Year 2 of the natural sciences course of the University of Cambridge (Part IB)

Use Einstein summation convention, writing ${\partial }_j=\partial /\partial x_j$. The contraction of two Levi-Civita symbols givesThe last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

The divergence theorem states that, for a continuously differentiable vector field $\mathbf{H}$ on a bounded region $V$ with piecewise smooth boundary $S$ and outward unit normal $\hat{\mathbf{n}}$,To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with $\mathbf{H}=\mathbf{F}\times \mathbf{G}$ now yields

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

For separation of variables, take $n=T(t)X(x)Y(y)Z(z)$ and divide the reaction-diffusion equation by this product wherever it is nonzero:Each spatial term depends on a different variable and must be constant. Write $X^{\prime \prime }=-k_x^{2}X$ and similarly for the other two factors. The Dirichlet boundary conditions give $X(0)=X(L)=0$. A zero separation constant permits only a linear function and hence the zero solution; a positive value of $X^{\prime \prime }/X$ likewise permits no nonzero function vanishing at both endpoints. The nontrivial choices are thereforeThe temporal ordinary differential equation isso, absorbing its constant multiplier into $a_{p,q,r}$,The superposition principle and completeness of the Fourier sine series give the stated threefold expansion with these factors. For an initial density $n_0$, orthogonality also determines the constants:

The fastest temporal growth occurs for the lowest spatial eigenvalue, with $p=q=r=1$. Its growth rate isFor a generic initial condition its coefficient is nonzero, so this term determines the long-time behavior. In fact, every nonzero, nonnegative initial density has $a_{111}>0$, since the product of the three first sine eigenfunctions is strictly positive in the interior. ConsequentlyFor $L<L_{\mathrm{crit}}$ every term decays. At equality the first term is stationary and all higher terms decay: equality does not produce exponential growth. This is a reaction-diffusion equation threshold arising from competition between production and loss through the Dirichlet boundary condition.

The separation of variables solution with Dirichlet boundary conditions on the cuboid isThe constants are fixed by the initial density through Fourier sine series. The fastest term again has $p=q=r=1$. When $L=2L_{\mathrm{crit}}$, its growth rate becomesThus the precise condition for no exponentially growing term isFor example, $\lambda =1/4$ gives ${\lambda }^{-2}+2\lambda =33/2>12$, while $\lambda =6$ gives $12+1/36>12$: either a sufficiently flat or a sufficiently elongated cuboid suppresses growth despite its unchanged volume. The function ${\lambda }^{-2}+2\lambda$ decreases to its minimum $3$ at $\lambda =1$, then increases without bound. Hence there are two positive threshold values, with growth only between them; equality leaves one stationary term and all other terms decaying.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.

Choose the convention $L_{x}G(x,s)=\delta (x-s)$, where $\delta$ is the Dirac delta distribution, together with $G(1/2,s)=G(1,s)=0$. The Green's function is continuous at $x=s$ and obeysThese conditions follow by integrating its differential equation across $s$; continuity prevents a derivative of the Dirac delta distribution. If $\ell$ and $r$ are homogeneous solutions satisfying the left and right boundary conditions, respectively, and $W=\ell r^{\prime }-{\ell }^{\prime }r$ is their Wronskian, thenFor $n\ne 0$ take $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$. Their Wronskian is $W(s)=n(2^n-1)s^{-n-1}$, and henceFor $n=0$ take $\ell (x)=\log (2x)$ and $r(x)=\log x$, giving $W(s)=\log 2/s$ andBoth formulas satisfy the endpoint conditions, continuity, and the unit jump directly. The unique solution of the boundary value problem isThe factor $s^{n+1}$ in the first formula matters: this Green's function is for the original monic differential operator, with ordinary measure $ds$, rather than for the operator multiplied by $x^{n+1}$.

For $n\ne 0$, split the Green's function integral at $s=x$. Writing $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$ givesFor $n\ne -1$, the two integrals areSubstitution and simplification yieldAt $x=1$ both contributions vanish. At $x=1/2$ the bracket is $-1/2+(2^n-1)/[2(2^n-1)]=0$. The differential operator kills $x^{-n}-1$ and sends $x/(n+1)$ to $1/x$, so this also verifies the inhomogeneous equation.

