03. Green's Functions

3.1 Properties of Green’s Function

Green's Function

Let $y_1$ and $y_2$ be linearly independent solutions to the Homogenous ODE $Ly=0$ where

$$Ly\equiv P_2{y}^{\prime \prime }+P_1{y}^{\prime }+P_{0}y=fa<x<b$$

with $y_1$ and $y_2$ satisfying one boundary condition each ($y_1(a)=0=y_2(b)$)

Let Green’s Function $g(x,\xi )$ be defined as

$$g(x,\xi )=\{\begin{array}{ll}\frac{y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}& a<\xi <x<b\\ \frac{y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}& a<x<\xi <b\end{array}$$

Note that it provides a kind of inverse to $Ly=f$

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Properties of Green's Function

Note that these are viewing $g(x,\xi )$ as a function of $x$

1. $g(x,\xi )$ satisfies the homogeneous ODE $(H)$ everywhere other than special point $x=\xi$ i.e.

$$L_{x}g=P_2(x)g_{xx}+P_1(x)g_x+P_0(x)g=0$$

in $a<x<\xi <b$ and in $a<\xi <x<b$
Note that the subscript $x$ indicates derivative are respect to $x$ and not $\xi$

2. $g(x,\xi )$ satisfies the same boundary conditions as $y$ i.e.

$$g(a,\xi )=g(b,\xi )=0$$

3. $g(x,\xi )$ is continuous at $x=\xi$ i.e.

$$\underset{x\to {\xi }^{+}}{\lim }g(x,\xi )=\underset{x\to {\xi }^{-}}{\lim }g(x,\xi )$$

However the first derivative of $g$ is discontinuous with a jump given by

$$\underset{x\to {\xi }^{+}}{\lim }{g}_x(x,\xi )-\underset{x\to {\xi }^{-}}{\lim }{g}_x(x,\xi )=\frac{1}{P_2(\xi )}$$

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3.2 Reverse-Engineering $g$

Finding Delta Function for $g$ from ODEs

Let

$$Ly=P_2{y}^{\prime \prime }+P_1{y}^{\prime }+P_{0}y=f\text{ for }a<x<b$$

Suppose $y(x)$ can be expressed by

$$y(x)={\int }_a^{b}g(x,\xi )f(\xi )d\xi$$

with boundary conditions

$$y(a)=0,y(b)=0$$

Using Boundary conditions then

$${\int }_a^{b}g(a,\xi )f(\xi )d\xi =0={\int }_a^{b}g(b,\xi )f(\xi )d\xi$$

for all functions $f(\xi )$

Suppose that $x$-derivative may be passed through integral sign so

$$L[y]=L\left[{\int }_a^{b}g(x,\xi )f(\xi )d\xi \right]={\int }_a^b{L}_x[g(x,\xi )]f(\xi )d\xi =f(x)$$

where $L$-subscript $x$ indicates derivative are respect to $x$ and not $\xi$

Thus

$$L_x[g(x,\xi )]=\delta (x-\xi )$$

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3.3 The Delta Function

3.3.1 Definition

Dirac Delta

Dirac Delta function $\delta (x)$ characterised by the following properties

$$\begin{array}{rl}\delta (x)& =0\text{ for all }x\ne 0\\ {\int }_{-\infty }^{\infty }\delta (x)dx& =1\end{array}$$

Note it is also known as Delta Function

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Sifting Property of Dirac Delta

For any function $\varphi (\xi )$

$${\int }_{-\infty }^{\infty }\delta (x-\xi )\varphi (\xi )d\xi =\varphi (x)$$

For $x\in (a,b)$ then

$${\int }_a^b\delta (x-\xi )\varphi (\xi )d\xi =\varphi (x)$$

Proof

$${\int }_a^b\delta (x-\xi )\varphi (\xi )d\xi ={\int }_{x-\xi }^{x+\xi }\delta (x-\xi )\varphi (\xi )d\xi$$

where $ϵ$ is an arbitrarily small positive parameter

So for smooth $\varphi (x)$ then

$${\int }_a^b\delta (x-\xi )\varphi (\xi )d\xi \sim [\varphi (x)+O(ϵ)]{\int }_{x-ϵ}^{x+ϵ}\delta (x-\xi )d\xi$$

and then as $ϵ\to 0$ we get equality to $\varphi (x)$

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3.3.2 Approximating the Delta Function

