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Assume $f : \mathbb{R} → \mathbb{R}$ is a $C^∞_c\mathbb{(R)}$ function. Use Fourier transform (in $x$) to obtain the Poisson integral formula solution to Laplace’s equation in the upper half-plane $u_{xx} + u_{yy} = 0$ for $− ∞ < x < ∞, y > 0, u(x, 0) = f(x), |u| $ bounded.

So If I write Fourier trnsform for $u$ and $f$,

$\displaystyle \hat{u}(k,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,y)e^{-ikx}dx$

$\displaystyle \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$

These imply that $\hat{u}(k,0)=\hat{f}(k)$.

Now I don't understand where to go. Could somebody please help me to proceed?

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    Assuming everything converges, what is the Fourier transform of $u_{xx} + u_{yy} = 0$ ? – 2017-02-13
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    With your notation, taking the FT in $x$ we get $-k^2 \hat{u}(k,y) + \partial_y^2 \hat{u}(k,y)= 0$ – 2017-02-13
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    So can I say now that $k^2u=u_{yy}$? – 2017-02-13
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    Then you can solve this ODE for $y$ and check it works. At the end, you can write $u(x,y)$ as a convolution of $f(x)$ with $h_y(x)$ (The Green function of the Laplace equation) – 2017-02-13
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    The solution of ODE is $A(k)e^{kt}+B(t)e^{-kt}$. But I don't understand how to involve Green's function. – 2017-02-13

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