The exceptional integer is $\boxed{n=-1}$, because then the integrals contain $s^{-1}$. In this caseand the Green's function integral becomesHere $L=d^2/d{x}^2$, and $u^{\prime \prime }=1/x$. Explicitly $u(1)=0$ and $u(1/2)=-(\log 2)/2+(\log 2)/2=0$. Thus both exceptional and nonexceptional formulas satisfy the required boundary conditions.

Assume first that $f$ and $g$ are integrable functions, so that Fubini's theorem justifies changing the order of the two integrals. In the defining Fourier transform of the convolution, put $y=x-x^{\prime }$:This proves the convolution theorem with the stated normalization:

Direct evaluation of the Fourier transform gives, for $k\ne 0$,At $k=0$ the defining integral is the interval length, so $\boxed{\tilde{g}(0)=2}$. This also equals the continuous limit of the displayed formula.

The convolution $(g\ast g)(x)$ is the length of the overlap of $(-1,1)$ with $(x-1,x+1)$. This overlap has length $2-|x|$ when $|x|<2$ and zero otherwise, so $u=g\ast g$. The convolution theorem now givesThe value at zero is also the area of the triangular function, providing a normalization check on its Fourier transform.

The Translation property of the Fourier transform givesFor the remaining integral, the same Translation property of the Fourier transform identifies $e^{20i{k}^{\prime }}\tilde{g}(k^{\prime })$ as the Fourier transform of $g(x+20)$. By the product form of the convolution theorem, with inverse normalization $1/(2\pi )$,But $g(x+20)$ is supported in $[-21,-19]$, whereas the four translated triangular functions making up $r$ are all supported in $[-8,8]$. Their product is identically zero. Therefore the requested integral is $\boxed{0}$ for every real $k$.

The matrix sending the standard basis vectors to the given columns isWith the Hermitian inner product, the squared column norms are $3/6+1/6+2/6=1$, $3/6+1/6+2/6=1$, and $4/6+2/6=1$. The off-diagonal products areThus the columns form an orthonormal basis, so $\boxed{U^{†}U=I}$ and $U$ is a unitary matrix. Here the operation maps the old basis vectors to the new vectors; the inverse coordinate change of basis uses $U^{†}$.

The conjugate transpose reverses the order of matrix multiplication. Since $A$ and $B$ are Hermitian matrices,Consequently $\boxed{AB\text{ is Hermitian}⟺AB=BA}$: the condition is exactly that the two matrices commute.

Take $\boxed{c=1}$. Indeed the conjugate transpose isso the anticommutator is a Hermitian matrix. The original PDF, p. 6, supplies the factor $BA$ that is corrupted in the local OCR.

Using the conjugate transpose and the Hermitian matrix assumptions,Thus $ABA$ is a Hermitian matrix, whether or not $A$ and $B$ commute.

Put $D=A-iB$, so $D^{†}=A+iB$. Because the Hermitian matrices $A,B$ commute,For every nonzero vector $v$,since $A$ is an invertible matrix. Thus $D$ and $D^{†}$ are invertible matrices, justifying the inverse in the definition of $U$. The conjugate transpose of $U=D^{†}{D}^{-1}$ is $(D^{†}{)}^{-1}D$. Since $D$ and $D^{†}$ commute, their inverses do too, andHence $U$ is a unitary matrix.

For the unitary matrix $C$, $C^{-1}=C^{†}$. Therefore $D=C^{-1}AC$ is a Hermitian matrix:Every positive integer power remains a Hermitian matrix, since $(D^p{)}^{†}=(D^{†}{)}^p=D^p$. Its eigenvalues are real by the finite-dimensional spectral theorem. Equivalently, matrix similarity giveswhose eigenvalues are $\boxed{{\lambda }_j^p\in \mathbb{R}}$, where the ${\lambda }_j$ are the real eigenvalues of $A$.

