Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

The azimuthal vector ${\mathbf{h}}_{\varphi }$ is automatically orthogonal to the other two. Their remaining dot product isThus orthogonal coordinates require $g{g}^{\prime }=\sinh \xi \cosh \xi$, which integrates to $g^2={\sinh }^2\xi +C$. The initial value gives $C=1$, and the smooth branch with $g(0)=1$ isThe corresponding coordinate-vector norms are $|{\mathbf{h}}_{\xi }|=|{\mathbf{h}}_{\theta }|=\sqrt{{\sinh }^2\xi +{\sin }^2\theta }$ and $|{\mathbf{h}}_{\varphi }|=\sinh \xi \sin \theta$, as in the hyperbolic parametrization of prolate spheroidal coordinates. Coordinate degeneracies on the axis and focal segment are understood.

Eliminating the angular coordinates on the requested surface givesIt is a prolate spheroid with long axis $z$, equatorial radius $\sinh 1$, polar semiaxis $\cosh 1$, and foci $(0,0,\pm 1)$.

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

Use separation of variables, $u=X(x)Y(y)$, and divide by $XY$ where nonzero. A real separation constant $\kappa$ givesso $X^{\prime \prime }=\kappa X$ and $Y^{\prime }=(\kappa -1-\sin y)Y$. Normalizing the second factor to one at zero givesThe spatial ordinary differential equation has three types of real solution:Products $X{Y}_{\kappa }$ are the separated eigenfunctions. The negative, zero and positive separation constants give trigonometric, linear and hyperbolic function forms respectively. There is no discrete quantization until boundary conditions on $x$ are supplied. Since $\cos y-1$ is bounded, the envelope of $Y_{\kappa }$ decays for $\kappa <1$, grows for $\kappa >1$, and is bounded and periodic for $\kappa =1$. The scalar prefactor can also be derived by removing a variable scalar reaction rate by an integrating factor.

The two homogeneous Dirichlet boundary conditions on $x$ force negative separation constants. The initial profile is an even function, so only the even sine-family eigenfunctions are needed:They obey $X_m(\pm 1)=0$ and ${\int }_{-1}^1{X}_m{X}_{n}dx={\delta }_{mn}$. Odd eigenfunctions have zero projection onto the even initial data. Hence the initial coefficients areThe desired eigenfunction expansion is thereforeAt $y=0$ this is the Fourier series of $1-x^2$ on the interval; its coefficients are absolutely summable. Each term satisfies the partial differential equation and side conditions. For $y>0$ the additional Gaussian decay in $m$ justifies termwise differentiation. Since $k_m^2+1>0$ and $e^{\cos y-1}\le 1$, the whole series decays uniformly as $y\to \infty$.

Set $L=d^2/d{t}^2-4$ and choose $L_{t}G(t,t^{\prime })=\delta (t-t^{\prime })$, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are $e^{\pm 2t}$. Decay at each end and continuity at $t=t^{\prime }$ therefore give $G=C{e}^{-2|t-t^{\prime }|}$. Integrating its equation across the source gives a unit jump in its first derivative:ThusThe solution of the boundary value problem is the convolutionThere is no extra homogeneous term: no nonzero combination of $e^{2t},e^{-2t}$ decays at both infinities. The forcing must of course permit the imposed decay; integrable $f$, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

For $t<0$, all forcing points satisfy $\tau >t$, so the Green's function formula becomesAn integration by parts gives $\int \tau e^{-2\tau }d\tau =-(\tau /2+1/4)e^{-2\tau }$. ConsequentlyThe exponential has the required decay as $t\to -\infty$.

For $t>1$, every forcing point has $\tau <t$. Hence the Green's function givesSince $\int \tau e^{2\tau }d\tau =(\tau /2-1/4)e^{2\tau }$, the remaining integral is $(e^2+1)/4$. ThereforeThis decays at positive infinity.

