Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 2

Original examination paper. The solutions below cover all ten questions, including every lettered and numbered part. The original PDF is authoritative for notation and matrices.

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts four distinct roots are $m=-2,0,2,4$. The separated polar solutions of the biharmonic equation therefore give the complete radial family in this angular sector,Equivalently, its radial equation is a fourth-order Cauchy-Euler equation; four distinct powers supply four linearly independent solutions, so no logarithmic terms are required.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. The latter depends on the approach direction even though it is bounded. The two boundary conditions at unit radius then give $C+D=0$ and $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the biharmonic streamfunction; no condition requiring $\psi$ itself to vanish at infinity was imposed.

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Definitions. In a punctured neighborhood of $z_0$, let the Laurent series be $f(z)={\sum }_{j=-\infty }^{\infty }{c}_j(z-z_0{)}^j$. The point is a pole of order $m\ge 1$ if $c_{-m}\ne 0$ and every coefficient with index less than $-m$ vanishes. Its residue is $\boxed{{\mathrm{Res}}_{z_0}f=c_{-1}}$. A simple pole has $m=1$.

Poles for $0<a<1$. Denote the bracket in the denominator by $D(z)$. Factoring givesPut $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite poles are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,At a nonzero simple root $r$ of $D$, the residue formula givesFor $r^2=q^2$, $D^{\prime }(r)=8ar$; for $r^2=q^{-2}$, $D^{\prime }(r)=-8ar$. Substitution yieldsThe two zeros of a function at $z=\pm i$ do not cause additional poles: there $D=-4$ and the numerator vanishes.

The case $a=1$. Simplify before using any formulas that divide by $1-a^2$:Thus its only finite pole isAs $a\uparrow 1$, the two inner poles merge with zero while the outer two escape to infinity. If one also considers the Riemann sphere, $f$ at $a=1$ has a simple pole at infinity because $f(z)\sim z/4$; the requested unit-circle calculation only uses finite poles.

Unit-circle integral. Set $z=e^{i\theta }$. ThenAt the two angles where the tangent is undefined, the real integrand has continuous limiting value zero. Thus the substitution is valid for the improper integral, or after that continuous extension. No pole lies on $|z|=1$ for any allowed $a$.

For $0<a<1$, the sum of the enclosed residues isThe residue theorem therefore givesAt $a=1$ the enclosed residue is $1/2$, giving $I=\pi$, exactly the same formula. The direct simplification $1/(1+{\tan }^2\theta )={\cos }^2\theta$ provides an independent check of that endpoint.

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.

Choose $\gamma >1/2$. The inverse integrand is ${\lambda }^{-s}/[2(s-1/2)]$. Because $-\log \lambda >0$, the exponential factor decays towards the left. Closing the Bromwich contour to the left and using the residue theorem picks up just the simple pole at $s=1/2$, with residue ${\lambda }^{-1/2}/2$. HenceIts defining integral checks both the answer and its convergence condition:

Now the inverse integrand can be writtenFor $0<\lambda <1/2$, the exponent has a positive real time parameter. Close the Bromwich contour to the left; the residue at $s=-1$ is $\frac{1}{2}(2\lambda )=\lambda$. For $1/2<\lambda <1$, close to the right, where there are no poles. The reversed orientation there still gives zero. ConsequentlyThe same result follows from the Laplace transform time-shift rule: $H(s)=\frac{1}{2}{e}^{-s\log 2}/(s+1)$ corresponds to $f(t)=e^{-t}\Theta (t-\log 2)$, where $\Theta$ is the Heaviside step function.

The original transform cannot fix a value at the single point $\lambda =1/2$. The Bromwich inversion formula chooses the average of the limiting values $1/2$ and zero, giving $\boxed{h(1/2)=1/4}$ under that convention. Indeed,regardless of the assigned value at the jump.

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite $p_{ij}=p{\delta }_{ij}+s_{ij}+{ϵ}_{ijk}{a}_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p{\delta }_{ij}+s_{ij}-{ϵ}_{ijk}{a}_k$. Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, $a_k=0$, hence $A_k=0$ and $\boxed{P_{ij}=P_{ji}}$. No equality between $\mu$ and $\nu$ is needed for that conclusion.

A normal mode is a fixed displacement pattern whose components share a common harmonic time dependence in the equations for small oscillations. Its normal frequency is the associated oscillation frequency; below, $\omega$ denotes the angular frequency, so the frequency in cycles per unit time is $\omega /(2\pi )$. A normal coordinate is the amplitude of one such pattern, chosen so that the linear equations decouple. For mass matrix $M$ and stiffness matrix $K$, the patterns solve $Kv={\omega }^{2}Mv$. Zero eigenvalues describe free rigid motions rather than periodic restoring oscillations.

