Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2

Seek a positive weight such that the differential operator becomes $-w^{-1}(p{y}^{\prime }{)}^{\prime }$. Matching the second derivative gives $p=w(1-x^2)$, while matching the first derivative requires $p^{\prime }=-xw$. ConsequentlyChoosing the multiplicative constant to be one givesThis is Sturm-Liouville form with zero potential term. Multiply the equations for $y_n,y_m$ by $y_m,y_n$, subtract, and integrate. The Lagrange identity for a Sturm-Liouville operator givesThe boundary expression vanishes because $p\to 0$ and the functions and their derivatives are finite. Since $n,m\ge 0$ are distinct, their eigenvalues $n^2,m^2$ are distinct, provingThe endpoint singularities of the weight are integrable, so these weighted inner products exist.

For $n=0$, a constant solution and its normalization give $y_0=1$. For $n=1$, substitute $y=ax+b$: the equation gives $b=0$, and the normalization gives $a=1$. For $n=2$, put $y=a{x}^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. ThusThese are the first three Chebyshev polynomials. For their squared weighted norms, use $x=\cos \theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:More generally, $y_n(\cos \theta )=\cos (n\theta )$ makes their orthogonality the usual cosine orthogonality.

Let $L=-(1-x^2)D^2+xD$, so $L{y}_n=n^2{y}_n$ and the forced equation is $({\gamma }^2-L)Z=e^{\epsilon x}$. The small-parameter Taylor expansion of the forcing isEach Chebyshev polynomial can be inverted separately under ${\gamma }^2-L$. None of the required denominators vanishes because $\gamma$ is a real noninteger. Thus the regular solution of a forced Chebyshev equation hasHere we select the same regular endpoint domain as in part (a), with finite function values and finite endpoint derivatives. This qualification is needed: the printed forced equation alone does not state endpoint conditions. Without this selection one can add $c_1\cos (\gamma \arccos x)+c_2\sin (\gamma \arccos x)$, which remain finite but generally have singular endpoint derivatives. With $x=\cos \theta$, regularity gives $Z_{\theta }(0)=Z_{\theta }(\pi )=0$; the corresponding homogeneous equation has only the zero solution for noninteger $\gamma$. This makes the displayed expansion unique.

Write the integral's multiplier in the Chebyshev polynomial basis:At ${\gamma }^2=5$, part (i) givesBy orthogonality, the $y_1$ term and all cross terms vanish. Using $I_0=\pi$ and $I_2=\pi /2$, the integral isThe remainder can be integrated against this integrable weight in the regular expansion, so it retains the stated order.

Put $\Psi =U(r)V(\varphi )$ in the Laplace equation in polar coordinates. After multiplication by $r^2/(UV)$, separation of variables givesThe angular function is $2\pi$-periodic and single-valued, so its nonconstant modes are $\cos (n\varphi ),\sin (n\varphi )$ with integer $n\ge 1$. Their radial Euler differential equation has solutions $r^n,r^{-n}$; finiteness at the origin excludes $r^{-n}$.

The zero angular mode has constant angular part and radial solutions $1,\log r$. The logarithmic term is again excluded at the origin. Superposition therefore yieldsThis is the regular harmonic expansion inside the disk.

On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant sine and cosine has zero average, soAll positive-power radial terms vanish at $r=0$, leaving $\Psi (0)=A_0$. HenceThis is the mean value property for harmonic functions. It is an average over arclength as well as over angle, since $ds=Rd\varphi$. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.

Write the boundary Fourier series as $f=A_0+{\sum }_{n\ge 1}(a_n\cos n\varphi +b_n\sin n\varphi )$. Since $R=2$, its coefficients obey $a_n=2^n{A}_n$ and $b_n=2^n{B}_n$. The mean isFor $n\ge 1$, the integral of $\cos n\varphi$ on the first semicircle is zero. On the second semicircle, use ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ and cosine orthogonality. This gives $a_2=1/4$ and every other $a_n=0$.

For the sine coefficients, every even $n$ gives zero on both semicircles. For odd $n$, product-to-sum formulas giveThus the requested radial-expansion coefficients are

For any interior point $\mathbf{p}$, choose a sufficiently small circle centred on $\mathbf{p}$ and wholly inside the disk. Translating the origin does not change the Laplace equation, so part (b) givesSuppose a real harmonic function, continuous on the closed disk, attains its global maximum $M$ at an interior point. Every value on each such circle is at most $M$, and their average is $M$. By continuity, every value on the circle must therefore equal $M$. Varying its radius shows that a neighbourhood of the point also has value $M$.

