Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

Let $V\subset {\mathbb{R}}^3$ be a bounded region with piecewise smooth closed boundary $\partial V$, let $\mathbf{n}$ be its outward unit normal vector, and let $\mathbf{A}$ be a continuously differentiable vector field on a neighbourhood of its closure. The divergence theorem statesHere $dV$ is the positive volume measure, $dS$ is the surface area measure, $\nabla \cdot \mathbf{A}={\partial }_x{A}_x+{\partial }_y{A}_y+{\partial }_z{A}_z$, and the right side is the total outward flux. All boundary components must be included, with their normals pointing away from $V$.

Differentiate the position vector with respect to the three coordinates:Their pairwise dot products vanish. For example, ${\mathbf{r}}_{\rho }\cdot {\mathbf{r}}_u=-\sinh \rho \cosh \rho \sin u\cos u+\sinh \rho \cosh \rho \sin u\cos u=0$. Thus these are orthogonal coordinates wherever the coordinate tangents are nonzero. Their scale factors of orthogonal coordinates areThe absolute Jacobian determinant is their product, givingThe ordered coordinate system is left handed: ${\mathbf{r}}_{\rho }\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)=-h_{\rho }{h}_u{h}_v$. The volume measure uses the absolute value. The axis and the focal circle contain coordinate degeneracies, as expected for oblate spheroidal coordinates. This is the hyperbolic parametrization of oblate spheroidal coordinates. The local TeX has an OCR substitution ${\sinh }^{2}u$ in the volume formula; the original PDF correctly prints ${\sin }^{2}u$.

Put $s^2=x^2+y^2$. For $\rho >0$, elimination of $u$ givesThese are oblate spheroids with equatorial radius $\cosh \rho$ and polar radius $\sinh \rho$, flattened along the $z$ axis. At $\rho =0$ the limiting surface is the disk $z=0$, $s\le 1$.

For $0<|u|<\pi /2$, elimination of $\rho$ instead givesThese are one-sheet hyperboloids of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho \ge 0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\ge 1$; at $u=\pm \pi /2$ it degenerates to the corresponding half of the $z$ axis.

Finally, fixed $v$ specifies a vertical half-plane through the $z$ axis, with horizontal direction $(\cos v,\sin v)$. The two values $v$ and $v+\pi$ together give a full plane. These surfaces form the mutually orthogonal families of the hyperbolic parametrization of oblate spheroidal coordinates.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

Seek a nonzero separated solution $\psi =T(t)R(r)\Theta (\theta )$. Substitution into the heat equation and division by the product givesThe left side depends only on time, so set it to $-k^2$. Multiplication by $r^2$ then separates the angular dependence, with separation constant $\ell (\ell +1)$. The three ordinary differential equations areThe temporal solution is $T=T_0{e}^{-\kappa k^{2}t}$. With $q=\cos \theta$, the angular equation is the Legendre differential equation; regularity at both angular endpoints selects $\ell =0,1,2,\dots$ and $\Theta \propto P_{\ell }(q)$, the Legendre polynomial. The radial equation has the Spherical Bessel functions $j_{\ell }(kr)$ and $y_{\ell }(kr)$ for $k>0$. Regularity at $r=0$, if imposed, excludes the singular second solution. For $k=0$ the radial powers are $r^{\ell }$ and $r^{-\ell -1}$. Without specified radial boundary conditions, no discrete set of $k$ is selected; $-k^2$ is a convention for the temporal separation constant.

For spherical symmetry the angular mode is $\ell =0$. Put $u(t,r)=r\psi (t,r)$. Direct differentiation givesso the radial heat equation in three dimensions becomes the ordinary one-dimensional heat equation $u_t=\kappa u_{rr}$. Since $r_{\pm }>0$, the zero endpoint temperatures become $u(t,r_{\pm })=0$.

