Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2

For the weighted inner product, the factor $w$ cancels the $w^{-1}$ in the Sturm-Liouville operator. Assuming the coefficients are real, two applications of integration by parts giveThis is the boundary form in the Lagrange identity for a Sturm-Liouville operator. The potential terms cancel; if $q$ were complex, an additional integral of $(q-q^{\ast })f^{\ast }g$ would remain, so suitable real coefficients are part of formal self-adjointness. The endpoint restrictions in part (a) eliminate the displayed form for every pair of functions in the domain.

The necessary boundary identity for a self-adjoint operator isAt regular endpoints, a standard full family of separated self-adjoint conditions iswith real coefficients and neither pair $({\alpha }_a,{\beta }_a)$, $({\alpha }_b,{\beta }_b)$ zero. For example, if ${\beta }_a\ne 0$, both functions obey $p{f}^{\prime }=-{\alpha }_{a}f/{\beta }_a$ at $a$, so $p{f}^{\prime \ast }g-p{f}^{\ast }{g}^{\prime }=0$ there. The cases ${\beta }_a=0$ and the endpoint $b$ follow directly. These include Dirichlet boundary conditions, Neumann boundary conditions and Robin boundary conditions.

Separate vanishing at each endpoint is sufficient but not necessary. Coupled conditions such as $y(b)=e^{i\chi }y(a)$ and $p(b)y^{\prime }(b)=e^{i\chi }p(a)y^{\prime }(a)$, with real $\chi$, make the two endpoint forms equal and hence their difference zero. More generally the allowed endpoint data must be a maximal subspace on which the boundary form vanishes; this ensures that the adjoint has the same domain. Merely imposing extra endpoint restrictions until the boundary term vanishes proves symmetry, but can overrestrict the domain and fail to give self-adjointness.

Let $L{y}_n={\lambda }_n{y}_n$. By self-adjointness,Taking $m=n$ first shows that each eigenvalue is real: $⟨y_n,y_n{⟩}_w>0$ for a nonzero eigenfunction, since $w>0$. ThereforeFor distinct eigenvalues this proves orthogonality of Sturm-Liouville eigenfunctions:The weight is essential: this need not be orthogonality under the unweighted integral.

The useful coordinate for this scale-dependent differential operator is $t=\ln x$. With $Y(t)=y(e^t)$, the chain rule givesThe interval becomes $0\le t\le 1$, and the endpoint values become $Y(0)=Y(1)=0$. This logarithmic change of variable in a Sturm-Liouville problem reduces the calculation to a constant-coefficient Dirichlet problem. Also $dx/x=dt$, so the transformation preserves the appropriate weighted inner product.

Divide the eigenvalue equation by $x$ and collect the first two derivative terms into a single derivative:Thus one choice of Sturm-Liouville form isIndeed $w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]=-x^2{y}^{\prime \prime }-x{y}^{\prime }-y$. Both $p$ and $w$ are positive and regular throughout this interval. The zero endpoint values make the boundary form vanish.

In logarithmic coordinates the eigenvalue problem becomes $-Y^{\prime \prime }=(\lambda +1)Y$. A nonzero function with zero values at both ends requires $\lambda +1>0$: multiplying by $Y^{\ast }$ and integrating gives $(\lambda +1){\int }_0^1|Y{|}^{2}dt={\int }_0^1|Y^{\prime }{|}^{2}dt>0$. Write $\kappa =\sqrt{\lambda +1}$. The first endpoint gives $Y=A\sin \kappa t$, and the second requires $\kappa =n\pi$, $n=1,2,\dots$.

For vector normalization, ${\int }_1^e|y{|}^{2}dx/x=|A{|}^2{\int }_0^1{\sin }^2(n\pi t)dt=|A{|}^2/2$. A positive real normalization constant therefore givesThe sine orthogonality identity also gives ${\int }_1^e{y}_m{y}_{n}dx/x={\delta }_{mn}$. Completeness is the usual completeness of the Fourier sine series on $(0,1)$ under the logarithmic coordinate transformation.

Expand the forcing and the solution in the normalized eigenfunctions: $1={\sum }_n{b}_n{y}_n$ and $y={\sum }_n{c}_n{y}_n$. Their coefficients areSince all ${\lambda }_n=n^2{\pi }^2-1$ are nonzero, the equation gives $c_n=b_n/{\lambda }_n$. Hence the required Sturm-Liouville eigenfunction expansion isThis is constant forcing in a logarithmic Dirichlet Sturm-Liouville problem. The coefficients are $O(n^{-3})$, so the solution series converges uniformly and has zero endpoint values. Applying $L$ is justified in the weighted square-integrable sense; the forcing sine series equals one at interior points, while its endpoint sums are zero. The equation is imposed in the interior, so this does not contradict the boundary conditions.

