Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 14B (Methods)

Use the Fourier transform convention $\tilde{h}(k)={\int }_{\mathbb{R}}{e}^{-ikx}h(x)dx$, with inverse $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\tilde{h}(k)dk$. Splitting at the origin, with $a>0$ to ensure convergence, givesThus

The original PDF has forcing $e^{-|x|}$; the supplied TeX loses the absolute-value bars. Set $q=|\omega |$. First assume $q>0$ and $q\ne 1$. The Fourier transform of a derivative, obtained by integration by parts with the boundary terms zero, is $\widetilde{u^{\prime \prime }}=-k^2\tilde{u}$. Applying it and the preceding transform givesUsing the inverse Fourier transform pairs from part (i),Both decay conditions hold. The first derivatives of the two exponential terms cancel at zero, so $u$ is continuously differentiable there; the second derivatives have matching limits and the differential equation holds at zero as well as on both half-lines. For uniqueness, a difference solves the homogeneous equation. Decay gives $A{e}^{-qx}$ on the positive half-line and $B{e}^{qx}$ on the negative half-line. Continuity gives $A=B$ and derivative continuity gives $-qA=qB$, forcing both to be zero.

The PDF's condition $\omega \ne \pm 1$ also allows $\omega =0$, but no decaying solution exists when $\omega =0$. Indeed on $x>0$, $-u^{\prime \prime }=e^{-x}$ and decay forces $u=-e^{-x}$; on $x<0$, decay forces $u=-e^x$. Their one-sided derivatives at zero are $1$ and $-1$, which cannot match. The missing condition $\omega \ne 0$ is therefore a genuine source omission, not a transcription repair. The transform expression at $q=0$ has a nonintegrable $k^{-2}$ singularity, consistently with this obstruction.

For smooth rapidly decaying initial data, apply the Fourier transform in $x$ and differentiate under the integral. The Fourier transform of a derivative turns the wave equation intoSolving this constant-coefficient ordinary differential equation givesAt $k=0$ the quotient means its continuous limit $t$, so $\tilde{v}(0,t)=\tilde{f}(0)+t\tilde{g}(0)$.

Since $\cos kt=(e^{ikt}+e^{-ikt})/2$, the Translation property of the Fourier transform inverts the first term as $(f(x+t)+f(x-t))/2$. Also $K_t(x)=\frac{1}{2}{1}_{[-t,t]}(x)$ has Fourier transform $\sin kt/k$, including its limiting value at zero. The convolution theorem therefore inverts the second term asThis proves the D'Alembert formulaThe formula itself extends beyond the rapidly decaying class: $f\in C^2$ and $g\in C^1$ suffice for a classical solution, verified by differentiation and the two initial conditions. The decay assumption above justifies the ordinary transform derivation without distributional qualifications.