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

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

For $u>0$, eliminating $v$ givesThus constant $u$ surfaces are confocal elliptic cylinders parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\le a$.

For generic $v$, eliminating $u$ givesThese are hyperbolic cylinders with the same focal lines. A specified angular value selects the half-branch with $\mathrm{sgn}x=\mathrm{sgn}(\cos v)$ and $\mathrm{sgn}y=\mathrm{sgn}(\sin v)$, because $u\ge 0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\ge a$ or $x\le -a$; at $v=\pi /2,3\pi /2$ they are $x=0$, $y\ge 0$ or $y\le 0$.

Finally, constant $z$ gives horizontal planes. The three surface families express the orthogonality of elliptic cylindrical coordinates.

The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:The given components are Cartesian components, so their ordinary dot product with this tangent isTherefore Stokes theorem givesA direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla \times \mathbf{F}{)}_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh 1\cosh 1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.

Write a nonzero separated solution as $T=X(x)Y(y)S(t)$. Division of the heat equation by this product givesThe two spatial factors solve independent constant-coefficient eigenvalue problems. Zero endpoint values in $x$ requireThe insulating conditions requireThe zero cosine mode is the nonzero constant function and must be retained; there is no nonzero zero-index sine mode. The temporal equation is $S^{\prime }=-\lambda {\pi }^2(n^2/a^2+m^2/b^2)S$. Superposing the complete sine-cosine basis yieldsThe constants are fixed by expansion of the initial profile. Orthogonality gives sine norm $a/2$; the cosine norm is $b$ at $m=0$ and $b/2$ otherwise. This is the heat equation on a rectangle with two insulated sides.

First decompose the initial $y$ dependence using ${\sin }^2(2\pi y/b)=[1-\cos (4\pi y/b)]/2$. Only the cosine indices $m=0$ and $m=4$ occur. Expand the remaining factor in a Fourier sine series:For $k_n=n\pi /a$, two integrations by parts giveThus even $n$ vanish, while $b_n=8a^2/(n^3{\pi }^3)$ for odd $n$. The nonzero double-series coefficients are $A_{n0}=b_n/2$ and $A_{n4}=-b_n/2$. Their separate heat-mode decay factors giveAt $t=0$ the bracket is $2{\sin }^2(2\pi y/b)$, which recovers the initial profile with the displayed sine coefficients. Each term satisfies the mixed Dirichlet boundary conditions and Neumann boundary conditions.

The slowest surviving mode has $n=1,m=0$. HenceIt is independent of $y$: insulation allows the constant transverse mode, while every nonconstant transverse mode decays faster. The next possible relative decay rates are $8\lambda {\pi }^2/a^2$ from $n=3,m=0$ and $16\lambda {\pi }^2/b^2$ from $n=1,m=4$.

For each source point $z$, construct the Green function from homogeneous solutions on either side:The endpoint conditions are then automatic. The equation $L_{x}G=\delta (x-z)$ requires continuity of $G$ and a unit jump in its first derivative: a discontinuity in $G$ would introduce an unwanted delta derivative, while integrating across the source fixes $G_x(z^{+},z)-G_x(z^{-},z)=1$. ThusSolving this two-by-two system, with the Wronskian $W=y_a{y}_b^{\prime }-y_b{y}_a^{\prime }\ne 0$, givesThis derivation also remains valid if one of the solution values at the matching point vanishes: only the Wronskian must be nonzero. For a regular nonsingular two-point problem it supplies the inverse kernel; at a singular endpoint, uniqueness requires separate checking, as part (d) illustrates.

The Green-function representation isThe delta identity gives $L{y}_p=f$ and the endpoint conditions follow from those of $G$. If a nonzero homogeneous solution also satisfies both endpoint conditions, it may be added; uniqueness is not automatic in that case.

For the Euler-Cauchy equation, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ isThe distinct roots giveThese solutions are linearly independent since their Wronskian is $2x^3\ne 0$ away from zero. Both nevertheless approach zero as $x\to 0$, which will make the endpoint conditions in the next part insufficient for uniqueness.

Choose $y_a=x$ and $y_b=x^3-x$, satisfying the left and right conditions respectively. Their Wronskian isPart (a) therefore supplies the valid kernelIt is continuous at the source, its first derivative has jump one, and it vanishes at both endpoints. However, it is not unique: $x^3-x$ is a nonzero homogeneous solution satisfying both stated endpoint conditions. Thus $G_0+(x^3-x)h(z)$ has the same defining properties for any suitable $h$.

This is a nonunique Green kernel at a singular endpoint with two vanishing modes. Choosing $y_a=x^3$ instead also gives a valid kernel, corresponding to the stronger selection $y(x)/x\to 0$ at the left endpoint. The original problem states no such selection, so the family must be acknowledged.

The original PDF shows $x^2$ in the nonzero source branch. Inserting that source into $G_0$ gives, for $x<1/2$,For $x\ge 1/2$, split the integral at $z=x$:Because of the homogeneous freedom established in part (i), the full answer to the stated endpoint problem isAt the interface, both branches have value $-5/96$ and derivative $-5/48$, so no extra delta or delta derivative occurs there. Direct substitution gives zero source on the left and $x^2$ on the right; the endpoint values are zero for every $C$. If the extra condition $y/x\to 0$ is imposed, it selects $C=-5/48$, but that is additional information absent from the printed question.

For the stated Fourier transform convention, Fourier inversion isFor example, the pointwise formula holds at continuity points under the usual integrability hypotheses; the same normalization applies to the $L^2$ or distributional inverse.

Assume initially that $f,g$ are integrable, so Fubini's theorem justifies interchanging the integrals. Transform the convolution and put $u=x-z$:This is the convolution theorem. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.

