Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2

The weight cancels the denominator in the Sturm-Liouville operator. Two integrations by parts, with real $p,q$, give the Lagrange identity for a Sturm-Liouville operatorThus the boundary form must vanish for every pair in the operator domain:Separated real Robin boundary conditions $A_{s}y(s)+B_{s}p(s)y^{\prime }(s)=0$ at $s=a,b$, with $(A_s,B_s)\ne (0,0)$, make each endpoint contribution vanish. This includes Dirichlet boundary conditions and Neumann boundary conditions. Coupled conditions can also work: periodic $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$ cancel the two endpoint contributions. For the regular problem, imposing a complete pair of such self-adjoint boundary conditions gives a self-adjoint differential operator; the displayed identity is the boundary-form criterion.

First, setting $f=g=y_n$ in the self-adjoint operator identity shows that ${\lambda }_n⟨y_n,y_n{⟩}_w$ is real, so every eigenvalue is real. For two eigenfunctions,Distinct eigenvalues therefore implyThis is orthogonality of Sturm-Liouville eigenfunctions.

Factor out $x$ from the differential expression:Comparing with $\mathcal{L}y=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ givesMultiplying all three by the same positive constant would give the same Sturm-Liouville operator.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and

Expand the solution and the forcing in the orthonormal Sturm-Liouville eigenfunction expansion. The forcing coefficients areEvery ${\lambda }_n$ is nonzero, so the solution coefficient is $b_n/{\lambda }_n$. ThusAs an independent check, solving $Y^{\prime \prime }+Y=-1$ directly with the two endpoint conditions givesIt satisfies both Dirichlet boundary conditions and the forced ordinary differential equation.

A single-valued potential must be $2\pi$-periodic in $\theta$. For a separated solution $\Phi =F(r)H(\theta )$, the Laplace equation in polar coordinates becomesThe periodic angular eigenvalue problem has $\mu =n^2$, $n\ge 0$: its nonconstant eigenfunctions are $\cos n\theta ,\sin n\theta$. Negative $\mu$ gives exponentials that cannot be periodic, and $\mu =0$ gives only a constant angular factor. For $n\ge 1$, the radial Cauchy-Euler equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ gives $F=A{r}^n+B{r}^{-n}$. At $n=0$ the two radial solutions are $1$ and $\log r$. Superposition therefore givesThe expansion is interpreted on an annulus where the harmonic function is defined; particular boundary conditions determine its constants and convergence region. Single-valuedness is essential: allowing a cut in the angular domain would also permit nonperiodic angular solutions.

Treat each angular Fourier mode independently. Boundedness at the axis excludes the negative radial powers and the logarithm inside; decay at infinity excludes the positive powers, constant and logarithm outside. For the cosine mode writeContinuity at $R$ gives $B_n=A_n{R}^{2n}$. The jump of the normal derivative is thenThus $A_n=a_n{R}^{1-n}/(2n)$ and $B_n=a_n{R}^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the potential of a charged cylindrical shell with angular Fourier data isThere is no angular mean term: the given surface charge density has zero average. The interior constant is zero because its boundary value must match the vanishing exterior mean. A nonzero mean charge per unit length would instead produce a logarithmic electrostatic potential, incompatible with the specified decay at infinity.

Apply the product rule to the vector field $\Phi \nabla \Psi -\Psi \nabla \Phi$:The gradient cross terms cancel. Integrating and applying the divergence theorem with outward unit normal $\mathbf{n}$ proves Green's second identity:Choose the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ with ${\Delta }_{\mathbf{r}}{G}_D=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ and $G_D=0$ on $S$. Taking $\Psi =G_D$ and using the Dirac delta distribution, the left side becomes $\Phi ({\mathbf{r}}^{\prime })$ while the second boundary term vanishes. HenceThe plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space fundamental solution is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about ${\mathbf{r}}^{\prime }$ before applying the smooth identity.

Put $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Away from the source, the radial Laplacian in three dimensions givesThus $\boxed{\Delta G_0=0\text{ for }\mathbf{r}\ne {\mathbf{r}}^{\prime }}$. At the source it is a distribution, not a smooth function; its outward gradient flux through a small sphere is $4\pi s^2/(4\pi s^2)=1$, giving $\Delta G_0=\delta$.

