Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.

Choose a complete orthonormal system of eigenfunctions ${\varphi }_n$ for the self-adjoint problem. Expand $f={\sum }_n{f}_n{\varphi }_n$ and $y={\sum }_n{c}_n{\varphi }_n$, withSubstitution into $Ly=f$ gives ${\lambda }_n{c}_n=f_n$. Since no eigenvalue is zero, the Sturm-Liouville eigenfunction expansion isFor eigenfunctions not normalized to one, divide each term additionally by $⟨{\varphi }_n,{\varphi }_n⟩$. The expansion is understood in the weighted function space and, with sufficient data regularity, in the operator domain. A zero eigenvalue would instead require its corresponding source coefficient to vanish and would leave an arbitrary homogeneous component.

Expanding the general Sturm-Liouville form givesMatching the second-derivative coefficients requires $p/w=x^{-4}$. Matching the first-derivative coefficients then gives $p^{\prime }/p=-2/x$. Up to a common positive normalization of $p,q,w$, one may therefore takeIndeed $-x^{-2}(x^{-2}{y}^{\prime }{)}^{\prime }-ay=-x^{-4}{y}^{\prime \prime }+2x^{-5}{y}^{\prime }-ay$. The left endpoint is singular in these coordinates. Its self-adjoint interpretation is supplied by the regular transformed problem in part (ii), rather than by assuming that $p(0)$ is finite.

Set $z=x^3/3$ and write $y(x)=Y(z)$. Thenso the two first-derivative contributions cancel exactly, leavingThe Dirichlet spectrum of $-d^2/d{z}^2$ on this interval is $(3n\pi {)}^2$, $n\ge 1$. If $\lambda +a\le 0$, the linear or hyperbolic solutions cannot satisfy both zero endpoints nontrivially, so no additional modes have been omitted. ThusTo verify normalization, use $dz=x^{2}dx$:This proves weighted orthonormality and also transfers completeness from the ordinary Fourier sine series. The construction is the cubic-coordinate reduction of a singular Sturm-Liouville operator. The positive constant $a$ shifts the entire spectrum down; a zero eigenvalue occurs if $a=9n^2{\pi }^2$, so the no-zero-eigenvalue proviso in part (a)(iii) must be imposed separately if that inverse formula is used here.

Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

For $\Phi =X(x)Y(y)$, separation of variables gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-(m^2+p^2)$. With the standard separation constants $m^2$ and $p^2$, the zero-endpoint solutions are sines. Thus, provided $mp\ne 0$,is nonzero, has zero value on all four sides, and satisfies $\Delta \Phi =-(m^2+p^2)\Phi$.

There is a missing qualification in the printed request for arbitrary integer $m,p$. The Dirichlet and Neumann spectra of a square show that a nonzero Dirichlet solution exists precisely whenWhen one of $m,p$ is zero, the displayed sine product vanishes identically, and a nonzero solution need not exist: for example $(m,p)=(0,0)$ or $(0,1)$ admits none. On the other hand $(0,5)$ does admit $\sin (3x)\sin (4y)$. Thus the natural intended case is two nonzero integer indices, with the stated spectral criterion covering the exceptional cases.

Choose the separated cosine solutions instead:Its derivatives are ${\Phi }_x=-m\sin (mx)\cos (py)$ and ${\Phi }_y=-p\cos (mx)\sin (py)$. They vanish on the appropriate opposite sides of the square for integer $m,p$, so the outward normal derivative is zero everywhere on the boundary. Also $\Delta \Phi =-(m^2+p^2)\Phi$.

This is nonzero for every pair of integer indices, including a zero index. For $m=p=0$ it is the constant solution, illustrating the zero mode allowed by Neumann boundary conditions. Negative indices give the same cosine modes. The construction is the Neumann part of the Dirichlet and Neumann spectra of a square.

For the real decaying Bessel form used here, take $k>0$. The original PDF has $K_m(r)$ as the last factor in its auxiliary differential equation, although both differentiated factors have argument $kr$; that last argument must also be $kr$ for the stated Modified Bessel differential equation.

Put $B(r)=K_m(kr)$. The corrected identity is $B^{\prime \prime }=-B^{\prime }/r+(k^2+m^2/r^2)B$. Since the field is radial, the radial Laplacian givesAway from the point source this must vanish. For $k>0$, a constant ratio of $r{B}^{\prime }$ to $B$ would make $B$ a pure power, incompatible with the $k^2{r}^2$ term of its equation. Therefore both displayed coefficients must vanish:The two orders give the same Modified Bessel function of the second kind: $K_{-1/2}(z)=K_{1/2}(z)=\sqrt{\pi /(2z)}{e}^{-z}$.^[ref]

The source describes $k$ merely as real, but the following displayed real square-root formula implicitly needs $k>0$. For the operator with arbitrary real $k$, its decaying fundamental solution instead uses $\kappa =|k|$; $k=0$ is obtained by a limit. The unmodified real $A{K}_m(kr)$ ansatz is not a real Bessel formula for all negative $k$.