Approximations for Delta Function

$\delta$ can be approximated by
Sequence of increasingly narrow functions with normalised area ${\delta }_n(x)$ where

$$\begin{array}{rl}{\int }_{-\infty }^{\infty }{\delta }_n(x)dx& =1\text{ for all }n=1,2,\cdots \\ \underset{n\to \infty }{\lim }{\delta }_n(x)& =0\text{ for all }x\ne 0\end{array}$$

One example is the hat function defined by

$${\delta }_n(x)=\{\begin{array}{ll}0& \text{ for }|x|>\frac{1}{n}\\ \frac{n}{2}& \text{ for }|x|\le \frac{1}{n}\end{array}$$

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3.3.3 Properties of Delta Function

Sifting Property from Approximations for Delta Function

Let $\varphi (x)$ be a smooth function
Let $\Phi (x)=\int \varphi (x)dx$ be the antiderivative

Using the hat approximating sequence then

$${\int }_{-\infty }^{\infty }{\delta }_n(x-a)\varphi (x)dx\to \varphi (a)\text{ as }n\to \infty$$

with

$${\int }_{-\infty }^{\infty }\delta (x-a)\varphi (x)dx=\varphi (a)$$

Proof

$$\begin{array}{rl}{\int }_{-\infty }^{\infty }{\partial }_n(x-a)\varphi (x)dx& ={\int }_{a-\frac{1}{n}}^{a+\frac{1}{n}}\left(\frac{n}{2}\right)\varphi (x)dx\\ & =\frac{n}{2}\left[\Phi \left(a+\frac{1}{n}\right)-\Phi \left(a-\frac{1}{n}\right)\right]\\ & =\frac{\left[\Phi \left(a+\frac{1}{n}\right)-\Phi \left(a-\frac{1}{n}\right)\right]}{\frac{2}{n}}\end{array}$$

Hence as $n\to \infty$

$${\int }_{-\infty }^{\infty }{\delta }_n(x-a)\varphi (x)dx\to {\Phi }^{\prime }(a)=\varphi (a)$$

Hence $\delta$ has the desired sifting property

$${\int }_{-\infty }^{\infty }\delta (x-a)\varphi (x)dx\equiv \varphi (a)$$

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Heaviside Function

Antiderivative of delta function is the Heaviside Function

$${\int }_{-\infty }^x\delta (s)ds=H(x):=\{\begin{array}{ll}0& x<0\\ 1& x>0\end{array}$$

Note that value of $H(x)$ at $x=0$ is indeterminate, sometimes taken to be $1$ and sometimes $\frac{1}{2}$

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3.4 Green’s Function via Delta Function

Finding Green's Function using Delta Function for ODEs

Consider

$$Ly\equiv P_2(x)y_{xx}(x)+P_1(x)y_x(x)+P_{0}y(x)=f(x)$$

with boundary conditions

$$y(a)=0,y(b)=0$$

Then

$$L_{x}g(x,\xi )=P_2(x)g_{xx}(x,\xi )+P_1(x)g_x(x,\xi )+P_{0}g(x,\xi )=\delta (x-\xi )$$

where this comes from Reverse Engineering Green’s Function

with

$$[g(x,\xi ){]}_{x={\xi }_{-}}^{x={\xi }_{+}}=0\text{ and }[g_x(x,\xi ){]}_{x={\xi }_{-}}^{x={\xi }_{+}}=\frac{1}{P_2(\xi )}$$

Proof - Lazy

Integrate $L_{x}g(x,\xi )$ from ${\xi }^{-}$ to ${\xi }^{+}$ taking into account that