Write $s=|{\mathbf{r}}^{\prime }|$ and ${\mathbf{r}}^{\prime \prime }=a^2{\mathbf{r}}^{\prime }/s^2$. If $|\mathbf{r}|=a$, expansion of squared Euclidean distances givesConsequently $|\mathbf{r}-{\mathbf{r}}^{\prime \prime }|=(a/s)|\mathbf{r}-{\mathbf{r}}^{\prime }|$, and the two terms in $H$ cancel. This method of images gives the Dirichlet Green function inside a sphere:The image point is outside the sphere, so only the physical source contributes a Dirac delta distribution inside. As $s\to 0$, the image position tends to infinity and its coefficient also diverges, but their ratio has the finite limitThusThe apparent singularity of the image construction at a central source is removable; the physical source singularity at $\mathbf{r}=0$ remains.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be the reflection of the source in the equatorial plane. For a source in the upper half-ball, the Dirichlet Green function inside a hemisphere isBoth terms vanish on the spherical boundary. On the flat boundary $z=0$, the two source distances are equal, and the distances to their spherical image points are also equal, so the difference vanishes there too. The reflected physical source is in the lower half-ball and both inverted image points are outside the sphere. Therefore the only source singularity in the desired region is the original one, with coefficient $+1$ in $\Delta G$. This combines reflection with the spherical method of images.

Reflecting repeatedly in the two parallel planes creates an infinite method of images. For ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$, introduceThe Dirichlet Green function between parallel planes isEach bracket is $O(n^{-2})$, so the paired series converges absolutely away from the sources. At $x=-1$, reflection pairs the positive image indexed by $n$ with the negative image indexed by $-n$. At $x=1$ it pairs that positive image with the negative image indexed by $1-n$. The distances agree in each pairing, making both boundary values zero. This can be checked with symmetric finite partial sums first; the unmatched end terms tend to zero, so no subtraction of separately divergent series is needed. Only ${\mathbf{r}}_0^{+}={\mathbf{r}}^{\prime }$ lies within the slab, so ${\Delta }_{\mathbf{r}}{G}_{\mathrm{slab}}=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. This is the usual Green function decaying at large transverse separation; such a condition selects the physical solution in the unbounded region.

A function is an analytic function at $z_0$ if it is holomorphic throughout some open neighbourhood of $z_0$, equivalently if it has a convergent power series there. In real components $g=u+iv$, continuous first partial derivatives in a neighbourhood together with the Cauchy-Riemann equationsthroughout that neighbourhood are sufficient, and for an analytic function they hold there. Satisfying these equations only at the single point is not enough to establish analyticity in a neighbourhood.

Factor the polynomial denominator as $(z-i{)}^2(z+i{)}^2$. The numerator has values $i$ and $-i$ at the two roots, so neither cancels. ThereforeThe factor $z^{-1/2}$ gives a branch point at zero, with local behaviour $f(z)\sim z^{-1/2}$. Infinity is also a branch point, since in the coordinate $w=1/z$ the leading behaviour is $f(1/w)\sim w^{5/2}$. A single-valued branch requires a branch cut joining these points. Away from that cut and the two poles, the selected branch is holomorphic. For the integral below, use $0<\arg z<2\pi$, putting the cut along the positive real axis.

Use a counterclockwise keyhole contour around the positive real branch cut, with small radius $\epsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\epsilon$, so its contribution also tends to $I$.

The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O({\epsilon }^{1/2})$, since $f=O({\epsilon }^{-1/2})$. Both vanish. HenceThe contour encloses both double poles $\pm i$ but excludes the branch point.

For a double pole at $z_0\in \{i,-i\}$, differentiate the remaining holomorphic factor:Since $z_0^2=-1$ and $1+z_0+z_0^2=z_0$, this simplifies toOn the selected branch $\sqrt{i}=e^{i\pi /4}$ and $\sqrt{-i}=e^{3i\pi /4}$, givingThe residue theorem therefore givesThe equality $C=2I$ uses the limiting contour integrals established in (c).

The signal is continuous at zero with $f(0)=1$. Its envelope is $e^{-\lambda |t|}$, but its angular frequency falls from $2$ on the incoming side to $1$ on the outgoing side. Its one-sided slopes are $f^{\prime }(0^{-})=\lambda$ and $f^{\prime }(0^{+})=-\lambda$, producing a cusp at the peak. Both tails decay to zero.

Use $\tilde{f}(\omega )={\int }_{\mathbb{R}}{e}^{-i\omega t}f(t)dt$. Expanding each cosine into two exponentials and integrating its half-line givesEquivalently,The positive $\lambda$ guarantees convergence of the Fourier transform integrals.

The Fourier transform of a derivative gives $(-{\omega }^2-1)\tilde{y}=-\tilde{f}$. For $\lambda =1$,This selects the solution decaying at both time infinities. Indeed, the inverse transform of $1/(1+{\omega }^2)$ is the one-dimensional modified Helmholtz Green function $K(t)=e^{-|t|}/2$, and the convolution theorem gives $y=K\ast f$. The printed equation has homogeneous solutions $e^t,e^{-t}$, so these two decay conditions matter; no causal initial-value condition was specified.

