# Applied linear operators and spectral methods/Greens functions 2

## Green's functions for linear differential operators

Consider the one-dimensional heat equation given by

${\displaystyle -{\cfrac {d^{2}u}{dx^{2}}}=f(x)\qquad x\in (0,1),~~u(0)=0,u(1)=0~.}$

This equation has Green's function ${\displaystyle g(x,y)}$ which satisfies

${\displaystyle -{\frac {\partial ^{2}g}{\partial x^{2}}}=\delta (x-y)\qquad x,y\in (0,1)}$

The Green's function is given by

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

or,

${\displaystyle g(x,y)=(1-y)~x+(y-x)~H(x-y)~.}$

Therefore,

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

Now,

${\displaystyle \left\langle x~\delta (x),\phi \right\rangle =\left\langle \delta (x),x~\phi \right\rangle =(x~\phi )(0)=0}$

Hence,

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

And,

${\displaystyle g_{xx}(x,y)=-\delta (x-y)}$

Note that the second derivative of ${\displaystyle g}$ 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 ${\displaystyle n}$th order linear differential operator we would want the ${\displaystyle n}$th derivative of ${\displaystyle g(x,y)}$ to be like a delta function. Thus the ${\displaystyle (n-1)}$th derivative of ${\displaystyle g(x,y)}$ should be like a Heaviside function and all lower derivatives should be continuous.

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

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

with

${\displaystyle {\mathcal {L}}=a_{n}(x)~{\cfrac {d^{n}}{dx^{n}}}+a_{n-1}(x)~{\cfrac {d^{(n-1)}}{dx^{(n-1)}}}+\dots a_{0}(x)}$

where ${\displaystyle a_{i}(x)\in C^{\infty }}$. If we integrate this equation across the point ${\displaystyle x=y}$ from ${\displaystyle x=y^{-1}}$ to ${\displaystyle x=y^{+}}$ we get

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

This condition is called a Jump condition.

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

${\displaystyle {\mathcal {L}}_{x}g(x,y)=\delta (x-y)}$

in the sense of distributions has the properties that

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

If the Green's function exists then

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

### Example

Let us consider a second order differentiable linear operator, i.e., ${\displaystyle n=2}$ on ${\displaystyle [a,b]}$ with separated boundary conditions,

{\displaystyle {\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 ${\displaystyle {\mathcal {B}}_{1}}$ and ${\displaystyle {\mathcal {B}}_{2}}$ are boundary operators.

The differential equation

${\displaystyle {\mathcal {L}}_{x}g(x,y)=0}$

and its required continuity conditions are satisfied is

${\displaystyle 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

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

where ${\displaystyle W(y)}$ is the Wronskian.

Therefore,

${\displaystyle A={\cfrac {1}{a_{2}(y)~W(y)}},~~~W(y)\neq 0}$

For the heat equation

${\displaystyle {\mathcal {L}}u=-u''}$

we can take

${\displaystyle u_{1}(x)=x,~~u_{2}(x)=(1-x),~~{\text{and}}~~W(y)=-1}$

That gives us ${\displaystyle A=1}$ and we recover the same ${\displaystyle g(x,y)}$ as before.

In the general case, the solution is

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