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My question is about part of a proof for the jump condition for a Green's function. First let $$LG(x,x')=-\{\frac{d^2}{d x^2}+q(x)\frac{d}{dx}+r(x) \}G(x,x')=\delta(x-x')$$ where $G$ is a green's function. We integrate both sides of this equation with respect to $x$ from $x'-\eta_1$ to $x'+\eta_2$ and then let $\eta_1\to0,\eta_2\to0$ which is written as $\int^{x'+}_{x'-}$, that is \begin{align} \int^{x'+}_{x'-} \{\frac{d^2}{d x^2}+q(x)\frac{d}{dx}+r(x) \}G(x,x') dx &= \int^{x'+}_{x'-}\frac{d^2G}{d x^2} dx\\ &=\frac{d G(x,x'^{+})}{dx}-\frac{d G(x,x'^{-})}{dx}\\ &=-\int^{x'+}_{x'-} \delta(x-x')dx=-1 \end{align}

From this we get the jump condition $\frac{d G(x,x'^{+})}{dx}-\frac{d G(x,x'^{-})}{dx}=-1$.

From the above my question is why do the terms $\{ q(x)\frac{d}{dx}+r(x) \}G(x,x') $ disappear in the integral.

AzJ
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1 Answers1

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Hint:

Note that $G(x,x^\prime)$ is continuous in both variables, and $q(x),r(x)$ are (usually) continuous in $x$. So, $r(x)G(x,x^\prime)$ is continuous as well, thus the anti-derivative $A_1(x,x^\prime)$ will also be continuous. Similarly, though more difficult to show, $q(x)\frac{dG(x,x^\prime)}{dx}$ has a continuous anti-derivative $A_2(x,x^\prime)$. If you can show this, then it is clear that

$\int_{x^{\prime -}}^{x^{\prime +}}\left[q(x)\frac{dG(x,x^\prime)}{dx} + r(x)G(x,x^\prime)\right]dx = \left[ A_2(x^{\prime+},x^\prime)-A_2(x^{\prime-},x^\prime) \right] + \left[ A_1(x^{\prime+},x^\prime)-A_1(x^{\prime-},x^\prime) \right] = 0 + 0 = 0$

Alex Jones
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