Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.

For Laplace's equation in polar coordinates, put $\Phi =R(r)\Theta (\theta )$. Separation of variables gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Periodicity selects integers $n\ge 0$. For $n>0$, the radial solutions are $r^n,r^{-n}$; for $n=0$, they are $1,\ln r$, while a term linear in $\theta$ is excluded by periodicity. Their superposition isInside the disk, boundedness at the origin removes the logarithm and all negative powers. The separately biased semicircular boundary values have mean $(V_1+V_2)/2$, no nonconstant cosine coefficients, and sine coefficients $2(V_1-V_2)/(\pi n)$ for odd $n$, zero for even $n$. The disk potential with two semicircular boundary values is thereforePut $w=(r/a)e^{i\theta }$. The odd logarithmic series sums to ${\sum }_{m\ge 0}{w}^{2m+1}/(2m+1)=\frac{1}{2}\log [(1+w)/(1-w)]$. The quotient has positive real part for $|w|<1$, so its argument is the principal arctangent of $2(r/a)\sin \theta /[1-(r/a{)}^2]$. Taking imaginary parts givesThe boundary limits away from the two voltage jumps are $V_1,V_2$. Along the diameter through a jump the value is the mean; incompatible adjacent boundary limits prevent a continuous extension at the junctions.

The denominator factors as $(z+1-i)(z+1+i)$ and neither factor cancels with the numerator. Thus there are simple poles at $z=-1\pm i$. At infinity the quotient tends to one and has a regular expansion in $1/z$, so it has a removable singularity at infinity, with extended value one.

The numerator is nonzero at $1$ and $-2/3$, so the finite singularities are a pole of order three at $1$ and a pole of order two at $-2/3$. The numerator degree exceeds the denominator degree by three, with leading behaviour $z^3/9$. Hence infinity is a pole of order three.

There are logarithmic branch points at $z=-3i$ and infinity. At zero, every local branch has $\log (3i)=\ln 3+i(\pi /2+2\pi k)\ne 0$, so division by $z^2$ gives a pole of order two at zero. Decay at infinity does not remove its branching: with $w=1/z$, the function is $w^2[\log (1+3iw)-\log w]$, which changes branch when $w$ encircles zero.

A holomorphic function is complex differentiable throughout an open region; equivalently it is locally represented by a convergent complex power series. For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$, $u_y=-v_x$. With continuous first partial derivatives, these equations imply holomorphicity and $f^{\prime }=u_x+i{v}_x$.

For $e^z=e^x\cos y+i{e}^x\sin y$, the equations hold and $u_x+i{v}_x=e^x(\cos y+i\sin y)=e^z$. For $\sin z=\sin x\cosh y+i\cos x\sinh y$, they also hold, and $u_x+i{v}_x=\cos x\cosh y-i\sin x\sinh y=\cos z$. Therefore

The Cauchy-Riemann equations give $v_x=2x$, $v_y=2-2y$. Integrating and matching both derivatives gives the harmonic conjugateThe real part of this polynomial is exactly the required $2x(1-y)$.

Here $v_x=-3x^2+3y^2+4y$ and $v_y=6xy+4x$. The harmonic conjugate and resulting polynomial areIndeed $\mathrm{Re}(-i{z}^3)=3x^{2}y-y^3$ and $\mathrm{Re}(2z^2)=2x^2-2y^2$.

For a pole of order $N$ at $z_0$, the residue is ${\mathrm{Res}}_{z_0}F=\frac{1}{(N-1)!}{\lim }_{z\to z_0}\frac{d^{N-1}}{d{z}^{N-1}}[(z-z_0{)}^{N}F(z)]$. For the rational integral, close the real-axis contour with an upper semicircle. The poles are double at $i$ and simple at $2i$, withThe arc integral vanishes as $O(R^{-5})$. By the residue theorem the whole real-axis integral is $\pi /9$, and evenness givesFor the second calculation, reciprocal expansion of the sine gives $1/\sin z=z^{-1}+z/6+O(z^3)$. Multiplying by $\cos z=1-z^2/2+O(z^4)$ yields $\cot z=z^{-1}-z/3+O(z^3)$. ThereforeAt every nonzero integer $n$, $\pi \cot \pi z$ has a simple pole with residue one, giving residue $1/n^2$ after division by $z^2$.

