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?
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 $x'\le x\le 1$.