Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 3

For the Green function of $D_x^2-1$, the homogeneous solutions satisfying the left and right endpoint conditions are $u(x)=\sinh x$ and $v(x)=\sinh (1-x)$. Their Wronskian isContinuity of the kernel at $x=\xi$ and integration of $G_{xx}-G=\delta (x-\xi )$ require $[G_x{]}_{\xi -}^{\xi +}=1$. Hence the Dirichlet modified Helmholtz Green function on an interval isHere $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The negative sign is essential for this operator convention. Integrating the forcing against the kernel givesThis satisfies both endpoint conditions. It solves the equation classically for continuous $f$, and in the corresponding weak sense for integrable forcing. A homogeneous difference satisfying both endpoint conditions must vanish, so the solution is unique.

Use the zero initial displacement and velocity given in the PDF. The characteristic roots of the homogeneous equation are $-1\pm i\mu$. Its causal Green function of a damped oscillator iswhere $H$ is the Heaviside step function. The positive-time kernel has $G(0+)=0$ and $G^{\prime }(0+)=1$, so its derivative jump generates a unit impulse. The zero-data forced solution is consequently the causal convolutionDifferentiating once gives $x^{\prime }(0)=0$; differentiating twice supplies the endpoint term $g(t)$ and verifies the differential equation. No homogeneous transient is required with these initial data.

The PDF bounds the absolute value of the forcing. Taking absolute values in the Green function integral and using $|\sin |\le 1$ yieldsFor $a\ne 1$, this givesThe ratio is nonnegative for either $a<1$ or $a>1$. The missing special case is $a=1$, for whichThis proves decay of a damped oscillator under exponentially decaying forcing. A one-sided upper bound on $g$ without an absolute value, as in the transcription, would not justify the conclusion.