Existence of Green's Function with Neumann Boundary Conditions 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.
 A: 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. 
