Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004

Compare the total differential $d\varphi ={\sum }_i{\partial }_{u_i}\varphi d{u}_i$ with $d\varphi =\nabla \varphi \cdot d\mathbf{r}$ and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{u}_i$. The orthonormality of the coordinate basis gives the gradient in orthogonal coordinates:To obtain the curl without neglecting the variation of the basis vectors, write $\varphi {\mathbf{e}}_1=(\varphi h_1)\nabla u_1$. The curl of a gradient vanishes, and the product rule givesLikewise $\nabla u_2\times \nabla u_3={\mathbf{e}}_1/(h_2{h}_3)$ because the system is right-handed. This cross product has zero divergence: $\nabla \cdot (\nabla u_2\times \nabla u_3)=\nabla u_3\cdot (\nabla \times \nabla u_2)-\nabla u_2\cdot (\nabla \times \nabla u_3)=0$. The divergence product rule applied to $(\varphi h_2{h}_3)(\nabla u_2\times \nabla u_3)$ now givesCyclic permutation and linearity give the general component formulas described by divergence and curl from orthogonal coordinate gradients:andIn cylindrical coordinates, $(h_r,h_{\theta },h_z)=(1,r,1)$. The specified field has no $z$ dependence or $z$ component, soIt is an irrotational vector field for every $r>0$ exactly when $\boxed{p=3,\alpha =4}$. This statement is local to an angular coordinate chart: the component proportional to $\theta$ is not single-valued under $\theta \mapsto \theta +2\pi$.

With constant positive thermal diffusivity $\nu$, no internal heat source and one spatial coordinate, the heat equation isSeparation of variables makes each spatial mode decay by $e^{-\nu k^{2}t}$. The endpoint conditions determine whether the initial triangular profile is expanded in a Fourier sine series or a Fourier cosine series.

The cooled endpoints impose Dirichlet boundary conditions $\theta (0,t)=\theta (L,t)=0$. The spatial modes are $\sin (n\pi x/L)$. Integrating the initial profile on its two straight segments givesThe heat flow from a triangular initial temperature with height $L/2$ is thereforeEvery nonzero mode decays, so ${\theta }_D\to 0$ as $t\to \infty$. There is no nonzero constant mode compatible with these endpoint temperatures.

By Fourier's law, insulation means zero outward heat flux. The Neumann boundary conditions are ${\theta }_x(0,t)={\theta }_x(L,t)=0$. The constant mode equals the initial integral average $L/4$. For $n\ge 1$, the cosine coefficients areOdd modes and modes divisible by four vanish; write the remaining indices as $n=2(2j+1)$. The heat flow from a triangular initial temperature with insulated endpoints givesThe nonconstant modes decay, so $\boxed{{\theta }_N\to L/4}$ uniformly as $t\to \infty$. Integrating the heat equation gives $d({\int }_0^L\theta dx)/dt=\nu [{\theta }_x{]}_0^L=0$, explaining the conserved mean. The cooled case tends instead to zero because heat leaves through its ends.

The initial triangle has nonzero endpoint slopes, so it is not compatible with the insulated condition at the corner $t=0$. The series nevertheless recovers the continuous initial temperature and satisfies the insulated condition for every $t>0$; one should not assert continuity of its spatial derivative at that initial corner.

Use the Fourier transform conventionThese hold directly under suitable integrability and inversion hypotheses, and extend to square-integrable functions through the Plancherel theorem. At a jump, ordinary Fourier inversion uses the mean of the one-sided limits under the usual piecewise smooth hypotheses.

For the two-sided exponential, split the integral at zero:For the interval indicator, the Fourier transform of an interval indicator givesAt $k=0$, the continuous limiting value is $4\pi$, the length of the interval. Changing the values at its two endpoints has no effect on the transform.

