Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 3

Let $L=d^2/d{x}^2+x^{-1}d/dx-x^{-2}$. A Green's function $G(x,\xi )$ satisfies $L_{x}G=\delta (x-\xi )$ and the chosen homogeneous boundary conditions in $x$. Away from $x=\xi$, it is a linear combination of $x$ and $x^{-1}$. It must be continuous there and obey the unit derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$, because the leading differential coefficient is one. Thensolves $Ly=f$ whenever the integral and boundary limits are valid. The Wronskian of $x,x^{-1}$ is $-2/x$. The Green kernels for a radial dipole Euler equation implement the two different choices of boundary conditions below.

The factor satisfying the left condition is $x$, while $x^{-1}$ satisfies decay at infinity. Matching the value and unit derivative jump givesFor the specified compactly supported forcing, integrating over $0<\xi <1$ givesFor example, on $0<x<1$, the two pieces contribute $-x^2/6$ and $-x(1-x)/2$. Both $y_1$ and $y_1^{\prime }$ match at one, and the function tends to zero at both ends. A homogeneous difference $ax+b/x$ with those two limits must vanish, establishing uniqueness.

The zero initial value and slope select a forward Green's function: set it to zero for $x<\xi$, and choose the right piece to start at zero with slope one. This givesFor $f=1$ on $(0,1)$ and zero elsewhere,Indeed $y_2={\int }_0^{\min (x,1)}(x-{\xi }^2/x)d\xi /2$. It and its first derivative match at one, while $y_2,y_2^{\prime }\to 0$ at zero. The only homogeneous solution with these two initial limits is zero, so this solution too is unique. Its growth at infinity is allowed by these boundary conditions.