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If we are given Poisson's equation in a space:

$$\nabla ^2 u=F$$

The solutions (those who admit Fourier transform) are given by:

$$u(x)=\int_\mathbb{R^n} G(x,y)F(y)dy$$

Where $G$ is the Green function (fundamental solution). In a distributional sense, the solution is:

$$u=G\ast F$$

Where the convolution $\ast$ is defined by:

$$\langle G\ast F,u\rangle=\langle G\otimes F,u ^\sharp\rangle$$

This seems to agree with wath we've seen.

But if instead of the whole space, we restric our equation to a domain $D$. With Dirichlet boundary conditions: $u\vert_{\partial D}=h$. The solution is:

$$u(x)=\int_D G(x,y)F(y)dV + \int_{\partial D} \frac{\partial G}{\partial n}(x,y) h(y)dS$$

How can this agree with the convolution? The surface integral doesn't seem right (I know the proof of this formula, just that this isn't completely clear).

Furthermore, does this mean that even if we know the Green function and the inhomogeneous term, we can't know the solution since we have to come up with the type of integral we need?

jinawee
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    Green's functions depend on boundary conditions. – paul garrett Jan 13 '15 at 18:17
  • @paulgarrett I know, but I expected $G$ to change, not the integral. – jinawee Jan 13 '15 at 18:18
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    Well, depending how you interpret things, I'll say the obvious: the without-boundary case allows $\mathbb R^n$ to operate as a group, and no boundary terms, either. The "domain" case wrecks the group structure, and loses the no-boundary property. (Most boundary conditions are not stable under translation...) Specifically, different sorts of boundary conditions must give different solution formulas... – paul garrett Jan 13 '15 at 18:25

2 Answers2

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Actually, $$ \int_D G(x,y)F(y) \, dV + \int_{\partial D} \frac{\partial G}{\partial n}(x,y) h(y) \,dS = G * (\mathbf{1}_D + \nabla\cdot (h n\,\sigma_{\partial D})), $$ where $\mathbf{1}_D$ is the indicator function of $D,$ $\sigma_{\delta D}$ is the area measure on $\partial D,$ and $n$ is the normal on $\partial D.$

Here it can be noted that $n\,\sigma_{\partial D} = -\nabla\mathbf{1}_D.$

md2perpe
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I would say you actually have the boundary terms in $R^N$ case, but it is just disappears because your data function $f$ is compact supported. Moreover, I suggest you work out the convolution solution for the case that $f$ is only integrable and Holder continuos, then you will have a better understanding how the "boundary" of $R^N$ disappears.

spatially
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