Differentiate the defining Fourier integral $n$ times with respect to $k$:For a nonnegative integer $n$, this givesIntegrability of $x^{n}f$ together with suitable lower moments justifies differentiation under the integral. The same multiplication-derivative identity holds more generally for tempered distributions.

Evenness means the negative half of the function is determined as well: this is the triangular function $q(x)=1-|x|$ on $|x|\le 1$, zero outside. Its transform is real and even:Integration by parts, or integrating the two terms separately, givesThe apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the Fourier transform of a triangular function.

Let $r(z)$ be the indicator of $[-1,1]$. The finite integral is exactly the convolution $p=q\ast r$. Its additional transform factor isApplying the convolution theorem and part (d),Equivalently this is $2(\sin k/k)[\sin (k/2)/(k/2){]}^2$. The value at zero also follows from $\int p=(\int q)(\int r)=1\times 2$.

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

An invertible matrix is a square matrix for which there is a matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I$. Write its assumed diagonalization as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. ThenIf the determinant is nonzero, every ${\lambda }_j$ is nonzero, so ${\Lambda }^{-1}=\mathrm{diag}({\lambda }_j^{-1})$ exists and $A^{-1}=P{\Lambda }^{-1}{P}^{-1}$. If the determinant is zero, at least one eigenvalue is zero; its nonzero eigenvector is mapped to zero, which an invertible map cannot do. HenceThe equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

For the positive integer $k$ in the nilpotence condition, diagonalize $B=P\Lambda P^{-1}$. ThenEvery diagonal entry must vanish, so each ${\lambda }_j=0$. Consequently $\Lambda =0$ and $\boxed{B=0}$. This proves that a diagonalizable nilpotent matrix is zero.

Without diagonalizability a counterexample isIts only eigenvalue is zero but its kernel has dimension one, so it cannot have a basis of eigenvectors. It is the simplest nontrivial nilpotent Jordan block.

By the spectral theorem, a Hermitian matrix has real eigenvalues and an orthonormal eigenbasis. Choose column vectors $u_j$ with $H{u}_j={\lambda }_j{u}_j$ and $u_i^{†}{u}_j={\delta }_{ij}$, and form $U=(u_1\cdots u_n)$. Then $U^{†}U=I$, so $U$ is unitary, and $HU=U\Lambda$ givesThe entries of $\Lambda$ are the real eigenvalues, repeated according to their multiplicities. Within a degenerate eigenspace, any orthonormal basis may be used. Equivalently $H=U\Lambda U^{†}$.

A real symmetric matrix has a real orthonormal eigenbasis. Put those eigenvectors into an orthogonal matrix $Q$, so $Q^{T}AQ=\mathrm{diag}(a_1,\dots ,a_n)$. With the change of coordinates $x=Q{x}^{\prime }$, or $x^{\prime }=Q^{T}x$,Thus the coefficients are the eigenvalues and the new coordinate directions are the eigenvectors. This orthogonal diagonalization preserves Euclidean lengths and angles while removing cross terms from the quadratic form.

The second coordinate direction is already an eigenvector with eigenvalue three. The first and third coordinates form a symmetric block, whose sum and difference directions giveFor example, the sum of the block's diagonal and off-diagonal entries is $(1+c)+(5-c)=6$, while their difference is $(1+c)-(5-c)=2c-4$. Equivalently the characteristic polynomial is $(\lambda -6)(\lambda -3)(\lambda -2c+4)$.

The displayed basis works for every real $c$. At $c=7/2$, the eigenvalue three has eigenspace spanned by the second and third listed vectors; at $c=5$, the eigenvalue six has eigenspace spanned by the first and third. Every nonzero vector in a repeated eigenspace is an eigenvector, and any orthonormal basis of it may replace the displayed choices.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.

For $f=u+iv$ analytic in $z=x+iy$, the Cauchy-Riemann equations areThey follow by computing the same complex derivative along the real and imaginary coordinate directions.

The Cauchy-Riemann equations giveAt a regular point, each level curve has tangent perpendicular to its nonzero gradient. The two gradients, and hence the two tangents, are orthogonal. Therefore regular level curves meet at right angles.

The qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }(z)|$, so at a critical point both gradients vanish and this argument gives no tangent direction. For example, for $f(z)=z^2$, the zero level curves of the real and imaginary parts are diagonals and coordinate axes; they meet at $45^{\circ }$ at zero. This is the regular-point statement of orthogonal level curves of harmonic conjugates.

Recognize that $z^2=(x^2-y^2)+2ixy$ and $e^{-z}=e^{-x}(\cos y-i\sin y)$. Their product has exactly the specified real part. Its imaginary part isOne can also recover it directly from the Cauchy-Riemann equations. Differentiating the specified real part givesIntegration of $v_y=u_x$ gives $v=v_0+h(x)$; the other equation $v_x=-u_y$ then forces $h^{\prime }(x)=0$. Thus the most general answer on a connected domain isIndeed the difference of any two analytic functions with the same real part has zero real part; its two Cauchy-Riemann equations force its imaginary part to be constant. This proves that no further analytic freedom is missing.

The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there areTo inspect infinity, put $w=1/z$:Hence infinity is a removable singularity at infinity, with limiting value zero. After this natural extension to the Riemann sphere it is also a simple zero. There are no other zeros or singularities.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.

Put $\zeta =z-3$ and apply the geometric series:ThusThe radius is the distance from the expansion centre $3$ to the nearest singularity $2i$, not the distance of that pole from the origin. On the boundary circle the series terms all have magnitude $1/\sqrt{13}$, so they do not approach zero: this particular series diverges at every boundary point.