Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024

For a continuously differentiable vector field $\mathbf{u}$ on a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth closed boundary $S=\partial V$, the divergence theorem statesHere $\nabla \cdot \mathbf{u}$ is its divergence, $dV$ is the volume element, $dS$ is the surface area element, $\hat{\mathbf{n}}$ is the outward unit normal, and $d\mathbf{S}=\hat{\mathbf{n}}dS$. Thus a volume integral of the divergence equals the outward flux through the boundary.

Since $\mathbf{a}$ is constant, the product rule gives $\nabla \cdot (\mathbf{a}\psi )=\mathbf{a}\cdot \nabla \psi$. Applying the divergence theorem therefore givesThis holds for every constant vector $\mathbf{a}$, so each component of the difference vanishes. Consequently the gradient has the boundary representation

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

Use Einstein summation convention and write ${\partial }_j={\nabla }_j$. Since ${\partial }_j{r}_i={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, the product rule givesThis proves $\boxed{P_i={\partial }_j(r_i{w}_j)-w_i}$. Now take $w_j={\partial }_j\psi$. Integrating this identity and applying the divergence theorem to the first term givesPart (b) changes the last volume integral of the gradient into a surface integral, yieldingHere $d{S}_i={\hat{n}}_{i}dS$; the Laplacian is ${\nabla }^2\psi ={\partial }_j{\partial }_j\psi$.

Apply the product rule to the tensor field, without assuming any symmetry of $\Sigma$:Integrating, the divergence theorem converts each of the two divergences into a boundary term:The repeated indices are summed according to Einstein summation convention, while $i$ labels the component of the resulting vector identity.

Choose the symmetric tensor fieldThe product rule and equality of mixed partial derivatives giveThis is the Euclidean scalar-gradient stress identity. Substitute it into part (e), and use ${\Sigma }_{ki}{\hat{n}}_k=({\partial }_n\psi ){\partial }_i\psi -\frac{1}{2}|\nabla \psi {|}^2{\hat{n}}_i$, where ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$. Reassembling the components gives the required pure surface integral:Sufficient smoothness of $\psi$ is understood for the displayed derivatives and applications of the divergence theorem.

Apply separation of variables with $\rho (x,t)=X(x)T(t)$ and temporal factor $T=e^{-\lambda t}$. The spatial eigenvalue equation isThe nontrivial solutions of these Dirichlet boundary conditions are sine eigenfunctions. Normalizing in the usual inner product on $(0,L)$ giveswith ${\int }_0^L{\psi }_n{\psi }_{m}dx={\delta }_{nm}$. The superposition principle and the Fourier sine series then giveThe constants are the initial-data coefficients $a_n={\int }_0^L\rho (x,0){\psi }_n(x)dx$. Here physical diffusion means $D>0$.

By orthogonality and the sampling property of the Dirac delta distribution,Consequently the reaction-diffusion equation solution isFor every $t>0$ this eigenfunction expansion is smoothed by the Gaussian decay in $n$. As $t\downarrow 0$, it recovers the specified initial condition in the sense of distributions.

The smallest decay-rate eigenvalue is ${\lambda }_1=D{\pi }^2/L^2-\gamma$, so the threshold isBecause $0<x_0<L$, $a_1>0$. For every fixed interior point $0<x<L$, the first sine eigenfunction is positive and the remaining terms have strictly faster decay relative to it. HenceOne can justify the relative-error limit directly: after factoring out the first exponential, the absolute remainder is bounded by $(2{\rho }_0/L){\sum }_{n\ge 2}{e}^{-D{\pi }^2(n^2-1)t/L^2}$, which tends to zero. Thus $\boxed{\rho \to 0\text{ if }\gamma <{\gamma }_c}$, and $\boxed{\rho (x,t)\to +\infty \text{ for each interior }x\text{ if }\gamma >{\gamma }_c}$. The boundary values stay zero. At equality the first term is stationary and the density approaches its positive sine profile.

Let $f(x)=\rho (x,0)$. The first normalized Fourier sine series coefficient isStrict positivity follows because $f\ge 0$, has positive total mass, and $\sin (\pi x/L)>0$ everywhere inside the interval. In particular, there is no cancellation that could remove the slowest term. This is the positivity of the first Dirichlet reaction-diffusion coefficient.

For finite initial mass $M={\int }_0^{L}f(x)dx$, all coefficients satisfy $|a_n|\le \sqrt{2/L}M$. The eigenfunction expansion therefore gives the uniform boundThe positive first profile thus controls the long-time behavior. We recover exactly the threshold $\boxed{{\gamma }_c=D{\pi }^2/L^2}$, with decay below it and unbounded positive growth at every interior point above it. The finite-mass condition is the usual meaning of an initial density profile here; standard positive-time smoothing gives the same conclusion in the usual broader admissible data classes.

For an exponential trial $y=e^{rx}$, the characteristic equation isIts distinct complex roots are $r=-a\pm ib$. Thus the general real solution of the homogeneous second-order linear differential equation isThe two displayed solutions have nonzero Wronskian $b{e}^{-2ax}$ and hence linear independence. The same formula permits complex solutions when $C_1,C_2$ are complex.

For the zero-data initial value problem, choose a causal Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ and $G=0$ for $x<\xi$. Continuity at $x=\xi$ prevents a derivative of the Dirac delta distribution, while integration across the source givesFor $x>\xi$, part (i) gives the shifted homogeneous solutions. The zero value at the join removes the cosine term, and the unit derivative jump fixes the sine coefficient. Therefore the causal Green function of a damped oscillator isEquivalently, multiply the first expression by the Heaviside step function $H(x-\xi )$. The required solution isIts initial value and first derivative vanish: the interval has zero length at $x=0$ and the kernel vanishes at its upper limit. Differentiating twice shows that the upper-limit contribution to $y^{\prime \prime }$ is precisely $g(x)$, while the remaining homogeneous kernel terms cancel in $Ly$.

