Use the Green's function for the half-plane to solve the problem
$$\begin{cases}
\Delta u(x_1,x_2) = 0 \ \ \text{in the half-plane} \ x_2 > 0\\
u(x_1,0) = g(x_1) \ \ \text{on the boundary} \ x_2 = 0
\end{cases}$$
where the boundary condition is
$$g(x_1) = \begin{cases}
\pi \ \ \text{if} \ |x_1| \leq 1\\
0 \ \ \text{if} \ |x_1| > 1\\
\end{cases}$$
Verify that the solution is, in fact, harmonic, and satisfies the given boundary condition.\
\noindent
Remark: In our theorems, we assumed that $g$ is continuous, which this one is not. What is happening at the two discontinuities?
Attempted (beginning) solution – The Green's function for the half-plane is given by
$$G(x,x_0) = \frac{1}{2\pi}\log |x – x_0| – \frac{1}{2\pi}\log |x – x^{*}_0|$$
where $x = x(x,y)$, $x_0 = (x_0,y_0)$, and $x^{*}_0 = (x_0,-y_0)$.
Now I am pretty confused with what to do next here, going through my professors notes I can seem to find anything related to Green's function in the half-plane. I had to search for it online to find it which is what is presented above.
The only part that seems remotely related in my notes is on the Half-space where we found the solution to Poisson's problem in the upper half place for $2$ dimensions which is
$$u(x_1,x_2) = -\frac{1}{4\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}\log \frac{(y_1 – y_2)^2 + (y_2 – x_2)^2}{(y_1 – x_1)^2 + (y_2 + x_2)^2}f(y_1,y_2)dy_2 dy_1 + \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x_2}{(y_1 – x_1)^2 + x_2^{2}}g(y_1)dy_1$$
My apologies ahead of time if this question is a bit messy. I just don't understand what I need to do here. Any suggestions are greatly appreciated.
Attempted solution (sort of) – The Green's function for the half-plane is given by
$$G(x,x_0) = \frac{1}{2\pi}\log |x – x_0| – \frac{1}{2\pi}\log |x – x^{*}_0|$$
where $x = x(x,y)$, $x_0 = (x_0,y_0)$, and $x^{*}_0 = (x_0,-y_0)$. In our case we have $$G(x_1,x_2) = \frac{1}{2\pi}\log|x_1 – x_2| – \frac{1}{2\pi}\log|x_1 – \tilde{x_2}|$$
Thus
$$\frac{\partial G}{\partial n} = \frac{1}{2\pi}\left(\frac{x_1 – \tilde{x_2}}{|\tilde{x_2} – x_1|^2}\right)\Bigg|_{\tilde{x_2} = 0} = \frac{1}{2\pi x_1}$$
Thus we have $$u(x_1,x_2) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{x_1}g(x_1)dx_1$$
Not sure if this is correct, any suggestions or comments are appreciated. If it is correct, is it possible to evaluate further? How do I show that the solution is harmonic?
Attempted solution 3 – The Green's function for the half-plane is given by
$$G(x,x_0) = \frac{1}{2\pi}\log |x – x_0| – \frac{1}{2\pi}\log |x – x^{*}_0|$$
where $x = x(x,y)$, $x_0 = (x_0,y_0)$, and $x^{*}_0 = (x_0,-y_0)$. In our case we have
\begin{align*}
G &= \frac{1}{2\pi}\log\left(\sqrt{(\tilde{x_1} – x_1)^2}\right) – \frac{1}{2\pi}\log\left(\sqrt{(\tilde{x_1}-x_1)^2 + (\tilde{x_2} + x_2)^2}\right)\\
&= -\frac{1}{4\pi}\left[\log\left((\tilde{x_1} – x_1)^2 + (\tilde{x_2} – x_2)^2\right) – \log\left((\tilde{x_1} – x_1)^2 + (\tilde{x_2} + x_2)^2\right)\right]
\end{align*}
Now,
\begin{align*}
\frac{\partial G}{\partial n}\Bigg|_{\tilde{x_2} = 0} &= -\frac{1}{4\pi}\left[\frac{2(\tilde{x_2} – x_2)}{(\tilde{x_1} – x_1)^2 + (\tilde{x_2} – x_2)^2} – \frac{2(\tilde{x_2} + x_2)}{(\tilde{x_1} – x_1)^2 + (\tilde{x_2} + x_2)^2} \right]\Bigg|_{\tilde{x_2} = 0}\\
&= \frac{x_2}{\pi((\tilde{x_1} – x_1)^2 + x_2^{2})}
\end{align*}
Thus we have
$$u(x_1,x_2) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x_2}{(\tilde{x_1} – x_1)^2 + x_2^{2}}g(x_1)dx_1$$
We see that $\Delta u(x_1,x_2) = 0$ since $g(x_1)$ is a constant function and taking the derivative with respect to $x_1$ will equal $0$. Therefore the solution $u(x_1,x_2)$ is harmonic.
I am not sure if this is correct or not. Also, how do we show that this solution satisfies the boundary condition?
Best Answer
HINT:
Using Green's Second Identity with $\nabla'^2 G(x_1,x_2|x_1',x_2')=-\delta(x_1-x_1')\delta(x_2-x_2')$ in distribution to write
$$\begin{align} u(x_1,x_2)&=\int_{x_2\ge 0}\left(G(x_1,x_2|x_1',x_2')\underbrace{\nabla'^2 u(x_1',x_2')}_{=0}-u(x_1',x_2')\underbrace{\nabla'^2 G(x_1,x_2|x_1',x_2')}_{=-\delta(x_1-x_1')\delta(x_2-x_2')}\right)\,dx_1'\,dx_2'\\\\ &=-\int_{-\infty}^\infty \left.\left(\underbrace{G(x_1,x_2|x_1',x_2')}_{=0\,\text{at}\,x_2'=0}\frac{\partial u(x_1',x_2')}{\partial x_2'}-\underbrace{u(x_1',x_2')}_{=g(x_1')\,\text{at}\,x_2'=0}\frac{\partial G(x_1,x_2|x_1',x_2')}{\partial x_2'}\right)\right|_{x_2'=0}\,dx_1'\\\\ &=\int_{-\infty}^\infty g(x_1')\left.\left(\frac{\partial G(x_1,x_2|x_1',x_2')}{\partial x_2'}\right)\right|_{x_2'=0}\,dx_1 \end{align}$$