Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2001 / Paper 1

These solutions cover all eleven questions and every printed subpart. The qualification needed at $\kappa =0$ in Question 8 and the limiting interpretation of the ray displacement in Question 10 are explained with their solutions.

Vector identity. Work in Cartesian coordinates and use the Einstein summation convention. Contraction of the two Levi-Civita symbols givesThus the convective acceleration identity isEach term is a geometrically defined vector field: the directional derivative of a vector along itself, the gradient of a scalar field, and a cross product with the curl. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate basis, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.

Spherical metric. LetDifferentiating gives the coordinate tangent vectorsTheir pairwise dot products vanish, proving that spherical coordinates are orthogonal coordinates. Their lengths are the scale factors of orthogonal coordinates:These formulas use the usual chart $r>0,0<\theta <\pi$; the origin and polar axis are coordinate singularities.

Azimuthal field. Write the speed as $U(r)$ to distinguish it from the velocity field $\mathbf{u}=U(r)e_{\varphi }$. Using the curl in spherical coordinates,Also $\frac{1}{2}\nabla |\mathbf{u}{|}^2=U{U}^{\prime }{e}_r$. Since $e_{\varphi }\times e_r=e_{\theta }$ and $e_{\varphi }\times e_{\theta }=-e_r$,The $U{U}^{\prime }$ terms cancel in the convective acceleration identity, givingThis centripetal acceleration of an azimuthal flow points toward the polar axis. A streamline is a circle of cylindrical radius $r\sin \theta$, so its acceleration is $-U^2{e}_{\rho }/(r\sin \theta )$, with $e_{\rho }=\sin \theta e_r+\cos \theta e_{\theta }$. This independent geometric argument recovers both nonzero components and explains why $U^{\prime }$ cannot appear: the particle travels at fixed $r$.

Transform pair. In the paper's symmetric normalization, the inverse Fourier transform isSplit the transform of $e^{-c|x|}$ at zero, with $c>0$:The inverse formula applied to this result givesBecause the kernel is even, replacing the transform variable by $x$ and the argument by $k$ yieldsThe factors of $\sqrt{2\pi }$ depend on the normalization and will also enter the convolution theorem.

Convolution theorem. For integrable functions define the convolutionSubstitute this into the Fourier transform, interchange the two absolutely integrable factors using Fubini's theorem, and set $z=x-y$:Therefore

Integral equation. Put $f_a(x)=1/(x^2+a^2)$. Since this is even, the left side is $f_a\ast g$. The transformed equation issoThe assumption $b>a$ makes the new scale $d=b-a$ positive. Inverting with the previously established transform pair givesThe Cauchy convolution semigroup gives a quick verification. With $K_c(x)=c/[\pi (x^2+c^2)]$, the formula is $g=(a/b)K_{b-a}$ and $f_a=(\pi /a)K_a$. Hence $f_a\ast g=(\pi /b)K_b=1/(x^2+b^2)$, as required. The solution is unique among integrable functions because the multiplier ${\tilde{f}}_a(k)$ never vanishes.

Laplace identities. The one-sided Laplace transform isfor $\mathrm{Re}p$ sufficiently large that the integral converges. For a piecewise smooth function of exponential order, integration by parts givesDifferentiating under the integral givesIteration also yields $\mathcal{L}[f^{\prime \prime }]=p^{2}F-pf(0)-f^{\prime }(0)$. For real $k$ and $p>0$, taking the imaginary part of $(p-ik{)}^{-1}$ gives

Transformed differential equation. Let $G=\mathcal{L}[g]$, using the initial values $g(0)=1,g^{\prime }(0)=0$. ThenSumming the three contributions givesFor $\alpha \ne 0$, the condition $G(p)\to 0$ as $p\to \infty$ fixes the integration constant:It is simpler to invert the derivative: $\mathcal{L}[tg]=-G^{\prime }=1/(p^2+{\alpha }^2)=\mathcal{L}[\sin (\alpha t)/\alpha ]$. ThusThis is the sinc function. Its Taylor series begins $1-{\alpha }^2{t}^2/6+O(t^4)$, verifying both initial values. For a direct differential check set $h=tg=\sin (\alpha t)/\alpha$. Then $h^{\prime \prime }+{\alpha }^{2}h=t{g}^{\prime \prime }+2g^{\prime }+{\alpha }^{2}tg=0$. For $\alpha =0$, take the continuous limit $\boxed{g(t)=1}$, which also follows from $G=1/p$.