The convolution is $(u\ast v)(x)={\int }_{-\infty }^{\infty }u(t)v(x-t)dt$. In this transform convention, the convolution theorem says $\widehat{u\ast v}=\hat{u}\hat{v}$, and Parseval's identity saysTo match the integral's scale exactly, set $u=f/2$ and $h(x)=\frac{1}{2}{1}_{[-1,1]}(x)=g(2\pi x)/2$. Then $\hat{u}=1/(1+k^2)$ and $\hat{h}=\sin k/k$, using the value one at zero. Their convolution is the box smoothing of a two-sided exponential with both scales equal to one:Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, Parseval's identity reduces the requested integral toUse ${\int }_0^1\cosh xdx=\sinh 1$, ${\int }_0^1{\cosh }^{2}xdx=1/2+\sinh 2/4$ and ${\int }_1^{\infty }{e}^{-2x}dx=e^{-2}/2$. Simplifying givesThe apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k{|}^{-6})$, so the integral converges.

A Hermitian matrix obeys $H^{†}=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne 0$, then $v^{†}Hv=\lambda v^{†}v$. This scalar is real because its complex conjugate equals $v^{†}{H}^{†}v=v^{†}Hv$. Since $v^{†}v>0$, the eigenvalue $\lambda$ is real.

For the Gram matrix, conjugate symmetry of the inner product givesThus $G$ is Hermitian. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.

Let $A{v}_1={\lambda }_1{v}_1$ and $A{v}_2={\lambda }_2{v}_2$, with both vectors nonzero and ${\lambda }_1\ne {\lambda }_2$. If $c_1{v}_1+c_2{v}_2=0$, apply $A-{\lambda }_{2}I$ to get $c_1({\lambda }_1-{\lambda }_2)v_1=0$. Hence $c_1=0$, and then $c_2=0$.

For the general independence of eigenvectors for distinct eigenvalues, use induction. In a relation ${\sum }_{j=1}^n{c}_j{v}_j=0$, applying $A-{\lambda }_{n}I$ gives ${\sum }_{j=1}^{n-1}{c}_j({\lambda }_j-{\lambda }_n)v_j=0$. Independence of the first $n-1$ vectors makes each coefficient vanish. Distinctness of the eigenvalues then gives $c_1=\cdots =c_{n-1}=0$, and the original relation gives $c_n=0$. No Hermitian assumption is required.

A real orthogonal matrix satisfies $Q^{T}Q=I$; a complex unitary matrix satisfies $U^{†}U=I$. Their columns are orthonormal in the corresponding real or complex inner product.

Adopt the convention that a complex inner product is linear in its second argument, conjugate symmetric, and positive definite: $⟨u,u⟩\ge 0$, with equality exactly when $u=0$. Conjugate symmetry and second-argument linearity then imply antilinearity in the first:For the specified real symmetric matrix use $b=a^2$. Its upper block is the outer product $(1,a{)}^T(1,a)$, so it has one zero eigenvalue and one eigenvalue $1+a^2$. A complete orthonormal eigenbasis isIf $c=1+b$, the last two eigenvalues coincide and their eigenspace is the span of $v_{+},v_c$; every nonzero vector in that span is an eigenvector. Since $c>0$, the zero eigenspace remains one-dimensional.

An ordinary point of the normalized second-order equation is a point where both coefficient functions $p(x)$ and $q(x)$ are analytic in a neighborhood. Ordinary continuity is enough for a local existence theorem, but analyticity is the condition used for a convergent local power series solution.

A regular singular point is a nonordinary point $x_0$ for which $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically across $x_0$. These pole restrictions permit the Frobenius method.

The Wronskian is $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. If constants $A,B$ obey $A{y}_1+B{y}_2=0$, differentiating gives $A{y}_1^{\prime }+B{y}_2^{\prime }=0$. At any point where $W\ne 0$, these form an invertible two-by-two system, forcing $A=B=0$. Thus the two solutions are linearly independent.

