Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2

A weighted Sturm-Liouville operator has the formwhere $p,q,w$ are real, $p>0$ and $w>0$ in the interval's interior, and the coefficients have the regularity required by the derivatives. The corresponding eigenvalue equation is $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Its natural weighted inner product is $⟨f,g{⟩}_w=\int wf\bar{g}dx$. Suitable endpoint conditions must also annihilate $[p(f{\bar{g}}^{\prime }-f^{\prime }\bar{g})]$ for a self-adjoint differential operator; the displayed differential expression alone does not specify those conditions.

Multiplication by a positive weight puts the differential equation into Sturm-Liouville form if $(xw{)}^{\prime }=(1-x)w$. Thus $x{w}^{\prime }=-xw$, and the constant factor can be chosen so thatThis fixes the otherwise arbitrary normalization of the weight consistently with the requested unit norms.

Write the polynomial as $p_n(x)={\sum }_{j=0}^n{c}_j{x}^j$. Comparing coefficients givesConsequently $c_j=(-1{)}^j(\genfrac{}{}{0ex}{}{n}{j})/j!$, identifying the solutions as Laguerre polynomials. In particular,Repeated integration by parts gives the moments ${\int }_0^{\infty }{e}^{-x}{x}^{j}dx=j!$. ThereforeFor later use, integration by parts in the weighted eigenvalue equations also givesThe endpoint expression vanishes because of the factor $x$ at zero and exponential decay at infinity. Hence distinct Laguerre polynomials are orthogonal, an instance of orthogonality of Sturm-Liouville eigenfunctions.

The optimal approximation is the orthogonal projection onto the linear span of $p_0,\dots ,p_N$. Their orthogonality givesCompleting each square proves both optimality and uniqueness:For real $f$, these are real coefficients. If complex-valued $f$ is allowed while the coefficients remain real, replace the numerator by its real part and use the modulus-squared norm. The question's real formulation corresponds to real $f$.

Here the first three squared norms are one. Using ${\int }_0^{\infty }{x}^j{e}^{-2x}dx=j!/2^{j+1}$,Thus

The coefficient formula from part (b) and the Leibniz rule yield Rodrigues' formula directly:The leading coefficient is $(-1{)}^n/n!$, so $p_n^{(n)}=(-1{)}^n$. Substitute this differentiated expression into one factor in the squared weighted norm and apply integration by parts $n$ times:Every endpoint term contains a derivative of $e^{-x}{x}^n$ of order less than $n$: it vanishes at zero, and its exponential decay beats every polynomial at infinity. This proves the normalization of the ordinary Laguerre polynomials for all $n$, including $n=0$ without any integrations by parts.

For a separated axisymmetric harmonic function $\Psi =R(r)\Theta (\theta )$, multiplying the Laplace equation by $r^2/(R\Theta )$ givesThe two sides depend on different variables, hence both equal a constant. Put $u=\cos \theta$ in the angular equation. Since $\sin \theta {\Theta }^{\prime }=-{\sin }^2\theta d\Theta /du$, it becomes the Legendre differential equationRegularity on both parts of the polar axis restricts $\lambda =\ell (\ell +1)$, with $\ell =0,1,\dots$, and selects the Legendre polynomial $P_{\ell }(u)$. The other angular solution is singular at an endpoint.

The radial Euler differential equation isTrying $R=r^s$ gives $s(s+1)=\ell (\ell +1)$, so $s=\ell$ or $s=-\ell -1$. Superposition of these separated harmonic functions givesThe Legendre polynomials form the angular expansion for regular data with rotational symmetry about the polar axis; the radial domain and the boundary conditions determine which coefficients survive and where the expansion converges.

With $u=\cos \theta$, the boundary data have the finite Legendre polynomial expansionInside the sphere, every $r^{-\ell -1}$ term is singular at the origin and must be absent. Matching at $r=a$ therefore givesIn the exterior, decay at infinity excludes all $r^{\ell }$ terms, including the constant term at $\ell =0$. ThusBoth harmonic functions have the prescribed Dirichlet boundary condition, and each is finite throughout its own domain. The interior expression is regular even at the origin because $r\cos \theta =z$ and $r^2(3{\cos }^2\theta -1)=2z^2-x^2-y^2$. Uniqueness of the Dirichlet problem identifies these as the respective solutions. Their normal derivatives need not match: the two Dirichlet problems do not require a single harmonic function across the bounding sphere.

