- #1
SqueeSpleen
- 141
- 5
Homework Statement
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 ] ##
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