For the given equation, the normalized coefficients are $p=(1-x)/(2x)$ and $q=-1/(4x)$, so zero is a regular singular point. Substitute a Frobenius series $y=x^r{\sum }_{n=0}^{\infty }{a}_n{x}^n$, $a_0\ne 0$. The lowest power gives the indicial equationThe coefficient of $x^{n+r}$ givesFor either root the common factor is nonzero for $n\ge 0$, soWith leading coefficient one, the two independent solutions areTheir analytic series factors both have infinite radius of convergence, because the ratio of consecutive terms tends to zero for every fixed $x$. The factor $x^{1/2}$ still gives the second solution a branch point at zero; it is not an entire function or an ordinary Taylor series there. On $x>0$, the general real solution is $C_1{y}_1+C_2{y}_2$. This is the $b=1/2$, $z=x/2$ case of the exponential reduction of a confluent hypergeometric equation with equal parameters.

For the given weighted inner product, a self-adjoint operator $L$ obeys $L=L^{†}$, including equality of its domain with the adjoint domain. In the differential-equation calculation this requiresfor all functions in a specified common domain, with endpoint conditions that remove the boundary form. Equality of these inner products by itself describes symmetry; the domain conditions complete the self-adjoint realization.

Divide the equation by $\sqrt{1-x^2}$ to obtain its Sturm-Liouville form:Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. For $L=-w^{-1}(pd/dx{)}^{\prime }$, the Lagrange identity for a Sturm-Liouville operator gives the endpoint form $[p({\bar{f}}^{\prime }g-\bar{f}{g}^{\prime }){]}_{-1}^1$. A natural self-adjoint choice isUnder $x=\cos \theta$, these become the ordinary zero-derivative conditions at $\theta =0,\pi$ for $-d^2/d{\theta }^2$. They therefore define the realization whose Neumann eigenfunctions of the Chebyshev differential operator are $T_n(x)=\cos (n\arccos x)$, with $n=0,1,\dots$.

Boundedness alone does not choose these endpoint conditions: $\sin (n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $p{y}^{\prime }$. Requiring bounded $y^{\prime }$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $z(x)=e^{-2x}u(x)$. It gives $u^{\prime \prime }+\lambda u=0$ with $u(0)=u(1)=0$. Positive $\lambda =k^2$ requires $\sin k=0$; zero or negative $\lambda$ yields only the trivial solution under both endpoint conditions. HenceMultiplication of the original equation by $e^{4x}$ gives $-(e^{4x}{z}^{\prime }{)}^{\prime }-4e^{4x}z=\lambda e^{4x}z$. Thus the appropriate weighted inner product has weight $e^{4x}$, andMultiplying each eigenfunction by $\sqrt{2}$ gives weighted norm one.

The homogeneous Cauchy-Euler equation has powers $x^r$ with $r(r-1)+2r-2=(r-1)(r+2)=0$. Boundedness at zero selects $x$, and boundedness at infinity selects $x^{-2}$. Put $G=Ax$ for $x<\zeta$ and $G=B{x}^{-2}$ for $x>\zeta$. Continuity gives $B=A{\zeta }^3$. Since the operator is $L=(x^{2}d/dx{)}^{\prime }-2$, the required delta-function normalization is ${\zeta }^2[G_x{]}_{{\zeta }^{-}}^{{\zeta }^{+}}=1$. Thus $-3A{\zeta }^2=1$, giving the bounded Green kernel for a radial Euler operator:In the requested notation, $\boxed{a=1,b=-2,c=1,d=-2,k_{-}=k_{+}=-1/3}$. The two expressions agree at $x=\zeta$.

With ordinary measure $d\zeta$, the solution isFor bounded forcing these integrals converge and give a bounded solution. It is unique because no nonzero combination of $x,x^{-2}$ is bounded at both ends. For a more general forcing, convergence and boundedness of the displayed integral must be checked.

