Dirichlet problem for the laplacian in the strip

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SUMMARY

The discussion focuses on solving the Dirichlet problem for the Laplacian in the strip defined as \( S = \{a < \operatorname{Re}(z) < b\} \). The author employs a conformal mapping from the strip to the unit disc and attempts to find harmonic functions that vanish on the boundary, ultimately concluding that the only solution is the zero function. The analysis involves separating variables and solving ordinary differential equations (ODEs) for \( \alpha(x) \) and \( \beta(y) \), leading to the realization that boundary conditions result in trivial solutions under certain constraints.

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pantboio
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I'm looking for all functions $u$ harmonic in $S$ and continuous in $\overline S$ such that
$$u(a,y)=u(b,y)=0,\forall y$$
and
$$\lim_{|y|\rightarrow +\infty} u(x,y)=0$$
where $S$ is the strip $\{a<\operatorname{Re}(z)<b\}$

My strategy is the following. I know that if $g$ is continuous on $\partial D$, with $D$ the unit disc,then
\begin{equation*}
u(z)= \left\{
\begin{array}{}
\frac{1}{2\pi}\int_0^{2\pi}g(e^{i\theta}) \frac{1-|z|^2}{|e^{i\theta}-z|^2}d\theta& \text{if } z\in D,\\
g(z)& \text{if } z\in\partial D\\
\end{array} \right.
\end{equation*}
is harmonic in $D$, continuous in $\overline D$ and $u=g$ on $\partial D$.

So i find a conformal map $\phi$ from the strip to the unit disc and i look for harmonic functions in the disc that vanish on the boundary. But in this case $g$ is the function identically equal to 0, hence the only harmonic functions i find is the zero one. In all this i have the strong sensation to have missed something fundamental from the theory, but i don't know what. Can someone give me a suggestion?
 
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pantboio said:
I'm looking for all functions $u$ harmonic in $S$ and continuous in $\overline S$ such that
$$u(a,y)=u(b,y)=0,\forall y$$
and
$$\lim_{|y|\rightarrow +\infty} u(x,y)=0$$
where $S$ is the strip $\{a<\operatorname{Re}(z)<b\}$

My strategy is the following. I know that if $g$ is continuous on $\partial D$, with $D$ the unit disc,then
\begin{equation*}
u(z)= \left\{
\begin{array}{}
\frac{1}{2\pi}\int_0^{2\pi}g(e^{i\theta}) \frac{1-|z|^2}{|e^{i\theta}-z|^2}d\theta& \text{if } z\in D,\\
g(z)& \text{if } z\in\partial D\\
\end{array} \right.
\end{equation*}
is harmonic in $D$, continuous in $\overline D$ and $u=g$ on $\partial D$.

So i find a conformal map $\phi$ from the strip to the unit disc and i look for harmonic functions in the disc that vanish on the boundary. But in this case $g$ is the function identically equal to 0, hence the only harmonic functions i find is the zero one. In all this i have the strong sensation to have missed something fundamental from the theory, but i don't know what. Can someone give me a suggestion?

As preliminary inspection we can follow the classical approach assuming that is $\displaystyle u(x,y)= \alpha(x)\ \beta(y)$ and that leads to the pair of ODE...$\displaystyle \alpha^{\ ''} + \lambda \alpha =0$ (1)

$\displaystyle \beta^{\ ''} - \lambda \beta =0$ (2)

... where $\lambda$ is a constant that will be better defined later.

For semplicity sake we suppose that $0 \le x \le 1$. Starting from (1) its [not identically equal to zero...] solution with the contour conditions $\displaystyle \alpha(0)= \alpha(1)=0$ is...

$\displaystyle \alpha= k\ \sin \sqrt{\lambda}\ x = k\ \sin \pi\ n\ x $ (3)

... where k is a constant and $\sqrt{\lambda}= \pi\ n$.

Now we observe (2), the solution of which is...

$\displaystyle \beta = c_1\ e^{\sqrt {\lambda}\ y} + c_{2}\ e^{- \sqrt{\lambda}\ y}$ (4)

... and soon a little problem appears. If the condition is $\displaystyle \lim_{|y| \rightarrow \infty} u(x,y)=0$ and S is defined in $0 \le x \le 1$ and $- \infty < y < + \infty$, then in (4) is $c_{1}=c_{2}=0$ and the Diriclet problem has the only solution is $u(x,y)=0$. If S, for example, is defined in $0 \le x \le 1$ and $0 \le y < + \infty$, then the solution is...

$\displaystyle u(x,y)= \sum_{n=1}^{\infty} k_{n}\ \sin (\pi\ n\ x)\ e^{- \pi\ n\ y}$ (5)

Kind regards

$\chi$ $\sigma$
 
Last edited:

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