Fourier transform to find an harmonic function (Strauss)

Question

I am trying to solve one of the problems of section 12.4 of the book "Partial Differential Equations" by Strauss. The problem says:

Use the Fourier transfor in the $x$ variable to find the harmonic function in the half plane ($y>0$) that satisfies the Neumann condition $u_y=h(x)$ on $y=0$.

So I did the fourier transform of the laplacian, but when I solve the new ODE I am missing one boundary condition. Is there any asymptotic condition that I should consider?

Thanks in advance!

Answer 1

Warning: All the formulas in this post are wrong. They all need $\pi$'s inserted in various places. (Partly I'm lazy. Partly we want to leave something for you to do. Mainly the problem is that every book uses a different definition of the Fourier transform, with the $\pi$'s in different places - no way I could possibly give FT formulas that will be valid with the FT in your book...)

Anyway, here's a basic fact about Poisson integrals versus the Fourier transform: If $f$ is a nice function on $\Bbb R$ (Lebesgue integrable is nice enough) and $$u(x,y)=\int e^{-y|\xi|}e^{ix\xi}\hat f(\xi)\,d\xi$$then $u$ is harmonic in the upper half plane and tends to $f$ (in various sense, depending on how nice $f$ is) on the boundary. Now it appears that $$u_y(x,y)=-\int e^{-y|\xi|}e^{ix\xi}|\xi|\hat f(\xi)\,d\xi.$$ So it would seem that you want $$\hat h(\xi)=-|\xi|\hat f(\xi).$$ So $f$ should be the inverse transform of $-\hat h(\xi)/|\xi|$.

Answer 2

There is some ambiguity in this problem, just as you pointed out in your question. If you start with the Fourier transform of your equation in $x$, which is given by $$ \hat{u}(s,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-isx}u(x,y)dx, $$ then the equation in $y$ is informally \begin{align} \frac{\partial^{2}}{\partial y^{2}}\hat{u}(s,y) & =-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-isx}\frac{\partial^{2}u}{\partial x^{2}}dx \\ & = -(is)^{2}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-isx}u(x,y)dx. \end{align} This gives rise to the equation $$ \frac{\partial^{2}}{\partial y^{2}}\hat{u}(s,y)-s^{2}\hat{u}(s,y)=0. $$ You were correct that there are two solutions $$ \hat{u}(s,y) = A(s)e^{y|s|}+B(s)e^{-y|s|}. $$ Therefore, $$ \hat{u}(s,0) = A(s)+B(s),\\ \hat{u}_{y}(s,0) = |s|A(s)-|s|B(s)=\hat{h}(s). $$ Because of the exponent, it is tempting to rule out a non-zero $A$, but you can't do that in general. For example, the following is a solution of Laplace's equation: $$ u(x,y) = e^{-(x^{2}-y^{2})/2}\cos(xy) = \Re e^{-(x+iy)^{2}/2} $$ Furthermore, $u_{y}(x,0)=0$. And this is okay because $A(s)$ in this case is the Fourier transform of a Gaussian, which tempers $e^{y|s|}$ because $A(s) = Ce^{-s^{2}}$, which leaves $e^{y|s|-s^{2}}$ absolutely integrable and square integrable in the variable $s$.

If you require $\int_{-\infty}^{\infty}|u(x,y)|^{2}dx \le M$ for some $M$ and all $y$, then you can rule out such problems. Regardless, you are correct that there is a missing piece that cannot be ignored without some further asymptotic or condition in $y$.