Write $C=A\sqrt{\pi /(2k)}$, so the radial field is $\Phi =C{e}^{-kr}/r$. Integrate its equation over a ball of radius $ϵ$ centred at the source. The divergence theorem givesThe second term is $4\pi k^{2}C{\int }_0^{ϵ}r{e}^{-kr}dr=O({ϵ}^2)$ and vanishes in the limit. MeanwhileThusThe minus sign follows from the convention $(\Delta -k^2)\Phi ={\delta }^{(3)}$, opposite to the positive static Yukawa Green function for $-\Delta +k^2$. For arbitrary real $k$, replace it in the exponential by $|k|$ to select decay at infinity. The $k=0$ limit is $-1/(4\pi r)$.

Reflect the source across the plane, defining ${\mathbf{r}}^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ for an interior point $z^{\prime }>0$. Superpose the direct Green function from part (b) and an image of opposite sign:The direct term supplies the delta source in the upper half-space; the image singularity lies outside that domain and contributes no source there. On $z=0$ the two distances are equal, so the terms cancel, giving homogeneous Dirichlet boundary conditions. Both decay at infinity, which is the additional condition selecting this half-space solution. This is the method of images realization of the Dirichlet image kernel for the three-dimensional modified Helmholtz equation. It includes the unscreened case $k=0$.

Let $f=u+iv$ be analytic on a neighbourhood of a bounded planar region $\Omega$ and its piecewise smooth boundary $C$, oriented with $\Omega$ on the left. The Cauchy integral theorem statesFor a region with holes, $C$ includes the clockwise inner boundaries as well as the counterclockwise outer one. In particular, the integral around a closed contour bounding a region throughout which $f$ is analytic is zero.

To prove this classical form, expand the line integral and apply Green's theorem to its real and imaginary parts:Both integrands vanish by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$. Analyticity supplies the continuous first partial derivatives needed by this Green's theorem proof. The requirement that the enclosed region contain no singularities is essential.

Put $s=1/n$, so $0<s<1$. For the complex exponentiation $z^s$, choose $-\pi /2<\arg z<3\pi /2$: the branch cut is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed simple pole is $z=ia$, with residueThe large arc is $O(R^{s-1})$ and the small arc is $O({ϵ}^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z{|}^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The residue theorem givesSince $1+e^{i\pi s}=2e^{i\pi s/2}\cos (\pi s/2)$,The endpoint estimates also prove convergence. More generally the same calculation is the fractional-power integral from an upper-half-plane contour, valid for $-1<s<1$.

Set $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\tan \theta =(z^2-1)/[i(z^2+1)]$, so the requested integral isThe denominator factors asFor $0<a<1$, its two inner roots are $z=\pm r$, $r=\sqrt{(1-a)/(1+a)}<1$, and the other two roots have modulus $1/r>1$. There is also a simple pole at zero. Their residues areThe latter values follow by differentiating the quadratic product, or from the supplied hint. The sum of the enclosed residues isThe residue theorem therefore givesAt the angles where the tangent is undefined, the original reciprocal has a removable zero limit. Its continuous extension gives the same integral; the transformed rational integrand has no pole on the unit circle.

With the stated Fourier transform convention, the exponential transform is a shifted Dirac delta distribution:Writing the sine as $(e^{i{\omega }_{0}t}-e^{-i{\omega }_{0}t})/(2i)$ then givesThese are tempered distributions, since the time-domain functions are not integrable on the full real line. The second formula also correctly vanishes when ${\omega }_0=0$.

By the definition of convolution and Fourier inversion,The second line uses $\nu =\omega -\eta$. The factor $2\pi$ comes from the inverse-transform normalization. This is the product form of the convolution theorem, initially justified for sufficiently decaying functions and extended when the indicated products and convolutions exist.

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform isusing the given Dirichlet integral and the sign of $t$. Also ${\mathcal{F}}^{-1}[\delta ](t)=1/(2\pi )$. The inverse of the proposed expression is consequently $(iA/2)\mathrm{sgn}t+B/(2\pi )$. Comparison with $H(t)=[1+\mathrm{sgn}t]/2$ for $t\ne 0$ givesThe value assigned at the single point $t=0$ does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating $1/\omega$ as an ordinary integral would miss the required prescription.