Write $L=r_{+}-r_{-}$. The two initial sine modes already satisfy these Dirichlet boundary conditions; their squared wave numbers are $(\pi /L{)}^2$ and $(2\pi /L{)}^2$. Each therefore evolves independently with its corresponding exponential decay. Dividing back by $r$ givesAt $t=0$ the initial profile is recovered, and each sine is zero at both endpoints. The first mode decays four times more slowly in its exponential rate than the second. This is the two-mode instance of the spherical-shell heat equation via radial rescaling; uniqueness follows, for example, by applying the heat equation energy identity to the difference of two solutions.

For a fixed source point $\xi \in (a,b)$, the Green function is defined bywhere $\delta$ is the Dirac delta distribution and the operator acts on $x$. The corresponding solution isApplying $L_x$ under the integral gives $f(x)$ by the defining delta identity, while the endpoint values vanish. With the stated continuous coefficients, one may define the normalized distributional equation $G_{xx}+(\beta /\alpha )G_x+(\gamma /\alpha )G=\delta (x-\xi )/\alpha (\xi )$; this avoids multiplying a general distributional second derivative by a merely continuous coefficient.

Existence of this inverse Green function additionally requires that the homogeneous two-point boundary value problem have only the zero solution. The assumptions printed in the question alone do not ensure this: $y^{\prime \prime }+y$ on $[0,\pi ]$ has the nonzero zero-endpoint solution $\sin x$. At such resonance there are solvability conditions and no ordinary inverse kernel for arbitrary $f$. The construction below is nonresonant.

Suppose the piecewise smooth Green function has jump $[G{]}_{\xi }=G({\xi }^{+};\xi )-G({\xi }^{-};\xi )$. Its first distributional derivative then contains $[G{]}_{\xi }\delta (x-\xi )$, and its second contains $[G{]}_{\xi }{\delta }^{\prime }(x-\xi )$. In the normalized equation from part (a), the first-derivative term can supply a delta but no delta derivative. The right side has only a delta. Thus its delta-derivative coefficient must vanish:Equivalently, the two homogeneous branches must join continuously. This is the continuity part of the jump condition for a second-order Green function.

Now $G$ is continuous, so $G_x$ has no delta contribution. Integrate its normalized differential equation over $(\xi -ϵ,\xi +ϵ)$. The bounded lower-order terms have integrals tending to zero, while the integral of $G_{xx}$ is the difference of the two one-sided first derivatives. Hence the jump condition isThe denominator is evaluated at the source point; it is not a variable $x$ left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of $L$.

The homogeneous characteristic equation is $r^2-2r+2=0$, with roots $1\pm i$. Choose the endpoint-adapted homogeneous solutionsso that $u(0)=0$ and $v(\pi /2)=0$. Their Wronskian iswhich never vanishes. A continuous Green function with the required endpoints has the form $C(\xi )u(x_{<})v(x_{>})$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Its derivative jump is $C(\xi )W(\xi )$. Because the leading coefficient is one, part (b) requires this to equal one. ThereforeIn particular, the branches are $-e^{x-\xi }\sin x\cos \xi$ for $x\le \xi$ and $-e^{x-\xi }\sin \xi \cos x$ for $x\ge \xi$. This is the Dirichlet Green function for a conjugated harmonic oscillator with interval length $\pi /2$, drift parameter one and oscillator wave number one.

Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:Using ${\sin }^{2}x+{\cos }^{2}x=1$, the answer isBoth endpoint values are zero. As a direct check, conjugating by $e^x$ sends $y^{\prime \prime }-2y^{\prime }+2y$ to $e^x(v^{\prime \prime }+v)$, and $v=1-\cos x-\sin x$ gives $v^{\prime \prime }+v=1$. Thus the differential equation and both boundary conditions are satisfied.

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

The $z$ direction is fixed, giving the eigenvalue one. The planar block has characteristic polynomialThus the full list of eigenvalues isThe last two are real precisely when $\sin \varphi =0$, so all three are real exactly for $\boxed{\varphi \in \pi \mathbb{Z}}$. At even multiples of $\pi$ the matrix is the identity and every vector is fixed. At odd multiples it is a half-turn: the axial eigenvalue remains $+1$, while the whole perpendicular plane has eigenvalue $-1$.