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

The first three output components equal $x_1,x_2,x_3$, while the fourth is $x_4-{\sum }_{k=1}^4{x}_k=-x_1-x_2-x_3$. HenceIts outputs have component sum zero, and applying $P$ to any such output leaves it fixed. ThusThe first three columns give a basis of the image of a linear map:They are linearly independent because their first three coordinates are the standard basis. Their span is precisely the three-dimensional hyperplane $y_1+y_2+y_3+y_4=0$.

Apply the Gram-Schmidt process. First $q_1=v_1/\sqrt{2}$. Subtracting the $q_1$ projection from $v_2$ gives $(-1,2,0,-1{)}^T/2$, of length $\sqrt{3/2}$. Subtracting both preceding projections from $v_3$ gives $(-1,-1,3,-1{)}^T/3$, of length $2/\sqrt{3}$. Therefore an orthonormal basis isTheir pairwise dot products vanish and each has length one. Although this image admits an orthonormal basis, $P$ itself is an oblique projection: its kernel is the fourth coordinate axis, which is not perpendicular to the image, and $P\ne P^T$.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Boundary form. For real functions and $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, integration by parts givesThe integral terms containing $p{y}_1^{\prime }{y}_2^{\prime }$ cancel, as do the $q$ terms. This is the Lagrange identity for a Sturm-Liouville operator. The boundary conditions must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous Dirichlet boundary conditions at both endpoints; homogeneous Neumann boundary conditions at both endpoints; one of each; or separated homogeneous Robin boundary conditionswhere each fixed real pair $(\alpha ,\beta )$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, which cancel the two endpoint contributions. These domains give the required self-adjoint differential operator in the regular setting.

Eigenfunctions. Set $x=\cos \theta$, $0\le \theta \le \pi$, and $Y(\theta )=y(\cos \theta )$. The chain rule givesThe Chebyshev differential equation therefore becomes $Y^{\prime \prime }+n^{2}Y=0$, with $Y(0)=Y(\pi )=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin (n\theta )$. ThusThere is no nonzero $n=0$ eigenfunction: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent eigenbasis.

Weight and normalization. Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into Sturm-Liouville form:Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. Under $x=\cos \theta$, the measure $w(x)dx$ becomes $-d\theta$, soThereforeThese endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $p{y}_n^{\prime }=-N_{n}n\cos (n\theta )$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the Dirichlet eigenfunctions of the Chebyshev differential operator, not the first-kind Chebyshev polynomials $\cos (n\arccos x)$, which have nonzero endpoint values.

Adapted homogeneous solutions. First assume real $\kappa \ne 0$. The functions $u(x)=\cosh (\kappa x)$ and $v(x)=\cosh (\kappa (1-x))$ solve $z^{\prime \prime }-{\kappa }^{2}z=0$ and obey $u^{\prime }(0)=v^{\prime }(1)=0$. Their Wronskian isIt is independent of $x$ and nonzero.

Green function. For $0<\xi <1$, continuity at $x=\xi$ and the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ follow by matching the Dirac delta function source. In particular a jump in $G$ itself would produce a derivative of the Dirac delta function, which is absent. Write $G=A(\xi )u(x)$ for $x<\xi$ and $G=B(\xi )v(x)$ for $x>\xi$. Solving the continuity and derivative-jump equations gives $A=v(\xi )/W$, $B=u(\xi )/W$. Hence the Neumann Green function for the modified Helmholtz equation on an interval isThe two pieces coincide at the source. Their derivative difference is $W/W=1$ and their endpoint derivatives vanish. The same expression has continuous one-sided limits as the source approaches an endpoint; the interior-source construction is what is needed in the integral solution.