$g$ is continuous (to prevent higher order singularities) so

$${\int }_{{\xi }^{-}}^{{\xi }^{+}}g(x)dx=0$$

(NOTE TO SELF) Maybe go through this another time

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3.5 Generalisation to Higher-Order Equations

Finding Green's Function for Higher-Order Equations

Let $L$ be a general linear differential operator of order $n\in N$ so

$$\begin{array}{rl}Ly(x)& \equiv P_n(x)y^{(n)}(x)+P_{n-1}{y}^{(n-1)}(x)+\cdots +P_1(x)y^{\prime }(x)+P_0(x)y(x)\\ & \equiv P_n(x)\frac{d^{n}y}{d{x}^n}+P_{n-1}(x)\frac{d^{n-1}y}{d{x}^{n-1}}+\cdots +P_1(x)\frac{dy}{dx}+P_0(x)y(x)\end{array}$$

for some coefficients $P_0,\cdots ,P_n$

Assume that all $P_i$ are continuous and coefficient $P_n$ of the highest derivative is nonzero

Define homogenous and inhomogeneous linear ODEs of order $n$ by

$$\begin{array}{rlrl}Ly& =0& & (H)\\ Ly& =f\not\equiv 0& & (N)\end{array}$$

with $n$ boundary conditions (linear combination of $y,\frac{dy}{dx},\cdots ,\frac{d^{n-1}y}{d{x}^{n-1}}$) at $x=a$ and $x=b$

$$B_{i}y{\mid }_{x=a,b}={\gamma }_i,i=1,2,\cdots ,n(BCN)$$

where ${\gamma }_i$ are constants and $B_i$ is in the form

$$B_{i}y=\sum _{j=1}^n({\alpha }_{ij}{y}^{(j-1)}(a)+{\beta }_{ij}{y}^{(j-1)}(b)b)$$

for some constants ${\alpha }_{ij},b_{ij}$

$(BCN)$ are homogeneous if ${\gamma }_i=0$ for all $i$ so

$$B_{i}y{\mid }_{x=a,b}=0i=1,2,\cdots ,n,(BCH)$$

Assume homogenous problem $(H+BCH)$ has non-trivial solutions then
By FAT, then expect inhomogeneous problem to have a unique solution

Let $u$ be the solution of problem $(H+BCN)$ thus

$$\begin{array}{rlrl}Lu(x)& =0& & a<x<b\\ B_{i}u{\mid }_{x=a,b}& ={\gamma }_i& & i=1,2,\cdots ,n\end{array}$$

Defining $\tilde{y}=y-u$
Thus $\tilde{y}$ satisfies inhomogeneous ODE $(N)$ but with homogenous boundary conditions $(BCH)$

Focusing on $(N+BCH)$ then

$$Ly(x)=\sum _{j=1}^n{P}_i(x)y^{(j-1)}(x)=f(x)\text{ for }a<x<b$$

with $n$ linearly independent homogeneous boundary conditions

$$B_{i}y{\mid }_{x=a,b}=\sum _{j=1}^n({\alpha }_{ij}{y}^{(j-1)}(a)+{\beta }_{ij}{y}^{(j-1)}(b))i=1,2,\cdots ,n$$

Then corresponding problem for Green’s Function $g$ is

$$L_{x}g(x,\xi )=\delta (x-\xi )\text{ for }a<x,\xi <b$$

with boundary conditions

$$B_{i}g(x,\xi ){\mid }_{x=a,b}=0,\text{ for }i=1,2,\cdots ,n$$

As $L$ is of order $n$ then there are $2n$ degrees of freedom and hence $n$ arbitrary integration constants
After applying $n$ independent boundary conditions then there are $n$ jump conditions at $x=\xi$
Integrating across $x=\xi$ then

$${\int }_{{\xi }^{-}}^{{\xi }^{+}}\left[P_n(x)\frac{{\partial }^n}{\partial x^n}g(x,\xi )+\cdots +P_0(x)g(x,\xi )\right]dx={\int }_{{\xi }^{-}}^{{\xi }^{+}}\delta (x-\xi )=1dx$$

