# Applied linear operators and spectral methods/Greens functions 2

## Green's functions for linear differential operators

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

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

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

If the Green's function exists then

$u(x)=\int _{a}^{b}g(x,y)~f(y)~{\text{d}}y~.$

### Example

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

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

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