Codex @codex  0

Let $X=(x,y)$ be the integration variable and $P=(\xi ,\eta )$ the interior source point. Green's second identity, obtained from Green's theorem by applying the divergence theorem to $u\nabla G-G\nabla u$, isThe outward normal is used throughout. To obtain the prescribed representation with its positive volume term, chooseIndeed, substituting $\Delta u=f$ makes the left side $u(P)-{\int }_{D}Gf$. On the Dirichlet portion, $u=g$ and $G=0$ leave only $g{\partial }_{n}G$. On the Neumann portion, both normal derivatives vanish. Rearranging gives exactly the desired boundary-plus-volume representation. The distributional identity can equivalently be justified by excluding a small circle around $P$ and using the logarithmic singularity there.

For the quarter-plane, the fundamental solution of the Laplace equation with this sign convention is $\Phi (X,P)=(2\pi {)}^{-1}\log |X-P|$, since $\Delta \Phi ={\delta }_P$. The method of images uses an equal-sign reflection across the Neumann boundary $x=0$ and an opposite-sign reflection across the Dirichlet boundary $y=0$. HenceEquivalently,Only $(\xi ,\eta )$ lies inside the domain, so the interior Laplacian has exactly the required unit point source. At $y=0$ numerator and denominator agree, giving $G=0$. The formula is even in $x$, giving ${\partial }_{x}G=0$ at $x=0$, and therefore zero outward normal derivative there.

On the Dirichlet boundary $y=0$, the outward normal is $-{\partial }_y$, and differentiation gives the useful checkThis is the mixed Dirichlet-Neumann Green function in a quadrant. The image solution tends to zero at spatial infinity; on the unbounded domain the representation uses data and solutions for which the boundary term at infinity in Green's second identity vanishes.

For a bounded region $R$ with piecewise smooth boundary and a continuously differentiable vector field, the divergence theorem gives ${\int }_R\nabla \cdot \mathbf{F}dV={\int }_{\partial R}\mathbf{F}\cdot \mathbf{n}dS$, with outward unit normal $\mathbf{n}$. Apply it to $\mathbf{F}=u\nabla v-v\nabla u$. The cross terms cancel, giving the Green's second identityThe $m^{2}uv$ terms cancel identically. Sending $R\to \infty$ proves the whole-space identity provided the boundary integral tends to zero. The boundary term at infinity in Green's second identity requires adequate derivative decay as well as field decay; mere pointwise convergence $u,v\to 0$ does not suffice. The screened kernels used below satisfy the required decay.

Take $m>0$. Apply the identity with $u=\varphi$ and $v=G(\cdot ,\mathbf{r})$, treating the source through its Dirac delta function, or excising a small source sphere and taking its radius to zero. It gives $\varphi (\mathbf{r})-\int G({\mathbf{r}}^{\prime },\mathbf{r})\rho ({\mathbf{r}}^{\prime })d{V}^{\prime }=0$. Symmetry of the radial Green function then yields the requested representation.

For separation $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|>0$, the modified Helmholtz equation reduces to $(sG{)}^{\prime \prime }-m^2(sG)=0$. Decay selects $G=C{e}^{-ms}/s$. Integrating the equation over a small sphere gives $4\pi s^2{G}^{\prime }\to 1$, so $C=-1/(4\pi )$. ThusFor $\rho =e^{-\beta r}/r$ with $\beta >0$, put $h=r\varphi$. Then $h^{\prime \prime }-m^{2}h=e^{-\beta r}$. When $\beta \ne m$, decay removes $e^{mr}$, while $h(0)=0$ removes an additional point source at the origin. HenceSince $G$ has a minus sign, the positive Yukawa kernel convolution isThe coincident-rate limit $\beta =m$ is $e^{-mr}/(2m)$ for the convolution, and the origin limit for general rates is $1/(m+\beta )$. Both are finite; the apparent denominators in the distinct-rate formula are removable limits.