My problem is to find the Green function $g(r;\tilde{r})$ of: $$ sg - \frac{1}{{{r^2}}}\frac{d}{{dr}}\left( {{r ^2}\frac{{dg}}{{dr}}} \right) = \delta (r - \tilde r){e^{ - s\tilde t }}$$ with $$ g( \pm 1;\tilde{r})=0$$
Below is my solution:
Note that $g'(r)$ is not continuous at $r=\tilde{r}$, so we need to solve the equation in two different domains: $[-1,\tilde{r})$ with $g(-1;\tilde{r})=0$, and $(\tilde{r},1]$ with $g(1;\tilde{r})=0$. Thus we have $$g(r;\tilde{r})=C_1\frac{\sinh\sqrt{s}(1+r)}{r}, r\in [-1,\tilde{r})$$ $$g(r;\tilde{r})=C_2\frac{\sinh\sqrt{s}(1-r)}{r}, r\in (\tilde{r},1]$$ Due to the continuity of $g(r;\tilde{r})$ at $r=\tilde{r}$, we have eq.1: $${C_1}\sinh \sqrt s (1 + \tilde r ) = {C_2}\sinh \sqrt s (1 - \tilde r )$$ Integrate the original differential equation by weight of $r^2$ from $(\tilde r-\epsilon,\tilde r+\epsilon)$ with respect to $r$, and then take $\epsilon\rightarrow0$, we obtain $$\left( {\frac{{dg}}{{dr}}} \right)_{{{\tilde r}^ - }}^{{{\tilde r }^ + }} = - {e^{ - s\tilde t }}$$ So we have eq.2: $${C_1}\cosh \sqrt s (1 + \tilde r ) + {C_2}\cosh \sqrt s (1 - \tilde r ) = \frac{{\tilde r {e^{ - s\tilde t }}}}{{\sqrt s }}$$ By eq.1 and eq.2 we get $${C_1} = \frac{{\tilde r \sinh \sqrt s (1 - \tilde r)}}{{\sqrt s \sinh 2\sqrt s }}{e^{ - s\tilde t }}$$ $${C_2} = \frac{{\tilde r \sinh \sqrt s (1 + \tilde r )}}{{\sqrt s \sinh 2\sqrt s }}{e^{ - s\tilde t }}$$ Finally, we obtain the solution to the problem: $$g(r;\tilde r) = \frac{{\tilde r}}{r}\frac{{\sinh \sqrt s (1 + r)\sinh \sqrt s (1 - \tilde r)}}{{\sqrt s \sinh 2\sqrt s }}{e^{ - s\tilde \tau }}, r<\tilde r$$ $$g(r;\tilde r) = \frac{{\tilde r}}{r}\frac{{\sinh \sqrt s (1 - r)\sinh \sqrt s (1 + \tilde r)}}{{\sqrt s \sinh 2\sqrt s }}{e^{ - s\tilde \tau }},r>\tilde r$$ Rewritten as $$g(r;\tilde r) = \frac{{\tilde r}}{r}\frac{{\cosh \sqrt s (2 - \left| {r - \tilde r} \right|) - \cosh\sqrt s (r + \tilde r)}}{{2\sqrt s \sinh 2\sqrt s }}{e^{ - s\tilde \tau }}$$ My question is, the Green function must be spatial symmetric, that means $g(r;\tilde r)=g(\tilde r;r)$, but my solution doesn't match that feature, so where am I wrong?
I finally know where I was wrong. The Green function for the differential equation is spatial symmetric only with weight of $r^2$, that is $r^2g(r;\tilde r)=\tilde r^2g(\tilde r,r)$.