# 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.

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.
• I'm confused at the notation $u(x,y)$. Is $u$ not variated by only one variable? – Felicio Grande Nov 16 '17 at 3:14
• I am using letter $u$ for Green's function, for additional confusion. – user357151 Nov 16 '17 at 3:27
• 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 $x\in(-1,1), x\neq 0$ – Felicio Grande Nov 16 '17 at 4:08
• @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 $x_0$. – user357151 Nov 17 '17 at 0:29