Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1

For a continuously differentiable vector field $\mathbf{G}$ on a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, the divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus total outward flux through a closed surface equals the integral of the local divergence inside it. A nonclosed surface first needs a suitable closure, with the added surface's flux accounted for separately.

In cylindrical coordinates, the surface has radius $r=\sqrt{2}z$. It is the lateral surface of a right circular cone, with vertex at the origin and axis along positive $z$. Its height is $h$ and the limiting upper rim has radius $\sqrt{2}h$; the horizontal disk at that rim is absent. The half-angle from the vertical satisfiesThe upper rim is excluded by $z<h$, which does not affect any convergent surface integral. The vertex likewise contributes no area.

Close the cone with the horizontal disk $D$ at height $h$, and let $V$ be the enclosed solid. The divergence isThe term $2y$ integrates to zero by reflection symmetry of each circular cross-section. Using the divergence theorem and cylindrical coordinates, the closed-surface flux isThe outward normal on $D$ is $+\hat{\mathbf{z}}$, so its flux is its area times $G_z=h$:Subtracting this cap contribution gives the required lateral flux:This is conical flux by closing with a cap. The normal convention matters: on the lateral cone the outward normal points radially outwards and partly downwards, so the $G_z$ contribution to the lateral flux is negative. To justify applying the theorem at the vertex, truncate a small tip and take its size to zero; the extra flux vanishes because $\mathbf{G}$ is regular there.

For $v=(t+1{)}^{2}u$, the derivatives areSubstitution cancels the two terms $2(t+1)u$, giving the diffusion equationThis is a scalar integrating factor for a reaction-diffusion equation, since $\exp [\int 2dt/(t+1)]=(t+1{)}^2$ up to a constant. We use $t\ge 0$, so the factor never vanishes. A classical solution at the initial corners additionally requires compatible data, $h(0)=0$ and $h(\pi )=f(0)$.

The transformed endpoint values are $u(0,t)=0$ and $u(\pi ,t)=3$. Apply boundary lifting with $u=3x/\pi +w$. The lifted profile is linear, so $w_t=w_{xx}$ with homogeneous Dirichlet boundary conditions andFor separation of variables, put $w=X(x)T(t)$. Then $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The endpoint conditions give nonzero spatial modes only for $\lambda =n^2$, $n=1,2,\dots$, with $X_n=\sin nx$ and $T_n=e^{-n^{2}t}$. Henceby orthogonality of the sine modes. Only the $n=2$ mode is present. Undoing both substitutions givesAt $t=0$ this has the prescribed initial profile, and at $x=\pi$ it is $3(t+1{)}^2$. Differentiating also verifies the original evolution equation directly.

The nonstationary part of $u$ decays exponentially, whereas the integrating factor grows quadratically. For fixed $x>0$,ConsequentlyAt $x=0$, both expressions are identically zero; the ratio formulation is understood for $x>0$. The diffusion modes have died away, leaving the lifted linear profile multiplied by the reaction-induced amplitude factor.

A causal Green function vanishes before the impulse. For $t>\tau$ it solves the homogeneous linear ordinary differential equation, whose characteristic roots are $-q\pm 2i$. Thus it has the formContinuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the jump condition for an impulse, $G_t({\tau }^{+},\tau )-G_t({\tau }^{-},\tau )=1$. The derivative from below is zero and the derivative from above is $2B$, givingHere $\Theta$ is the Heaviside step function. The zero initial data exclude any additional homogeneous response. By the superposition principle, the output for a general locally integrable input isThe upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.

The pulse height is the constant $t_0$, not the time variable. In the undamped case the convolution therefore reduces toIntegration gives the continuous displacement and velocityThe post-pulse expression is equivalently $x(t)=(t_0/2)\sin t_0\sin (2t-t_0)$. Since $t_0>0$, it vanishes for every subsequent time precisely when $\sin t_0=0$. ThusThese are integer multiples of the oscillator's natural period $\pi$. At such switch-off times both $x(t_0)$ and $x^{\prime }(t_0)$ are zero. Vanishing displacement alone would not suffice: a nonzero switch-off velocity would restart an oscillation. This is rectangular-pulse cancellation of an undamped oscillator.

For the stated forward Fourier transform, the inverse has the opposite exponential sign and a factor $1/(2\pi )$:This holds under the usual hypotheses for Fourier inversion, or in the sense of distributions when appropriate.

Suppose initially that $f,g$ are absolutely integrable, so Fubini's theorem justifies exchanging the integrations. The transform of their convolution isSet $s=z-t$, so $z=s+t$ and $dz=ds$. The exponential separates:Therefore the convolution theorem isNo extra $2\pi$ occurs in this direction with the specified normalization.

Since $\gamma >0$, the causal damped sine has an absolutely convergent Fourier transform for real $\omega$. Resolving the sine into complex exponentials givesCombining the two fractions yieldsThe formula also includes $p=0$, for which both the original function and its transform vanish. Its sign reverses when the real parameter $p$ reverses.

For an integrable applied voltage, the convolution theorem givesThe integrated electric current is the zero-frequency transform, $Q=\tilde{I}(0)$. Since $\tilde{V}(0)$ can be arbitrary, zero total electric charge for every such input is equivalent toFor nonzero $p_1,p_2$, this can be writtenBecause both amplitudes are positive, nonzero $p_1,p_2$ must have opposite signs. If exactly one $p_i$ is zero there is no cancellation with positive amplitudes; if both vanish the kernel is identically zero and any amplitudes work. In particular, if both frequency parameters were additionally required to be positive, no such nontrivial system would exist. The PDF requires positivity of $a_i,{\gamma }_i$, but allows real fixed $p_i$.