Use the equilibrium coordinates specified in the paper; their distances from the origin are one and their mutual separation is $\sqrt{3}$. Let $u_i=(x_i,y_i)$ be the small displacements. Each initially unstretched spring has, to first order, extension equal to the relative displacement projected along its original bond. Choose the three oriented extensionsThe kinetic energy is exact in these displacement coordinates, while the potential energy needed for small oscillations is its quadratic part:The factor $\sqrt{3}$ in the equilibrium bond length cancels when forming the unit bond directions. Applying the Euler-Lagrange equation, or differentiating $V$ for the force on each mass, givesThe total force vanishes componentwise, as expected for three masses coupled only by internal spring forces.

Order coordinates as $q=(x_1,x_2,x_3;y_1,y_2,y_3)$. For either uniform displacement, $t_x=(1,1,1;0,0,0)$ or $t_y=(0,0,0;1,1,1)$, every relative displacement and every extension vanishes. These normal modes have zero normal frequency and represent two independent planar translations.

For $r=(-1,-1,2;\sqrt{3},-\sqrt{3},0)$, substitution givesThus $Kr=0$ too. A small rigid rotation through angle $\alpha$ moves equilibrium position $(X_i,Y_i)$ by $\alpha (-Y_i,X_i)$. With the specified positions this displacement is $-\alpha r/2$, identifying the third vector as infinitesimal rigid rotation. HenceA zero-frequency normal coordinate obeys $\ddot{Q}=0$, so $Q=A+Bt$. This permits a constant displacement or a uniform velocity at the level of the linearized equations. In particular, the rotational pattern preserves lengths to first order; finite-angle rotation follows a curved path outside a single linear displacement pattern.

For the normal modes of an equilateral spring triangle, the complement of the three rigid motions has dimension three. One convenient mutually orthogonal basis isThese vectors are also orthogonal to $t_x,t_y,r$, and each has squared length $12$. Their geometry comes from applying symmetric linear deformations to the equilibrium triangle: $b$ is uniform expansion, $d$ stretches one axis while compressing the other, and $s$ is the second independent shear pattern. Their extensions areInsert each row into the force equations. For example, $b$ gives $m{\ddot{x}}_1=3\sqrt{3}k=-3k{b}_{x_1}$ and $m{\ddot{y}}_3=-6k=-3k{b}_{y_3}$, with the same proportionality in every component. The other two rows similarly give force $-3k/2$ times their displacement vector. Thusand the remaining normal modes and normal frequencies areAny orthogonal pair spanning the same two-dimensional shear subspace is equally valid, because its eigenvalue is repeated. The six displayed vectors are nonzero and mutually orthogonal, so they supply all six independent normal modes without solving a characteristic polynomial.

For explicit normal coordinates, normalize $t_x,t_y$ by $\sqrt{3}$ and $r,b,d,s$ by $\sqrt{12}$, and write $q={\sum }_j{Q}_j{\hat{v}}_j$. Since all masses are equal,Positive-frequency coordinates have $Q_j=A_j\cos ({\omega }_{j}t)+B_j\sin ({\omega }_{j}t)$; the three rigid coordinates have the linear form found above.

Cosets and counting. A left coset of a subgroup $H$ in a group $G$ is $gH=\{gh:h\in H\}$. The left cosets partition $G$ and each has $|H|$ elements. Lagrange's theorem states that for a finite group, $\boxed{|G|=[G:H]|H|}$; in particular, the subgroup order divides the group order.

The matrix group. Work throughout in the finite field ${\mathbb{F}}_5$, so every nonzero entry has a multiplicative inverse. Denote a matrix by $A(a,b,c)$, with $a,c\ne 0$. Ordinary matrix multiplication, interpreted using modular arithmetic, givesIts diagonal entries remain nonzero, proving closure. Associativity is inherited from matrix multiplication, $A(1,0,1)$ is the identity element, andThis lies in the same set and multiplication checks the inverse in either order. Thus these matrices form a group. There are four choices each for $a$ and $c$ and five for $b$, giving $\boxed{|G|=4\cdot 5\cdot 4=80}$.

The subgroup. Write $h(a,c)=A(a,a-c,c)$. ThenIt contains the identity $h(1,1)$, is closed, and has inverse $h(a^{-1},c^{-1})$. Interchanging the two factors leaves the product unchanged because multiplication in ${\mathbb{F}}_5$ is commutative. Therefore $H$ is an abelian subgroup, withEquivalently $H$ is isomorphic to ${\mathbb{F}}_5^{\times }\times {\mathbb{F}}_5^{\times }$ by $h(a,c)\mapsto (a,c)$, which also explains its abelian property. Its order divides $80$, verifying Lagrange's theorem.

There are five distinct left cosets. Representatives can be taken as $U_t=A(1,t,1)$, $t\in {\mathbb{F}}_5$: indeedEvery matrix lies in exactly the coset specified by this unique $t$.

Elements whose square is the identity. Squaring givesThus $a,c\in \{1,-1\}=\{1,4\}$ and $b(a+c)=0$. For equal signs, $a+c\ne 0$ in ${\mathbb{F}}_5$, so $b=0$. For opposite signs, $a+c=0$ and all five $b$ values work. The complete set is thereforeIt contains $2+5+5=12$ elements, including the identity; the other eleven are involutions. Since $12$ does not divide $80$, Lagrange's theorem proves that this set is not a subgroup.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.