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The Dirichlet problem I read is as follows:

If $f$ is an integrable function, find a function $u$ such that for $x \in \mathbb{R}, y>0$ \begin{align} u_{xx} + u_{yy} & =0 \\ \lim_{y \to 0^+} u(x,y) &= f(x) \text{ almost everywhere} \end{align}

Does the method listed in this Find the solution of the Dirichlet problem in the half-plane y>0. also work if $u$ and $u_x$ are not required to vanish as $|x| \to \infty$ and $u$ is bounded? Since if we discard these conditions we can not exclude the case $\lambda>0$.

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  • $\begingroup$ You are correct that some asymptotic or integrability conditions are necessary for this to be a nicely-defined problem. Yes, there are some non-uniqueness possibilities if none of the (various possible) constraints are imposed. Do you have a particular instance that you care about? $\endgroup$ Aug 17 '17 at 1:46
  • $\begingroup$ No, I'm asked to prove the general case, and if $f$ is continuous , then the solution is unique and the limit is uniform. But I'm confused since there are different versions of this problem, like if $f$ is bounded continuous, then the solution is unique, or if $u$ is continuous the solution is unique. Im wondering which version is correct. $\endgroup$ Aug 17 '17 at 2:04
  • $\begingroup$ I made some slight edits to your post, let me know if you are unhappy with them. $\endgroup$
    – mattos
    Aug 17 '17 at 2:57
  • $\begingroup$ Let $G_y(x) = \frac{y}{\pi (x^2+y^2)} = \frac{d}{dx}\frac{\log(x-iy)-\log(x+iy)}{2i\pi}$. Then $G_y(x)$ is harmonic for $y >0$ as well as $G_y \ast f(x)$, and integrating by parts $$\lim_{y \to 0^+} G_y \ast f(x) = \lim_{y \to 0^+}\int_{-\infty}^\infty G_y(t) f(x-t)dt = \int_{-\infty}^\infty 1_{t < 0} f'(x-t)dt = f(0)$$ at least if $f' \in L^1$ $\endgroup$
    – reuns
    Aug 17 '17 at 8:41
  • $\begingroup$ If you require $u$ to be uniformly bounded in the upper half-plane, then $f$ will have to be essentially bounded. So that requirement on $u$ is too strong in general for a general $f\in L^1(\mathbb{R})$. $\endgroup$ Aug 19 '17 at 17:11
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In short, consider the Green function $$G_y(x) = \frac{y}{\pi (x^2+y^2)} = \frac{d}{dx}\frac{\log(x-iy)-\log(x+iy)}{2i\pi}$$ Then $G$ is harmonic for $y >0$ as well as $G_. \ast f$.

Integrating by parts and using that $$\lim_{y \to 0^+}\log(x\pm iy)- \log|x|=\pm i\pi 1_{x < 0}$$ the convergence being locally uniform away from $x=0$, you'll obtain $$\lim_{y \to 0^+} G_y \ast f(x) = \lim_{y \to 0^+}\int_{-\infty}^\infty G_y(t) f(x-t)dt = \int_{-\infty}^\infty 1_{t < 0} f'(x-t)dt = f(x)$$ at least if $f,f' \in L^1$

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The final Poisson form of the solution will give you a non-unique solution: $$ u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-x')^2+y^2}f(x')dx'. $$ The non-uniqueness is easily verified because $v(x,y)=y$ is a solution of the Laplace equation that vanishes identically on the real line. So $u(x,y)+Cy$ is another solution, for any constant $C$.

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  • $\begingroup$ I know this result, but could u tell me how to derive it? Does the method mentioned in the pages I post still work for the case that $u, u_x$ are not required to vanish at infinity and $u$ is not necessary bounded for $|y|\to\infty$? $\endgroup$ Aug 17 '17 at 7:49
  • $\begingroup$ @RussellHiwayor I posted a derivation to the problem you linked to. The final convolution form can be seen to be the real part of a holomorphic function if, for example, $f$ is absolutely integrable. And it can be verified that the boundary condition is met almost everywhere for such $f$. $\endgroup$ Aug 17 '17 at 7:57
  • $\begingroup$ Are you sure this is $\int_{-\pi}^\pi$ and not $\int_{-\infty}^\infty$ ? Because it is a convolution equation, the solution is of the form $u(x,y) = G_y \ast f(x)$ where $G_y$ is the Green function, ie. the solution for $f = \delta$ $\endgroup$
    – reuns
    Aug 17 '17 at 8:15
  • $\begingroup$ @reuns : Yes, that was a typo. Gotten started with 1/pi and kept using pi. $\endgroup$ Aug 17 '17 at 14:51
  • $\begingroup$ ok but I don't agree with this answer using the Fourier transform, see mine $\endgroup$
    – reuns
    Aug 18 '17 at 19:13

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