Integrate the slope in part (i), using $y(-\alpha )=0$. On the left,Its value at the interface is $y(0)=\frac{3}{2}[(1+\alpha {)}^{2/3}-1]$. On the right, integrate $(1+x{)}^{-1/2}$ from the interface and enforce continuity. ThereforeBoth $y$ and $y^{\prime }$ are continuous at zero. The ray steepens as it approaches the interface from the left and flattens as it leaves to the right.

The integrand has no explicit $y$ dependence. Thus the Euler-Lagrange equation gives the conserved transverse optical momentumAt $x=0$, $\mu (0)=\sqrt{2}$ and the crossing slope is one, so $p=1$. Solving for the positive slope gives $y^{\prime }=1/\sqrt{{\mu }^2-1}$, henceThe slopes agree at zero. The same conserved momentum holds across the interface because the refractive index is continuous and there is no point source in the variational equation.

Here a least-time ray means the minimum between the source and the receiver point selected by this ray. For those fixed endpoints, ${\partial }^2[\mu \sqrt{1+y^{\prime 2}}]/\partial y^{\prime 2}=\mu (1+y^{\prime 2}{)}^{-3/2}>0$ proves strict convexity in the slope, so the stationary ray is the unique minimum among graph paths. If the receiver could instead be freely placed anywhere on the line and its height were also minimized over, the horizontal ray would be faster; that different problem is incompatible with the prescribed crossing slope.

The reduced forcing is $f(\xi )=\xi 1_{[1,\infty )}(\xi )$. Its exponential-weighted integrals converge, even though the resulting reduced function is not bounded. For $0\le x<1$, the Green's function formula givesFor $x\ge 1$ it givesUse $\int \xi \sinh \xi d\xi =\xi \cosh \xi -\sinh \xi$ and ${\int }_x^{\infty }\xi e^{-\xi }d\xi =(x+1)e^{-x}$ to simplify this to $y(x)=-x+e^{-x-1}$. Dividing by $x$ yieldsThus $a(x)=-(2/e)\sinh x/x$ with its continuous value at zero. At $x=1$, both pieces give $u=-1+e^{-2}$ and $u^{\prime }=-2e^{-2}$, so neither a Dirac delta distribution nor its derivative is introduced at the forcing jump. The solution is bounded at both ends, with $u\to -1$ at infinity. Its reduced function $y=xu\sim -x$ confirms the growth qualification in part (i).

For $x>0$, differentiation of $u=y/x$ givesThus the radial Laplacian equation reduces toBoundedness of $u$ at zero implies $y(0)=0$. Conversely, a regular reduced solution with $y(0)=0$ gives finite $u(0)=y^{\prime }(0)$.

There is an important qualification to the reduction in the printed question: bounded $u$ at infinity requires $y=O(x)$, not bounded $y$. The example in part (ii) actually has $y\sim -x$, so it cannot be the bounded solution described in part (a). The differential equation and origin condition reduce as asserted, and the same Green's function can still be used with the correct growth condition. This is the radial reduction and growth at infinity distinction.

For constant diffusivity, the trial Rayleigh quotient simplifies toIts derivative is $1-2/q^2$, so its only positive critical point is $q=\sqrt{2}$. The second derivative is $4/q^3>0$; also the quotient diverges as $q\downarrow 0$ or $q\to \infty$. Therefore this is the global optimum within the trial family:The two requested profiles areThe initial radial graph decreases from one at the centre to zero at the rim, has horizontal tangent at the centre, and is concave downward for $0<r<1$. The central temperature decreases exponentially from one to zero, with initial time derivative $-(3+2\sqrt{2})$. These are trial profiles, not exact solutions of the full heat equation. Although the radial trial function is not twice differentiable at the centre, it has finite weighted energy and belongs to the variational trial class used by the Rayleigh-Ritz variational principle.

For the proposed trial profile, $R^{\prime }=-q{r}^{q-1}$. Its weighted energy isThe squared weighted norm isBoth are finite for $q>0$ and $n\ge 0$, and $R(1)=0$. Therefore the Rayleigh-Ritz variational principle givesWhen a fundamental diffusion mode exists, its decay rate is the infimum and is bounded above by this expression.

The existence qualification matters for the full range of $n$ printed in the question. For $n>2$, trial functions concentrated within radius $\epsilon$ have quotient proportional to ${\epsilon }^{n-2}$, proving the loss of a spectral gap under strongly degenerate radial diffusion. The infimum is zero and is not attained, because zero energy together with $R(1)=0$ forces the zero function. Thus there is no positive lowest-mode decay rate in that case. Even at $n=2$, the mode equation is $r^2{R}^{\prime \prime }+3r{R}^{\prime }+\nu R=0$: its characteristic powers are $r^{-1\pm \sqrt{1-\nu }}$, and for $\nu >0$ every nonzero solution is unbounded at the centre. Hence the asserted bounded fundamental mode cannot be assumed for all $n\ge 0$. The trial quotient remains valid, and the next part's $n=0$ case has the usual fundamental mode.

The domain is the three-dimensional upper half-space. Reflect the source point to ${\mathbf{r}}^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images places an opposite source at this reflected point, giving the Dirichlet image kernel for the three-dimensional modified Helmholtz equationOn $z=0$, the two Euclidean distances are equal, so the terms cancel exactly. The image lies outside the domain, so its contribution satisfies the homogeneous modified Helmholtz equation everywhere inside; only the original source contributes a Dirac delta. Both terms decay at infinity, including for $k=0$, and the homogeneous Dirichlet boundary condition follows.