For other angles, the only real eigenvector direction is the rotation axis; the perpendicular plane is rotated rather than preserving an individual real direction. Its complex eigenvectors can be taken as $(1,-i,0{)}^T$ and $(1,i,0{)}^T$ for $e^{i\varphi }$ and $e^{-i\varphi }$, respectively. These represent the two complex conjugate planar eigenvalues.

With the same right-handed active rotation convention as $R_z$, the $y$ direction is fixed, ${\mathbf{e}}_z$ rotates towards ${\mathbf{e}}_x$, and ${\mathbf{e}}_x$ towards $-{\mathbf{e}}_z$. ThereforeIts columns give the images of the three Cartesian unit vectors, which also makes the signs of this rotation matrix explicit.

A product of orthogonal matrices is orthogonal. In fact,Thus $\boxed{R_1\text{ is orthogonal for every }\theta ,\varphi .}$ Its determinant is also one, since each factor has determinant one, so it remains a proper rotation matrix.

Both factors and their product are real, so the Hermitian conjugate equals the transpose. Part (i) therefore givesHence $\boxed{R_1\text{ is unitary for every }\theta ,\varphi .}$ A real orthogonal matrix is always a unitary matrix.

A real matrix is a Hermitian matrix exactly when it is a symmetric matrix. Since $R_1$ is orthogonal, this is equivalent to $R_1=R_1^{-1}$, or $R_1^2=I$. A generic rotation does not satisfy this; for example $\theta =0$, $\varphi =\pi /2$ gives a nonsymmetric quarter-turn. Thus $\boxed{R_1\text{ is not Hermitian in general.}}$

The exceptions are the identity and half-turns, as characterized by real rotation matrices that are Hermitian. For this product, a half-turn occurs when either $\theta$ or $\varphi$ is an odd multiple of $\pi$; the identity occurs when both are even multiples. For example, the half-turn condition follows from $\mathrm{tr}{R}_1=(1+\cos \theta )(1+\cos \varphi )-1=-1$.

The Hermitian conjugate of a real matrix is its transpose, soTherefore $\boxed{M\text{ is anti-Hermitian for all }\theta ,\varphi .}$ It is also a real skew-symmetric matrix. It is Hermitian as well only in the exceptional case $M=0$, precisely when $R_1$ is symmetric. This includes the identity and half-turn cases from part (d), so the zero-matrix overlap should not be excluded.

Applying the rightmost rotation matrix first givesEquality with ${\mathbf{e}}_y$ requires $\cos \theta =0$, so $\theta =\pi /2$ in the specified range. Then $\sin \varphi =1$ and $\cos \varphi =0$ giveIn the reverse order, $R_z$ first fixes ${\mathbf{e}}_z$, so $R_2{\mathbf{e}}_z=R_y(\pi /2){\mathbf{e}}_z={\mathbf{e}}_x$. Therefore $\boxed{R_2\mathbf{v}\ne {\mathbf{v}}^{\prime }}$. This explicitly demonstrates that rotations about different axes do not generally commute.

At the angles in part (f),Let the two diagonal matrices be $D=\mathrm{diag}(d_1,d_2,d_3)$ and $D^{\prime }=\mathrm{diag}(e_1,e_2,e_3)$. Multiplication givesThe vector condition is exactly $c{e}_2{d}_3=b$. For $c\ne 0$, the general form is thereforeEvery pair $p,q$ is attainable, for example with $D=I$ and $D^{\prime }=\mathrm{diag}(-p,b/c,-q)$. The transformation need not be orthogonal or invertible; the question asks only for diagonal rescalings with this one vector image. If $c=0$ and $b\ne 0$, no linear map can satisfy the condition. If $c=b=0$, the $(2,3)$ entry is arbitrary as well.

Let $A=A^{†}$ be a Hermitian matrix and $Av=\lambda v$ with $v\ne 0$. The scalar $v^{†}Av$ equals its own complex conjugate, becauseSince $v^{†}Av=\lambda v^{†}v$ and the squared norm $v^{†}v$ is positive, $\boxed{\lambda \in \mathbb{R}}$.