This is zero-frequency cancellation of a causal response kernel. The phrase arbitrary voltage is understood within a class for which the net-charge integrals exist; a permanent DC input need not have finite total charge.

An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

Using $(AB{)}^{†}=B^{†}{A}^{†}$, the conjugate transpose of the proposed matrix isThusThe commutator of two Hermitian matrices is skew-Hermitian, and multiplication by $i$ changes it into a Hermitian matrix. No simplicity assumption on the spectra is needed for this part.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

Choose the common orthonormal eigenbasis established in part (ii), and let $U$ be the unitary matrix of its columns. ThenConsequentlyThus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $N{N}^{†}=N^{†}N=A^2+B^2$, so $N$ is a normal matrix, as also follows from unitary diagonalization of a normal matrix.

Since $C$ is unitary, $C^{-1}=C^{†}$. Hence $B=C^{-1}AC=C^{†}AC$ is Hermitian, and so is its positive integer power:It follows that $B^p$ has real eigenvalues. More explicitly, cancellation of adjacent $C{C}^{-1}$ factors gives the similarity transformationIf the real eigenvalues of $A$ are ${\lambda }_j$, the eigenvalues in question are thereforeFor even $p$ they are nonnegative; for odd $p$ their signs can be either. The similarity argument actually works for every invertible $C$, although the Hermiticity argument uses its unitarity.

The Cauchy-Riemann equations give $u_x=v_y$ and $u_y=-v_x$. ThereforeTaking the nonnegative square root yieldsThe two gradients are also orthogonal, since $\nabla u\cdot \nabla v=v_y{v}_x-v_x{v}_y=0$, although equality of their lengths alone is the requested result.

Differentiate the prescribed imaginary part with respect to $y$:The Cauchy-Riemann equation $u_x=v_y$ integrates toIts $y$ derivative is $u_y=-e^x[(x+1)\sin y+y\cos y]+F^{\prime }(y)$. On the other hand, $-v_x=-e^x[(x+1)\sin y+y\cos y]$. The second Cauchy-Riemann equation therefore requires $F^{\prime }(y)=0$. Combining the real and imaginary parts,Thus the most general answer on a connected domain isA real constant preserves the given imaginary part. Conversely, the difference of any two such analytic functions has zero imaginary part, and its Cauchy-Riemann derivatives force it to be constant. This proves completeness, rather than only exhibiting one possible harmonic conjugate.

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.

The origin is an ordinary point of the Hermite differential equation, so put $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$. Substitution and matching powers givesThus the linear recurrence relation isEven and odd coefficients evolve independently from arbitrary $a_0,a_1$. The general power series solution iswith an empty product equal to one. Its first terms areBoth series converge at every finite $x$, since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree $N$, the highest coefficient in the differential equation requires $\lambda =2N$. Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. HenceHere $H_N$ is a physicists' Hermite polynomial. Two explicit choices areDirect differentiation verifies both. These are polynomial solutions up to an arbitrary overall factor. Without a polynomial requirement or additional boundary conditions, the equation also has nonpolynomial solutions for other real $\lambda$; discreteness here comes from termination.

Insert the Frobenius solution and its differentiated series into the equation. Collecting the terms before shifting an index givesThe coefficient of the lowest power $x^{\sigma -1}$ is $a_0\sigma (\sigma +\gamma -1)$. Because $a_0\ne 0$, the indicial equation givesFor the coefficient of $x^{n+\sigma -1}$ with $n\ge 1$, shift the index in the second sum. The hypergeometric Frobenius recurrence isWhenever the left coefficient is nonzero, division yieldsFor noninteger $\gamma$, the two independent local solutions can therefore be identified aswith a consistent local branch for the fractional power. The hypergeometric function coefficients follow by iterating the recurrence using rising factorials.

There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an undetermined coefficient at Frobenius resonance, use the undivided coefficient equation to test compatibility. For example, $\gamma =0,\sigma =0,n=1$ requires $0=\alpha \beta a_0$, so a pure exponent-zero series with $a_0\ne 0$ is impossible if $\alpha \beta \ne 0$. When $\gamma =1$ the two indicial roots coincide, and a second independent solution generally needs a logarithmic term. The indicial roots and undivided relation remain the valid coefficient conclusions in these exceptional cases.

Define $h=f-y^{\prime }{f}_{y^{\prime }}$. For a density without explicit $x$ dependence, the chain rule givesThus the Euler-Lagrange equation implies the Beltrami identityWhere $y^{\prime }\ne 0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

The monotonic horizontal coordinate allows the path to be written as a graph. Its arc length element is $ds=\sqrt{1+y^{\prime 2}}dx$, so the variational density is $f(y,y^{\prime })=g(y)\sqrt{1+y^{\prime 2}}$. It has no explicit horizontal-coordinate dependence. The Beltrami identity becomesConsequently every smooth minimizing path satisfiesThis is a first integral of the weighted arc-length problem; fixed endpoint values remove the endpoint variation terms.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

The path rises from $(-1,1)$ to its maximum height $\sqrt{2}$ at $x=0$, then falls symmetrically to $(1,1)$. Its radius is $\sqrt{2}$, and its endpoint polar angles are $3\pi /4$ and $\pi /4$, giving angular span $\pi /2$. Hence its ordinary arc length isEquivalently, $\sqrt{1+y^{\prime 2}}=\sqrt{2}/\sqrt{2-x^2}$ and integrating over $[-1,1]$ gives the same value. This Euclidean length differs from the weighted destruction integral $D_{\min }\simeq 1.76275$: the slightly longer route saves undergrowth by travelling at greater height.

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is