Complete the square in the Hermitian matrix polynomial:For every nonzero vector $v$,Thus $H$ is a positive-definite matrix, so $\boxed{0\text{ is not an eigenvalue of }H}$. In finite-dimensional spectral theorem terms, each eigenvalue isIf ${\alpha }^2\ge 4\beta$, the scalar polynomial has real roots of a polynomialAn allowed nonzero root can be an eigenvalue of $A$, a Hermitian matrix which is also an invertible matrix, in which case $H$ has a zero eigenvalue. For example, $\alpha =2$, $\beta =1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha =\beta =0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.

For every vector $v$, the Hermitian inner product givesThe norm vanishes only on the zero vector, so $Cv=0$ for every $v$ and $\boxed{C=0}$. In contrast,is a nonzero nilpotent matrix. Multiplying a matrix by its conjugate transpose detects column norms; multiplying it by itself has no corresponding positivity property.

The cyclic property of the trace impliesTaking the matrix trace of the asserted equality therefore gives $nc=0$. For a positive integer $n$ over the complex numbers this requires $\boxed{c=0}$. It is also sufficient: choose $M=N=0$, or any two commuting matrices. Thus a finite-dimensional operator commutator cannot be a nonzero scalar multiple of the identity matrix.

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.

Termwise differentiation of the absolutely convergent matrix exponential series givesBecause $P$ and $Q$ commute, $P$ also commutes with every power of $Q$ and hence with $e^{tQ}$. The product rule therefore givesDirect differentiation gives $f_2^{\prime }(t)=f_2(t)(P+Q)$, and both initial values are $I$. The uniqueness theorem for ordinary differential equations, applied to the finitely many matrix entries, implies $f_1=f_2$. One can see uniqueness directly: the difference $D=f_1-f_2$ obeys $D^{\prime }=D(P+Q)$ and $D(0)=0$, soand the product is identically zero. At $t=1$ this yields

Let $K(t)=e^{tR}S{e}^{-tR}$. The product rule and differentiation of the matrix exponential givewhere brackets are the operator commutator. Evaluating at zero gives $K(0)=S$, $K^{\prime }(0)=[R,S]$ and $K^{\prime \prime }(0)=[R,[R,S]]$. The Taylor series at zero is consequentlyIn particular $[R,[R,S]]=R^{2}S-2RSR+S{R}^2$, which also follows by multiplying the two matrix exponential series through quadratic order. For fixed finite matrices the remainder has the stated Big O notation in any matrix norm.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

The Cauchy-Riemann equations determine the harmonic conjugate. Firstso $v_y=u_x$ integrates with respect to $y$ toNowand the second Cauchy-Riemann equation, $v_x=-u_y$, forces $h^{\prime }(x)=0$. HenceThe normalization at zero gives $C=0$, so the required entire function is

For the first function, the complex exponential has no zeros, while the denominator has three distinct simple zeros. Thus there are $\boxed{\text{simple poles at }z=0,2i,-2i;\text{no finite zeros}}$. To examine infinity, substitute $z=1/w$:The factor $e^{1/w}$ has an essential singularity at zero, and multiplication by a nonzero finite-order factor cannot remove it. Therefore $\boxed{\infty \text{ is an isolated essential singularity}}$ of the first function.

For the second function, $\sin z$ has simple zeros at $z=n\pi$. At zero, its Taylor series givesso $\boxed{z=0\text{ is removable, with extended value }1}$. Every other $z=n\pi$ is a double pole, since the numerator is nonzero there. There are no finite zeros, including after removal at zero.

At infinity the double poles $n\pi$ accumulate. Equivalently, after $z=1/w$, the function $1/[w^2{\sin }^2(1/w)]$ has double poles at $w=1/(n\pi )$ tending to zero. Hence $\boxed{\infty \text{ is a non-isolated singularity}}$. The removable/pole/essential trichotomy for isolated singularities does not apply there; calling this an isolated essential singularity would be incorrect.

Put $w=z-1$ and $a=1-i$. Partial fraction decomposition givesBoth terms admit a geometric series at $w=0$. The local Laurent series is actually a Taylor series, since the center is regular:The nearest pole is $z=i$, a distance $|1-i|=\sqrt{2}$ from the center; the other pole is at distance $4$. Thus the requested local radius of convergence is $\boxed{R=\sqrt{2}}$.