I am trying to show, with the following PDE: $$\begin{cases}\Delta u=f &\text{on}\>\>\Omega\\ u=g &\text{ on }\partial\Omega \end{cases}$$ that for all $x_0\in\Omega$, we have: $$u(x_0) = \iiint_{\Omega}G(x,x_0)f(x) dx+\iint_{\partial\Omega}g(x)\frac{\partial}{\partial \vec n}G(x,x_0)dS$$ Where $\vec n$ is the normal vector to $\partial\Omega$. Now, we have defined a Green's Function to be: $$G(x,x_0) = H(x)+\Phi(x)$$ Where $\Phi(x)$ is the Fundamental Solution in the respective Dimension. We also have the characteristics of $G$ and $H$, $H$ is a smooth Harmonic function in all of $\Omega$, and $G(x,x_0)=0$ on $\partial\Omega$.

For simplicity let us assume we are in Dimension $3$. Then, I believe we can solve this by Superposition. We had previously solved: $$\begin{cases}\Delta u=0 &\text{on}\>\>\Omega\\ u=g &\text{ on }\partial\Omega \end{cases}$$ The solution was given by: $$u(x_0)=\iint_{\partial\Omega}g(x)\frac{\partial}{\partial\vec n}G(x,x_0)dS$$ Good. This is half of what we need to show. Now, by Superposition I need to find the answer to: $$\begin{cases}\Delta u=f &\text{on}\>\>\Omega\\ u=0 &\text{ on }\partial\Omega \end{cases}$$ I started by applying Green's Second Identity to $H(x)$ and $u$ on $\Omega$: $$\iiint_{\Omega}H\Delta u-\Delta H u dx = \iint_{\partial\Omega}H\frac{\partial u}{\partial\vec n} dS_x-\iint_{\partial\Omega}\frac{\partial H}{\partial \vec n} u dS_x$$ Now, we know that $\Delta H$ is zero on $\Omega$. I also know that $u$ and all its derivatives are $0$ on $\partial\Omega$ by Superposition. So I am left with: $$\iiint_{\Omega}H\Delta u dx=\iiint_{\Omega}G\Delta u dx -\iiint_{\Omega}\Phi(x) \Delta u dx$$ Now I know above that $\Delta u = f$. All I need to show is that the above second term on the right hand side goes to $0$. But how do I show this??


2 Answers 2


You say that you know that $u$ is zero on the boundary $\partial \Omega$ and that you want to show that:

$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d}x = 0$$

This is just a classic application of Green's billionth identity. We have:

$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d} x = \int_\Omega u(x) \Delta \Phi \ \mathrm{d} x+ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) - u (\nabla \Phi \cdot \mathbf{\hat{n}}) \ \mathrm{d} S$$

Since $\Delta \Phi = 0$ in $\Omega$ and $u = 0$ on $\partial \Omega$, this leaves us with the following integral:

$$ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) \ \mathrm{d} S $$

Note that we have the following identity by the product rule:

$$ \nabla(\Phi u) = \Phi \nabla u + u \nabla \Phi $$

So that $\Phi \nabla u = \nabla(\Phi u ) - u\nabla \Phi $. If we now plug this into our integral we get:

$$ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) \ \mathrm{d} S = \int_{\partial\Omega} \nabla(\Phi u ) \cdot \mathbf{\hat{n}} \ \mathrm{d} S - \int_{\partial\Omega} u \nabla\Phi \cdot \mathbf{\hat{n}} \ \mathrm{d} S = 0$$

The first integral vanishes, since we are integrating the conservative vector field $\nabla (\Phi u)$ over a closed curve $\partial\Omega$, and the second integral vanishes since $u = 0$ on $\partial \Omega$. The conclusion is then:

$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d}x = 0$$

There you go! This is exactly what you wanted to show.


I strongly advise you have a close look at the following lecture notes taylor made for Green's functions. It will certainly solve your problem. I fact having read this I realized you did not take care of singularities occurred by the fundamental solution that is why you problem to show that the term $$\iiint_{\Omega}G\Delta u dx -\iiint_{\Omega}\Phi(x) \Delta u dx$$ cancel out. All details about that issues is given therein


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