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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?

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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

$$\begin{align} u(x_1,x_2)&=\int_{x_2\ge 0}\left(G(x_1,x_2|x_1',x_2')\underbrace{\nabla'^2 u(x_1',x_2')}_{=0}-u(x_1',x_2')\underbrace{\nabla'^2 G(x_1,x_2|x_1',x_2')}_{=-\delta(x_1-x_1')\delta(x_2-x_2')}\right)\,dx_1'\,dx_2'\\\\ &=-\int_{-\infty}^\infty \left.\left(\underbrace{G(x_1,x_2|x_1',x_2')}_{=0\,\text{at}\,x_2'=0}\frac{\partial u(x_1',x_2')}{\partial x_2'}-\underbrace{u(x_1',x_2')}_{=g(x_1')\,\text{at}\,x_2'=0}\frac{\partial G(x_1,x_2|x_1',x_2')}{\partial x_2'}\right)\right|_{x_2'=0}\,dx_1'\\\\ &=\int_{-\infty}^\infty g(x_1')\left.\left(\frac{\partial G(x_1,x_2|x_1',x_2')}{\partial x_2'}\right)\right|_{x_2'=0}\,dx_1 \end{align}$$

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  • $\begingroup$ I am not sure if I follow from my book: Partial Differential Equations by Walter A. Strauss, Green's second identity is given by $$\iiint_{D}(u\Delta v - v\Delta u)dx = \iint_{\partial D}\left(u\frac{\partial v}{\partial n} - v\frac{\partial u}{\partial n}\right)dS$$ $\endgroup$ – justanewb Jul 19 '17 at 23:18
  • $\begingroup$ I probably just don't understand the formula or how to apply it to this problem. $\endgroup$ – justanewb Jul 19 '17 at 23:19
  • $\begingroup$ You have the correct formula for $\mathbb{R}^3$ if $v=G$; for $\mathbb{R}^2$ reduce the 3-D integral over $D$ to a 2-D integral over the upper-half plane and the integral over $\partial D$ to a contour integral that is comprised of the real line and a semicircle of "infinite radius." The integral over the "infinite semicircle" is $0$ given the behavior of $G$. The left-hand side of the one you wrote becomes $u(x_1,x_2)$. The right-hand side becomes $- \int_{-\infty}^\infty g(x_1')\frac{\partial G}{\partial x_2'}\,dx_1'$ as expected. $\endgroup$ – Mark Viola Jul 19 '17 at 23:32
  • $\begingroup$ My apologies, my intellectual inferiority does not comprehend. I should have mentioned the formula is for $\mathbb{R}^3$. But I just don't understand how we use it for this particular problem it just doesn't make sense to me on where I begin given what I have. $\endgroup$ – justanewb Jul 19 '17 at 23:35
  • $\begingroup$ The solution I posted is all you need. It gives you the answer. Just use the form of $G$ you wrote in your OP, take its derivative with respect to $x_2'$, then after that let $x_2'=0$. You will get an integral representation for $u$ $\endgroup$ – Mark Viola Jul 19 '17 at 23:39

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