Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 8

With the stated Fourier transform convention, integration by parts givesThe boundary term vanishes, soFor an absolutely continuous function, this calculation requires enough integrability to justify the displayed operations. Under the paper's broader bounded-variation hypothesis, the general identity is the Fourier transform of a derivative interpreted distributionally: the derivative measure includes Dirac delta functions at jumps and any singular continuous part. If one instead differentiates only the smooth pieces and discards jumps, the formula needs the corresponding jump corrections. Absolute integrability and bounded variation ensure the limits of $f$ at both infinities are zero, but do not by themselves make its ordinary derivative capture those singular parts.

For real $f$, complex conjugation commutes with its absolutely convergent Fourier integral:ThusThis Hermitian symmetry pairs the positive- and negative-frequency components of a real signal; its real part is even and its imaginary part is odd.

Transform in time, leaving $r$ as a parameter. Applying the derivative identity twice gives $\widetilde{{\partial }_t^2\sigma }=-{\omega }^2\tilde{\sigma }$. ThereforeThese manipulations can be justified for sufficiently decaying smooth fields, or with distributional Fourier transforms when ordinary integrability is unavailable.

At fixed frequency, the transformed equation has characteristic roots $\pm ik$, whereFor a consistent explicit branch chooseThis branch is even in $\omega$. Away from the two thresholds, the spatial solution is $\tilde{\sigma }=A(\omega )e^{ikr}+B(\omega )e^{-ikr}$, so Fourier inversion givesAt $|\omega |=\mu$, the spatial equation has a double zero characteristic root and its full solution is $C+Dr$. The two exponentials at $k=0$ represent only its constant part. This isolated-frequency issue does not change an integral with ordinary integrable amplitudes; threshold-supported distributional modes require separate treatment.

Reality requires $\tilde{\sigma }(\omega ,r)=\tilde{\sigma }(-\omega ,r{)}^{\ast }$. The way it constrains the amplitudes depends on whether the spatial wave number is real or imaginary.

For $|\omega |>\mu$, $k$ is real and even. Complex conjugation interchanges the two exponentials, so comparing independent spatial factors givesThe companion equation $B(\omega )=A(-\omega {)}^{\ast }$ is the same constraint with frequency reversed and conjugated. Neither oscillatory spatial branch grows as $r\to \infty$.

For $|\omega |<\mu$, put $k=i\kappa$ with $\kappa >0$. ThenBoundedness excludes the growing exponential, and reality then pairs the remaining amplitudes:These are the two independent low-frequency requirements. At threshold boundedness eliminates the separate linear-in-$r$ solution discussed in (ii). The constraints are understood for a Fourier transform representation in which spatial mode amplitudes are well defined; sufficient amplitude integrability also ensures the spectral integral itself is bounded.

Split the inverse transform into the intervals $|\omega |<\mu$ and $|\omega |>\mu$, and in each case pair the positive-frequency integral with its complex conjugate at negative frequency. The low-frequency pair becomessince $B=0$ and the decaying spatial factor is real. In the high-frequency positive half, replace $A(\omega )$ by $B(-\omega {)}^{\ast }$. The negative half is again its conjugate. Combining the two pieces givesEvery integrand satisfies the massive wave equation because ${\omega }^2=k^2+{\mu }^2$. Linearity, and taking the real part, preserve the equation whenever differentiation under the integrals is justified, or distributionally under the corresponding spectral assumptions. The construction is exactly the real bounded Fourier modes of the Klein-Gordon equation decomposition.

The displayed expression uses ordinary spectral amplitudes. If distributions supported at $\omega =\pm \mu$ are admitted, a further bounded threshold contribution $c_1\cos \mu t+c_2\sin \mu t$, independent of $r$, is possible. It is not a missing propagating or evanescent branch of the open frequency intervals.