Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V (A10142W1)

Source: released examination PDF, bundle pages 230–233.

Solutions to all seven questions, including every optional question and Section C. In Question 6(b), the printed transcendental equation excludes valid separated solutions at common poles of the tangents; the complete characteristic equation and an explicit counterexample are given below.

For an oriented surface $\Sigma$ with piecewise smooth closed boundary $\partial \Sigma$, and a vector field $\mathbf{F}$ that is continuously differentiable on an open neighborhood of $\Sigma$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector. The orientation of $\partial \Sigma$ follows the right-hand rule: with the right thumb pointing along $\mathbf{n}$, the fingers curl in the positive direction. Equivalently, walking along the boundary with one's head along $\mathbf{n}$ leaves the surface on the left. Each boundary component receives this induced orientation.

Write $r=\sqrt{x^2+y^2}>0$. The cylindrical coordinates give ${\mathbf{e}}_{\theta }=(-y/r,x/r,0)$ and ${\mathbf{e}}_r=(x/r,y/r,0)$, so the vector components of the vector field areThe first two components of the curl, calculated in Cartesian coordinates, are $-x{f}^{\prime }(z)/r$ and $-y{f}^{\prime }(z)/r$. For the third component, the chain rule givesand henceCombining these vector components in the cylindrical basis yieldsThe restriction $r>0$ matters because the cylindrical basis is undefined on the axis.

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of the closure of $V$, the divergence theorem statesHere $\mathbf{n}$ is the outward unit normal vector; the right side is the outward flux through the boundary.

Apply the product rule to each vector component and sum:Thus the divergence of a scalar field times a vector field contains a divergence term and a dot product with the gradient of the scalar field.

Choose $\mathbf{F}=\nabla \psi$ in the preceding product rule and apply the divergence theorem. Since $\nabla \cdot \nabla \psi =\Delta \psi$, this gives the first of Green's identities:Here ${\partial }_n\psi =\mathbf{n}\cdot \nabla \psi$ is the outward normal derivative. Interchange $\varphi$ and $\psi$, then subtract. The gradient dot products cancel, yielding the second of Green's identities:For example, $\varphi ,\psi \in C^2(\overline{V})$ with the needed extensions suffice for these applications of the divergence theorem.

False. The constant function $u=1$ is a harmonic function and has zero normal derivative on every smooth boundary, but it is nonzero in $V$.

In fact, the first of Green's identities givesHence $\nabla u=0$; path connectedness makes $u$ constant throughout $V$. Thus this homogeneous Neumann boundary condition determines a harmonic function only up to an additive constant.

True. The first of Green's identities and the Robin boundary condition ${\partial }_{n}u=-u$ giveBoth integrands are nonnegative, so $\nabla u=0$ in $V$ and $u=0$ on the boundary. By path connectedness, $u$ is a constant function, and its boundary value makes that constant zero. Therefore $\boxed{u=0}$.

False. Take $V=\{x^2+y^2+z^2<1\}$, the open unit ball, and $u(x,y,z)=x$. This is a nonzero harmonic function. On the unit sphere, the outward unit normal vector is $\mathbf{n}=(x,y,z)$, soThus the prescribed Robin boundary condition holds. Unlike the preceding case, Green's identities now give ${\int }_V|\nabla u{|}^2={\int }_{\partial V}{u}^2$, which permits nonzero solutions of the differential equation.

True. Two applications of Green's identities, or two integrations by parts, giveThe biharmonic equation makes the left side zero, while the two clamped boundary conditions $u={\partial }_{n}u=0$ annihilate the boundary integral. Consequentlyso $\Delta u=0$. Apply the first of Green's identities once more with $\varphi =\psi =u$: because $u=0$ on the boundary, it gives ${\int }_V|\nabla u{|}^2=0$. By path connectedness, $u$ is constant, and the zero boundary value implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed integrations by parts.

For the given wrapped hyperbolic cosine curve,so $\boxed{C\subset S}$. The curve makes one full turn around the circular cylinder, descending from $(-1,0,\cosh \pi )$ to $(1,0,1)$ and ascending to its initial point. Its two parameter endpoints coincide; their tangent vectors $(0,-1,-\sinh \pi )$ and $(0,-1,\sinh \pi )$ do not agree, so the resulting closed curve has a corner there.

