Green's Function: Solving 1D Laplace Equation

[squenshl]

Homework Statement

I am trying to find the Green's function in one space dimension. The Green's function is G(x,y) = $$\Phi(x-y)$$ - $$\phi(x,y)$$ where $$\phi(x,y)$$ is the solution to the Laplace problem (x fixed): $$\Delta$$_y$$\phi$$ = 0 in $$\Omega$$ with $$\phi(x,\sigma)$$ = $$\Phi(x-\sigma)$$ for $$\sigma$$ on $$\delta\Omega$$. I have $$\Phi(x)$$ = -|x|/2.

Homework Equations

The Attempt at a Solution

The Laplace equation in one dimension is just $$\phi''$$ = 0 so solving this is trivial, $$\phi$$ = ax + b but how go I get the constants from $$\phi(x,\sigma)$$ = $$\Phi(x-\sigma)$$ = -|x-$$\sigma$$|/2

 

[HallsofIvy]

The Green's function for this problem is a function, G(x,t), such that:
1: It satisifies the equation $d^2 G/dx^2= 0$ for all t.
2: It satisifies the boundary conditions,
3: It is continuous at x= t.
4: There is a jump of 1 in the derivative at x= t: $\lim_{x\to t^-} dG/dx- \lim_{x\to t^+} dG/dx= 1$

From the first condition, it follows that G(x,t) must be of the form
G(x,t)= Ax+ B for $x\le t$
G(x,t)= Cx+ D for $t\le x$
and the "constants", A, B, D, and D, may depend on t.

From the fourth condition, C- A= 1. From the third condition, At+ B= Ct+ D
But you haven't given any boundary conditions. What are your boundary conditions? On what interval are you solving this?

 

[squenshl]

We only have $$\phi(x,\sigma$$) as the only condition. I think that we only have to apply this to the solution$$\phi$$ = c₁y + c₂. But how do we find the constants if we only have one condition.