Use the sign convention in which a fundamental solution of the Laplace equation satisfiesin the sense of distributions. It is a harmonic function away from the source. Writing $\rho =|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial equation is $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. The divergence theorem on a small circle about the source fixes its total flux:ConsequentlyThe arbitrary constant is a harmonic function. For the alternative convention $-\Delta G={\delta }^{(2)}$, the logarithm has the opposite sign; subsequent Green's function formulas must use that convention consistently.

A Dirichlet Green function satisfies ${\Delta }_{\mathbf{x}}{G}_D(\mathbf{x},\mathbf{y})={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ in the domain and $G_D=0$ when $\mathbf{x}$ lies on its boundary. For a source $\mathbf{y}\ne 0$ in the unit disc, put its opposite image at ${\mathbf{y}}^{\ast }=\mathbf{y}/|\mathbf{y}{|}^2$. The method of images givesThe image lies outside the unit disc, so its logarithm is a harmonic function inside. The original logarithm has precisely the unit source strength found in part (a). On $|\mathbf{x}|=1$,so the boundary value is zero. Equivalently, in complex coordinates, the Dirichlet Green function inside the unit disk isThis also covers a source at the origin, where $G_D(z,0)=(2\pi {)}^{-1}\log |z|$.

Fix the observation point $\mathbf{x}$ and apply Green's second identity in the source variable $\mathbf{y}$. Since $\Delta u=0$, ${\Delta }_{\mathbf{y}}{G}_D={\delta }_{\mathbf{x}}$ and $G_D=0$ on the boundary, it givesThe identity can be justified by deleting a small circle about $\mathbf{x}$ and letting its radius tend to zero; the logarithmic flux contributes $u(\mathbf{x})$.

Set $\mathbf{x}=(r,\varphi )$ and $\mathbf{y}=(\rho ,{\varphi }^{\prime })$, and write $c=\cos (\varphi -{\varphi }^{\prime })$. The symmetric expression for the Green's function isAt the boundary, the outward normal derivative is ${\partial }_{\rho }$ at $\rho =1$, and henceSince $ds=d{\varphi }^{\prime }$ on the unit circle, the Poisson integral on the unit disk follows, with Poisson kernel

At the origin, the Poisson kernel is one, so the mean value property for harmonic functions givesFor $0<r<1$, the denominator in the Poisson kernel satisfies $1-2r\cos \vartheta +r^2\ge (1-r{)}^2$. ThusThe nonnegativity of $f$ allows this upper bound to be inserted under the Poisson integral on the unit disk:This is the upper bound in the Harnack inequality on the unit disk. The same argument with the minimum of the Poisson kernel gives the corresponding lower bound $(1-r)u(0)/(1+r)$.

In the Laurent seriesabout a pole of order $m$, the residue is the coefficientwhere the circle is traversed anticlockwise. Equivalently, the residue is $\frac{1}{(m-1)!}{\frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]|}_{z=z_0}$.

The Taylor series of the entire function $g$ givesThe coefficient of $(z-z_0{)}^{-1}$, and therefore the residue, isIf $g(z_0)=0$, the apparent pole is a removable singularity and the residue is zero.

For the simple-contour form of the residue theorem, let $C$ be a positively oriented, simple, closed, piecewise smooth contour. Assume $f$ is holomorphic on a neighbourhood of $C$ and its interior except at finitely many isolated singularities $z_1,\dots ,z_s$, all strictly inside $C$, with none on $C$. ThenThe more general version weights each residue by the contour's winding number, with the usual requirement that the contour be homologous to zero in the domain of analyticity containing it and the nonsingular region.

For the requested application, assume $f$ is holomorphic on a neighbourhood of the contour and its closed interior, and $z_0$ is strictly inside it. Its Taylor series showsOnly the second term contributes a residue. The residue theorem therefore proves the derivative form of the Cauchy integral formula:

Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives

Write $a=s+ib$ with real $s,b$, and first translate the real integration variable by $s$. It remains to integrate $e^{-z^2}$ along the horizontal line $\mathrm{Im}z=b$.

For $b>0$, apply the Cauchy integral theorem to the rectangle with vertices $-R,R,R+ib,-R+ib$. The integrand is an entire function, sois the negative of the two vertical-edge integrals. On either vertical edge, $z=\pm R+iy$ andTheir combined absolute value is at most $2b{e}^{-R^2+b^2}$ and tends to zero. The rectangle below the real axis gives the same estimate with $|b|$ when $b<0$; $b=0$ needs no deformation. Absolute convergence follows from $|e^{-(x+ib{)}^2}|=e^{b^2}{e}^{-x^2}$. Taking $R\to \infty$ proves the Gaussian contour translation identity

Take $\lambda >0$ and $t>0$, the positive diffusivity requirement for the forward heat kernel. The PDF does not explicitly specify the sign of $\lambda$; the stated real Gaussian function formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the Fourier transform, its inverse and the following integrations are justified. The resulting heat-kernel convolution extends to wider classes of initial data.