Solution for a general source. By superposition, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$. Splitting the integral at $\xi =x$ givesNo nonzero homogeneous solution satisfies both endpoint conditions when $\kappa \ne 0$: the condition at zero eliminates $\sinh (\kappa x)$, and the condition at one then eliminates $\cosh (\kappa x)$. Thus the Green function gives the unique solution.

The printed zero-parameter case. The statement permits real $\kappa$, but $\kappa =0$ is exceptional. Integrating a putative equation $G^{\prime \prime }=\delta (x-\xi )$ would give $G^{\prime }(1)-G^{\prime }(0)=1$, contrary to the two zero derivatives. Thus this Green function does not exist at zero. For $y^{\prime \prime }=f$ with homogeneous Neumann boundary conditions, solvability instead requires ${\int }_0^{1}f=0$, and a compatible solution has an arbitrary additive constant. This is the Fredholm solvability condition associated with the constant zero mode.

Explicit forcing. For $f(x)=x$ and $\kappa =1$, the particular solution is $-x$. Write $y=C\cosh x+D\sinh x-x$. Then $y^{\prime }(0)=0$ gives $D=1$ and $y^{\prime }(1)=0$ gives $C=(1-\cosh 1)/\sinh 1=-\tanh (1/2)$. ThereforeDirect differentiation gives $y^{\prime }=\cosh (x-1/2)/\cosh (1/2)-1$, which vanishes at both endpoints, and $y^{\prime \prime }-y=x$, verifying the solution independently of the Green function integration.

Variational formulation. Take real $p,q,w$ with $p,w>0$ in the interval, and impose homogeneous Dirichlet boundary conditions, or other suitable conditions that cancel the boundary terms. The functionalhas Euler-Lagrange equationwhich is equivalent to $-(p{y}^{\prime }{)}^{\prime }+qy-\lambda wy=0$. The same construction is constrained stationarity of the energy $\int [p(y^{\prime }{)}^2+q{y}^2]$ subject to $\int w{y}^2=1$, with $\lambda$ a Lagrange multiplier. For fixed zero endpoint values, the variation vanishes at both ends; homogeneous zero-flux conditions also cancel the boundary term $[2p{y}^{\prime }\delta y{]}_a^b$.

Multiplying the differential equation by $y$ and using integration by parts gives the Rayleigh quotientFor a nonzero admissible trial function, $R[y]\ge {\lambda }_0$ by the Rayleigh-Ritz variational principle. The Rayleigh-Ritz method chooses trial functions satisfying the endpoint conditions, often with adjustable coefficients, and minimizes this quotient over that trial family. Expanding $y=\sum c_n{y}_n$ in a normalized eigenbasis shows why the lower bound holds: $R[y]=\sum {\lambda }_n{c}_n^2/\sum c_n^2$, a weighted average of eigenvalues no smaller than the least one.

Radial quadratic trial. The radial problem becomeswhere the printed endpoint notation $y$ refers to $\psi$. Thus $p=w=r^2$ and $q=0$. At the origin the vanishing $r^2$ and regularity of $\psi$ make the boundary term vanish; the weight must be retained in both numerator and denominator.

For a quadratic $\psi =A+Br+C{r}^2$, the derivative condition sets $B=0$ and the outer condition gives $C=-A$. Overall amplitude cancels in the Rayleigh quotient, so take $\tilde{\psi }=1-r^2$. ThenwhileConsequently

Why the estimate is higher. For a regular radial function put $u=r\psi$. Then $u(0)=u(1)=0$, and direct differentiation transforms the equation to $u^{\prime \prime }+\lambda u=0$. Its first mode is $u=\sin (\pi r)$, corresponding to $\psi =\sin (\pi r)/r$ with its continuous value at the origin. Thus ${\lambda }_0={\pi }^2$. The quadratic trial is not proportional to that Spherical Bessel function, so it has nonzero components in higher modes. Its Rayleigh quotient is therefore strictly larger than ${\pi }^2$, in agreement with $21/2>{\pi }^2$.

