I'm trying to solve a question, but I can't figure it out. The question is:

Does there exist a Green's function in $1$D for $\Delta$ on $(-1,1)$ with Neumann Boundary Conditions?

I know that the Fundamental Solution in 1D is given by: $$\Phi(x-x_0) =\frac{|x-x_0|}{2}$$ I found the Green's function with the Dirichlet Boundary Conditions (i.e. $u(-1)=u(1)=0$)

But I dont know how to solve it when $u'(-1)=u'(1)=0$? If anyone would help me get started on this question it would be appreciated.


The boundary conditions $u'(-1)=u'(1)=0$ are incompatible with the property $\Delta_x u(x,y) = \delta_{x=y}$, so if this property is required, Green's function does not exist. Indeed, $$ 0 = u'(1)-u'(-1) = \int_{-1}^1 \Delta u = 1 \tag1$$ is impossible.

But if one still wants to solve the Neumann boundary problem using Green's function (a natural thing to want), the way out of the difficulty is to require $$\Delta_x u(x,y) = \delta_{x=y} - \frac12\tag2$$ instead. Then the integral $\int_{-1}^1 \Delta u$ is zero. A function satisfying (2) with Neumann boundary conditions can be found: $$u(x,y) = \frac{|x-y|}{2} - \frac{x^2+y^2}{4} \tag3$$ One can use (3) to solve the Neumann problem $\Delta w = f$ provided $\int_{-1}^1 f=0$ (a condition necessary for existence of solution), in the usual way: $$ w(x) = \int_{-1}^1 u(x,y)f(y)\,dy $$ This works because $$ \Delta w(x) = \int_{-1}^1 (\Delta_x u(x,y)) f(y)\,dy =\int_{-1}^1 (\delta_{x=y} - 1/2) f(y)\,dy = f(x) $$ since the integral of $(1/2)f$ vanishes.

So, depending on one's understanding of Green's function, the answer is no or yes.

  • $\begingroup$ I'm confused at the notation $u(x,y)$. Is $u$ not variated by only one variable? $\endgroup$ – Felicio Grande Nov 16 '17 at 3:14
  • $\begingroup$ I am using letter $u$ for Green's function, for additional confusion. $\endgroup$ – user357151 Nov 16 '17 at 3:27
  • $\begingroup$ Can't we just say that because $G(x,x_0)=\Phi(x-x_0)+H(x)$ where $G$ is Greens Function, $\Phi$ is the Fundamental Solution and $H(x)$ is a harmonic smooth function on the interval $(-1,1)$, and $\Phi$ is not differentiable that we cannot find $G'(x,x_0)$, therefore there does not exist a Green's Function which satisfies the Neumann Boundary Conditions on $(-1,1)$ (it does exist for all $x\in(-1,1), x\neq 0$ $\endgroup$ – Felicio Grande Nov 16 '17 at 4:08
  • 1
    $\begingroup$ @FelicioGrande No we can't just say that. That argument seems to say that there are no Green's functions at all, for any boundary conditions. The derivative of $G$ exists everywhere except at $x_0$. $\endgroup$ – user357151 Nov 17 '17 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.