The velocity vector of the parametric curve isIts speed iswhere $\cosh t>0$ fixes the sign of the square root. Hence the arc length isThe coincidence of the endpoints does not duplicate any positive arc length.

Parametrize the finite cylindrical surface patch byThe surface area element is $|{\mathbf{R}}_t\times {\mathbf{R}}_z|dtdz=dtdz$. The seam $t=\pm \pi$ has zero surface area, so it causes no overcounting. ThereforeUnrolling the unit circular cylinder preserves surface area, which explains the equality with the area under the hyperbolic cosine in the figure. The patch consists of the cylindrical surface alone; no flat base is included.

At a fixed $z\in [-1,1]$, both cylinder inequalities becomeThus the cross section is a square of side $2\sqrt{1-z^2}$ and area $4(1-z^2)$. Integrating these cross-sectional areas gives the volume of this bicylinder:

For the surface area of a tricylinder, permutation symmetry shows that each of the three cylindrical surfaces contributes the same surface area to the boundary. On the face $x^2+y^2=1$, put $x=\cos \theta$, $y=\sin \theta$. The remaining two inequalities requireso $|z|\le \min (|\sin \theta |,|\cos \theta |)$. The surface area element on this unit circular cylinder is $d\theta dz$. There are eight congruent angular sectors, with the minimum equal to $\sin \theta$ for $0\le \theta \le \pi /4$. Thus the area of this face isEvery boundary point lies on at least one of these three cylindrical surfaces; their pairwise intersections are curves of zero surface area. Adding the three faces therefore gives

The Fourier cosine series on $[0,L]$ isIt is the Fourier series of the even extension to $[-L,L]$, repeated with period $2L$. The Fourier sine series isIt comes from the odd extension and its $2L$-periodic extension. These Fourier coefficient formulas follow by Fourier orthogonality, since every nonconstant sine or cosine has squared integral $L/2$ on $[0,L]$, while the constant function has squared integral $L$.

Under the stated piecewise smooth hypotheses, the Dirichlet convergence theorem for Fourier series gives the average of the two one-sided limits of the corresponding periodic extension. The even extension is continuous at both endpoints, so the Fourier cosine series converges toThe odd extension agrees with $f$ on $(0,L)$, but its left and right limits are $-f(0),f(0)$ at zero and $f(L),-f(L)$ at the join $L$. Therefore the Fourier sine series converges toIndeed every partial sum of the Fourier sine series is already zero at both endpoints. The endpoint values agree with $f$ only when the corresponding boundary values vanish.

Set $u=x/L$ and $q=n\pi$. The required integrals follow fromConsequently the Fourier coefficients areThus the Fourier cosine series isand its sum is $e^{x/L}$ everywhere on $[0,L]$. The Fourier sine series isand its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.

First, the Fourier coefficients of the Fourier cosine series decay as $O(n^{-2})$, whereas those of the Fourier sine series are of order $n^{-1}$. In particular, the absolute convergence bound for the Fourier cosine series gives a uniform tail estimateThus its tail is absolutely controlled using fewer Fourier harmonics. The Fourier sine series does not have an absolutely summable sequence of Fourier coefficients.

Second, the even extension is continuous, so its Fourier cosine series converges uniformly on the closed interval and approximates the nonzero endpoint limits correctly. The odd periodic extension has jumps, producing the Gibbs phenomenon in its Fourier sine series near zero and $L$. Every Fourier sine series partial sum is zero at both endpoints, so by approaching $L$ from inside $(0,L)$ one obtainsHence the Fourier sine series approximation cannot converge uniformly even on the whole open interval, despite pointwise convergence at each interior point. The Fourier cosine series approximation has faster coefficient decay and avoids the persistent errors near the endpoints.

Take $\boxed{p(x)=x(L-x)}$. It vanishes at both endpoints. Its odd periodic extension is continuously differentiable across the joins, whereas its even extension has corners because $p^{\prime }(0)=L$ and $p^{\prime }(L)=-L$ are nonzero.

