# Examples of Greens functions for Laplace's equation with Neumann boundary conditions.

The von Neumann boundary problem is a PDE in $\Omega$ \begin{cases}\Delta u＝0\\\frac{\partial u}{\partial \vec n}\rvert_{\Gamma}＝g\end{cases}where $g$ is assumed to be smooth on $\Gamma＝\partial\Omega$ $\vec n$ is the unit normal vector. $\Omega$ is bounded.

The green function of this problem is $G(p,q)$ such that\begin{cases} \Delta_p G(p,q)＝\delta(p-q) \\ \frac{\partial_p G}{\partial \vec n}＝0\end{cases}

I want an explicit formula as an example. when the domain$\Omega$ is special, such as a sphere, a cube, etc.

----------05/10----------

According to Kenny Wong's answer, the conditions of $g$ should satisfy $\int_\Gamma gdS＝0$, and the boundary condition for $G(p,q)$ should be changed to $\frac{\partial G}{\partial\vec n}\lvert_\Gamma＝constant$ such.that the integral of the constant is one on $\Gamma$.

I'm afraid that what I'm about to say is not quite you what to hear: your Neumann PDE does not have a solution for arbitrary choices of $g$.

To see why, let's compute the integral of $g$ over the surface $\Gamma$: $$\int_\Gamma g \ dS = \int_\Gamma \frac{\partial u}{\partial n} \ dS=\int_\Omega \nabla^2 u \ dV =\int_\Omega 0\ dV = 0.$$ Therefore, the average value of $g$ on the boundary surface $S$ must be zero. If $g$ fails to satisfy this condition, your PDE cannot be solved.

For a similar reason, your definition of the Green's function must be modified. Indeed, if we integrate $\frac{\partial G}{\partial n}$ over the surface $\Gamma$, we find that $$\int_\Gamma \frac{\partial G(\vec r, \vec r')}{\partial n_{\vec r}}dS(\vec r) = \int_\Omega \nabla_{\vec r}^2 G(\vec r,\vec r') \ dV(\vec r) = \int_V \delta(\vec r-\vec r') \ dV(\vec r) = 1.$$ So it is inconsistent to set $\frac{\partial G}{\partial n}$ equal to zero on the boundary $\Gamma$.

Fortunately, all is not lost! We can redefine the Green's function $G$ so that it satisfies $$\begin{cases} \nabla_{\vec r}^2 G(\vec r,\vec r') = \delta(\vec r - \vec r') &{\rm \ \ on \ \ } \Omega , \\ \frac{\partial G(\vec r, \vec r')}{\partial n_{\vec r}}=\frac 1 A & {\rm \ \ on \ \ } \Gamma, \end{cases}$$ where $A = \int_\Gamma dS$ is the area of the boundary surface $\Gamma$.

Now, Green's identity states that \begin{multline}\int_\Omega \left( u(\vec r) \nabla_{\vec r}^2 G(\vec r , \vec r') - G(\vec r,\vec r') \nabla_{\vec r} u(\vec r) \right) \ dV(\vec r) \\ = \int_\Gamma \left( u(\vec r) \frac{\partial G(\vec r,\vec r')}{\partial n_{\vec r}} - G(\vec r,\vec r') \frac{\partial u(\vec r)}{\partial n_{\vec r}}\right) dS(\vec r).\end{multline} Plugging in my new definition of $G(\vec r, \vec r')$, we see that any $u(\vec r)$ satisfying your PDE must satisfy we can immediately see that any solution to the PDE must satisfy $$u(\vec r') = - \int_\Gamma G(\vec r,\vec r')g(\vec r) \ dS(\vec r) + c, \ \ \ \ \ \ (\ast )$$ where the constant $c$ is equal to $\frac 1 A \int_\Gamma u(\vec r) dS(\vec r)$, the average value of $u$ on $\Gamma$.

Having redefined the Green's function, I'll give you an explicit expression in the case where $\Omega$ is a two-dimensional circular disk of radius $1$. Here it is: $$G(\vec r, \vec r') = \tfrac 1 {2\pi} \ln | \vec r -\vec r' | + \tfrac 1 {2\pi} \ln |\vec r - \vec r''|,$$ where $\vec r'' = \vec r'/|\vec r'|^2$ is the image of $r'$ under an inversion about the unit circle. It is similar to the Dirichlet Green's function, expect that we have a plus sign in front of the "image" term instead of a minus sign.

I believe that if you plug this Green's function into $(\ast )$ (with the constant $c$ chosen arbitrarily), you do indeed get a solution to your PDE, provided that $g$ obeys the consistency condition $\int_\Gamma g \ dS = 0$. This is stated in these lecture notes from my university, and also in Riley, Hobson and Bence, though I would love to see a rigorous proof! I wonder if anyone knows a good reference?

• I know of a book by Duffy that is entirely devoted to Green's functions. It has been mentioned to me on this site, BTW. Maybe it might help. Also, good answer! Commented May 9, 2017 at 22:40
• @GiuseppeNegro Thanks! Do you know if the formula ($\ast$) is ALWAYS a solution to the Neumann problem, given suitable conditions? I know that, if $u$ is a solution, then $u$ must satisfy ($\ast$); the problem is that I can't quite see how you prove the converse! Maybe this is explained in Duffy's book? Both the sources I quoted are written by physicists, which makes me nervous. I looked in Evan's PDEs book, but he only deals with the Dirichlet case. Commented May 9, 2017 at 22:43
• Duffy's book is not written by a mathematician either. :-D I bet that what you assert is true, namely, that (*) always defines a solution to the Neumann problem. The proof must boil down to integrating by parts via the Green's identity, exactly as in Dirichlet's case. Commented May 10, 2017 at 7:40