As an independent check, solving $-Y^{\prime \prime }-Y=1$ directly givesIt has both zero endpoint values and satisfies $Ly=1$. The invertibility of the Dirichlet operator makes it the same solution as the expansion.

Use separation of variables, writing $\Psi =F(r)\Theta (\theta )$. After multiplication by $r^2/(F\Theta )$, the radial and angular terms are functions of independent variables, so setWith $x=\cos \theta$, the angular equation isRegularity at both poles selects $\lambda =\ell (\ell +1)$, $\ell =0,1,\dots$, and the Legendre polynomials $P_{\ell }(x)$; the other angular branch is singular at a pole. The radial Euler-Cauchy equation is $r^2{F}^{\prime \prime }+2r{F}^{\prime }-\ell (\ell +1)F=0$. A power $r^s$ gives $(s-\ell )(s+\ell +1)=0$, hence $F=a_{\ell }{r}^{\ell }+b_{\ell }{r}^{-\ell -1}$.

Superposing these axisymmetric spherical harmonics givesThe coefficients must give convergence appropriate to the domain. In a region containing the origin, regularity removes the negative powers; in an exterior region with zero potential at infinity, it removes the growing powers and the constant. No such radial restriction is needed for a general annular region.

Resolve the angular surface charge density into Legendre polynomials:For a single degree $\ell$, write the common surface potential as $V_{\ell }{P}_{\ell }$. Regularity and decay require radial factors $(r/R{)}^{\ell }$ inside and $(R/r{)}^{\ell +1}$ outside. Their exterior-minus-interior radial derivative at $R$ is $-(2\ell +1)V_{\ell }{P}_{\ell }/R$. Thus spherical harmonic matching with a derivative jump gives $V_{\ell }=R{\sigma }_{\ell }/(2\ell +1)$.

Only degrees zero and two occur, yieldingBoth expressions have surface value $2AR/3-2AR{P}_2/15$, and their derivative difference is $-2A/3+(2A/3)P_2=-A{\sin }^2\theta$, checking both interface conditions. The interior quadratic is regular at the origin, and the exterior potential tends to zero.

The leading far-field term is the monopole term. Indeed the total charge is $Q=R^2\int \sigma d\Omega =8\pi A{R}^2/3$, and the normalization of the specified jump gives $Q/(4\pi r)=2A{R}^2/(3r)$. No extra permittivity factor is inserted into the supplied normalization. Regularity here means finite piecewise smooth potential; its normal derivative necessarily jumps at the charged surface.

For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

Fix the evaluation point $r^{\prime }$ inside $V$ and put $\psi (r)=G(r,r^{\prime })$ in Green's second identity. The two $k^2$ terms cancel because ${\nabla }^2\varphi =-k^2\varphi$ and ${\nabla }^{2}G={\delta }^{(3)}-k^{2}G$. The remaining volume integral is $\int \varphi (r){\delta }^{(3)}(r-r^{\prime })dV=\varphi (r^{\prime })$. ThusThe Dirac delta has a positive sign in this problem's Green-function equation, fixing the order of the two terms. All boundary derivatives act on the integration variable $r$.

Impose a homogeneous Dirichlet boundary condition on the Green function in its first variable, $G(r,r^{\prime })=0$ for $r\in S$. The second boundary term then vanishes, givingIn a bounded domain, a Dirichlet Green function requires $k^2$ not to be a Dirichlet eigenvalue of $-\Delta$; otherwise a resonance can obstruct its existence. The representation assumes such a Green function exists.

In all space, put $\rho =|r-r^{\prime }|$ and try $G=A{e}^{ik\rho }/\rho$. For $\rho >0$, direct differentiation gives $({\rho }^2{G}^{\prime }{)}^{\prime }+k^2{\rho }^{2}G=0$, so the Helmholtz equation holds away from the source. Integrate its inhomogeneous version over a small source-centred ball. The $k^{2}G$ volume integral tends to zero, while the flux isIt must equal the integrated delta, one. ThereforeThis is the negative of the convention used in the outgoing Green function for the three-dimensional Helmholtz equation, where the source is $-\delta$. The sign here follows the present $+\delta$ convention. The factor $e^{ik\rho }$ chooses the outgoing solution, with time convention $e^{-i\omega t}$ and the Sommerfeld radiation condition; one could add homogeneous solutions without a radiation prescription.

