= Solution
\b[Transform convention and derivatives.] Use the <Fourier transform> pair
$$
\boxed{\widetilde f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx,\qquad f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}\widetilde f(k)e^{ikx}\,dk.}
$$
For sufficiently decaying $f,f'$ with integrable second <derivative>, two applications of <integration by parts> give
$$
\begin{aligned}
\widetilde{f''}(k)&=[f'e^{-ikx}]_{-\infty}^{\infty}+ik\int_{-\infty}^{\infty}f'e^{-ikx}\,dx\\
&=ik[f e^{-ikx}]_{-\infty}^{\infty}+(ik)^2\widetilde f(k)=\boxed{-k^2\widetilde f(k)}.
\end{aligned}
$$
The identity also holds for <tempered distributions>, which is the setting needed for the <Dirac delta function> boundary data.
\b[Upper half-plane solution.] Take the <Fourier transform> in $x$ only, writing $\Phi(k,y)=\widetilde\phi(k,y)$. <Laplace's equation> becomes $\Phi_{yy}-k^2\Phi=0$. For $k\ne0$, discard the exponentially growing solution and use $\Phi(k,0)=1$, the <Fourier transform> of the <Dirac delta function>. This gives $\Phi(k,y)=e^{-|k|y}$. The value at $k=0$ can be set to its continuous limit one. Pointwise decay of $\phi$ as $y\to\infty$ does not require this individual transformed value to decay: the total integral in $x$ remains one while the profile spreads.
The inverse <Fourier transform> now yields, for $y>0$,
$$
\phi(x,y)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{-|k|y}e^{ikx}\,dk=\frac1\pi\int_0^\infty e^{-ky}\cos(kx)\,dk.
$$
Evaluate the last <integral> as the real part of $\int_0^\infty e^{-(y-ix)k}\,dk=(y-ix)^{-1}$:
$$
\boxed{\phi(x,y)=\frac{y}{\pi(x^2+y^2)},\qquad y>0.}
$$
This is the <Poisson kernel for the upper half-plane>. Its boundary condition is a <distributional convergence>, rather than an ordinary value at $(0,0)$: for a smooth compactly supported test <function> $g$,
$$
\int_{-\infty}^{\infty}\phi(x,y)g(x)\,dx=\frac1\pi\int_{-\infty}^{\infty}\frac{g(yt)}{1+t^2}\,dt\longrightarrow g(0).
$$
The <dominated convergence theorem> applies because $g$ is bounded and $(1+t^2)^{-1}$ is integrable. Also $0\leq\phi(x,y)\leq1/(\pi y)$, proving the required decay as $y\to\infty$.
\b[Direct verification.] Differentiating the rational expression gives
$$
\phi_{xx}=\frac{2y(3x^2-y^2)}{\pi(x^2+y^2)^3},\qquad\phi_{yy}=\frac{2y(y^2-3x^2)}{\pi(x^2+y^2)^3}.
$$
Their sum is zero throughout $y>0$, so the solution is a <harmonic function> there.
Back to article page