By integration by parts on the first term then

$$\begin{array}{r}{\int }_{{\xi }^{-}}^{{\xi }^{+}}\left[(P_{n-1}(x)-P_n^{\prime }(x))\frac{{\partial }^{(n-1)}}{\partial x^{(n-1)}}g(x,\xi )+\cdots +P_0(x)g(x,\xi )\right]dx\\ +{\left[P_n(x)\frac{{\partial }^{(n-1)}}{\partial x^{(n-1)}}g(x,\xi )\right]}_{x={\xi }^{-}}^{x={\xi }^{+}}=1\end{array}$$

Equation is balanced by setting jump condition on $(n-1)$th derivative

$${\left[\frac{{\partial }^{(n-1)}}{\partial x^{(n-1)}}g(x,\xi )\right]}_{x={\xi }^{-}}^{x={\xi }^{+}}=\frac{1}{P_n(\xi )}$$

and ensuring all lower derivatives are continuous across $x=\xi$

$$[g(x,\xi ){]}_{x={\xi }^{-}}^{x={\xi }^{+}}=[g_x(x,\xi ){]}_{x={\xi }^{-}}^{x={\xi }^{+}}=\cdots ={\left[\frac{{\partial }^{(n-2)}}{\partial x^{(n-2)}}g(x,\xi )\right]}_{x={\xi }^{-}}^{x={\xi }^{+}}=0$$

Hence Green’s Function can be determined where the solution to BVP is given by

$$y(x)={\int }_a^{b}g(x,\xi )f(\xi )d\xi$$

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3.6 Green’s Function in terms of Adjoint

Finding Green's Function from Adjoint

Suppose

$$Ly(x)=\sum _{j=1}^n{P}_i(x)y^{(j-1)}(x)=f(x)\text{ for }a<x<b$$

with $n$ linearly independent homogeneous boundary conditions

$$B_{i}y{\mid }_{x=a,b}=\sum _{j=1}^n({\alpha }_{ij}{y}^{(j-1)}(a)+{\beta }_{ij}{y}^{(j-1)}(b))i=1,2,\cdots ,n$$

If $g(x,\xi )$ satisfies $L_{x}g(x,\xi )=\delta (x-\xi )$ with homogenous boundary conditions $(BC)$ then

1. $g(\xi ,x)$ satisfies corresponding adjoint equation $L_x^{\ast }g(\xi ,x)=\delta (x,\xi )$ and $(B{C}^{\ast })$

2. If $(L+BC)$ is fully self-adjoint then $g$ is symmetric so $g(x,\xi )\equiv g(\xi ,x)$

Proof priori unknown function $G(x,\xi )$ For $Ly=f$ then with respect to $x$

Consider

$$⟨Ly,G⟩=⟨f(x),G(x,\xi )⟩={\int }_a^{b}G(x,\xi )f(x)dx$$

Using adjoint

$$⟨Ly,G⟩=⟨y,L^{\ast }G⟩={\int }_a^{b}y(x)L_x^{\ast }G(x,\xi )dx$$

where $G$ needs to satisfy adjoint boundary conditions corresponding to $BC$ imposed on $y$

If $G$ satisfies (in order to isolate $y$)

$$L_x^{\ast }G(x,\xi )=\delta (x-\xi )$$

Then

$${\int }_a^{b}y(x)L_x^{\ast }G(x,\xi )dx=y(\xi )={\int }_a^{b}G(x,\xi )f(x)dx$$

Hence

$$y(x)={\int }_a^{b}G(\xi ,x)f(\xi )d\xi ={\int }_a^{b}g(x,\xi )f(\xi )d\xi$$

Thus $G(\xi ,x)\equiv g(x,\xi )$ so $G$ is the transpose of $g$

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