Now take $Au=\lambda u$ and $Av=\mu v$. The inner product satisfiesSubtracting gives $(\mu -\lambda )u^{†}v=0$. Thus distinct eigenvalues imply $\boxed{u^{†}v=0}$, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.

A complex square matrix $U$ is a unitary matrix if $\boxed{U^{†}U=U{U}^{†}=I}$. Equivalently, it preserves the complex inner product and hence the norm. If $Uv=\mu v$ with $v\ne 0$, thenDivision by the positive squared norm gives $\boxed{|\mu |=1}$. Thus every eigenvalue lies on the unit circle.

Part (a) makes each ${\lambda }_j$ real. In the eigenvector basis,and $|{\lambda }_j+ia{|}^2={\lambda }_j^2+a^2>0$. Hence $\boxed{A+iaI\text{ is invertible}}$.

All factors that are rational functions of $A$ commute. Taking the Hermitian conjugate reverses product order and givesThereforeso $\boxed{V\text{ is unitary}}$. This is a Cayley transform of a Hermitian matrix, with scale $a$ and the convention that an infinite real eigenvalue maps to one.

Applying it to each eigenvector yieldsThe eigenvectors are unchanged, and $|{\nu }_j|=1$ because numerator and denominator are complex conjugates. No ${\nu }_j$ equals one when $a>0$. The scalar map is injective on real $\lambda$, so the eigenspaces, including any degeneracies, are preserved exactly.

Use $U^{†}=U^{-1}$ and take the Hermitian conjugate in the reverse product order:The factorization $I-U^{-1}=-U^{-1}(I-U)$ gives $(I-U^{-1}{)}^{-1}=-(I-U{)}^{-1}U$, while $I+U^{-1}=U^{-1}(I+U)$. ConsequentlyThe last equality uses the fact that $I+U$ commutes with $I-U$ and its inverse. Thus $B$ is a Hermitian matrix. The exclusion of the eigenvalue one is essential to define the inverse. Spectrally, an eigenvalue $e^{i\vartheta }\ne 1$ is sent to $-\cot (\vartheta /2)$, a real number. This is the inverse Cayley transform of a unitary matrix without eigenvalue one.

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

Insert the specified real part into the polar Cauchy-Riemann equations. The first gives $v_r=n{r}^{n-1}\sin (n\theta )$, and integration with respect to $r$ givesThe second equation then requires $v_{\theta }=n{r}^n\cos (n\theta )$, so $C^{\prime }(\theta )=0$. The remaining imaginary additive constant is fixed by $f(0)=0$. ThusThe resulting polynomial is holomorphic at the origin too, consistently extending the calculation made in polar coordinates with $r>0$.

Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the zeros are exactlyAt each point ${\cos }^{\prime }(z_n)=-\sin z_n\ne 0$, so the cosine has a simple zero. Cubing it givesThe order of a zero of a holomorphic function is therefore three at every zero. There are no finite singularities because this is an entire function.

To classify infinity, examine ${\cos }^3(1/w)$ at $w=0$. The identity ${\cos }^{3}z=(3\cos z+\cos 3z)/4$ shows that its Laurent series contains the nonzero negative-power coefficientsThere are infinitely many terms in the principal part, the criterion for an essential singularity. Therefore $\boxed{\infty \text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial entire function at infinity.

If $z_0\ne 0$, factor out $z_0$ and use the geometric series:ThusThe radius of convergence is the distance to the nearest singularity, the simple pole at $z=-z_0$. The case $z_0=0$ must be treated separately: the function is $1/z$, has no Taylor series about zero and is already its own Laurent series on $0<|z|<\infty$.

For $|z|<1$, multiply the absolutely convergent Taylor series of the sine by the geometric series:The product is odd, so all even coefficients vanish. Collecting the coefficient of $z^{2m+1}$ givesFor example, the first terms are $z+5z^3/6+101z^5/120+\cdots$. The denominator vanishes at $z=\pm 1$, and the numerator is nonzero there, so these are genuine simple poles. Therefore $\boxed{R=1}$ is the radius of convergence.