Find the solution of the Dirichlet problem in the half-plane y>0.

Find the solution of the Dirichlet problem in the half-plane $y>0$.
$${u_y}_y +{u_x}_x=0, -\infty<x<\infty,y>0$$
$$u(x,0)=f(x),-\infty<x<\infty$$
$u$ and $u_x$ vanish as
$$ \lvert x\rvert\rightarrow\infty$$
and $u$ is bounded as $$y\rightarrow\infty$$
I’ve tried hard for this,but my answer is not matching.I’ve gone through by applying infinite fourier transformation as my text book suggests,but I’m not even getting the intermediate equations correct.

Please help in getting over it.

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The method that would be used by Fourier is to separate variables, discard all of the resulting separated functions that are unbounded in the upper half plane, and then form integral linear combinations of the others.

Start by separating variables with $u(x,y)=X(x)Y(y)$; to do this, plug into the equation and divide by $XY$:
$$
\frac{X”}{X}+\frac{Y”}{Y} = 0 \\
\frac{X”}{X} = \lambda = -\frac{Y”}{Y}.
$$
The values of $\lambda$ will always be real in well-posed problems of this type. If $\lambda > 0$ then $X(x) = Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x}$, and these are either unbounded at $x=+\infty$ or at $x=-\infty$. So you, you can replace $\lambda$ with $-\mu^2$ to simplify matters. Then the separated solutions that are bounded have the form
$$
u_{\lambda}(x,y) = C(\mu)e^{i\mu x}e^{-|\mu|y},\;\;\; -\infty < \mu < \infty.
$$
There are other ways to write these by looking at $\mu > 0$ and $\mu < 0$ separately, but this form is convenient for use with the Fourier transform. You end up with a proposed solution
$$
u(x,y) = \int_{-\infty}^{\infty}C(\mu)e^{i\mu x}e^{-|\mu|y}d\mu.
$$
In order to match the condition at $y=0$,
$$
f(x) = u(x,0) = \int_{-\infty}^{\infty}C(\mu)e^{i\mu x}d\mu \\
\implies C(\mu) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\mu x}dx
= \frac{1}{\sqrt{2\pi}}\hat{f}(\mu).
$$
Therefore,
$$
u(x,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(\mu)e^{-|\mu|y}e^{i\mu x}d\mu
$$
That’s basically the solution. However, you can reduce to an integral involving $f$ instead of one involving its Fourier transform.

You can get rid of the transform for $f$ and rewrite as a convolution integral using the convolution theorem for Fourier transforms. This is done by writing $e^{-|\mu|y}=\hat{g_y}(\mu)$, which is achieved by inverse Fourier transforming $e^{-|\mu|y}$ with respect to $\mu$:
\begin{align}
g_{y}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|\mu|y}e^{i\mu x}d\mu \\
& = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\mu(ix+y)}d\mu+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{\mu(ix-y)}d\mu \\
& = \frac{1}{\sqrt{2\pi}}\left(\frac{1}{y+ix}+\frac{1}{ix-y}\right)\\
& = \frac{2y}{\sqrt{2\pi}(x^2+y^2)}
\end{align}
Then you apply the convolution theorem to obtain
$$
u(x,y)= \frac{1}{\pi}\int_{-\infty}^{\infty}f(x’)\frac{y}{(x’-x)^2+y^2}dx’.
$$
This is the Poisson integral for the solution.