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I am currently dealing with Poisson's equation $- \Delta u= f $ on some open domain $U$ and $u =g$ on the boundary $\partial U.$

Now a fundamental solution is a solution to $- \Delta u(x) = \delta(x)$ on the whole $\mathbb{R}^n$. Is this correct?

A Green's function is now rather a construct that is supposed to satisfy Poisson's equation on some given domain with Dirichlet boundary conditions. So we should have $-\Delta_x G(x,y) = 0$ on $U$ and $G(x,y) = 0$ on $\partial U.$

Now, Evans constructs the Green function by saying that $G(x,y) = \phi(x-y) - \psi^{x}(y)$ where $\psi^{x}(y)$ satisfies Laplace's equation on $U$ and $\psi^{x}(y) = \phi(y-x)$ on $\partial U$. Unfortunately, he only defines the function for $x \neq y$. Apparently because $\phi(0)$ is -infinity.

My first question is: Why is this not a problem, if we don't define the Green's function for $x=y$?

and my second question is: Evans says that $-\Delta_y G(x,y) = \delta(x)$ in $U$ and I don't see how this follows from the definition? In particular, many references like mathworld claim that the Green's function actually satisfies $-\Delta_y G(x,y) = \delta(x-y)$?

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For the second question, note that for each $v\in C_0^\infty(\mathbb{R}^N)$ \begin{eqnarray} \int_{\mathbb{R}^N}\nabla\phi(x-y)\nabla v(y)dy &=& \int_{B_R}\nabla \phi(y)\nabla v(x-y)dy, \nonumber \\ &=& \int_{\partial B_{\delta}}\frac{\partial \phi(y)}{\partial \eta}v(x-y)dy-\int_{B_\delta}\nabla \phi(y)\nabla v(x-y)dy, \nonumber \end{eqnarray}

where $B_R, B_\delta$ are balls with center in the origin, $R$ is big so that $B_R$ contains the support of $v$ and $\delta$ is small, $\eta$ stands for normal direction. By letting $\delta \to 0$ in the above equality and using the definitions, we get that $$\langle -\Delta _y G(x,y),v\rangle=v(x)=\langle \delta_x, v\rangle.$$

With respect to your first question: that's the nature of the Green function and as you observed, it is like this because the fundamental solution blows up in the origin.

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  • $\begingroup$ thank you. and where does the equation $-\Delta_x G(x,y) = \delta(x-y)$ similar like here mathworld.wolfram.com/GreensFunction.html come from? $\endgroup$ – RealAnalysis May 6 '15 at 0:27
  • $\begingroup$ sorry, not sure I understand this. would you mind adding this to your answer ( I mean, why mathworld gets a $\delta(x-y)$ out of the Green's function?), cause so far I only see $ - \Delta_y G(x,y) = \delta(x)$ according to you answer and not where a $\delta(x-y)$ might eventually come from? $\endgroup$ – RealAnalysis May 6 '15 at 0:53
  • $\begingroup$ For me $\delta_x$ means the distribution, which applied in a smooth function with compact support $v$, gives $v(x)$. Now the question is, what is the meaning of $\delta(x-y)$? $\endgroup$ – Tomás May 6 '15 at 12:22

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