Currently I am studying for an exam about partial differential equations. While looking through some of the exercises concerning Green's function on the plane, I came across a rather impenetrable-seeming integral connected to a Dirichlet BVP. As the textbook we are using (Partial Differential Equations, Peter Olver) does not provide clear examples on this topic, I figured that the internet would be a fine next step towards a solution. I also think that some experienced mathematicians out there might enjoy an exercise, and explaining it. Let me be clear: this exercise will not be examined. The problem is as follows:

Solve for $u$ where \begin{align} -\Delta u (x,y)= \frac{1}{1+y}\text{ on }\{(x,y):y>0\}\text{ and }u(x,0) = 0\text{ for all }x\in \mathbb{R}. \end{align}

Now, we are asked to use the Green's function for the upper half plane to find a solution. Using the method of images, one finds such a function quite easily: \begin{align} G(x,y;\theta,\eta) = \frac{1}{4\pi}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right). \end{align} By Green's representation formula, the solution is then given by \begin{align}u(x,y) = \frac{1}{4\pi}\int_{-\infty}^{\infty}\int_0^{\infty}\frac{1}{1+\eta}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right)\,\mathrm{d}\eta\,\mathrm{d\theta}. \end{align} I am wondering, did the author perhaps choose this example problem because the resulting integral would be solvable? If so, could someone shed light on the problem? And if not, is there a better way to solve it?

My thanks.


1 Answer 1


I'll assume you made a typo in the equation, since the Green's function you have is for $\nabla$, not $-\nabla$

The inhomogeneous part only depends on $y$, so you can guess a solution of the form $u(x,y) = g(y)$, where

$$ -g''(y) = \frac{1}{1+y}, \ g(0) = 0 $$

Then integrating twice gives $$ g(y) = (1+y)\ln(1+y) - y + cy $$

where $c$ is some arbitrary constant.

Change the order of integration to.

$$ \frac{1}{4\pi}\int_0^\infty \frac{1}{1+\eta}\int_{-\infty}^\infty \ln \left(\frac{(\theta-x)^2 + (\eta-y)^2}{(\theta-x)^2 + (\eta+y)^2}\right)\ d\theta\ d\eta $$

Substitute $u = \theta-x$ and rewrite $a = |\eta-y|$, $b=|\eta+y|$. The inner integral can be evaluated as

\begin{align} \int_{-\infty}^\infty \ln\left(\frac{u^2+a^2}{u^2+b^2}\right) du &= u\ln\left(\frac{u^2+a^2}{u^2+b^2}\right)\Bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \left(\frac{2u^2}{u^2+a^2} - \frac{2u^2}{u^2+b^2}\right)du \\ &= 2\int_{-\infty}^\infty \left(\frac{a^2}{u^2+a^2} - \frac{b^2}{u^2+b^2}\right) du \\ &= 2\left(a\arctan\frac{u}{a} - b\arctan\frac{u}{b}\right)\Bigg|_{\infty}^\infty \\ &= 2\pi (a-b) \end{align}

So we're left with

\begin{align} \frac12 \int_0^\infty \frac{1}{1+\eta}\big(|\eta-y|-|\eta+y|\big) d\eta &= \int_0^y \frac{1}{1+\eta}(-\eta)d\eta + \int_y^\infty \frac{1}{1+\eta}(-y)d\eta \\ &= -\int_0^y \left(1 - \frac{1}{1+\eta}\right)d\eta - y\int_y^\infty \frac{1}{1+\eta}d\eta \\ &= -y + \ln(1+y)+y\ln(1+y) + y\lim_{b\to\infty}\ln(1+b) \end{align}

You may notice that the last term of the integral does not converge. This is because the solution is not unique. However, if you add a second boundary condition such that $u_y(x,b) = g'(b) = 0$, then the solution is indeed

$$ u(x,y) = (1+y)\ln(1+y) - y + y\ln(1+b), \ y \in (0,b) $$

for any $b$ arbitrarily large. This matches the solution obtained through integration as above.

  • $\begingroup$ Thank you! I could not have hoped for a more rigorous answer, and it is such a relief to see an example of how to approach the problem. Indeed I forgot to add a minus sign. I'm sure this will be of help to me. The last bit was a little confusing, however. In general, how does one recognize that the Poisson equation has a unique solution? Did we already know this in advance? Might help with guessing solutions. $\endgroup$
    – IAnemaet
    Commented May 23, 2018 at 12:28

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