Past exam of the differential equations 2 course of the University of Oxford / 2014 / 2 / b

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.