Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2011

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

The substitution separates the uniform equilibrium from the decaying perturbation. Direct differentiation givesConsequently the reaction-diffusion equation reduces to the heat equation $v_t=v_{xx}$. Its data become

When the three constants coincide, the Dirichlet boundary conditions for $v$ are homogeneous. Separation of variables gives $X^{\prime \prime }+n^{2}X=0$ and $T^{\prime }=-n^{2}T$, selecting the normal modes $\sin (nx)e^{-n^{2}t}$ for $n\ge 1$. ThusTwo integrations by parts give ${\int }_0^{\pi }x(\pi -x)\sin (nx)dx=2[1-(-1{)}^n]/n^3$, while the constant term uses ${\int }_0^{\pi }\sin (nx)dx=[1-(-1{)}^n]/n$. Therefore the Fourier sine series isFor $u_0\ne 2$, its first coefficient is nonzero. Multiplying by $e^{-\beta t}$ and retaining the slowest-decaying mode givesThe remainder estimate holds uniformly in $x$ for sufficiently large positive time. If $u_0=2$, the fundamental vanishes and the next possible mode has decay rate $\beta +9$, explaining the exclusion.

There is an initial endpoint compatibility issue unless $u_0=0$: the given polynomial profile vanishes at the endpoints, whereas the boundary values equal $u_0$. The series realizes the initial data in L2 space and pointwise in the open interval, imposing the prescribed boundary values for $t>0$. For nonzero $u_0$ it cannot also be continuous at both initial endpoint corners.

Trying $y=x^m$ in the Cauchy-Euler equation givesso the homogeneous solution isContinuity with zero value at the singular endpoint excludes the second branch. Use the left and right adapted solutions $\ell (x)=x$, $r(x)=x^{-4}-x$, whose Wronskian isThe Dirichlet Green kernel for an Euler operator with exponents one and minus four is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and has first-derivative jumpThus $L_{x}G=\delta (x-\xi )$ with ordinary source measure $d\xi$. For bounded forcing the Green function representation isThe apparent singular prefactor in the first term is canceled by the integral's decay at zero. The only regular homogeneous solution vanishing at $1$ is zero, so this solution is unique in the stated regular class.

For the piecewise source, split the integrals at $1/2$. If $x<1/2$, the needed integrals are ${\int }_0^x{\xi }^{6}d\xi =x^7/7$ and ${\int }_x^{1/2}({\xi }^6-\xi )d\xi$. If $x>1/2$, only ${\int }_0^{1/2}{\xi }^{6}d\xi =1/896$ remains. Simplification yieldsAt the interface both expressions have value $-31/8960$ and derivative $129/4480$, so no unintended Dirac delta source appears. Their second derivatives have the allowed jump associated with the discontinuous forcing.

With the stated Fourier transform convention, the inverse Fourier transform and the deterministic autocorrelation of a square-integrable function areAssume initially sufficient decay and smoothness, for example a Schwartz function, to justify interchanging integrations and Fourier inversion. With $s=x+y$, the cross-correlation identity isEvaluating $h$ directly and by its inverse transform at zero proves the Parseval identity:Approximation extends the identity to L2 space, as the Plancherel theorem.

For the compact cosine profile, the Fourier transform of a truncated cosine follows from $\cos x=(e^{ix}+e^{-ix})/2$:At $k=\pm 1$ the apparent singularities are removable, with value $\pi /2$. The direct squared norm is ${\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx=\pi /2$. Hence Parseval identity and evenness givePutting $t=\pi k/2$ makes the left side ${\pi }^3/8$ times the requested integral. ThereforeAt $t=\pi /2$ the integrand's continuous extension is $1/{\pi }^2$, so no principal-value prescription is required.

A square matrix is invertible precisely when its determinant is nonzero. Equivalent conditions are full rank, a trivial kernel, and zero not being an eigenvalue.

For a lower triangular matrix, the first row has only the possibly nonzero entry $A_{11}$. A Laplace expansion along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yieldsThis proof includes singular triangular matrices; it makes no division by diagonal entries.