Two integrations by parts, with $q=n\pi /L$, giveThe Fourier coefficients are thereforeThe Fourier sine series has a uniform absolute tail $O(N^{-2})$, and each of its partial sums satisfies both zero boundary values exactly. The Fourier cosine series has an absolute tail bound $O(N^{-1})$; at $x=0$ its omitted even-index Fourier coefficients have the same sign and yield an error of order $N^{-1}$. This verifies a strictly faster uniform error bound for the sine approximation in this example.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.

Use the same difference $w$ and wave energy as in part (i). The damped wave equation now gives the energy dissipation identity for a linearly damped stringBecause $E(0)=0$ and $E(t)\ge 0$, the energy remains zero. Hence $w_t=w_x=0$, and the homogeneous Dirichlet boundary conditions force $w=0$. Therefore the solution of a differential equation is unique also for $\boxed{\alpha >0}$; positive damping can only decrease this energy.

Substitute the separated solution $y=e^{-\alpha t}F(t)\sin (nx)$ into the damped wave equation. Its time derivatives givewhereas $y_{xx}=-n^2{e}^{-\alpha t}F\sin (nx)$. ThusThe last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin (nx)$. For $0<\alpha <n$, let $\nu =\sqrt{n^2-{\alpha }^2}>0$. The simple harmonic motion equation has solution of a differential equationThis is an underdamped normal mode: its oscillations have the decaying factor $e^{-\alpha t}$.

When $\alpha =n$, the reduced ordinary differential equation is $F^{\prime \prime }=0$. The initial conditions $F(0)=0$, $F^{\prime }(0)=1$ giveThis is the critically damped normal mode, also obtained by taking $\nu \to 0$ in $\sin (\nu t)/\nu$.

When $\alpha >n$, set $\mu =\sqrt{{\alpha }^2-n^2}>0$. The reduced ordinary differential equation is $F^{\prime \prime }-{\mu }^{2}F=0$. Enforcing the same initial conditions givesThe corresponding overdamped normal mode can also be writtenBoth rates in the exponential functions are negative because $0<\mu <\alpha$, so the hyperbolic sine factor does not prevent decay.

For $\alpha =3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n^{\prime }(0)=1$. By superposition the Fourier sine series solution isAt $t=0$, its initial velocity is ${\sum }_{n\ge 1}{b}_n\sin (nx)={\sin }^{2}x$. Multiply by $\sin (mx)$, integrate on $[0,\pi ]$, and use the permitted interchange and Fourier orthogonality. This yields the requested integral expressionUnder $x\mapsto \pi -x$, ${\sin }^{2}x$ is unchanged and $\sin (n(\pi -x))=(-1{)}^{n+1}\sin (nx)$. Hence every even-index Fourier coefficient is zero, including $b_2$.

For odd $n$, use ${\sin }^{2}x=(1-\cos 2x)/2$ and the product-to-sum formula:ThusIn particular $b_1=8/(3\pi )$ and $b_3=-8/(15\pi )$. An explicit version of the solution of a differential equation isOnly the first, overdamped normal mode, the third, critically damped normal mode, and the odd underdamped normal modes of index at least five occur. No odd-index Fourier coefficient vanishes.

Conservation of thermal energy and Fourier's law give the heat equationsThe density times the specific heat capacity, ${\rho }_i{c}_i$, is the volumetric heat capacity, and $k_i$ is the thermal conductivity.

The condition $T_1(-L,t)=0$ holds the left exterior face at a fixed zero reference temperature; $T_2(L,t)=0$ does the same for the right exterior face. The condition $T_1(0_{-},t)=T_2(0_{+},t)$ expresses perfect thermal contact, with no temperature jump at the interface. Finally,expresses continuity of the signed heat flux in the positive $x$ direction: the interface neither stores nor produces heat.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

The space ${\mathbb{R}}^3$ is simply connected, and the electric field has zero curl. Hence it is a conservative vector field. More explicitly, Stokes theorem gives zero circulation around any closed loop, since such a loop bounds a spanning surface in ${\mathbb{R}}^3$. Therefore the line integral between two fixed endpoints is independent of the chosen path.

Fix a reference point ${\mathbf{r}}_0$ and define the electric potentialTaking a short straight terminal segment in each coordinate direction shows ${\partial }_j\phi =-E_j$. ConsequentlyThe electric potential is determined up to an additive constant; the later condition at infinity fixes that constant for the finite charge distribution in part (b).

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.