Question

Determine the Green, s function and express the solution as a definite integral

-y^''=f(x), y(0)=0, y'(1)=0

Answer 1

Answer:-

Given IVP:

$$-y''(x)=f(x),\quad y(0)=0,\quad y(1)=0$$

The general solution of homogeneous equation

$$y(x)=c_1+c_2x$$

Hence,

$$G(x,s)=\left\{\begin{matrix} c_1+c_2x,\quad 0\leq x\leq s, \\ c_3+c_4x,\ \quad s< x\leq 1. \end{matrix}\right.$$

Initial conditions give

$$G(0,s)=c_1=0,\quad G(1,s)=c_4=-c_3$$

So

$$G(x,s)=\left\{\begin{matrix} c_2x,\quad 0\leq x\leq s, \\ c_3(1-x),\ \quad s< x\leq 1. \end{matrix}\right.$$

The continuity of Green function gives

$$G(x,s)=\left\{\begin{matrix} x(s-1),\quad 0\leq x\leq s, \\ s(x-1),\ \quad s< x\leq 1. \end{matrix}\right.$$

The solution of DE

$$y(x)=\int_0^1G(x,s)f(s)ds$$ $$=\int_0^xx(s-1)f(s)ds+\int_x^1s(x-1)f(s)ds$$