1. ## Green's function

green's functions are new to me so i'm struggling with these

how does one use green's method to solve:

-(K(x)u')'=f(x), 0<x<1; u(0)=u(1)=0; K(x)>0

thank you

2. Originally Posted by morganfor
green's functions are new to me so i'm struggling with these

how does one use green's method to solve:

-(K(x)u')'=f(x), 0<x<1; u(0)=u(1)=0; K(x)>0

thank you
To find a green's function first solve the homogenous problem

$\frac{d}{dx}\left(k(x)u' \right)=0 \iff k(x)u'=c \iff u(x)=c\int_{0}^{x}\frac{dt}{k(t)}+d$

Now using this we need to find functions that satisfy the boundary conditions.

$u_0(x)=\int_{0}^{x}\frac{dt}{k(t)} \implies u_0(0)=0$

and

$u_1(x)=\int_{0}^{x}\frac{dt}{k(t)}-\int_{0}^{1}\frac{dt}{k(t)} \implies u_1(1)=0$

Now we take the Wronskian to get times k(x)

$k(x)W(x)=k(x)\begin{vmatrix} \int_{0}^{x}\frac{dt}{k(t)} & \int_{0}^{x}\frac{dt}{k(t)}-\int_{0}^{1}\frac{dt}{k(t)} \\ \frac{1}{k(x)} & \frac{1}{k(x)} \end{vmatrix}=\int_{0}^{1}\frac{dt}{k(t)}$

So our Greens function is

$g(x,s)=\begin{cases} -\frac{\int_{0}^{x}\frac{dt}{k(t)}-\int_{0}^{1}\frac{dt}{k(t)}}{\int_{0}^{1}\frac{dt} {k(t)}} \left( \int_{0}^{s}\frac{dt}{k(t)} \right), \quad 0 < s < x \\ -\frac{\int_{0}^{x}\frac{dt}{k(t)}}{\int_{0}^{1} \frac{dt}{k(t)} }\left(\int_{0}^{s}\frac{dt}{k(t)}-\int_{0}^{1}\frac{dt}{k(t)}\right) , \quad 0 < x < s\end{cases}$

Can you finish from here?