First integral. If $F$ has no explicit $x$ dependence, the chain rule givesThe Euler-Lagrange equation makes this zero without requiring division by $y^{\prime }$. Hence the Beltrami identity is

Optical paths. By Fermat principle, a ray makes the optical path length stationary. Within the strip $0<y\le 1$, its functional for a graph $y(x)$ isThe Beltrami identity givesFor a nonvertical segment $k>0$, write $R=1/k$. Then $(y^{\prime }{)}^2=(R^2-y^2)/y^2$, soThus the optical rays in a reciprocal-height refractive index satisfyrestricted to the part in the strip. They are arcs of circles centered on $y=0$. Vertical lines are the additional solutions, obtained as a limiting case or directly from the constant horizontal optical momentum. The same curves are geodesics of the Poincaré half-plane model, whose length element is $ds/y$.

Horizontal displacement. Let the ray enter at $(x_0,1)$, travelling to the right and downward at angle $0<\theta <\pi /2$ to the vertical. The refractive index is continuous at $y=1$, so there is no instantaneous angular jump at entry. The tangent has $(dx/ds,dy/ds)=(\sin \theta ,-\cos \theta )$; hence $y^{\prime }(x_0)=-\cot \theta$. The first integral gives $k=\sin \theta$, or $R=\csc \theta$. The circle derivative $y^{\prime }=-(x-c)/y$ then fixes $c=x_0-\cot \theta$.

As $y\downarrow 0$, the rightward ray approaches $x=c+R$. The total limiting horizontal distance is thereforeFor vertical incidence, $D=0$ by continuity. For a leftward ray the displacement has the opposite sign, with the same magnitude. Since $\tan (\theta /2)$ increases on $[0,\pi /2]$,If entry is required to have a strictly downward component, this is a supremum rather than an attained maximum. A horizontal tangent at entry gives the limiting unit-radius arc, which immediately bends downward in the medium.

The lower boundary is a limit, rather than a point reached in finite optical travel time. On a nonvertical circular ray, $ds/dy=R/\sqrt{R^2-y^2}\to 1$ as $y\downarrow 0$. Thus the optical path length contains a divergent integral proportional to $\int dy/y$. The horizontal distance stays finite even though propagation toward $y=0$ takes infinite time.

Local errors of the Euler schemes. Let $h=\Delta t>0$, and evaluate $f$ and its partial derivatives at an exact starting point $(y(t),t)$. DefineThe exact Taylor expansion is $y(t+h)=y+hf+\frac{1}{2}{h}^{2}A+O(h^3)$. The forward Euler method retains only $y+hf$, so its one-step defect, exact minus computed value, isFor the Backward Euler method, substitution of the exact solution into the scheme gives the residualbecause $f(y(t+h),t+h)=f+hA+O(h^2)$. Under local smoothness, solving the implicit step changes this residual by a factor $1+O(h)$, so the one-step defect has the same leading term:These are unnormalized local truncation errors, as in the question. A convention dividing by $h$ instead gives $O(h)$. Both schemes have first-order global error under the usual smoothness and stability hypotheses.

Euler stability. For the Dahlquist test equation $y^{\prime }=-\Lambda y$, put $z=h\Lambda$, with $\mathrm{Re}z>0$. The numerical amplification factors areStrict decay requires $|R|<1$; the boundary $|R|=1$ gives bounded but generally nondecaying values. For Forward Euler method,For positive real $\Lambda$, this reduces to $0<h\Lambda <2$. The absolute stability region in the $z$ plane is the disk $|z-1|\le 1$. For Backward Euler method, $|1+z|>1$ whenever $\mathrm{Re}z>0$, so it decays for every step size. It is A-stable and also L-stable, since $R_{BE}(z)\to 0$ for large $z$ in this half-plane. Here $z$ has the opposite sign from the alternative convention $z=h\lambda$ for $y^{\prime }=\lambda y$.