Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 8

The homogeneous Cauchy-Euler equation has powers $x^r$ with $r(r-1)+2r-2=(r-1)(r+2)=0$. Boundedness at zero selects $x$, and boundedness at infinity selects $x^{-2}$. Put $G=Ax$ for $x<\zeta$ and $G=B{x}^{-2}$ for $x>\zeta$. Continuity gives $B=A{\zeta }^3$. Since the operator is $L=(x^{2}d/dx{)}^{\prime }-2$, the required delta-function normalization is ${\zeta }^2[G_x{]}_{{\zeta }^{-}}^{{\zeta }^{+}}=1$. Thus $-3A{\zeta }^2=1$, giving the bounded Green kernel for a radial Euler operator:In the requested notation, $\boxed{a=1,b=-2,c=1,d=-2,k_{-}=k_{+}=-1/3}$. The two expressions agree at $x=\zeta$.

With ordinary measure $d\zeta$, the solution isFor bounded forcing these integrals converge and give a bounded solution. It is unique because no nonzero combination of $x,x^{-2}$ is bounded at both ends. For a more general forcing, convergence and boundedness of the displayed integral must be checked.

The Green's function is nonpositive, so $f\ge 0$ gives $y\le 0$. If also $f\le m$, thenThe integral solution therefore givesBoth bounds are attained, by the constant forcings $f=m$ and $f=0$, respectively. In particular, boundedness of the solution does not require it to approach zero.

Differentiate the integral formula using the Leibniz integral rule. Its two moving-endpoint terms cancel, leavingEach term before subtraction lies between zero and $m/(3x)$. ThereforeThe bounds are sharp at any fixed positive $x$: put $f=m$ below that point and zero above it for the upper bound, or reverse those choices for the lower bound. If continuity of the forcing is required, smooth approximations approach the same bounds.