MHB Dirichlet problem for the laplacian in the strip

  • Thread starter Thread starter pantboio
  • Start date Start date
  • Tags Tags
    Laplacian
Click For Summary
The discussion focuses on finding harmonic functions in the strip S defined by the conditions u(a,y)=u(b,y)=0 and the limit condition as |y| approaches infinity. The initial approach involves using a conformal map to relate the strip to the unit disc and applying the Poisson integral formula, which leads to the conclusion that the only harmonic function satisfying the boundary conditions is the zero function. Further exploration using separation of variables reveals that while solutions exist for certain bounded conditions, they ultimately lead back to the trivial solution under the specified limit condition. The participants express concern about potentially overlooking fundamental aspects of the theory that could yield non-trivial solutions.
pantboio
Messages
43
Reaction score
0
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?
 
Physics news on Phys.org
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:
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
951
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K