# Thread: [SOLVED] Solve initial value problem with Green's function

1. ## [SOLVED] Solve initial value problem with Green's function

I need to solve the following intial value problem using Green's Function:

(Ay)(x) = h(x), Ay:=d^2y/dx^2 + dy/dx, y(0) = y'(0) = 0

where h is defined as the following:

h(x) := { 0, when 0<=x<1
{ 1, when x >= 1

Can someone help, im not really sure where to start

2. Firstly, I must admit I cannot see why on earth someone would want to solve anything (...other than a two-point BVP...) by using the Green's function method.

Ok, so we are looking for a symmetric function $G(x,s)$ to satisfy

$\int_0^2G(x,s)h(s)ds=\begin{cases} 0, \ x<1 \\
y(x), \ x\geq 1\end{cases}$

where I took the liberty of setting [0,2] as the domain, but this is inessential as we can modify the Green's function and get the domain as large as we like.

For $x<1$, we have $h(x)=0$, so $y(x)=0$ solves the IVP. and we have the liberty to define G any way we like.
For $x>1$, the IVP $y''+y'=1, \ y(0)=y'(0)=0$ has the solution $y(x)={\rm e}^{-x}+x-1$. So we demand

${\rm e}^{-x}+x-1=\int_1^2G(x,s)ds$ (1)

Since G has to be symmetric, write ${\rm e}^{-x}+x-1=\int_1^2{\rm e}^{-x}\phi(s)+(x-1)\psi(s)ds$

and observe that (1) is satisfied for $\phi(s)=\frac{{\rm e}^{-s}}{{\rm e}^{-1}-{\rm e}^{-2}}, \ \psi(s)=2(s-1)$. Define G this way on $0\leq x\leq 2, \ 0\leq s\leq 2$. So we get the solution of the IVP as

$y(x)=\int_0^2\left[\frac{{\rm e}^{-s-x}}{{\rm e}^{-1}-{\rm e}^{-2}} + 2(x-1)(s-1)\right]h(s) ds$.

### green function IVP

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