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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 $Ag(x,y)$?

  • 0
    The definition doesn't discuss properties of the function $a(y)$. As long as $a(y)$ is sufficiently nice then it seems possible to define the jump to be 1. Perhaps I am missing something in your question.2011-07-25
  • 0
    @Alice: No, it doesn't. It only mentioned that $a$ is sufficiently smooth.2011-08-04

2 Answers 2