Greens Function

Green’s Function #

L(y)=f(x)

A function G(x,t) defined for $a\le x\le t$, $t\le x \le b$ is called green’s function of L(y) if it satisfies:

1. G(x,t) has derivatives upto order n.
   1. $$ \frac{\partial ^{k}G}{\partial x^{k}}|_{t^{-}}^{t^{+}}=0,k=0,1,2,…n-2 $$
   2. $$\frac{\partial ^{k}G}{\partial x^{k}}|_{t^{-}}^{t^{+}}= 1/c_0(t) ,k=n-1$$
2. G(x,t) considered as a function of x, is a solution of L(y)=0. L(G)=0 on each of the intervals. It satisfies linear homogeneous boundary conditions

Theorem of unique green function #

If y(x)=0 is the only solution of the BVP $L(y)=0,V_{k}(y)=0$ Then, L(y) has one and only one Green function

Let the basis solutions be $y_{1}…y_{n}$

$$ \begin{align*} G(x,t)&= a_{1}y_{1}+a_{2}y_{2}+…a_{n}y_{n},a<x<t\\ G(x,t)&= b_{1}y_{1}+b_{2}y_{2}+…b_{n}y_{n},t<x<b\\ \end{align*} $$

The derivative conditions are used to create n equations with n unknowns $p_{k}=b_{k}-a_{k}$. Find $p_{k}$ using cramer’s rule

Applying boundary conditions $$ \begin{align*} V_{k}(y)&= A_{k}(y)+B_{k}(y)\\ V_{k}(G)&= 0\\ (a_{1}A_{k}(y_{1})+a_{2}A_{k}(y_{2})+…+a_{n}A_{k}(y_{n})) +(b_{1}B_{k}(y_{1})+b_{2}B_{k}(y_{2})+…+b_{n}B_{k}(y_{n}))&= 0 \end{align*} $$ Substitute $a_{k}=b_{k}-p_{k}$ and solve n equations for $b_{k}$ using cramer’s rule

Therefore, we have the unique Green’s function G(x,t) of L(y).

Construction of Green’s Function for second order equations #

$$ \begin{align*} L(y)&= c_{0}(x) y’’+c_{1}(x) y’+c_{2}(x)y=0\\ G(x,t)&= \begin{cases}G_{1}(x,t),a\le x<t\G_{2}(x,t),t<x\le b\end{cases}\\ G_{2}|{x=t^{+}}-G{2}|{x=t^{-}}&= 0\\ \frac{\partial G{2}}{\partial x}|{x=t^{+}}-\frac{\partial G{1}}{\partial x}|{x=t^{-}}&= \frac{1}{c{0}(t)} \end{align*} $$

Let $y_{1},y_{2}$ be two linearly independent solution of 2nd order DE

Let $G_{1}(x,t)= a_{1}y_{1}(x),G_{2}(x,t)=a_{2}y_{2}(x)$

$$ \begin{align*} a_{2}y_{2}(t)-a_{1}y_{1}(t)&= 0\\ a_{2} \frac{dy_{2}}{dt}- a_{1} \frac{dy_{1}}{dt}&= \frac{1}{c_{0}(t)}\\ a_{1},a_{2}\text{ satisfying this would be}\\ a_{1}&= \frac{-y_{2}(t)}{c_{0}(t)W(t)}\\ a_{2}&= \frac{-y_{1}(t)}{c_{0}(t)W(t)}\\ \end{align*} $$

where W is the Wronskian of (y1,y2)

Therefore, G1, G2 have been found and therefore the green function.

Could also be given by

$$ G(x,t)= \frac{1}{c_{0}(t)W(t)}\begin{vmatrix}y_{1}(t) & y_{2}(t)\y_{1}(x) & y_{2}(x)\end{vmatrix} $$

In general, the first n-1 rows will contain upto (n-2) derivatives of $y_{k}(t)$

Theorem of particular solution #

General solution of $L(y)=f(x)$

$$ \begin{align*} y(x)=a_{1}y_{1}(x)+a_{2}y_{2}(x)+…+a_{n}y_{n}(x)+\int_{x_{0}}^{x}G(x,t)f(t)dt \end{align*} $$