For the diagonal example, put $N=n(n+1)/2$. Its determinant is $b^{1+2+\cdots +n}=b^N$, and invertibility requires $b\ne 0$. Write the complex number as $b=r{e}^{i\theta }$ with $r>0$. A nonzero purely imaginary determinant requires $\cos (N\theta )=0$. ThusEquivalently $b^N\in i\mathbb{R}\setminus \{0\}$; the angle is understood modulo $2\pi$.

For skew-Hermitian matrices, reversing products under the conjugate transpose gives $(CD{)}^{†}=D^{†}{C}^{†}=DC$. ConsequentlyThe latter condition is vanishing of the anticommutator. A universal choice for the last request isIndeed $(CD-DC{)}^{†}=DC-CD=-(CD-DC)$, so the commutator of two skew-Hermitian matrices is skew-Hermitian.

Over the chosen scalar field, an $n$-by-$n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. Repeated eigenvalues are allowed provided all eigenspaces together supply a basis.

For the given matrix, the characteristic polynomial is ${\lambda }^2-4\lambda -5=(\lambda -5)(\lambda +1)$. One may chooseTaking these as columns gives the change of basisThis is not a rotation matrix. The PDF's terminology cannot hold literally: $A$ is real and nonsymmetric, whereas a real matrix orthogonally similar to a real diagonal matrix is symmetric. Also $V_1\cdot V_2=-4\ne 0$, so scaling these eigenvectors cannot make an orthonormal rotation basis. A general invertible similarity is the valid construction.

Let $D=P^{-1}AP$. Inserting $P{P}^{-1}$ between successive factors gives $A^n=P{D}^n{P}^{-1}$, and henceTo write $A^n={\alpha }_{n}A+{\beta }_{n}I$, match its action on the two eigenspaces:Solving gives the matrix powers for a diagonalizable matrix with two eigenvaluesThe cases $n=0,1$ recover $I,A$, respectively. Equivalently the spectral projections of a diagonalizable matrix are $(A+I)/6$ and $(5I-A)/6$, and the power multiplies them by $5^n$ and $(-1{)}^n$.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

For endpoint-preserving variations of a functional, the Euler-Lagrange equation isAlong an extremal, the chain rule gives $d(F-y^{\prime }{F}_{y^{\prime }})/dx=F_x+y^{\prime }[F_y-d{F}_{y^{\prime }}/dx]=F_x$. Thus when $F$ has no explicit $x$ dependence, the Beltrami identity isFor a positive speed profile, the elapsed-time integrand in Fermat principle is $F=\sqrt{1+(y^{\prime }{)}^2}/c(y)$. Its first integral givesIntegrating each monotone-height branch yieldsReal branches require $Ac(y)\le 1$. Turning points are joined in the original graph $y(x)$, even though the inverse derivative is singular there.

For the inverse-height speed profile, this becomes $dx/dy=\pm a/\sqrt{y^2-a^2}$, with $a=A>0$. Integration gives $x-b=\pm a\mathrm{arcosh}(y/a)$, so the light rays are catenariesEqual endpoint heights at opposite $x$ values force $b=0$, since the real hyperbolic cosine is even and strictly increases with the absolute value of its argument. The remaining endpoint condition is $a\cosh (1/a)=\cosh 1$. One exact solution is $a=1$, givingHere $ds/dx=\cosh x$ and $1/c(y)=y=\cosh x$ account for the squared hyperbolic cosine in the time integral.