Thus the set of points attaining $M$ is open in the disk, and it is closed by continuity. The disk is connected, so this set must be the whole disk. The function is constant and has the same maximum on the boundary. Otherwise the maximum cannot be interior. Compactness of the closed disk guarantees existence of a maximum, proving the maximum principle for harmonic functions:Continuity up to the boundary is the standard hypothesis for this attained-maximum formulation.

Use the convention $\Delta G=+{\delta }^{(3)}$. The Dirichlet Green function for this Poisson equation satisfies, for each source point ${\mathbf{r}}^{\prime }\in V$,It is a harmonic function away from its source, with a unit delta singularity there. Its boundary condition is homogeneous even when the original potential has nonzero boundary values; those values enter through the boundary integral. This sign convention is the one giving the plus signs in the formula requested in part (b).

Apply Green's second identity to $\Phi (\mathbf{r})$ and $G(\mathbf{r},{\mathbf{r}}^{\prime })$, differentiating in $\mathbf{r}$:The volume terms become $\Phi ({\mathbf{r}}^{\prime })-{\int }_{V}G\rho dV$. On the boundary $G=0$ and $\Phi =f$, soHere $\partial /\partial n$ is the outward normal derivative in the integration variable. One may justify the identity at the singularity by excising a small sphere around ${\mathbf{r}}^{\prime }$; its unit flux supplies the evaluation $\Phi ({\mathbf{r}}^{\prime })$.

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be reflection of the interior source across the equatorial plane. Using the ball Green's function from part (c), formBoth terms vanish on the unit sphere. On the flat disk $z=0$, distances to reflected points coincide, and the same is true of their inversion images, so the difference is zero there as well. Only ${\mathbf{r}}^{\prime }$ is inside the upper half-ball; the reflected source is in the lower half-ball, and both inversion images lie outside the full ball. Thus there is exactly one interior delta source, of the required strength.

Explicitly, with $s=|{\mathbf{r}}^{\prime }|$ and stars denoting inversion in the unit sphere,This is the Dirichlet Green function inside a hemisphere, constructed from four image-source terms.

A sufficient contour setting for both results is a positively oriented, simple, closed, piecewise smooth contour $C$, with the function holomorphic on an open neighbourhood of $C$ and its entire interior. The Cauchy integral theorem then statesIf $z_0$ is strictly inside $C$, the Cauchy integral formula statesThe contour must not pass through $z_0$. Positive orientation means anticlockwise; reversing it reverses the sign of the contour integral. The holomorphy assumption throughout the interior excludes hidden singularities there.

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore

A sufficient assumption is that $f$ is absolutely continuous, $f,f^{\prime }\in L^1(\mathbb{R})$, and $f(t)\to 0$ at both infinities. Integration by parts in the Fourier transform givesThe decay assumption removes the boundary term, and integrability makes both transforms ordinary convergent integrals. The same derivative identity extends to tempered distributions.

Apply the Fourier transform of a derivative twice to the damped harmonic oscillator equation. Under the corresponding convergence assumptions,ThusSince $\kappa ,\Omega >0$, this denominator has no real zero: its imaginary part could vanish only at $\omega =0$, where its real part is positive. The transform therefore defines a regular response on the real frequency axis.

Let $\beta =\sqrt{{\Omega }^2-{\kappa }^2}$, interpreted as $i\sqrt{{\kappa }^2-{\Omega }^2}$ in the overdamped case. Factoring the denominator givesFourier inversion and the convolution theorem givewhereThus the requested numerator isFor all positive $\kappa ,\Omega$, both poles lie in the upper half-plane. In the overdamped case their imaginary parts are $\kappa \pm \sqrt{{\kappa }^2-{\Omega }^2}$, both positive. At equality they merge into a double pole.

For $t>0$, close the inverse-transform contour in the upper half-plane. There $|e^{i\omega t}|\le 1$; the rational factor is $O(|\omega {|}^{-2})$, so the arc integral vanishes. For $\Omega >\kappa$, write $\beta =\sqrt{{\Omega }^2-{\kappa }^2}>0$. The two simple poles are $i\kappa \pm \beta$, and the residue theorem yieldsThis causal Green function of a damped oscillator has damped oscillations with envelope $e^{-\kappa t}$ and tends to zero as $t\to \infty$.