Using the stated Fourier transform convention and integration by parts,The transformed ordinary differential equation has solution $\tilde{u}(k,t)=e^{-\lambda k^{2}t}{\tilde{u}}_0(k)$. Fourier inversion then givesComplete the square:Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining Gaussian integral even though its shift is imaginary. ThereforeThe heat kernel has integral one and is an approximate identity as $t\downarrow 0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.

Put $s=\lambda t>0$. In the heat-kernel convolution, complete the real square asThe Gaussian integral therefore givesAt $t=0$ the same formula gives $\beta =\alpha$. Equivalently, ${\beta }^{-2}={\alpha }^{-2}+4\lambda t$: the variance of this normalized Gaussian function grows from $1/(2{\alpha }^2)$ to $1/(2{\alpha }^2)+2\lambda t$.

The initial profiles have unit integral and concentrate at zero as $\alpha \to \infty$. Precisely, for every smooth compactly supported test function $\phi$,Thus the limit is the Dirac delta distribution, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,The limiting evolved profile is the heat kernel for a unit point source.

An order-$n$ Cartesian tensor has $3^n$ components whose values in two orthonormal Cartesian frames satisfywith repeated indices summed as in Einstein notation. It is an isotropic tensor when its components are unchanged by every proper rotation, $R\in SO(3)$.

For rank two, a half-turn about any coordinate axis changes the sign of each component having exactly one index along that axis. Invariance therefore makes every off-diagonal component zero. Quarter-turns interchanging two coordinate axes then require $T_{11}=T_{22}=T_{33}$. Thus, without assuming symmetry in advance,Conversely, the Kronecker delta is invariant because $R_{ik}{R}_{jk}={\delta }_{ij}$, so every tensor of this form is isotropic.

For rank three the general proper-rotation invariant form iswhere ${ϵ}_{ijk}$ is the Levi-Civita symbol. The convention matters: the isotropic rank-three tensor and reflections distinction says that invariance of a polar tensor under all $R\in O(3)$ would force $b=0$. A nonzero ${ϵ}_{ijk}$ is invariant under proper rotations, and is a pseudotensor under the full orthogonal group. The shell integrals below are invariant under reflections as well, so their odd moments vanish.

Introduce the second moment tensor $M_{ij}={\int }_V{x}_i{x}_j{d}^{3}x$. The spherical shell is preserved by every rotation, so this is an isotropic tensor, $M_{ij}=A{\delta }_{ij}$. Its trace determines the constant:Contracting with the components of the two unit vectors givesThis is the second-order case of radial tensor moments in three dimensions.

Under inversion $\mathbf{x}\mapsto -\mathbf{x}$, the spherical shell and its volume element are unchanged, while the product of three scalar products changes sign. Its integral is therefore its own negative:Equivalently, its rank-three moment tensor is completely symmetric and cannot have a nonzero multiple of the antisymmetric Levi-Civita symbol.

Let $M_{ijkl}={\int }_V{x}_i{x}_j{x}_k{x}_l{d}^{3}x$. It is an isotropic tensor and is invariant under every permutation of its four indices. Thus the three coefficients in the general rank-four isotropic form must coincide:A double tensor contraction givesHere the three contracted Kronecker delta products contribute $9$, $3$ and $3$. Contracting instead with the four unit vectors yieldsThis agrees with subtracting two instances of the isotropic fourth moment of a ball.

For an ordinary conservative mechanical system, choose generalized coordinates $q_i$ measured from a static equilibrium point. Assume the kinetic energy is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the potential energy vanishes. Retaining only terms of total degree two gives the small oscillation LagrangianAn irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a stable equilibrium. The Euler-Lagrange equations areA normal mode has a fixed mode shape $\mathbf{v}$ and one time factor: $\mathbf{q}=\mathbf{v}Q(t)$. Its angular frequency $\omega$ solves the generalized eigenvalue problem $K\mathbf{v}={\omega }^{2}M\mathbf{v}$, with $\ddot{Q}+{\omega }^{2}Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega /(2\pi )$. A zero mode satisfies $K\mathbf{v}=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.

Label the equal masses $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi /3,4\pi /3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe first is rigid rotation of all three beads: every arc length stays fixed, giving a zero mode. In the second, the equal masses move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal masses move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero normal modes have zero total angular momentum. These are the normal modes of three beads connected by springs on a circular hoop with $M=2m$.