Applied linear operators and spectral methods/Greens functions 2

Green's functions for linear differential operators

Consider the one-dimensional heat equation given by

$-\frac{d^{2}u}{d{x}^2}=f(x)x\in (0,1),u(0)=0,u(1)=0.$

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

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

The Green's function is given by

$g(x,y)=\{\begin{array}{cc}(1-y)x& 0\le x<y\\ (1-x)y& y<x\le 1\end{array}$

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,

$⟨x\delta (x),\varphi ⟩=⟨\delta (x),x\varphi ⟩=(x\varphi )(0)=0$

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 $ℒ$ acting on $g$ such that

$ℒg=\delta (x-y)$

with

$ℒ=a_n(x)\frac{d^n}{d{x}^n}+a_{n-1}(x)\frac{d^{(n-1)}}{d{x}^{(n-1)}}+\dots a_0(x)$

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

${a_n(x)\frac{d^{n-1}g}{d{x}^{n-1}}|}_{x=y^{-}}^{x=y^{+}}=\underset{y^{-1}}{\overset{y^{+}}{\int }}\delta (x)\text{d}x=1$

This condition is called a Jump condition.

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

${ℒ}_{x}g(x,y)=\delta (x-y)$

in the sense of distributions has the properties that

1. ${ℒ}_{x}g(x,y)=0$ for all $x\ne y$.
2. $\frac{d^{k}g(x,y)}{d{x}^k}$ is continuous at $x=y$ for $k=0,1,\dots ,n-2$.
3. ${\frac{d^{n-1}g(x,y)}{d{x}^{n-1}}|}_{x=y^{-}}^{x=y^{+}}=\frac{1}{a_n(y)}$.
4. $g(x,y)$ must satisfy all appropriate homogeneous boundary conditions.

If the Green's function exists then

$u(x)=\underset{a}{\overset{b}{\int }}g(x,y)f(y)\text{d}y.$

Example

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

$\begin{array}{cc}{\alpha }_{1}u(a)+{\beta }_1{u}^{\prime }(a)& =0⟹{ℬ}_1[u(a)]=0\\ {\alpha }_{2}u(b)+{\beta }_2{u}^{\prime }(b)& =0⟹{ℬ}_2[u(b)]=0\end{array}$

where ${ℬ}_1$ and ${ℬ}_2$ are boundary operators.

The differential equation

${ℒ}_{x}g(x,y)=0$

and its required continuity conditions are satisfied is

$g(x,y)=\{\begin{array}{cc}{A}_1{u}_1(x)u_2(y)& a\le y<x\\ A_2{u}_2(x)u_1(y)& y<x\le b\end{array}$

The first two terms above are to satisfy the boundary conditions while the third terms gives us continuity.

The jump condition is that

${\frac{dg(x,y)}{dx}|}_{x=y^{-}}^{x=y^{+}}=\frac{1}{a_2(y)}=A[{u_2}^{\prime }(y)u_1(y)-{u_1}^{\prime }(y)u_2(y)]=AW(y)$

where $W(y)$ is the Wronskian.

Therefore,

$A=\frac{1}{a_2(y)W(y)},W(y)\ne 0$

For the heat equation

$ℒu=-u^{″}$

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

$u(x)=\underset{a}{\overset{x}{\int }}\frac{u_2(x)u_1(y)f(y)}{a_2(y)W(y)}\text{d}y+\underset{x}{\overset{b}{\int }}\frac{u_1(x)u_2(y)f(y)}{a_2(y)W(y)}\text{d}y$