The original PDF has the absolute-value hypothesis $|g(\xi )|<C{e}^{-s\xi }$, which is missing from the local OCR. From the Green's function formula and $|\sin |\le 1$,Because $s>a>0$, the upper bound tends to zero. The squeeze theorem proves $\boxed{y(x)\to 0\text{ as }x\to \infty }$.

With the paper's Fourier transform convention, the Fourier inversion theorem givesThis holds at continuity points under the usual inversion hypotheses; at a jump with one-sided limits it gives their mean. The factor $1/(2\pi )$ and the sign change in the exponential are part of this normalization.

Insert the inverse Fourier transforms of $f$ and its complex conjugate, initially under enough decay to justify interchanging integrals:The middle step uses the stated Dirac delta distribution identity, and the last uses its sampling property. This is the Parseval identity for the chosen normalization. Approximation extends the result to square-integrable functions, including the function used below.

The real odd function has a purely imaginary odd Fourier transform. Setting $A=\pi /2$ and using a product-to-sum identity,The apparent singularities are removable. Direct evaluation or limits giveIn particular, the transform stays finite where its numerator and denominator both vanish.

The squared norm of the original function isApplying the Parseval identity to part (c), using the evenness of the squared transform, givesThus the last integral equals ${\pi }^2/8$. Now set $u=\pi k/2$. Sincewe obtainThe apparent singularity at $u=\pi /2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.

Let $Av=\lambda v$ with $v\ne 0$. Taking the Hermitian inner product with $v$ gives $v^{†}Av=\lambda \Vert v{\Vert }^2$. For an anti-Hermitian matrix,Consequently $\overline{\lambda }\Vert v{\Vert }^2=-\lambda \Vert v{\Vert }^2$. As $\Vert v{\Vert }^2>0$, the eigenvalue satisfies $\boxed{\overline{\lambda }=-\lambda }$, or $\boxed{\lambda =i\theta \text{ with }\theta \in \mathbb{R}}$. Zero is included as a purely imaginary eigenvalue.

If $I+A$ were a singular matrix, a nonzero vector $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an eigenvalue of the anti-Hermitian matrix, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an invertible matrix, which will also be used below.

Taking the conjugate transpose term by term in the convergent matrix exponential givesThe matrices $A$ and $-A$ commute, so the matrix exponential product law gives $U^{†}U=e^{-A}{e}^A=I$ and likewise $U{U}^{†}=I$. Hence $\boxed{U\text{ is unitary}}$.

For $V$, use the inverse and product rules for the conjugate transpose:The factors $I\pm A$ commute, as do their inverses, since they are expressions in the same matrix. Consequentlyand similarly $V{V}^{†}=I$. Therefore $\boxed{V\text{ is unitary}}$. This rational construction is the skew-Hermitian form of the Cayley transform of a Hermitian matrix, obtained by writing $A=iH$ with $H$ a Hermitian matrix.

If $Av=\lambda v$, then $A^{n}v={\lambda }^{n}v$, and the matrix exponential series gives $Uv=e^{\lambda }v$. Also $(I+A{)}^{-1}v=(1+\lambda {)}^{-1}v$, since $1+\lambda \ne 0$. ThusThe assumed diagonalization ensures that these account for all eigenvalues, including multiplicities. In particular, for $\lambda =i\theta$,both of modulus one, in agreement with the unitary matrix property.

Here $A^2=-I$. Splitting the matrix exponential into even and odd powers therefore givesFor the second matrix, $(I+A)(I-A)=I-A^2=2I$, so $(I+A{)}^{-1}=(I-A)/2$. HenceBoth are real orthogonal matrices, and hence also unitary matrices.

Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

In the orthonormal basis of part (b), the three quadric surfaces have the following forms. For $\alpha =1$,a two-sheeted hyperboloid with axis $u$, vertices $(\pm 1/\sqrt{6},0,0)$, and one sheet on each side of the plane $u=0$. For $\alpha =0$,an elliptic cone with vertex at the origin and the same axis. For $\alpha =-1$,a one-sheet hyperboloid with axis $u$ and a waist ellipse at $u=0$, whose semiaxes are $1/\sqrt{3}$ along $v$ and $1/2$ along $w$. The sections at fixed $u$ enlarge away from the waist.

The orthogonal change of coordinates preserves distance to the origin. On the required quadric surface,Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm 1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates areSubstitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.

Write $z=x+iy$ and square the modulus equation. It becomesFor $P\ne 1$, completing the square gives the Apollonius circleHenceThe printed statement needs the exception $\boxed{P=1⟹y=0}$: equal distances from $ib$ and $-ib$ give their perpendicular bisector, a straight line rather than a finite-radius circle. For $P>1$ the center lies in the upper half-plane; this is the case needed in part (ii). The original PDF confirms the numerator $z+ib$, correcting the duplicated, corrupted OCR formula.

For the upper Apollonius circle, let its constant ratio be $P>1$. Matching the center and radius from part (i) givesSubtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. AlsoThe relevant inverse hyperbolic cosine is positive. Now $\log |(z+ib)/(z-ib)|$ is the real part of a local holomorphic logarithm, hence a harmonic function away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.

On $y=0$ the numerator and denominator have equal modulus, so the logarithm is zero. On the circular boundary its value is $\log P=\mathrm{arcosh}(y_0/R)$. Therefore the harmonic potential between a grounded line and a circular boundary isThis verifies Laplace's equation and both prescribed Dirichlet boundary conditions. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.