For $\kappa >\Omega$, put $\eta =\sqrt{{\kappa }^2-{\Omega }^2}$, so $0<\eta <\kappa$ and the poles are $i(\kappa \pm \eta )$. Closing in the upper half-plane for $t>0$ again includes both simple poles. The residue theorem givesThis overdamped response has no oscillation. Its slow tail is $e^{-(\kappa -\eta )t}/(2\eta )$, and it tends to zero because $\eta <\kappa$.

At $\Omega =\kappa$, the inverse-transform denominator is $-(\omega -i\kappa {)}^2$. For $t>0$, its upper-half-plane double-pole residue is the derivative of the exponential:It is also the continuous limit of both answers above as $\beta \to 0$ or $\eta \to 0$. This critically damped response tends to zero with a linear factor multiplying the exponential.

In all three cases, for $t<0$ close the contour in the lower half-plane, where $e^{i\omega t}$ decays. There are no poles inside, soAt $t=0$ its continuous value is zero and $G^{\prime }(0^{+})=1$, producing the unit derivative jump required by $(D_t^2+2\kappa D_t+{\Omega }^2)G=\delta (t)$. Thus all three formulas are causal Green's functions and all decay to zero at late times.

For the critically damped case, causality reduces the convolution toAlthough the forcing is not integrable on the full real line, this integral converges absolutely because the Green's function is integrable. Thus the calculation is valid directly; the corresponding Fourier transform can be interpreted as a tempered distribution.

Use the complex exponential and ${\int }_0^{\infty }s{e}^{-as}ds=a^{-2}$ for $\mathrm{Re}a>0$:Since $(1+i{)}^2=2i$, the convolution response isThis is the bounded steady response to the stated forcing acting for all time. Without an initial condition or the convolution selection, the general solution additionally contains $(C_1+C_{2}t)e^{-\kappa t}$. No switch-on at $t=0$ was specified, so zero initial displacement and velocity should not be imposed on this all-time forcing.

Split the Cartesian second-rank tensor into symmetric and antisymmetric parts:Define $u_k=\frac{1}{2}{ϵ}_{kij}{T}_{ij}$. The symmetric part contributes zero to this contraction. The epsilon-delta identity then givesThereforeThe symmetric part and the three independent antisymmetric components determine these quantities uniquely. Under proper rotations $u$ is a vector; under reflections it is an axial vector.

Apply the symmetric gradient and curl decomposition separately to the electric field and the magnetic field:The corresponding axial parts areThe second equalities use the vacuum Maxwell equations. Their divergence constraints also show that both symmetric gradient tensors are traceless.

Differentiate ${\mathbf{u}}^E=\frac{1}{2}\nabla \times \mathbf{E}$ with respect to time. The vacuum Maxwell equations give ${\partial }_t\mathbf{E}=c^2\nabla \times \mathbf{B}$, soThe curl of the curl identity is $\nabla \times (\nabla \times \mathbf{B})=\nabla (\nabla \cdot \mathbf{B})-{\nabla }^2\mathbf{B}$. Since $\nabla \cdot \mathbf{B}=0$,

The same Maxwell equations givewhere the last equality uses $\nabla \cdot \mathbf{E}=0$. Combining with part (ii),The divergences of the symmetric gradients satisfyThus the local identity is ${\partial }_t(u_i^E+u_i^B)={\partial }_j(S_{ij}^E-c^2{S}_{ij}^B)$. Integrate over the fixed volume and apply the divergence theorem componentwise:Here $d{A}_j=n_{j}dA$ uses the outward normal vector.

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

Choose positive $\varphi$ so the bob is displaced to the right relative to the cart. Its Cartesian coordinates areDifferentiation gives ${\dot{X}}^2+{\dot{Y}}^2={\dot{x}}^2+2L\cos \varphi \dot{x}\dot{\varphi }+L^2{\dot{\varphi }}^2$. Thus the exact kinetic energy and potential energy areRetain quadratic terms, using $\cos \varphi =1+O({\varphi }^2)$ in the kinetic cross term and $1-\cos \varphi ={\varphi }^2/2+O({\varphi }^4)$ in the potential. The Lagrangian isIts Euler-Lagrange equations giveThe kinetic cross term is essential: acceleration of the support couples the pendulum and cart. These equations define the normal modes of a spring-mounted pendulum cart.