For $t<0$, evaluate Fourier inversion by closing the contour in the lower half-plane, where $e^{i\omega t}$ decays. The contour is clockwise. The enclosed simple poles and residues of $\tilde{y}$ areThus inversion equals $-i$ times the sum of the residues of $e^{i\omega t}\tilde{y}$, givingAs a check, applying $d^2/d{t}^2-1$ gives $-e^t\cos 2t$. The convolution also gives $y(t)=e^{-t}[\frac{1}{8}+\frac{1}{5}\cos t+\frac{2}{5}\sin t]$ for $t\ge 0$; the two pieces have matching values $13/40$ and matching first derivatives $3/40$ at zero.

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

Use three components of the proposed isotropic tensor form. Since $T_{1122}=0$, one gets $\alpha =0$. Since $T_{1212}={ϵ}_{123}^2=1$, one gets $\beta =1$. Finally $T_{1221}={ϵ}_{123}{ϵ}_{213}=-1$ gives $\gamma =-1$. Hence contraction of two Levi-Civita symbols gives

The integration domain is invariant under every rotation matrix, and its volume element is unchanged. The substitution $\mathbf{y}=R\mathbf{x}$ therefore provesThus $I$ is an isotropic tensor. It is also completely symmetric in its four indices, because the coordinate factors commute. In the general isotropic form, exchanging indices $j,k$ equates $\alpha$ and $\beta$, and exchanging $j,l$ equates $\alpha$ and $\gamma$. Therefore

The contraction in the PDF is $I_{iijj}$. By Einstein notation, its integrand is $({\sum }_i{x}_i^2)({\sum }_j{x}_j^2)=r^4$. In spherical coordinates,

Contract the isotropic expression from (i). Since ${\delta }_{ii}=3$ and ${\delta }_{ij}{\delta }_{ij}=3$,The value in (ii) gives $A=4\pi /105$, so the isotropic fourth moment of a ball isFor example ${\int }_B{x}_1^{4}dV=4\pi /35$ and ${\int }_B{x}_1^2{x}_2^{2}dV=4\pi /105$, illustrating the factor of three between these two fourth moments.

Write the inverse tensor asInsert this into $({\delta }_{ik}+\lambda X_{ik})Y_{kj}={\delta }_{ij}$ and equate powers of $\lambda$. One obtains successivelyThus the Neumann series givesEvery contraction is an ordinary ordered matrix product. No transpose or symmetry assumption is used. Multiplying the truncated series on either side by $I+\lambda X$ leaves $I-{\lambda }^4{X}^4$, which verifies the error order. For a fixed finite matrix this expansion converges whenever $|\lambda |\Vert X\Vert <1$ in a submultiplicative matrix norm.

For the stated square side length $2a$, the volume is $4a^{2}h$ and the uniform mass density is ${\rho }_1=m_1/(4a^{2}h)$. Direct integration about its central vertical axis gives the moment of inertia of a uniform square plate:Thus the PDF's coefficient $1/6$ is inconsistent with its stated side length. The coefficient $1/6$ would apply to a square of side $a$.

For the disc, use cylindrical coordinates and $m_2={\rho }_2\pi a^{2}h$:For the sphere, use spherical coordinates and $m_3=4\pi {\rho }_3{a}^3/3$:These agree with the moment of inertia of a uniform disc and the moment of inertia of a uniform solid sphere. The square's thickness does not enter its axial moment of inertia once the mass is fixed.

Measure each generalized coordinate ${\theta }_i$ from an unstressed configuration. The diagram has one torsional spring joining the frame to object 1, one joining objects 1 and 2, and one joining objects 2 and 3. The rotational kinetic energy and elastic potential energy areEach spring contribution follows by integrating its restoring torque: ${\int }_0^{\Delta \theta }{k}_{j}sds=k_j(\Delta \theta {)}^2/2$. ThusThe Euler-Lagrange equations giveThese hold for either interpretation of the square's moment of inertia; the geometry-based value from (a) is used first below.

For the stated geometry and numerical masses, the inertia and stiffness matrices areA normal mode $\mathbf{\theta }=\mathbf{v}{e}^{i\omega t}$ requires $(K-{\omega }^{2}M)\mathbf{v}=0$. Writing $z={\omega }^2$, its characteristic equation isAt the claimed $z=6$, this determinant is $-27/50\ne 0$. Hence the printed claim that $\sqrt{6}$ is a normal frequency is false for the stated data, and there is no corresponding nonzero normal mode.

The complete geometry-based answer is obtained from the three positive roots $z_j$:An exact closed form for the squared angular frequencies isThey are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding eigenvector can be chosen asThe last two rows of $(K-zM)\mathbf{v}=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring potential energy ensures that all squared angular frequencies are positive.

If instead one takes the PDF's printed $I_1=m_1{a}^2/6$ as a stipulated inertia, then $M=\mathrm{diag}(1/2,1/4,2/25)$ andThe corresponding angular frequencies are approximatelyThe same formula for $\mathbf{v}(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne 0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.