The Green's function is nonpositive, so $f\ge 0$ gives $y\le 0$. If also $f\le m$, thenThe integral solution therefore givesBoth bounds are attained, by the constant forcings $f=m$ and $f=0$, respectively. In particular, boundedness of the solution does not require it to approach zero.

Differentiate the integral formula using the Leibniz integral rule. Its two moving-endpoint terms cancel, leavingEach term before subtraction lies between zero and $m/(3x)$. ThereforeThe bounds are sharp at any fixed positive $x$: put $f=m$ below that point and zero above it for the upper bound, or reverse those choices for the lower bound. If continuity of the forcing is required, smooth approximations approach the same bounds.

If the integrand has no explicit $x$ dependence, differentiate $f-y^{\prime }{f}_{y^{\prime }}$ along a stationary path:The last equality uses $f_x=0$ and the Euler-Lagrange equation. This proves the Beltrami identity, the first integral $\boxed{f-y^{\prime }{f}_{y^{\prime }}=C}$.

If $f$ has no explicit $y$ dependence, the Euler-Lagrange equation becomes $d(f_{y^{\prime }})/dx=0$. ThusThe variable $y$ is then a cyclic coordinate. This is the case used for the azimuth in the ramp calculation.

For variations $y+ϵ\eta$ with $\eta (a)=\eta (b)=0$, integration by parts gives $\delta I={\int }_a^b[f_y-d(f_{y^{\prime }})/dx]\eta dx$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationFor the ramp, use conservation of energy from rest at the rim. The vertical drop is $R\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. The spherical surface arc length is $ds=R\sqrt{d{\theta }^2+{\sin }^2\theta d{\varphi }^2}$. On a monotone segment described by $\varphi (\theta )$, the travel time is consequentlyFor increasing $\theta$ this has the printed limits ${\theta }_0$ to ${\theta }_1$; for a descent, reverse the decreasing angular limits to keep time positive. The azimuth is a cyclic coordinate of this integrand. Its conserved derivative gives the first integral for the brachistochrone on a hemispherical ramp:Solving for its square yieldsFor a regular stationary path with finite $C$, this tends to zero as the rim $\theta \to \pi /2$ is approached. The ratio of azimuthal to meridional displacements is $\sin \theta {\varphi }^{\prime }$, which also tends to zero. Hence the initial direction is along a meridian toward the bottom. This conclusion concerns the initial heading; the later route can turn before reaching the bottom when its rim endpoints require a nonzero azimuthal displacement.

For example, impose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$, and use variations vanishing there. Homogeneous Neumann boundary conditions $p{y}^{\prime }=0$ are another choice removing the boundary terms. Put $N[y]={\int }_a^b(p{y}^{\prime 2}+q{y}^2)dx$, $D[y]={\int }_a^{b}w{y}^{2}dx>0$, and $I=N/D$. At a stationary function let $\lambda =I[y]$. The quotient rule gives $\delta I=(\delta N-\lambda \delta D)/D$. After integration by parts, stationarity under every allowed variation givesso the fundamental lemma of the calculus of variations yields the stated Sturm-Liouville problem. Conversely an eigenfunction with those boundary conditions has $N=\lambda D$ and makes the quotient stationary. The lowest eigenvalue minimizes the Rayleigh quotient under the usual self-adjoint, lower-bounded hypotheses; higher eigenvalues are stationary values with orthogonality constraints when minimizing successively.

The hydrogen equation becomes $-(r^2{\psi }^{\prime }{)}^{\prime }-2r\psi =2E{r}^2\psi$. Its radial energy quotient is thereforeRegularity at zero and decay at infinity remove the boundary term $r^2\psi {\psi }^{\prime }$. The condition $\psi (0)=1$ fixes an arbitrary amplitude for these trial functions; it is not their three-dimensional probability normalization, which cancels from the quotient.

