Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 3

For $x\ne \xi$, the homogeneous solutions are $e^x,e^{-x}$. The left boundary condition selects $\sinh x$, while boundedness at infinity selects $e^{-x}$ on the right. Thus write $G=A(\xi )\sinh x$ for $x<\xi$ and $G=B(\xi )e^{-x}$ for $x>\xi$.

A unit Dirac delta distribution in the second derivative requires continuity of $G$ and a derivative jump of one:The second equation and the first give $-A{e}^{\xi }=1$, so the Dirichlet half-line Green function for d2 minus 1 isThe second expression is also a method of images construction. Applying the Green's function to the forcing givesFor bounded forcing this is bounded, vanishes at zero and solves the equation; a bounded homogeneous solution vanishing at zero is zero, proving uniqueness. More general forcing is allowed when the integrals converge and the resulting function has the required growth. Arbitrary unbounded forcing need not admit a bounded solution; this distinction is relevant in part (b).

For $x>0$, differentiation of $u=y/x$ givesThus the radial Laplacian equation reduces toBoundedness of $u$ at zero implies $y(0)=0$. Conversely, a regular reduced solution with $y(0)=0$ gives finite $u(0)=y^{\prime }(0)$.

There is an important qualification to the reduction in the printed question: bounded $u$ at infinity requires $y=O(x)$, not bounded $y$. The example in part (ii) actually has $y\sim -x$, so it cannot be the bounded solution described in part (a). The differential equation and origin condition reduce as asserted, and the same Green's function can still be used with the correct growth condition. This is the radial reduction and growth at infinity distinction.

The reduced forcing is $f(\xi )=\xi 1_{[1,\infty )}(\xi )$. Its exponential-weighted integrals converge, even though the resulting reduced function is not bounded. For $0\le x<1$, the Green's function formula givesFor $x\ge 1$ it givesUse $\int \xi \sinh \xi d\xi =\xi \cosh \xi -\sinh \xi$ and ${\int }_x^{\infty }\xi e^{-\xi }d\xi =(x+1)e^{-x}$ to simplify this to $y(x)=-x+e^{-x-1}$. Dividing by $x$ yieldsThus $a(x)=-(2/e)\sinh x/x$ with its continuous value at zero. At $x=1$, both pieces give $u=-1+e^{-2}$ and $u^{\prime }=-2e^{-2}$, so neither a Dirac delta distribution nor its derivative is introduced at the forcing jump. The solution is bounded at both ends, with $u\to -1$ at infinity. Its reduced function $y=xu\sim -x$ confirms the growth qualification in part (i).