Dirichlet problem for the half plane

Question

In solving this problem, for a separated solution, set $u(x,y)=X(x)Y(y).$ Then $$\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}=-\lambda,$$ yields two ODEs: (with $\lambda=\beta^2$ for $\beta >0$) $$X''(x)+\lambda X(x)=0 \implies X(x)=A \cos \beta x + B \sin \beta x,$$ $$Y''(y)-\lambda Y(y)=0 \implies Y(y)=C \cosh \beta y + D \sinh \beta y.$$ Combining these two and form the sum: $$u(x,y)=\sum_{n=1}^{\infty} ( A_n \cos \beta_n x + B_n \sin \beta_n x ) \cdot ( C_n \cosh \beta_n y + D_n \sinh \beta_n y ).$$

Since the problem is defined for $y >0,$ we need $u(x,y) \rightarrow 0$ as $y \rightarrow -\infty.$ Still, I don't know how to continue here on ward and completing the problem. Any help is much appreciated.

Answer 1

The first problem with your approach is that the problem is not in a compact domain, so we should not expect there to be a discrete countable set of eigenvalues. Therefore the solution has to be an integral instead of a sum. It is also easier to use exponentials than trig. functions for this case.

Separating variables, you have the correct equations, but let's write the separation constant as $\lambda=k^2$, and then we have solutions $$ X_k(x) = A(k) e^{ikx} + B(k) e^{-ikx}, \quad Y_k(y) = C(k) e^{ky} + D(k) e^{-ky}, $$ where the dependence on the eigenvalue has been made explicit. Now, to keep $u$ bounded as $y \to \infty$, we need to also take the $Y_k$ bounded as $y \to \infty$: i.e., $e^{-\lvert k \rvert y}$; we can discard the coefficient since it plays no further rôle that cannot be taken up by those in $X_k$.

The total solution is then $$ u(x,y) = \int_{-\infty}^{\infty} e^{ikx-\lvert k \rvert y} A(k) \, dk. $$ Of course, now we have to find $A(k)$ using the boundary conditions $u(x,0)=f(x)$, or $$ f(x) = \int_{-\infty}^{\infty} e^{ikx} A(k) \, dk. $$ This is a Fourier transform, so we proceed in the same way as Fourier series, by multiplying by $e^{-imx}$ and integrating with respect to $x$ (we assume that this integral makes sense, or this approach doesn't work; the final answer can still be obtained this way, but by a more complicated limiting process): $$ \int_{-\infty}^{\infty} e^{-imx} f(x)\, dx = \int_{-\infty}^{\infty} e^{-imx} \int_{-\infty}^{\infty} e^{ikx} A(k) \, dk \, dx. $$ Changing the order of integration on the right, $$ \int_{-\infty}^{\infty} e^{-imx} f(x) \, dx = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{ix(k-m)} \, dx \right) A(k) \, dk. $$ At this point we use the orthogonality relation for the Fourier transform, which is $$ \int_{-\infty}^{\infty} e^{ix(k-m)} \, dx = 2\pi \delta(k-m), $$ which may be shown in a number of different ways, such as using a cutoff to make the integral finite and taking the limit. Hence we find, using the property $ \int_{-\infty}^{\infty} \delta(k-m) A(k) \, dk = A(m) $, $$ A(m) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-imx} f(x) \, dx, $$ and so we have $$ u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx-\lvert k \rvert y} \left( \int_{-\infty}^{\infty} e^{-ikx'} f(x') \, dx' \right) dk. $$

In this case, we can simplify this further: change the order of integration again to get $$ u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x') \left( \int_{-\infty}^{\infty} e^{ik(x-x')-\lvert k \rvert y} \, dk \right) dx'. $$ The inside integral is straightforward, and can be done by splitting it at zero: we find that $$ \int_{-\infty}^{\infty} e^{ik(x-x')-\lvert k \rvert y} \, dk = \frac{2y}{(x-x')^2+y^2}, $$ so the neatest answer is $$ u(x,y) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y}{(x-x')^2+y^2} f(x') dx'. $$

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Another way to derive this involves calculating the Green's function for Laplace's equation on the upper half-space, and using Green's third identity to get this integral more directly; you haven't given enough context to determine which is the intended method.