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I'm not sure if this question is for physics forum, but my book's title is "Green's Functions in Quantum Physics", so I ask here.

The book says that the Green's function defined as

$$ (z-L( \mathbf{r}))G(\mathbf{r},\mathbf{r'};z)=\delta(\mathbf{r-r'}) $$ solves the inhomogeneous differential equation $$ (z-L( \mathbf{r}))u(\mathbf{r})=f(\mathbf{r}) $$ with the "same boundary condition" and it gives the solution $$ u(\mathbf{r})=\int d\mathbf{r'}G(\mathbf{r},\mathbf{r'};z)f(\mathbf{r'}). $$

I think that the third equation is just from by multiplying $f{(\mathbf{r})}$ to first equation and integrating it.

But then, why do we need the statement "with same boundary condition"? I don't see where to use that condition. My book does not prove this and wikipedia link Green function also states this, but no proof. Is there any simple explanation or proof for this?

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  • $\begingroup$ Better on Maths SE $\endgroup$ – Rob Jeffries Jan 26 '15 at 2:22
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Suppose for simplicity that $u(\boldsymbol{r})$ satisfies the boundary condition $u(\boldsymbol{r}_0)=0$ for $\boldsymbol{r}_0$ in the boundary then the integral on the right hand side of your last equation should satisfy

\begin{equation} \int{d\boldsymbol{r}'G(\boldsymbol{r}_0,\boldsymbol{r}';z)f(\boldsymbol{r}')}=0 \end{equation}

and then $G(\boldsymbol{r}_0,\boldsymbol{r}';z)=0$ it is, $G$ satisfies the same boundary condition as $u$. The same is true for Neumman boundary conditions.

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