Laplacian on the straight circle I would like to well understand the notion of Green's function. 
I know that a Green's function, $G(x,s)$, of a linear differential operator $L = L(x)$ acting on distributions over a subset of the Euclidean space $ℝ^n$, at a point s, is any solution of
$$LG(x,s)=\delta(s-x),$$
where $δ$ is the Dirac delta function.
Find the Green's function on the unit interval $[0,1]$.
We have to solve the linear equation $\frac{d^2f(x)}{dx}=g(x)$ with boundary condition $f(0)=a$ and $f(1)=b$. 
So the Green's function satisfies $\frac{d^2G(x,x')}{dx}=\delta(x-x')$. What will be the form that $G$ will take and why?
I think our linear differential operator $L$ could be the Laplacian operator, but I don't know how to solve it. Is anyone could explain to me how applicate this function at an interval on the real line?
 A: You are making this much to difficult because you are ignoring what you were told.  You were told that the differential operator was  $d^2/dx^2$, the "Laplacian", but only in one dimension.
The general solution to $d^2y/dx^2= 0$ is the linear function $y= ax+ b$.   The solution to $d^2G/dx^2= \delta(x- x')$, then, is a "broken line: $G(x, x')= px+ q$  for $0\le x \le x'$ and $G(x, x')= rx+ s$ for $x'\le x\le 1$.
It must satisfy the boundary conditons: G(0, x')= a.  Since x' must be between $0$ and $1$, $x= 0\le x'$ this is $G(0, x')= p(0)+ q= q= a$. And $G(1, x')= b$.  Since $x'$ must be between $0$ and $1$, $x'\le x= 1$, this is $G(1, x')= r(1)+ s= r+ s= b$.
So far we have $G(x, x')= px+ a$ for $0\le x\le x'$ and $G(x, x')= rx+ (b- r)$ for $x'\le x\le 1$.
The Green's function must be continuous so the two linear parts must be equal for $x= x'$:  $px'+ a= rx'+ (b- r)$.  $px'= rx'+ b- r- a$ so $p= (rx'+ b- r- a)/x'$.  
The Green's function is of the form $G(x, x')= (rx'+ b- r- a)(x/x')+ a$ for $0\le x\le x'$ and $G(x, x')= rx+ b- r$ for $x'\le x \le 1$.
The derivative of the Green's function has a step discontinuity of $1$ at $x= x'$.  The derivative of $(rx'+ b- r- a)(x/x')$ is $(rx'+ b- r- a)/x'$ and the derivative of $rx+ b- r$ is $r$.  We must have $(rx'+ b- r- a)/x'- r= (b- r- a)/x'= 1$.  So $b- r- a= x'$ and $b= x'+ a+ r$.  
Now $G(x,x')= (rx'+ x')(x/x')+ a= (r+ 1)x+ a$ for $0\le x\le x'$ and $G(x, x')= rx+ x'+ a+ r- r= rx+ x'+ a$ for $x'\le x\le 1$.