The cotangent contour summation uses the positively oriented square of half-side $R=N+1/2$. Its length is $8R$ and $|z|\ge R$ on its boundary. The supplied cotangent bound implies $|\oint Fdz|\le 8\pi C/R\to 0$. Its enclosed residues sum to $-{\pi }^2/3+2{\sum }_{n=1}^N{n}^{-2}$. The residue theorem therefore proves the Basel problem value

The one-sided Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$. Its Bromwich inversion formula is $f(t)=(2\pi i{)}^{-1}{\int }_{\gamma -i\infty }^{\gamma +i\infty }{e}^{pt}\bar{f}(p)dp$, with an upward vertical contour to the right of all singularities and within the appropriate convergence half-plane.

To derive the convolution theorem for Laplace transforms, transform $x(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. With $u=t-\tau \ge 0$, the triangular integration region becomes a quadrant. Fubini's theorem gives $\bar{x}(p)={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-p(u+\tau )}f(\tau )g(u)dud\tau =\bar{f}(p)\bar{g}(p)$, under the usual absolute convergence hypotheses. Inversion gives the required convolution formula.

For the differential equation, the transformed initial conditions give $(p^2-1)\bar{y}=\bar{g}+p+1$, hence $\bar{y}=\bar{g}/(p^2-1)+1/(p-1)$. Since $\mathcal{L}[\sinh t]=1/(p^2-1)$ for $\mathrm{Re}p>1$, the convolution yieldsThe integral and its first derivative vanish at zero, so the homogeneous term supplies both initial values. Differentiating twice gives the forcing through the unit derivative of the hyperbolic-sine kernel at zero.

For a Cartesian tensor under a proper orthogonal change $R$, the rank-$n$ law is $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, with repeated indices summed. An isotropic tensor has $T^{\prime }=T$ for every such rotation.

For rank two, $T=RT{R}^T$. A half-turn about a coordinate axis, obtained by applying a quarter-turn twice, reverses each off-diagonal component with exactly one index on that axis. Invariance forces every off-diagonal entry to vanish. A quarter-turn about the third axis exchanges $T_{11}$ and $T_{22}$; quarter-turns about the other axes equate them with $T_{33}$. Thus the isotropic second-rank tensor isConversely $R(\lambda I)R^T=\lambda I$, proving sufficiency.

The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are Cartesian tensors: transposing $T^{\prime }=RT{R}^T$ gives $(T^{\prime }{)}^T=R{T}^T{R}^T$, so $S^{\prime }=RS{R}^T$ and $A^{\prime }=RA{R}^T$. Hence this is a decomposition into a symmetric tensor and an antisymmetric second-rank tensor, rather than just a decomposition of numerical arrays.

For the axial vector of an antisymmetric second-rank tensor, contract the two Levi-Civita symbols:ThusWith this convention $Ax=x\times a$, so the displayed matrix is the negative of the cross-product matrix for $a\times x$.

For proper rotations the Levi-Civita symbol is invariant, so contracting the two rank-one tensors ${\partial }_j$ and $v_k$ gives $c_i={ϵ}_{ijk}{\partial }_j{v}_k$ with transformation $c_i^{\prime }=R_{i\ell }{c}_{\ell }$. Thus the curl is a vector under proper rotations. Under an orientation-reversing orthogonal change it acquires an additional determinant sign: it is an axial vector when $v$ is a polar vector.

Commuting smooth partial derivatives and using epsilon contraction gives the curl of the curl identity:Now take the curl of Faraday's law and use the vacuum Ampère-Maxwell equation: $\nabla \times \nabla \times E=-{\partial }_t(\nabla \times B)=-c^{-2}{E}_{tt}$. Since $\nabla \cdot E=0$, the identity gives $\Delta E-c^{-2}{E}_{tt}=0$. Taking the curl of the Ampère-Maxwell equation gives the same result for $B$, using $\nabla \cdot B=0$. Hence both satisfy the electromagnetic wave equation

For small oscillations with quadratic kinetic and potential energies, a normal mode has a fixed spatial amplitude pattern and one angular normal frequency. Its amplitude is a normal coordinate; choosing all modes diagonalizes the two quadratic forms and decouples their equations. A zero frequency describes a free symmetry motion.