Reflect $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images gives the Dirichlet image kernel for the three-dimensional Helmholtz equation:The two distances coincide on the plane, so its boundary value is zero; the image source is outside the domain. It therefore has the required positive delta source inside and the selected outgoing behaviour.

Let $h(R)=e^{ikR}/R$. At a boundary point $r=(\xi ,\eta ,0)$, the distances to the real and reflected sources have $z$ derivatives $-z^{\prime }/R$ and $+z^{\prime }/R$. ThusThe outward unit normal of the upper half-space is $-e_z$, and $z^{\prime }/R=\cos \theta$. HenceUsing the decay assumption to discard the distant hemisphere in the boundary representation givesThe qualification at infinity concerns the whole Green boundary term, not just the pointwise limit of $\varphi$. Sufficient decay of the field and its derivatives, as assumed, makes it vanish; compatible radiation conditions can also do so. As a sign check, taking $k\to 0$ in the kernel gives $z^{\prime }/(2\pi R^3)$, the positive Poisson kernel for the upper half-space.

Put $d=\sqrt{a^2-b^2}>0$. The two roots areSince $(a-d)(a+d)=b^2$, their moduli are $b/(a+d)<1$ and $(a+d)/b>1$. Thus exactly one root is inside the unit circle, and neither lies on it.

For the real integral use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The change of variables givesThe contour is counterclockwise. At its single interior simple pole, the residue isBy the residue theorem, the reciprocal affine sine integral is thereforeThe condition $a>b>0$ also makes the real denominator strictly positive, consistent with the positive answer and the absence of contour poles on the unit circle.

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.

The Fourier transform of $u_{xx}$ is $-k^2\tilde{u}$: two integrations by parts have no boundary terms under the stated decay assumption. Time differentiation commutes with the spatial transform. ThereforeFor every fixed spatial wavenumber $k$, this is a constant-coefficient ordinary differential equation in time.

The characteristic polynomial is $(r+\gamma {)}^2+c^2{k}^2$, with roots $r=-\gamma \pm ick$. For $ck\ne 0$, the transformed general solution isThe repeated-root case $ck=0$ is $e^{-\gamma t}[A_0+B_{0}t]$, rather than two independent copies of the same exponential. An alternative basis regular at $k=0$ is $e^{-\gamma t}[C(k)\cos (ckt)+D(k)\sin (ckt)/(ck)]$, using the limit $t$ for the last factor at zero.

For $c\ne 0$, the Fourier shift theorem turns the two exponential factors into translations. Equivalently, set $v=e^{\gamma t}u$; then $v_{tt}=c^2{v}_{xx}$. The D'Alembert formula gives the general formwhere $F,G$ are arbitrary functions with the required regularity and decay. The arbitrary profiles can accommodate a spatially transformed zero-frequency velocity by their integral components; the isolated repeated root must not be used to discard that degree of freedom. If $c=0$, the equation instead has $u=e^{-\gamma t}[F(x)+tG(x)]$.

This is the exponentially conjugated wave equation. Positive $\gamma$ produces damping, but the algebra also covers negative $\gamma$, which gives exponential amplification. The sign of $c$ is immaterial because the differential equation contains only $c^2$.

The initial profile is $f(x)=e^{-|x|}$, with the absolute value confirmed in the original PDF. Its Fourier transform isUsing ${\tilde{u}}_t(k,0)=0$ in the regular cosine-sine form giveswith the quotient interpreted as $t$ when $ck=0$. Put $v_0=|c|>0$. Inverting, or applying the D'Alembert formula to $e^{\gamma t}u$, whose initial velocity is $\gamma f$, givesFor an explicit expression at $t\ge 0$, let $r=|x|$ and $s=v_{0}t$. The integral is $2e^{-r}\sinh s$ for $r\ge s$ and $2(1-e^{-s}\cosh r)$ for $r\le s$. ThereforeThe branches agree at $r=v_{0}t$. At zero time, the integral vanishes and the translated profiles coincide, recovering $f$. Their time derivatives cancel, while the integral contributes $\gamma f$, cancelling the derivative of the prefactor, so $u_t(x,0)=0$.

For $c=0$ the limiting answer is $u=(1+\gamma t)e^{-\gamma t}{e}^{-|x|}$. The cusp in $f$ means the solution is understood as a weak solution, classically smooth away from the characteristic fronts $x=\pm v_{0}t$; it is not globally twice differentiable. No assumption that $\gamma$ is positive is needed for this initial-value solution.

