Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 3

Set $L=d^2/d{t}^2-4$ and choose $L_{t}G(t,t^{\prime })=\delta (t-t^{\prime })$, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are $e^{\pm 2t}$. Decay at each end and continuity at $t=t^{\prime }$ therefore give $G=C{e}^{-2|t-t^{\prime }|}$. Integrating its equation across the source gives a unit jump in its first derivative:ThusThe solution of the boundary value problem is the convolutionThere is no extra homogeneous term: no nonzero combination of $e^{2t},e^{-2t}$ decays at both infinities. The forcing must of course permit the imposed decay; integrable $f$, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

For $t<0$, all forcing points satisfy $\tau >t$, so the Green's function formula becomesAn integration by parts gives $\int \tau e^{-2\tau }d\tau =-(\tau /2+1/4)e^{-2\tau }$. ConsequentlyThe exponential has the required decay as $t\to -\infty$.

For $t>1$, every forcing point has $\tau <t$. Hence the Green's function givesSince $\int \tau e^{2\tau }d\tau =(\tau /2-1/4)e^{2\tau }$, the remaining integral is $(e^2+1)/4$. ThereforeThis decays at positive infinity.

For $0<t<1$, split the Green's function integral where the absolute value changes sign:The two terms inside the bracket are respectivelyAdding them yieldsThe affine term gives $x^{\prime \prime }-4x=t$, while both exponential terms are homogeneous solutions. The values at the joins are $x(0)=(3e^{-2}-1)/16$ and $x(1)=-(1+e^{-2})/16$, agreeing with parts (i) and (ii). Their first derivatives also match: $x^{\prime }(0)=(3e^{-2}-1)/8$ and $x^{\prime }(1)=(1+e^{-2})/8$. Thus the pieces introduce no unwanted Dirac delta distributions.