For $0<t<1$, split the Green's function integral where the absolute value changes sign:The two terms inside the bracket are respectivelyAdding them yieldsThe affine term gives $x^{\prime \prime }-4x=t$, while both exponential terms are homogeneous solutions. The values at the joins are $x(0)=(3e^{-2}-1)/16$ and $x(1)=-(1+e^{-2})/16$, agreeing with parts (i) and (ii). Their first derivatives also match: $x^{\prime }(0)=(3e^{-2}-1)/8$ and $x^{\prime }(1)=(1+e^{-2})/8$. Thus the pieces introduce no unwanted Dirac delta distributions.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

By the two Fourier transform rules in part (b),Consequently transforming the differential equation givesorThe inverse Fourier transform gives the reverse implication under the same regularity hypotheses. This expresses the Fourier symmetry of the quantum harmonic oscillator operator.

Part (b) shows that the transform of $g_n$ is $q_n(k)e^{-k^2/2}$ with $q_n$ of degree $n$. Part (c) shows that it solves the same eigenvalue equation as $g_n$, with the same ${\mu }_n$. The assumed uniqueness of the polynomial solution of each degree therefore forces $q_n=c_n{p}_n$; hence ${\tilde{g}}_n=c_n{g}_n$.

The constants can also be determined. Substituting $g_n=p_n{e}^{-x^2/2}$ into the equation gives the Hermite differential equationComparison of the leading coefficient yields ${\mu }_n=-(2n+1)$. If the leading coefficient of $p_n$ is $a_n$, part (b) gives leading coefficient $\sqrt{2\pi }(-i{)}^n{a}_n$ for $q_n$. ThusThus Hermite functions are Fourier eigenfunctions with the paper's unnormalized Fourier transform; dividing that transform by $\sqrt{2\pi }$ gives the unitary eigenvalues $(-i{)}^n$.

Since $M^T=-M$, $(M^2{)}^T=(M^T{)}^2=M^2$, so $M^2$ is a real symmetric matrix. Moreover,proving that it is a negative semidefinite matrix.

For odd $n$, determinant invariance under transpose givesso $\det M=0$. Hence $M$ and $M^2$ have a zero eigenvalue. Finally, if $M^{2}w=0$, then $0=w^T{M}^{2}w=-\Vert Mw{\Vert }^2$, so $Mw=0$. Thuswhere equality of these kernels includes the reverse implication, which is immediate.

Because $M$ commutes with $M^2$,Its squared Euclidean norm isA scalar equals its transpose, while $M^T=-M$, so $u_1^{T}M{u}_1=-u_1^{T}M{u}_1=0$. ConsequentlyThus each negative eigenvalue has at least this pair of orthonormal eigenvectors.

The plane spanned by $u_1,v_1$ is an invariant subspace for $M$, sinceIts orthogonal complement is also invariant: for $w$ orthogonal to both, $u_1^{T}Mw=-(M{u}_1{)}^{T}w=0$ and $v_1^{T}Mw=-(M{v}_1{)}^{T}w=0$. The restriction of the real symmetric matrix $M^2$ to this nonempty complement therefore has a unit eigenvector $u_2$, by the spectral theorem for real symmetric matrices.

If its eigenvalue is $-{\mu }_2^2<0$, the argument in part (b) shows that $v_2=M{u}_2/{\mu }_2$ is a unit eigenvector orthogonal to $u_2$. It also lies in the same complement; explicitly,Repeating the construction decomposes every negative eigenspace into orthogonal pairs. The zero eigenspace has dimension $n-2r$ for some number $r$ of pairs, and therefore has the same parity as $n$. This is the paired eigenspaces of the square of a real skew-symmetric matrix property.

For $n=2m$, there can be $m$ distinct negative eigenvalues and no zero, or at most $m-1$ negative ones plus zero, because a nonempty zero eigenspace has dimension at least two. For $n=2m+1$, there are at most $m$ distinct negative ones plus the mandatory zero. HenceThe bound is attained by skew two-by-two blocks with distinct positive parameters, adding one zero direction when $n$ is odd.

The required actions have already been identified:In three dimensions choose a unit vector $w$ perpendicular to both. Its one-dimensional orthogonal complement to the pair is invariant under $M$. On a one-dimensional real space, a skew-symmetric matrix must act as zero: $Mw=\eta w$ would imply $\eta =w^{T}Mw=0$. Thus $Mw=0$.