Multiply $H$ by the exponential form of the sine and use the modulation property of the Fourier transform:Inserting part (iii) givesThe reciprocal terms are real principal-value distributions and the delta terms are imaginary. Equivalently, exponential damping $H(t)e^{-ϵt}$ gives ${\omega }_0/[(ϵ+i\omega {)}^2+{\omega }_0^2]$, whose distributional limit is the boxed expression. The shifted-difference formula remains valid at ${\omega }_0=0$, where all terms cancel.

Transform only the time variable. Two integrations by parts give ${\mathcal{F}}_t[y_{tt}]=(i\omega {)}^2\tilde{y}=-{\omega }^2\tilde{y}$, when the ordinary transform boundary terms vanish; the same derivative rule holds for tempered distributions. Spatial derivatives and the fixed spatial delta commute with the transform. ThusorThe spatial Dirichlet boundary conditions transform to $\tilde{y}(\omega ,0)=\tilde{y}(\omega ,L)=0$.

Multiply the transformed equation by $\sin (k_{n}x)$ and integrate from zero to $L$. Two integrations by parts, using the zero endpoint values of both $\tilde{y}$ and the sine, giveThe delta source contributes $\tilde{F}(\omega )\sin (k_n{x}_0)$. Multiplying by $2/L$ and using the coefficient definition therefore yieldsAway from the resonant frequencies this givesAt $\omega =\pm k_n$ the quotient must be interpreted with a pole prescription if the source excites that mode. In distributional form, arbitrary delta terms at those frequencies may also be added; they are the free homogeneous oscillations. This is the frequency-space issue addressed explicitly in part (iii).

Expand in the complete Fourier sine series $\tilde{y}(\omega ,x)={\sum }_{n\ge 1}{b}_n(\omega )\sin (k_{n}x)$ and apply Fourier inversion. Substitution of part (ii) givesTo check the source normalization, apply ${\partial }_t^2-{\partial }_x^2$ mode by mode. It cancels the denominator; the remaining inverse transform gives $2\pi F(t)$, and completeness givesThe endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.

The displayed real-axis integrals in the paper are not ordinary convergent integrals at their poles. One definite choice is the Cauchy principal value at each $\omega =\pm k_n$. If a causal solution is wanted, use $[k_n^2-(\omega -i0{)}^2{]}^{-1}$ instead. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, so the causal solution iswhen the past integral exists. The principal-value choice instead uses the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the causal Fourier resolvent of a point-driven fixed string.

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of $L$. Thus a tensor of order $n$ transforms asRepeated old-basis indices are summed from one to three. Orthogonality of the new basis gives $L{L}^T=I$. Taking determinants yields $(\det L{)}^2=1$, hence $\boxed{\det L=+1\text{ or }-1}$. The positive case preserves orientation and the negative case reverses it.

A pseudotensor has an additional determinant factor under an orthogonal transformation. Thus its rank-three transformation law isThe contracted expression $L_{ia}{L}_{jb}{L}_{kc}{ϵ}_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of ${ϵ}_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the determinant expansion of $L$, fixing that scalar to $\det L$. ThereforeThe Levi-Civita symbol consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three tensor transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.

Integrating the Dirac delta distribution in $x_3$ reduces the inertia integral to the planar lamina. Use $u=x_1-x_2$, $v=x_2$, so $x_1=u+v$ and the parallelogram becomes the square $-D\le u,v\le D$, with unit Jacobian determinant. The mass is $M=4\sigma D^2$, and symmetry gives a centre of mass at the origin.

Odd rectangular integrals vanish. The required second moments areThe mixed moments involving $x_3$ vanish. Hence the inertia tensor isThe off-diagonal entries are minus the product moment, not the product moment itself. This is the unit-shear case of the inertia tensor of a sheared rectangular lamina.

Put $J_0=4\sigma D^4/3=M{D}^2/3$. The normal to the lamina is already an eigenvector, with moment $3J_0$. The planar block has characteristic polynomialso the principal moments of inertia areA corresponding orthonormal set of principal axes isThe sign of each axis vector is immaterial. The smaller-moment axis makes angle $\frac{1}{2}\arctan (2)$ with the positive $x_1$ direction, approximately $31.7^{\circ }$; the other planar axis is perpendicular. As a check, $I_{-}+I_{+}=3J_0=I_3$, the perpendicular axis theorem for a planar lamina.

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.

Write $d=2k-m{\omega }^2$. The characteristic determinant of the tridiagonal matrix isThe three values of $d$ are $\sqrt{2}k$, $0$ and $-\sqrt{2}k$. Therefore the normal mode frequencies and convenient Euclidean-normalized generalized eigenvectors areEach pair satisfies $V\mathbf{u}={\omega }^{2}T\mathbf{u}$. The first mode moves all masses in phase, the middle mode leaves the central mass fixed, and the last has the central mass moving opposite to the outer masses. All squared frequencies are positive, consistent with the positive spring potential and fixed endpoints.