[Math] Neumann problem for Laplace equation on Balls by using Green function

harmonic-analysislaplacianpartial differential equations

It is well known that for Dirichlet problem for Laplace equation on balls or half-space, we could use the green function to construct a solution based on the boundary data. For instance, one could find a nice proof in Evans PDE book, chapter 2.2, it is called the Poisson's formula.

Now, it comes to my attention that could we study the Neumann problem for Laplace equation on Balls or half-space by using the same method? i.e., try to find the corresponding Green functions and write down the Poisson's formula as for Dirichlet problem. I worked out the case for half-space, but I meet difficulties in balls case.

For example, we are trying to solve
\begin{cases}
-\Delta u =0 &x\in B(0,1)\\
\partial_\gamma u =g &x\in \partial B(0,1)
\end{cases}

We also know that we need to require such that $\int_B g=0$ in order to solve the equation, and I think this instance has something to do to write down the Poisson's formula, but didn't really get it…

I believe that this problem has been well studied, but I can not find it online…. So, if you know the answer, please just write done for the Poisson's formula for balls case for Neumann problem, or kindly directly me to where I can find a solution.

By the way, here is what I found for half-space case

Best Answer

In a halfspace, the construction of Neumann-Green function proceeds by reflection similar to the construction of Dirichlet-Green function; the difference is that one uses even reflection instead of odd reflection, thus achieving zero derivative instead of zero value. So, in $n\ge 3$ dimensions the function is (according to one convention) $$G(x;y) = -\frac{c_n}{|x-y|^{n-2}}-\frac{c_n}{|x-\xi|^{n-2}}$$ where $\xi$ is the reflection of $y$ in the boundary of the halfspace.

On a bounded domain such as a ball, the Neumann-Green function has to be defined a bit differently from the Dirichlet-Green function, because having $\Delta_x G(x;y)=\delta_y$ is incompatible with the Neumann boundary condition. Instead, the requirement on the Laplacian is $$\Delta_x G(x;y)=\delta_y-k,\quad \text{ where } \ k=|\Omega|^{-1}\tag{1}$$ Some authors prefer to put Laplacian on the second variable here; but when normalized by condition $\int G(x,y)\,dx=0$, the function is symmetric.

The requirement (1) makes the integral $\int_\Omega \Delta_x G(x;y)\,dx$ equal to zero, which is consistent with the boundary condition $\frac{\partial }{\partial n_x}G(x;y)=0$ on $\partial\Omega$. Note that the usage of Neumann-Green function is not affected by this $-k$; given any reasonable function $f$ with $\int_\Omega f=0$, we can solve the Neumann problem for the Poisson equation $\Delta u=f$ as $$ u(x)=\int_\Omega G(x;y) f(y)\,dy $$ since $$\Delta u(x)=\int_\Omega \Delta_x G(x;y) f(y)\,dy = f(x)- \int_\Omega k f(y)\,dy = f(x)$$

Explicit formulas

The singular part of the Laplacian in (1) is contributed by the fundamental solution; the constant part can be contributed by a quadratic polynomial. The hard part is to find a harmonic function to cancel out the normal derivative. A step toward finding it is to perform even reflection as for half-plane; in two dimensions this is enough but in three dimensions one needs further correction.

In the book Partial Differential Equations by Emmanuele DiBenedetto, pages 106-107, one can find the following formulas for $G(x;y)$ in the ball of radius $R$. (NB: DiBenedetto uses the convention $\Delta_y G(x;y)=-\delta_x+k$.)

Two dimensions $$G(x;y) = -\frac{1}{2\pi}\left( \ln|\xi-y|\frac{|x|}{R} +\ln|x-y|\right) - \frac{1}{4\pi R^2} |y|^2$$

where $\xi=\frac{R^2}{|x|^2}x$ is the reflection of $x$ across the boundary of the ball.

Three dimensions

$$G(x;y) = \frac{1}{4\pi}\left( \frac{1}{|x-y|} + \frac{R}{|x|} \frac{1}{|\xi-y|}\right) + \frac{1}{4\pi}\ln\left( (\xi-y)\cdot \frac{x}{|x|} +|\xi-y|\right) -\frac{1}{8\pi R^3} |y|^2 $$

Unlike the Dirichlet case, the author does not give a formula for $n$ dimensions, which suggests there isn't a simple one.

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