: Jul 9, 2017
2. Jul 9, 2017

### Orodruin

Staff Emeritus
There is no constant in the series. Your boundary condition at $x = L$ prohibits a non-trivial solution with $\lambda = 0$. You need to expand the constant in the functions that are part of the series.

3. Jul 9, 2017

### SqueeSpleen

First of all, sorry for the poor redaction. I have re-read what I wrote and when I meant "as the zero eigenvalue is discarded" I wrote "as the constant eigenvalue is discarded" which makes no sense at all.

Yes, that's why I said that I have solution only for the constant equal to zero, I discarded the zero eigenvalue, and the cosine series cannot represent the constant value that's not zero, so I think that using that series representation I will only obtain the zero function, that's obviously not a solution if $V \neq 0$. So, I tried to find a new series representation.

I know how to expand the constant in a sine series, but if I'm with cosines I think it's impossible, as they're always orthogonal to constants even in $[0, \pi ]$.
My reasoning: I know that the Fourier series with cosines, sines and the constant function can represent any function in $L^{2} [- \pi, \pi ]$. So, as far as I know, if you want to represent a function in $[0, \pi]$ you can always use the sine series to find the fourier series of the odd extension of that function on $[-\pi, \pi]$ and use it in $[0, \pi ]$. Same reasoning but with even extension and you will use cosine series. The thing is, the cosine series needs the constant to represent constant terms, as the even extension of the constant is still a constant, and they're orthogononal to sines and cosines in $[ - \pi , \pi ]$.
Most of what I know about trigonometric series is in $[ - \pi , \pi ]$ so I always try to relate things to that interval.

Last edited: Jul 9, 2017
4. Jul 9, 2017

### Orodruin

Staff Emeritus
Yes it can - you can approximate it arbitrarily well with the actual eigenfunctions. The function will have contributions from all eigenfunctions.

There is absolutely no difference between the sine series you proposed and the cosine series. The only difference is in the coordinate.

No, this is true only if your SL operator has Neumann boundary conditions on both boundaries, which makes the constant function an eigenfunction. The constant function is definitely not orthogonal to this set of eigenvectors - just check the inner product!

5. Jul 9, 2017

### SqueeSpleen

I may be understanding something bad, but:
$$\int_{0}^{\pi} \cos (nx) dx = \dfrac{1}{n} \sin (nx) \bigg|_{0}^{\pi} = 0, \qquad \forall n \in \mathbb{N}$$
I checked with calculator to see if I was making a mistake integrating and I still have this equal to zero, at least for n=1,2,3 and 4, but I don't think it's going to change.
What inner product should I use to compute the coefficients?

6. Jul 9, 2017

### SqueeSpleen

Oh I think that I got it. I should use a series of cosines like
$\cos ( (n+ \dfrac{1}{2} ) \dfrac{ \pi }{L} x )$
In which case, it's the same as the series I used. I'm not sure how to do a reasoning to arrive this cosine series, but perhaps doubling the interval so I'm changing $\dfrac{ \partial u }{ \partial x } (0,y)$ for $u(-L,y)$ might work. And the expression will be simplier.

Well, what I tried to say when I said that I discarded that cosine series is that the series with $\lambda = \left( \dfrac{n \pi}{L} \right)^{2}$ wasn't going to work, so that approach was flawed.

7. Jul 9, 2017

### Orodruin

Staff Emeritus
How would you do it if you had any other boundary conditions? Do it the same way! The boundary conditions will tell you what the appropriate eigenfunctions are.

8. Jul 9, 2017

### SqueeSpleen

Sorry, I probably should have waited after reading Sturm - Liouville theory before approaching these exercises. I only had done exercises without reaching that chapter, but today I will read about this to be able to solve the problems in a more systematic way instead of having to try in this way.