Let $q_j=R[{\theta }_j-j\pi /2]$ be the small signed arc displacements, with cyclic indices $j=0,1,2,3$. The springs follow arcs, so their extensions are $q_{j+1}-q_j$. The Lagrangian of the normal modes of a four-bead spring ring isThe stiffness matrix divided by $k$ is the cycle graph Laplacian, with diagonal entries two, adjacent entries minus one, and zero opposite entries. An orthonormal basis of mode vectors and their frequencies isThe normal mode coordinates are $Q_{\alpha }=u_{\alpha }\cdot q$, with equations ${\ddot{Q}}_{\alpha }+{\omega }_{\alpha }^2{Q}_{\alpha }=0$. The zero mode rotates all beads together without stretching a spring and has $Q_0=A+Bt$. In each of the two degenerate middle modes, opposite beads have opposite signed arc displacements while the other pair remains stationary. Any fixed linear combination is another mode of that frequency. The last mode alternates the signed displacement of adjacent beads. The radius cancels from the frequencies because arc length supplies the same factor $R^2$ to the angular kinetic and spring energies.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

In a group of order six, element orders divide six. If there were no element of order three or six, every nonidentity element would have order two, making the group abelian by the preceding argument. Choose $a\ne e$ and $b\notin \{e,a\}$. Then $\{e,a,b,ab\}$ is a subgroup of order four: its elements are distinct, it is closed because $a,b$ commute and square to $e$, and each element is its own inverse. This contradicts Lagrange's theorem, since four does not divide six. Hence an element of order three or six exists.

Suppose $x,y$ have order three but $x\ne y,y^2$. Their cyclic subgroups have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^i{y}^j$, $0\le i,j<3$, are distinct. Indeed $x^i{y}^j=x^k{y}^{\ell }$ implies $x^{i-k}=y^{\ell -j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.

If an element has order six, it generates $G$, so $G\cong C_6$. Otherwise choose an element $x$ of order three. The preceding result excludes any further order-three elements outside $\{e,x,x^2\}$, and there are no elements of order six by assumption. Each of the other three elements must therefore have order two. ThusThe noncyclic case is the symmetric group on three points, as the classification of a group of order six further identifies.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$. A group homomorphism preserves multiplication, $\Phi (ab)=\Phi (a)\Phi (b)$. It follows that $\Phi (e_1)=e_2$ and $\Phi (g^{-1})=\Phi (g{)}^{-1}$.

The kernel of a group homomorphism $K=\{g:\Phi (g)=e_2\}$ contains the identity and satisfies $a{b}^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi (gk{g}^{-1})=\Phi (g)e_2\Phi (g{)}^{-1}=e_2$. Thus $gK{g}^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a normal subgroup.

The product $f={\prod }_{i<j}(x_i-x_j)$ is a Vandermonde determinant up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. ThereforeIf a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1{)}^{r}f$ and $(-1{)}^{s}f$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus all decompositions have the same parity, and an even decomposition forces every decomposition to be even. This defines the sign of a permutation; its kernel is the alternating group, linking the parity argument to the normal-kernel result above.

An invariant subspace $W$ satisfies $D(g)W\subseteq W$ for every $g\in G$; applying the inverse gives equality. The representation is irreducible when the only invariant subspaces are $0$ and the full space.

The inverse and conjugate transpose each reverse multiplication order, so their combination preserves it:Its values are invertible, so this is the conjugate contragredient representation.

If $W$ is invariant under $D_2$, take $w\in W$ and $v\in W^{\perp }$. For the standard Hermitian inner product,since $D_2(g^{-1})w\in W$. Hence $D_1(g)v\in W^{\perp }$, proving its invariance. If $D_1$ is irreducible, $W^{\perp }$ is either zero or the whole space; finite-dimensional orthogonal complementation makes $W$ the whole space or zero, respectively. Thus $D_2$ is irreducible whenever $D_1$ is irreducible.