I learn the classical definition of Green's function from Hunter's Applied Analysis.
Consider the second-order ordinary differential operators $A$ of the form $Au=au''+bu'+cu,$ where $a,b$ and $c$ are sufficiently smooth functions on $[0,1]$.
Think about the Dirichlet boundary value problem for the second-order differential operator $A$ defined above: $Au=f,\qquad u(0)=u(1)=0,\qquad \tag{10.9}$ where $f:[0,1]\to{\mathbb C}$ is a given continuous function.
The author gives a heuristic discussion using Dirac delta function: the Green's function $g(x,y)$ associated with the boundary value problem in (10.9) is the solution of the following problem: $Ag(x,y)=\delta(x-y),\qquad g(0,y)=g(1,y)=0.\qquad \tag{10.13}$ He reformulates $(10.13)$ in classical, pointwise terms. The book said that we want $g(x,y)$ to satisfy the homogeneous ODE(as a function of $x$) when $x\neq y$, and we want the jump in $a(x)g_x(x,y)$ across $x=y$ to equal one in order to obtain a delta function after taking a second $x$-derivative. We therefore make the following definition:†
† A function $g:[0,1]\times[0,1]\to{\mathbb C}$ is a Green's function for (10.9) if it satisfies the following conditions.
(a) The function $g(x,y)$ is continuous on the square $0\leq x,y\leq 1$, and twice continuously differentiable with respect to $x$ on the triangles $0\leq x\leq y\leq 1$ and $0\leq y\leq x\leq 1$, meaning that the partial derivatives exist in the interiors of the triangles and extend to continuous functions on the closures. The left and right limits of the partial derivatives on $x=y$ are not equal, however.
(b) The function $g(x,y)$ satisfies the ODE with respect to $x$ and the boundary conditions: $\begin{align} Ag=0\qquad \text{in}~0
(c) The jump in $g_x$ across the line $x=y$ is given by $g_x(y^+,y)-g_x(y^-,y)=\frac{1}{a(y)}$ where the subscript $x$ denotes a partial derivative with respect to the first variable in $g(x,y)$, and $g_x(y^+,y)=\lim_{x\to y^+}g_x(x,y),\qquad g_x(y^-,y)=\lim_{x\to y^-}g_x(x,y).$
The words in bold---
...we want the jump in $a(x)g_x(x,y)$ across $x=y$ to equal one in order to obtain a delta function after taking a second $x$-derivative
refer to condition (c) in the definition above.
Here is my question:
How can one get $Ag(x,y)=\delta(x-y)$ from $g_x(y^+,y)-g_x(y^-,y)=\frac{1}{a(y)}?$ Added:
The confusion is that I don't know what the words in bold mean. Finally, we want $Ag(x,y)=\delta(x-y),$ but what's the relation between "taking a second $x$-derivative" of $a(x)g_x(x,y)$ and $Ag(x,y)$?