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Math Help - Green's function

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by morganfor View Post
    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?
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