Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 15F (Complex Methods)

For an integer $n\ge 0$ and $\mathrm{Re}s>0$, the definition of the Laplace transform gives $F_0(s)={\int }_0^{\infty }{e}^{-st}dt=1/s$. For $n\ge 1$, integration by parts givesThe endpoint term vanishes because exponential decay dominates every fixed power of $t$. Induction therefore gives $F_n(s)=n!/s^{n+1}$. Multiplication by $e^{at}$ replaces $e^{-st}$ by $e^{-(s-a)t}$ in the defining integral, soThe second formula holds for $\mathrm{Re}(s-a)>0$ and is the Laplace transform shift theorem.

By the Bromwich inversion formula, for $t>0$ and $\gamma >0$,Close the upward vertical segment by a left semicircle of radius $R$ centred at $\gamma$. On that arc $\mathrm{Re}s\le \gamma$, so $|e^{st}|\le e^{\gamma t}$, while $|s^2(s+1{)}^2|$ is bounded below by a constant times $R^4$ for large $R$. Its length is $\pi R$, hence the arc integral tends to zero as $O(R^{-3})$. The contour is positively oriented and encloses the double poles at $0$ and $-1$.

Their residues areandThe residue theorem therefore yieldsThe value at zero follows by continuity and is $f(0)=0$.

Part (i) gives $\mathcal{L}(t)=1/s^2$ and $\mathcal{L}(t{e}^{-t})=1/(s+1{)}^2$. The convolution theorem for Laplace transforms therefore givesRepeated integration by parts yieldsSubstitution gives $f(t)=t-2+(t+2)e^{-t}$, agreeing with the contour integration result.

Write $F=\mathcal{L}f$. Using the Laplace transform of a derivative and the initial data givesThe transformed equation is $(s^4+2s^3+s^2)F=1$, so $F=1/[s^2(s+1{)}^2]$. The preceding inverse Laplace transform therefore givesFor a direct check, $f^{\prime }=1-(t+1)e^{-t}$, $f^{\prime \prime }=t{e}^{-t}$, $f^{(3)}=(1-t)e^{-t}$, and $f^{(4)}=(t-2)e^{-t}$. These satisfy both the ordinary differential equation and all four initial conditions.