The columns of $W=(u_1{v}_{1}w)$ form an orthonormal basis, so $W$ is an orthogonal matrix. The displayed actions giveand thereforeIf $M=0$, take ${\mu }_1=0$ and any orthogonal matrix $W$. Otherwise its nonzero two-dimensional block supplies the pair used above.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

The quadratic form has real symmetric matrixIts unit eigenvectors can be chosen as $e_u=(1,1,0)/\sqrt{2}$ with eigenvalue $1$, and $e_v=(1,-1,0)/\sqrt{2}$, $e_w=(0,0,1)$ with eigenvalue $9$. SetThe orthogonal diagonalization turns the surface equation intoIt is an ellipsoid, specifically a prolate spheroid centered at the origin. Its long principal axis is the line in direction $\boxed{(1,1,0)/\sqrt{2}}$, with semiaxis length $\boxed{3}$. Each direction in the plane perpendicular to that line has semiaxis length $\boxed{1}$; the displayed $e_v,e_w$ give one orthonormal choice of the two short principal axes. Their nonuniqueness reflects the repeated eigenvalue $9$ and rotational symmetry about the long axis.

The orthogonal change of coordinates preserves the sphere equation, giving $u^2+v^2+w^2=4$. Subtract it from the spheroid equation in part (b):Thus the intersection consists of the two circlesTheir common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates givesTheir planes have equations $x_1+x_2=\pm 3\sqrt{3}/2$, each with normal vector $(1,1,0)$. The difference ${\mathbf{c}}_{+}-{\mathbf{c}}_{-}$ is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.

For an analytic function, the complex differentiability limit is independent of the direction of approach. A real increment gives $f^{\prime }(z)=u_x+i{v}_x$, whereas an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if $u,v$ have continuous first partial derivatives near a point and obey these equations, their real differentiability gives, for a complex increment $h$,Dividing by $h$ proves the complex differentiability there. Continuous first partial derivatives are a sufficient regularity hypothesis for this converse; agreement along the two coordinate directions alone is only a necessary condition.

Using the complex cosine identity,we findIts real part is exactly the prescribed function. Thus a harmonic conjugate is $v=y\cos x\cosh y-x\sin x\sinh y$, and the general answer isTo check the only freedom, two analytic functions with this real part differ by a function with zero real part. The Cauchy-Riemann equations then force both partial derivatives of its imaginary part to vanish, so it is constant on the connected domain. No normalization was supplied to fix $C$.

Multiply the exponential power series by the two prefactors:Collecting powers yields the Laurent seriesIts first terms are $z+1+3/(2z)+7/(6z^2)+13/(24z^3)+\cdots$. Infinitely many negative-power coefficients are nonzero, so $\boxed{z=0\text{ is an essential singularity}}$. The residue is the coefficient of $z^{-1}$:The phrase "power series" here refers to a Laurent series; no ordinary Taylor series about an essential singularity exists.

The denominator vanishes at $z=\pm i\sqrt{2}$, giving two simple poles. Their distances from the center are both $|1\mp i\sqrt{2}|=\sqrt{3}$. There are no nearer singularities, so the Taylor series has radius of convergence

Apply the Cauchy-Hadamard theorem to the power series coefficients $a_n=[n/(n+\alpha ){]}^{n^2}$ for $n\ge 1$:For example, taking logarithms gives $-n\log (1+\alpha /n)\to -\alpha$. ThereforeAny convention for the isolated $n=0$ coefficient has no effect on the radius of convergence.

Away from zeros of the leading coefficient, the normal-form differential equation isThe only finite singular points are $z=0,i,-i$. At zero, $zP(z)=3/(1+z^2{)}^2$ and $z^{2}Q(z)=1/(1+z^2{)}^2$ are analytic, so zero is a regular singular point. At $z=\pm i$, the coefficient $P$ has a genuine double pole, so $(z\mp i)P$ is not analytic; these are irregular singular points. ThusIf infinity is included, set $t=1/z$ and $Y(t)=y(1/t)$. The transformed coefficients are $\tilde{P}=2/t-3t^3/(1+t^2{)}^2$ and $\tilde{Q}=t^2/(1+t^2{)}^2$. They satisfy the regular singular point at infinity criterion, so $\boxed{\infty \text{ is regular singular}}$ as well.

Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.