Can someone give me hints/directions how to figure this problem out? Thanks
$\frac{d^2G}{dx^2} = \delta(x-\bar{x})$
$G(0) = G(L) = 0$
Can someone give me hints/directions how to figure this problem out? Thanks
$\frac{d^2G}{dx^2} = \delta(x-\bar{x})$
$G(0) = G(L) = 0$
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It's obvious that $G(x)$ is continuous.
Let $G(\bar x)=g$. We can solve two boundary value problems
$G''(x)=0, G(0)=0, G(\bar x)=g$
and
$G''(x)=0, G(\bar x)=g, G(L)=0$
and then find such $g$ that $G'(\bar x+0)-G'(\bar x-0)=1$.