Past exam of the applied partial differential equations course of the University of Oxford / 2023 / 1

The supplied positive free-space kernel satisfies $-\Delta H=\delta$, so use the same sign convention for the Dirichlet Green function. For an interior source $Q=(\xi ,\eta ,\zeta )$, requireAway from $Q$, the function is harmonic, and $G-H$ is regular near $Q$. The method of images reflects the source in each coordinate plane, reversing its sign on each reflection. ThereforeOn either face, the terms cancel in pairs. Only the original source lies in the domain, so the other three terms are harmonic there; the required singularity and decay also follow. This is the Dirichlet Green function for a three-dimensional quarter-space. Using $\Delta G=\delta$ instead would change the sign of the kernel and the representation formula together, as in sign conventions for a Dirichlet Green function.

For the preceding sign convention, Green's second identity gives the harmonic representationprovided the far-boundary term vanishes or the bounded-solution representation applies. The face $x=0$ contributes nothing because $g_1=0$. On $\zeta =0$, the outward normal is $-e_{\zeta }$, so the remaining kernel is $-{\partial }_{n_Q}G={\partial }_{\zeta }G$. Differentiating the image sum givesThus the requested constants areThe Poisson kernel of a three-dimensional quarter-space is the difference of the upper-half-space kernel and its reflection in $x=0$. For bounded, suitably continuous data it produces the bounded harmonic solution, taking the prescribed traces away from the edge. Continuous data at the common edge must be compatible. Without any growth restriction at infinity, boundary values alone do not give uniqueness: a multiple of the harmonic function $xz$ can be added while leaving both face values zero.

Introduce the Euler differential operator $D=x{\partial }_x$. The fourth-order spatial expression is $D^{4}u$. The moving boundary has similarity coordinate $\xi =1$, so its two conditions giveHence $\alpha =0$ and $v^{\prime }(1)=1$. With $D_{\xi }=\xi {\partial }_{\xi }$, substitution in the equation yieldsThis is a Cauchy-Euler equation; putting $r=\log \xi$ turns it into the constant-coefficient equation $V^{(4)}-\beta V^{\prime }=0$. Its characteristic polynomial is $n(n^3-\beta )$.

If $\beta >0$, write $a={\beta }^{1/3}$. The roots are $0,a,-a/2\pm i\sqrt{3}a/2$. Decay excludes the constant and growing modes, leavingThe boundary conditions force $A=0$ and $B=2/(\sqrt{3}a)$. If $\beta <0$, writing $\beta =-a^3$ leaves only one decaying mode, $C{\xi }^{-a}$; $v(1)=0$ then forces the zero function, which fails the derivative condition. If $\beta =0$, the solutions are polynomials of degree at most three in $\log \xi$, and none can both decay and satisfy $v^{\prime }(1)=1$.

Consequently all real self-similar solutions are the one-parameter familyThe decaying profile for a fourth-order dilation equation oscillates in logarithmic distance. No positivity condition on $u$ was imposed, so these sign changes are permitted.