By translation and rotation invariance, use a radial Green function $G(R)$, where $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$ is the Euclidean distance. Away from the source, the radial modified Helmholtz equation isWriting $u=RG$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$, where $\kappa =|k|$. For $\kappa >0$, decay eliminates $e^{\kappa R}/R$, leaving $G=A{e}^{-\kappa R}/R$. For $\kappa =0$, the two radial branches are a constant and $1/R$; decay again leaves $A/R$.

To fix the singular normalization, integrate the defining equation over a ball of radius $\epsilon$ about ${\mathbf{r}}^{\prime }$. The divergence theorem givesThe volume term is $O({\epsilon }^2)$, whereasConsequentlySource qualification. The printed $e^{-kR}$ is the decaying branch only when $k\ge 0$, although the source declares $k$ real. For negative $k$ it grows, so it cannot be the requested decaying Green function. The absolute value above covers every real $k$ and reduces to the printed form for nonnegative $k$.

Set $\kappa =|k|$. The differential operator in the modified Helmholtz equation depends only on $k^2={\kappa }^2$. Let $G$ be its decaying Green function. Acting on the proposed integral with the operator, in the sense of distributions, givesThe last step is the sifting property of the Dirac delta. ThusFor a regular rapidly decaying source, the $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ singularity of the kernel found in part (ii) is locally integrable in three dimensions, and its convolution with the source tends to zero at infinity. This verifies both the equation and the decay condition. The sign of the kernel below corresponds to ${\nabla }^2-k^2$; reversing the sign of the entire operator reverses the kernel.

Only two coefficients in the Sturm-Liouville eigenfunction expansion are nonzero: $a_0=2$ and $a_1=2-\alpha$. Since $\alpha \ne 2$, the result of part (i) simplifies toUsing Parseval's identity, the prescribed weighted norm becomesThus $1-\alpha =\pm 2$, so $\alpha =-1$ or $3$. Positivity selectsThe standing nonresonance assumption still applies: this answer requires that $3$ is not one of the unspecified higher eigenvalues. Their actual values are not supplied. If a higher eigenvalue were $3$, the stated exclusion would fail and the data would not have an admissible $\alpha$ under that exclusion.

Write $Y={\sum }_{n\ge 0}{b}_n{y}_n$ in the complete Sturm-Liouville eigenfunction expansion. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The Sturm-Liouville operator has the symmetry identitybecause both functions obey the homogeneous Dirichlet boundary conditions. Consequentlyso that $({\lambda }_m-\alpha )b_m=a_m$. The stipulated exclusion of every eigenvalue makes all denominators nonzero, and givesThis is the resolvent operator applied to $f$ in the weighted space. The same projection shows uniqueness: a homogeneous solution would have $({\lambda }_m-\alpha )b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.

The path rises from $(-1,1)$ to its maximum height $\sqrt{2}$ at $x=0$, then falls symmetrically to $(1,1)$. Its radius is $\sqrt{2}$, and its endpoint polar angles are $3\pi /4$ and $\pi /4$, giving angular span $\pi /2$. Hence its ordinary arc length isEquivalently, $\sqrt{1+y^{\prime 2}}=\sqrt{2}/\sqrt{2-x^2}$ and integrating over $[-1,1]$ gives the same value. This Euclidean length differs from the weighted destruction integral $D_{\min }\simeq 1.76275$: the slightly longer route saves undergrowth by travelling at greater height.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

The monotonic horizontal coordinate allows the path to be written as a graph. Its arc length element is $ds=\sqrt{1+y^{\prime 2}}dx$, so the variational density is $f(y,y^{\prime })=g(y)\sqrt{1+y^{\prime 2}}$. It has no explicit horizontal-coordinate dependence. The Beltrami identity becomesConsequently every smooth minimizing path satisfiesThis is a first integral of the weighted arc-length problem; fixed endpoint values remove the endpoint variation terms.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Choose the common orthonormal eigenbasis established in part (ii), and let $U$ be the unitary matrix of its columns. ThenConsequentlyThus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $N{N}^{†}=N^{†}N=A^2+B^2$, so $N$ is a normal matrix, as also follows from unitary diagonalization of a normal matrix.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

Using $(AB{)}^{†}=B^{†}{A}^{†}$, the conjugate transpose of the proposed matrix isThusThe commutator of two Hermitian matrices is skew-Hermitian, and multiplication by $i$ changes it into a Hermitian matrix. No simplicity assumption on the spectra is needed for this part.

Use the branch from part (i) and close the real-axis path with a large upper semicircle, indenting clockwise around $-1$ and $1$. On the segment between them, take the upper-bank value of the branch cut. This is an instance of square-root endpoint integrals by an upper-half-plane contour.

The contributions can be controlled separately. Since $g(z)=O(z^{-2})$, the large arc contributes $O(R^{-1})\to 0$. Near either branch point, $g=O({\epsilon }^{-1/2})$, so an arc of length $O(\epsilon )$ contributes $O(\sqrt{\epsilon })\to 0$. On the upper bank of $(-1,1)$,which is odd in $x$, so its integral is zero. On the exterior real intervals $h(-x)=-h(x)$, while the rational numerator is odd. Hence $g(-x)=g(x)$ and the two exterior contributions sum to $2J$.

The enclosed residues use $h(i)=i\sqrt{2}$ and $h(2i)=i\sqrt{5}$:The residue theorem gives $2J=2\pi i[-i/(3\sqrt{2})-i/(6\sqrt{5})]$. ThereforeAs an independent check, $u=\sqrt{x^2-1}$ converts the integral towhich yields the same value. This check also confirms the branch signs and the factor of two in the contour argument.