1

Today is class we were looking at Green's functions for second-order operators. I am having trouble understanding one example and I was hoping someone could shed some light on it. We saw issues arise we attempted to find the Green's function for \begin{equation*} \left\{ \begin{array}{l} &Av''=u \\ &v'(0)=v'(1)=0 \end{array} \right. \end{equation*} To get around this we considered the Green's function which satisfies \begin{equation*} \left\{ \begin{array}{l} -AJ_{xx}= \delta_y - 1 \\ J_x(0)=J_x(1)=0 \\ \int_0^1 J(x,y) dx = 0 \end{array} \right. \end{equation*} If we let $L$ be the differential operator, then \begin{equation*} \int L[J] u dx = u(y)-\int_0^1 u(x) dx \end{equation*} This was supposed to remove the issue we had for the first equation. I have a few questions.

1) When solving for the Green's function in the first equation (by solving the equation on each interval (0,y) and (y,1) and then assuming cty of the function and a jump discont for the first derivative) I see that we can't solve it. I was wondering what the underlying reason for this was so that I can see in general when a Green's function will exist. I would assume it has to do with the lack of uniqueness for solutions in this case, but I was wondering if someone could give me a more complete reason. Does it have anything to do with the eigenvalue for this equation also being a root of the characteristic equation?

2) I tried solving this second equation, but it appears to have infinitely many solutions. I will post my work for the second equation below and perhaps someone would be kind enough to tell me what I'm doing wrong.

$A J_{xx}(x,y) = \delta(x-y)-1 \Rightarrow J_{left} = c_1 + c_2 x - \frac{x^2}{2A}, \; J_{right} = d_1 + d_2 x - \frac{x^2}{2A}$. The boundary conditions yield $c_2 = 0$ and $d_2 = \frac{1}{A}$. Thus $J_{left} = c_1 - \frac{x^2}{2A}, \; J_{right} = d_1 + \frac{1}{A} x - \frac{x^2}{2A}$. Using the continuity of $J$ we get that $c_1=d_1 + \frac{y}{A}$. The jump discontinuity of $J_x$ is always satisfied and so we get a family of solutions \begin{equation*} J(x,y) = \left\{ \begin{array}{ll} \left(d_1 + \frac{y}{A} \right) - \frac{x^2}{2A} & x < y \\ d_1 + \frac{x}{A} - \frac{x^2}{2A} & x \geq y \end{array} \right. \end{equation*}

3) What is the underlying trick that is being utilizing in this method? It seems like we could find a Green's function when the forcing term has zero average and this trick appears to actually create a function with zero average on the right side. Is there more to it than this?

Sorry for so many questions. Thank you in advance for the help.

John
  • 463
  • 3
  • 10
  • There are no solutions to the Poisson equation with a forcing which does not have zero average. The physical interpretation of this is quite simple. The solution of the Poisson equation is a stationary solution of the corresponding heat equation. With no forcing, the heat equation keeps the average constant, so the time derivative of the average in a forced heat equation is the average of the forcing. So you can never have a stationary solution unless the average of the forcing is zero. The proof in this particular case follows the physical intuition pretty nicely. – Ian Mar 25 '16 at 02:43
  • Thanks for the insight with the physical interpretation. That was very helpful. – John Mar 25 '16 at 14:45
  • Small correction: with no forcing, the heat equation with homogeneous Neumann boundary conditions keeps the average constant. Any other boundary condition can be interpreted as basically "forcing at the boundary"...which is in fact exactly what it looks like when you write down the weak formulation. – Ian Mar 25 '16 at 14:52
  • ...and another correction: with Neumann or "Neumann-like" conditions (where you are told the derivative on the boundary, possibly in terms of the solution itself), the boundary condition looks like a forcing in the weak formulation. With Dirichlet conditions it doesn't, so to extend that analogy to the Dirichlet context, you have to identify Dirichlet boundary conditions as a limit of Neumann-like boundary conditions. Specifically the solution with $u(0)=u_0$ can be realized by considering $u'(0)=k(u(0)-u_0)$ and sending $k \to \infty$. – Ian Mar 25 '16 at 15:06

0 Answers0