For a constant orthogonal Cartesian change of coordinates $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law for order $n$ isRepeated indices are summed. In Cartesian orthonormal coordinates the metric identifies upper and lower indices, so the same orthogonal matrix acts on every index.

An isotropic tensor has the same numerical components after every proper orthogonal change of axes: $Q^{\otimes n}T=T$ for every $Q$ with $Q^{T}Q=I$ and $\det Q=1$. For example, ${\delta }_{ij}$ is isotropic because $Q_{ik}{Q}_{jk}={\delta }_{ij}$. Some conventions require invariance under all orthogonal transformations, including reflections; this stronger convention must be distinguished for odd-order tensors, notably the Levi-Civita symbol, which is invariant under proper rotations but changes sign under its ordinary three-index transformation for a reflection.

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.

An axial vector transforms under an orthogonal matrix $Q$ asFor a proper rotation, $\det Q=1$, so this is the usual polar vector rule. Under an orientation-reversing change, the extra sign distinguishes the two. In particular, spatial inversion leaves an axial vector unchanged while reversing a polar vector.

Let $C_i=(\nabla \times u{)}_i={ϵ}_{ijk}{\partial }_j{u}_k$. The determinant identity for the Levi-Civita symbol isIt follows by multiplying the three-index determinant identity by $Q_{ir}$ and using orthogonality. The derivative transformation from part (ii) then givesHence the curl of a polar vector field is an axial vector field. The Levi-Civita determinant factor is exactly the extra sign required under reflections.

Write $A_{ij}={\partial }_j{u}_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the scalar, symmetric-traceless and axial tensor decomposition:Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Its associated axial vector isIndeed the contraction identity ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$ gives ${ϵ}_{ijk}{\omega }_k=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Adding the three parts reconstructs $A_{ij}$.

The minus sign is required by the index order $A_{ij}={\partial }_j{u}_i$. The symmetric gradient and curl decomposition written instead for ${\partial }_i{u}_j$ has the opposite axial sign. The divergence supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.

Direct differentiation givesThe curl is $(0,0,b-a)$, so the preceding convention givesPut $h=(a+b)/2$. The normalized principal axes and eigenvalues of the symmetric tensor $s$ areThe planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne 0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.

The kinetic energy is $T=m{\sum }_i{\dot{z}}_i^2/2$. Differentiating the quadratic potential energy gives the equations for small oscillations,Put $z=v\cos (\omega t+\phi )$ and $\lambda =m\ell {\omega }^2/\tau$. The eigenvalue problem is $Kv=\lambda v$. Its characteristic polynomial isThus a convenient set of normal modes isEach shape can be multiplied by an arbitrary sinusoidal time amplitude. In the lowest mode all beads move in phase. In the middle mode the central bead is fixed and the outer beads move oppositely. In the highest mode the two outer beads move together against the centre. These vectors are mutually orthogonal and span the three displacement coordinates.

These are the three-mass case of the normal modes of a uniformly loaded finite string, with discrete sine shapes $v_i^{(j)}=\sin (ij\pi /4)$. The sketches use straight string segments between masses and fixed zero displacements at the walls. The small-slope and constant-tension assumptions justify the quadratic energy and linear equations.

Resolve the initial displacement into the mode shapes from part (i):There is no component along $(1,0,-1)$: the initial displacement is invariant under exchanging beads 1 and 3, whereas that mode is antisymmetric. Zero initial velocities select cosine time dependence. ThereforeHere ${\omega }_1=\sqrt{\tau (2-\sqrt{2})/(m\ell )}$ and ${\omega }_3=\sqrt{\tau (2+\sqrt{2})/(m\ell )}$. At zero time these expressions give $(0,a,0)$ and all three velocities vanish.

The motion is a superposition of the two symmetric normal modes, with opposite signed modal amplitudes. The outer beads always move together. Equivalently, $z_2=a\cos [({\omega }_1+{\omega }_3)t/2]\cos [({\omega }_3-{\omega }_1)t/2]$, displaying the modulation from the two frequencies. The result describes the linearized string while the initial amplitude and subsequent slopes remain small compared with the bead spacing.

The identity commutes with every group element, so it lies in the center of a group. If $z,w$ are central and $g\in G$, then $gw=wg$ implies $g{w}^{-1}=w^{-1}g$. ConsequentlyThus $z{w}^{-1}$ is central. The subgroup test provesEquivalently, the centre is nonempty, closed under products, and closed under inverses. This proof does not require finiteness. In fact the centre is an Abelian subgroup, and it is normal because $gz{g}^{-1}=z$ for each central element.