Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 11H / Solution

Substituting $y=x^p$ in the homogeneous ordinary differential equation gives $p(p-1)-2=(p-2)(p+1)=0$. ThusFor a Green's function decaying at the two ends, use $u(x)=x^2$ on the side of zero and $v(x)=x^{-1}$ on the side of infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-3$. The function must be continuous at $x=\xi$, or its second derivative would contain a derivative of a Dirac delta. Its derivative jump must be one, to give the specified positive delta source. These two conditions yield the boundary-decaying Green kernel for an inverse-square differential operator:Its derivatives at the join are $-2/3$ on the left and $1/3$ on the right, confirming the jump one. Away from the join each branch solves the homogeneous equation, and both endpoint conditions hold.

Integrate this Green's function against the forcing, whose support is between zero and one. For $0<x\le 1$,For $x\ge 1$, only the right branch contributes. ThereforeOn the first branch, $y^{\prime }=(2x/3)\log x+x/9$ and $y^{\prime \prime }=(2/3)\log x+7/9$, so $y^{\prime \prime }-2y/x^2=1$. The second branch is a multiple of the homogeneous solution $x^{-1}$, so its forcing is zero. At $x=1$, both $y$ and $y^{\prime }$ agree, with values $-1/9$ and $1/9$, so there is no additional delta source. Finally $x^2\log x\to 0$ as $x\downarrow 0$ and $x^{-1}\to 0$ as $x\to \infty$. A homogeneous solution obeying both boundary conditions has $C=D=0$, establishing uniqueness.