Consider the one-dimensional heat equation given by

$-{\cfrac {d^{2}u}{dx^{2}}}=f(x)\qquad x\in (0,1),~~u(0)=0,u(1)=0~.$ This equation has Green's function $g(x,y)$ which satisfies

$-{\frac {\partial ^{2}g}{\partial x^{2}}}=\delta (x-y)\qquad x,y\in (0,1)$ The Green's function is given by

$g(x,y)={\begin{cases}(1-y)~x&0\leq x<y\\(1-x)~y&y<x\leq 1\end{cases}}$ or,

$g(x,y)=(1-y)~x+(y-x)~H(x-y)~.$ Therefore,

$g_{x}(x,y)=(1-y)-H(x-y)+(y-x){\frac {\partial }{\partial x}}[H(x-y)]=(1-y)-H(x-y)+(y-x)\delta (x-y)$ Now,

$\left\langle x~\delta (x),\phi \right\rangle =\left\langle \delta (x),x~\phi \right\rangle =(x~\phi )(0)=0$ Hence,

$g_{x}(x,y)=(1-y)-H(x-y)$ And,

$g_{xx}(x,y)=-\delta (x-y)$ Note that the second derivative of $g$ is a delta function.

We can use this observation to arrive at a more general description of the
Green's function for a particular differential equation. That is, for a
$n$ th order linear differential operator we would want the $n$ th derivative of
$g(x,y)$ to be like a delta function. Thus the $(n-1)$ th derivative of
$g(x,y)$ should be like a Heaviside function and all lower derivatives should
be continuous.

In particular, consider the operator ${\mathcal {L}}$ acting on $g$ such that

${\mathcal {L}}~g=\delta (x-y)$ with

${\mathcal {L}}=a_{n}(x)~{\cfrac {d^{n}}{dx^{n}}}+a_{n-1}(x)~{\cfrac {d^{(n-1)}}{dx^{(n-1)}}}+\dots a_{0}(x)$ where $a_{i}(x)\in C^{\infty }$ . If we integrate this equation across the
point $x=y$ from $x=y^{-1}$ to $x=y^{+}$ we get

$\left.a_{n}(x){\cfrac {d^{n-1}g}{dx^{n-1}}}\right|_{x=y^{-}}^{x=y^{+}}=\int _{y^{-1}}^{y^{+}}\delta (x)~{\text{d}}x=1$ This condition is called a Jump condition .

This suggests that the Green's function $g(x,y)$ satisfying

${\mathcal {L}}_{x}g(x,y)=\delta (x-y)$ in the sense of distributions has the properties that

${\mathcal {L}}_{x}g(x,y)=0$ for all $x\neq y$ .
${\cfrac {d^{k}g(x,y)}{dx^{k}}}$ is continuous at $x=y$ for $k=0,1,\dots ,n-2$ .
$\left.{\cfrac {d^{n-1}g(x,y)}{dx^{n-1}}}\right|_{x=y^{-}}^{x=y^{+}}={\cfrac {1}{a_{n}(y)}}$ .
$g(x,y)$ must satisfy all appropriate homogeneous boundary conditions.
If the Green's function exists then

$u(x)=\int _{a}^{b}g(x,y)~f(y)~{\text{d}}y~.$
Let us consider a second order differentiable linear operator, i.e., $n=2$ on
$[a,b]$ with separated boundary conditions,

${\begin{aligned}\alpha _{1}~u(a)+\beta _{1}~u'(a)&=0\qquad \implies {\mathcal {B}}_{1}[u(a)]=0\\\alpha _{2}~u(b)+\beta _{2}~u'(b)&=0\qquad \implies {\mathcal {B}}_{2}[u(b)]=0\end{aligned}}$ where ${\mathcal {B}}_{1}$ and ${\mathcal {B}}_{2}$ are boundary operators .

The differential equation

${\mathcal {L}}_{x}g(x,y)=0$ and its required continuity conditions are satisfied is

$g(x,y)={\begin{cases}A_{1}~u_{1}(x)~u_{2}(y)&a\leq y<x\\A_{2}~u_{2}(x)~u_{1}(y)&y<x\leq b\end{cases}}$ The first two terms above are to satisfy the boundary conditions while the
third terms gives us continuity.

The jump condition is that

$\left.{\cfrac {dg(x,y)}{dx}}\right|_{x=y^{-}}^{x=y^{+}}={\cfrac {1}{a_{2}(y)}}=A[u_{2}'(y)~u_{1}(y)-u_{1}'(y)~u_{2}(y)]=A~W(y)$ where $W(y)$ is the Wronskian .

Therefore,

$A={\cfrac {1}{a_{2}(y)~W(y)}},~~~W(y)\neq 0$ For the heat equation

${\mathcal {L}}u=-u''$ we can take

$u_{1}(x)=x,~~u_{2}(x)=(1-x),~~{\text{and}}~~W(y)=-1$ That gives us $A=1$ and we recover the same $g(x,y)$ as before.

In the general case, the solution is

$u(x)=\int _{a}^{x}{\cfrac {u_{2}(x)~u_{1}(y)~f(y)}{a_{2}(y)~W(y)}}~{\text{d}}y+\int _{x}^{b}{\cfrac {u_{1}(x)~u_{2}(y)~f(y)}{a_{2}(y)~W(y)}}~{\text{d}}y$ Template:Lectures