2
$\begingroup$

Consider the Laplacian $-\Delta$ on (in a suitable sense) twice differentiable functions subject to homogeneous Dirichlet boundary conditions $\mathscr{H}=\{f : f(0)=f(1)=0\}$. We can identify the Green's function of $-\Delta$ to be

$$\mathcal{G}(s, t) = \min(s, t) - st.$$

This Green's function will of course work for solving the Poisson equation on any linear subspace of $\mathscr{H}$, but it is the distributional form of the Green's function—we may be able to find a functional form of the Green's function that seem more appropriate for a given subspace.

For example, if we consider the subspace of functions $f\in\mathscr{H}$ such that $\int_0^1 f(t)\,\mathrm{d}t =0$, we can find a functional form of the Green's function on this subspace that obeys the integral condition:

$$\mathcal{G}(s, t) = \min(s, t) - st− 3(1 − s)s(1 − t)t.$$

Now for my question, if we consider the Green's function for the Laplacian on the functions $f\in\mathscr{H}$ which are also subject to homogeneous Neumann boundary conditions $f'(0)=f'(1)=0$, I cannot find a functional form of the Green's function that obeys both types of conditions. Does such a form exist?

Through some applications, I have an indication that the extra Neumann conditions are non-restricting in some sense, but it is all very unclear to me what is going on. Is this something that any of you have had experience with, or can any of you refer me to some relevant work?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I have no idea why you would "have an indication that the extra Neumann conditions are non-restricting in some sense." If you find any solution $g(t)$ to your inhomogeneous equation, then a solution satisfying your conditions will have to be of the form $g(t)-f(t)$ for some homoegeneous solution $f(t)$. All of $f'(0)=g'(0)$, $f'(1)=g'(1)$ and $\int_0^1 [f(t)-g(t)]\, dt = 0$ are independent affine conditions on the 2-dimensional space of homogeneous solutions. You will not be able to find an $f(t)$ to satisfy all three conditions unless you are lucky and $g(t)$ already satisfies some of them. $\endgroup$
    – Igor Khavkine
    Commented Sep 28, 2015 at 10:07
  • $\begingroup$ @IgorKhavkine: I think we are talking about different things. When I restrict to a subspace of functions $f\in\mathscr{H}$, I assume we are only considering $h$s such that the equation $-\Delta f=h$ can be solved. I have several examples where I add extra subspace conditions and then find a form of the Green's function that obeys these constraints. The trick I have used so far is to use the extra conditions to identify functions you can add to the Green's function without affecting the distributional definition. As the above example shows, these may be more complex than homogeneous solutions. $\endgroup$
    – Lars Lau Raket
    Commented Sep 29, 2015 at 8:01

0

Reset to default

You must log in to answer this question.