The stated endpoints also allow a second stationary ray. Set $t=1/a$. The function $\cosh t/t$ decreases and then increases: its derivative has the sign of $t\sinh t-\cosh t$, whose derivative is $t\cosh t>0$. Its unique minimum occurs where $t\tanh t=1$, at $t\simeq 1.199679$. Since the known root $t=1$ lies below this minimum's location, there is exactly one further root, $t\simeq 1.4242930655$. Thus the two stationary catenary rays in an inverse-height speed profile also includeNo extra condition in the question selects uniquely between these rays. The minimum height is a minimum along the chosen ray, not a global travel-time assertion. Indeed, competitors descending vertically to height $\eta >0$, crossing horizontally, and rising again have time ${\cosh }^2(1)-{\eta }^2+2\eta$. As $\eta \downarrow 0$ this tends to ${\cosh }^2(1)\simeq 2.381098$, below both stationary times. Their corners can be smoothed, so Fermat principle here must be understood as stationarity rather than unrestricted global minimization.

Use a Lagrange multiplier $\mu$ for the norm constraint. For real variations $u+\delta \eta$ with $\eta (0)=\eta (\pi )=0$, integration by parts givesThus stationarity is equivalent to the Sturm-Liouville problemMultiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the eigenvalues, and the normalized stationary functions are their eigenfunctions. Conversely the eigenvalue equation makes the constrained first variation vanish. The self-adjoint operator has weight one.

For $ϵ=0$, zero and negative eigenvalues yield no nonzero solution with both Dirichlet boundary conditions. Positive eigenvalues have solutions $u=A\sin (\sqrt{\lambda }x)$ after the left condition, and the right condition gives $\sqrt{\lambda }=n$, $n\ge 1$. Hence the lowest eigenpair iswhere the amplitude follows from ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$.

The Rayleigh-Ritz variational principle bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi /2$ givesThus the sine trial bound for a linear potential on an interval isFor $ϵ>0$ the bound is strict, since $-Y_1^{\prime \prime }+ϵx{Y}_1=(1+ϵx)Y_1$ is not a constant multiple of $Y_1$. Also ${\lambda }_{ϵ}\ge 1$ by the Poincaré inequality and the nonnegative potential. Equality in the upper bound holds at $ϵ=0$.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

The Dirichlet uniqueness for the Poisson equation follows from a positive integral. Suppose $T_a,T_b$ satisfy the same data, and put $w=T_a-T_b$. Then $w$ is a harmonic function in $V$ and vanishes on $S$. Green's first identity, applied with both arguments $w$, givesThus $\nabla w=0$, and $w$ is constant on the connected body. Its zero boundary value makes that constant zero. This proves at most one temperature field; it does not require an independent existence construction. As usual, the argument assumes a bounded body and sufficient regularity for Green's first identity, or the corresponding finite-energy weak formulation.

For the composite body, let $w_{-}$ and $w_{+}$ be the differences in $V_1$ and $V_2$. They are harmonic functions, have a common interface value $w$, and obeyThe unit normal vector $n$ points out of $V_1$; the outward normal of $V_2$ on its inner boundary is $-n$. Keeping that sign in Green's first identity givesThe outer boundary term in $E_2$ is zero. Weight these equations by the positive coefficients that match the flux condition:Both Dirichlet energies must vanish. Hence both differences are constants; the outer data set the constant in $V_2$ to zero, and interface continuity sets the constant in $V_1$ to zero. This establishes uniqueness for a harmonic transmission problem. Physically $\beta$ and $\alpha$ play the roles of the inner and outer thermal conductivities; continuity of temperature and normal heat flux is what cancels the interface contribution.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

For the grounded wall, use an image charge $-q$ at $(2b,0)$. Define the Euclidean distances $r=\sqrt{x^2+y^2}$ and $r_{\ast }=\sqrt{(x-2b{)}^2+y^2}$. The method of images givesThe real source has the required Dirac delta function coefficient, and the image is outside the domain. On $x=b$, $r=r_{\ast }$, so $V_D=0$ exactly. This verifies the Dirichlet boundary condition rather than only its derivative.

For the insulating wall, the image instead has the same strength $q$. With the distances defined above,It has the required source in the physical domain. Its normal derivative isAt $x=b$ the terms are $b/(b^2+y^2)$ and $-b/(b^2+y^2)$, so they cancel. The Neumann boundary condition leaves the additive constant $C$ free.