3
$\begingroup$

I am reading through PDE book by Evans to get a better understanding of what a Green function is. In my understanding the Green function enables an explicit representation of the solution of certain PDE (I think mainly linear ones).

Looking at the derivation is, as always, insightful.

The author specifically derives Green function for the Laplace equation

$$ \left\{ \begin{array}{ll} -\Delta u = f & \text{in $U$} \\ u = g & \text{on $\partial U$} \end{array} \right. $$

I quote the derivation (which isn't long) and I'll put my questions in the middle.

Suppose first of all $u \in C^2(\overline{U})$ is an arbitrary function. Fix $x \in U$ choose $\epsilon > 0$ so small that $B(x,\epsilon) \in U$, and apply the Green's formula from C.2 on the region $V_{\epsilon} = U - B(x,\epsilon)$ to $u(y)$ and $\Psi(y - x)$. We thereby compute $$ \int_{V_{\epsilon}} u(y) \Delta \Psi(y - x) - \Psi(y - x) \Delta u(y) dy = \int_{\partial V_{\epsilon}} u(y) \frac{\partial \Psi}{\partial \nu}(y - x) - \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y), $$ $\nu$ denoting the outer unit normal vector in $\partial V_{\epsilon}$. Recall next $\Delta \Psi (x - y) = 0$ for $x \neq y$. We observe also $$ \left|\int_{\partial B(x,\epsilon)} \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y) \right| \leq C\epsilon^{n-1} \max_{\partial B(0,\epsilon)} \left| \Psi \right| = o(1) $$ as $\epsilon \to 0$. Futhermore the calculations in the proof of Theorem 1 show $$ \int_{\partial B(x,\epsilon)} u(y) \frac{\partial \Psi}{\partial \nu} dS(y) = \frac{1}{\left| \partial B(x,\epsilon) \right|} \int u(y)dS(y) = u(x) $$ as $\epsilon \to 0$. Hence sending $\epsilon \to 0$ yields the formula $$ u(x) = \int_{\partial U} \Psi(y - x)\frac{\partial u}{\partial \nu}(y) - u(y) \frac{\partial \Psi}{\partial \nu}(y - x) dS(y) - \int_{U} \Psi(y - x) \Delta u(y) dy \hspace{10mm} (25) $$

My observation here is that most of the calculation being done so far they're independent from the use of the ball $B(x,\epsilon)$. My question here is: Do we need such a ball so we can apply the mean value property? Apart from that specific calculation I don't see any other use of the ball.

The observation made by the author is that formula (25) would fully define $u$ if we knew the normal derivative at the boundary (which we don't, we only know the function at the boundary but not it's normal derivative). Therefore a corrective term $\phi^x(y)$ for fixed $x$ is added. Such a corrective term is defined as the solution of the following boundary problem

$$ \left\{ \begin{array}{ll} \Delta \phi^x = 0 & \text{in $U$} \\ \phi^x = \Psi(y - x) & \text{on $\partial U$} \end{array} \right. $$

And applying the Green's formula once more yields $$ -\int_U \phi^x(y) \Delta u(y) dy = \int_{\partial U} u(y) \frac{\partial \phi^x}{\partial \nu}(y) - \Psi(y - x) \frac{\partial u}{\partial \nu}(y) dS(y) \hspace{10mm} (27) $$

The Green function is then defined as $$ G(x,y) := \Psi(y - x) - \phi^x(y) \hspace{10mm} (x, y \in U, x \neq y) $$ Adding (25) and (27) together gives

$$ u(x) = - \int_{\partial U} u(y) \frac{\partial G}{\partial \nu}(x,y) dS(y) - \int_{U} G(x,y) \Delta u(y) dy $$

Finally a remark at the end

Remark. Fix $u \in U$. Then regarding $G$ as a function of $y$, we may simbolically write $$ \left\{ \begin{array}{ll} -\Delta G = \delta_x & y \in U \\ G = 0 & y \in \partial U \end{array} \right. $$

Question : Why is the remark true? why can we simbolically write a PDE with the Dirac delta as input?

Question : This derivation requires the use of a boundary term. However I wonder what changes would we need to make if the domain is unbounded and we search for solutions (for example) in $C^2(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, does the boundary term just disappear? So we only endup with:

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

$\endgroup$
5
  • $\begingroup$ The problem is that Evans totally ignores any rigorous discussion of distributions (or generalized functions) except for a few short heuristic remarks on delta function. In any discussion of Green's functions it is much more profitable to spend some time on this theory, so that the remark your are mentioning becomes the definition. Here is also my older answer on a somewhat more elementary question: link Finally, for your last question, look in the same book for the... $\endgroup$
    – Artem
    Jun 2, 2023 at 2:41
  • $\begingroup$ ...fundamental solution of the Laplace equation. $\endgroup$
    – Artem
    Jun 2, 2023 at 2:41
  • $\begingroup$ @Artem , since I wrote this question I have been doing some research. In my understanding the "fundamental solution" is "the\a" solution to the problem without boundary conditions while the "Green Function" is used for boundary values problem. Can you confirm if this understanding is correct? There's a relationship maybe highlighted by Evans in his definition of Green function (which is the fundamental solution + some corrective term). About distributiuons... I am a bit familiar with distribution theory maybe I'll do some research in that sense as well. $\endgroup$ Jun 2, 2023 at 5:02
  • $\begingroup$ In general you are correct, but you should be careful: there exists no universally accepted terminology. E.g., the fundamental solutions to Laplace equation are sometimes called "free space Green's functions." There is also another class of "fundamental solutions" for PDEs, when the delta function is assumed to be in the initial conditions (see, e.g., fundamental solution to the heat equation). Also, there are Green's functions for initial value problem. So, it is a little confusing, especially if a student wants 100% precise definitions (as in, e.g., group theory), but they do not exist. $\endgroup$
    – Artem
    Jun 2, 2023 at 13:15
  • $\begingroup$ What about my first question? $\endgroup$ Jun 2, 2023 at 21:55

0

You must log in to answer this question.