For $\psi =e^{-\alpha r}$, elementary exponential integrals give $\int r^2{\psi }^{2}dr=1/(4{\alpha }^3)$, $\int r^2{\psi }^{\prime 2}dr=1/(4\alpha )$, and $\int r{\psi }^{2}dr=1/(4{\alpha }^2)$. HenceFor the compactly supported linear trial for hydrogen, the norm, gradient and Coulomb integrals are ${\beta }^3/30$, $\beta /3$, and ${\beta }^2/12$, respectively. ThusThe derivative changes from negative to positive at four, so this is the minimum over positive $\beta$. The quantum variational principle gives upper bounds on the ground-state energy; the exponential trial is better, and in fact $e^{-r}$ satisfies the exact differential equation at $E=-1/2$. It is the nodeless hydrogen ground state in the given units. The linear trial is admissible for the energy quadratic form despite its derivative jump at $r=\beta$; it need not possess a classical second derivative at that point.

For a separated harmonic function $R(r)\Theta (\theta )$ in polar coordinates, single-valuedness forces integer angular frequencies. The radial factors for frequency $m\ge 1$ are $r^m,r^{-m}$, while frequency zero has a constant and $\ln r$. Finiteness at the origin excludes $r^{-m}$ and $\ln r$.

The source's angular factor is $\cos \theta$ in every term, as printed, rather than $\cos (n\theta )$. The power-law angular mode for the planar Poisson equation givesHence, assuming the series permits termwise differentiation, the general finite, single-valued solution isThe arbitrary harmonic series must itself converge to a finite function on the prescribed disk. If ${\alpha }_0\ne 0$, the source has a direction-dependent limit at the origin; the displayed potential still extends finitely there, but the equation at that point is understood weakly rather than as a smooth classical source.

For the special source, a particular potential is $r^3\cos \theta /8$. Its boundary cosine term is canceled by $-a^{2}r\cos \theta /8$, and the boundary sine term is supplied by $(r/a{)}^2\sin 2\theta$. The resulting Dirichlet boundary conditions has solutionIt is finite at zero, has the required boundary values, and its Laplacian is $r\cos \theta$.

The far radial derivative selects the growing frequency-one term $Ur\cos \theta$. The matching decaying harmonic is $B\cos \theta /r$. The no-penetration condition gives ${\partial }_r\varphi =(U-B/r^2)\cos \theta =0$ at $r=a$, so $B=U{a}^2$. Thus the potential flow around a circular cylinder isThe remaining single-valued exterior harmonics are excluded by the zero radial derivative at the cylinder and the prescribed derivative at infinity, apart from the arbitrary additive constant. In particular the logarithmic radial term would have a nonzero radial derivative on the cylinder. A circulation term proportional to $\theta$ is excluded by single-valuedness of the potential.

Use Green's second identity in the source coordinate:Take $u({\mathbf{x}}^{\prime })=\psi ({\mathbf{x}}^{\prime })$ and $v({\mathbf{x}}^{\prime })=-1/[4\pi |\mathbf{x}-{\mathbf{x}}^{\prime }|]$, the fundamental solution of the Laplace equation satisfying ${\Delta }^{\prime }v=\delta ({\mathbf{x}}^{\prime }-\mathbf{x})$. With localized charge and the decaying potential, the boundary integral tends to zero on expanding spheres. The volume identity then gives $\psi (\mathbf{x})+(1/{ϵ}_0)\int vqd{V}^{\prime }=0$, hence the Coulomb potentialConvergence of this integral and a condition fixing the potential at infinity are needed: mere pointwise decay of $q$ is not enough, and a harmonic function could otherwise be added. A bounded, compactly supported charge distribution is a sufficient setting for this derivation.

For the disk put $\sigma =Q/(\pi a^2)$. On its axis, every ring of radius $\rho$ is at distance $\sqrt{{\rho }^2+z^2}$, soFor $z>0$ subtract $z$; for $z<0$ add $z$. This is the electrostatic potential of a uniformly charged disk with zero potential at infinity. The absolute value is essential for reflection symmetry and the correct behavior on the lower axis.