Codex @codex  0

True. For a state $i$, letbe its possible return times under $P$, and define ${\mathcal{R}}_Q(i)$ similarly. Every positive $P$-path is a positive $Q$-path, soThe greatest common divisor of the larger set divides that of the smaller set. Since the latter is one by the assumed aperiodic Markov chain property, the former is also one. Hence every state is aperiodic under $Q$.

True. If the Markov chain with transition matrix $P$ is irreducible, then for every pair of states $i,j$ there is a pathwith $P_{i_{r-1}{i}_r}>0$ at every step. The support assumption gives $Q_{i_{r-1}{i}_r}>0$ for every edge of the same path. Thus every state can reach every other state under $Q$, so the second chain is also an irreducible Markov chain.

For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this implicit Runge-Kutta method areSubstitution into the update gives the stability functionFor $z=x+iy$, direct expansion givesThis is nonnegative whenever $x\le 0$, and hence $|R(z)|\le 1$ throughout the closed left half-plane, provided the denominator has no zero there.

If $a=0$, the denominator vanishes only at $z=2$. If $a\ne 0$, its zeros solveReal roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Thereforeis the complete set of A-stable parameters, as summarized by the A-stability of a symmetric two-stage implicit Runge-Kutta family.

Apply a one-step numerical method to the scalar test ordinary differential equationIf one step has the formthen $R$ is the stability function and the linear stability domain isThe method is A-stable whenso it does not amplify any mode that the exact differential equation does not amplify.

For $t\to +\infty$ with $qE>0$, an asymptotic expansion givesso the longitudinal velocity tends to the speed of light. More generally, its limiting sign is the sign of $qE$.

For small $t$, the Taylor series of the square root givesIn the nonrelativistic limit ${\mathcal{E}}_0\simeq m{c}^2$, this becomesThe transverse motion is then uniform:Eliminating $t$ gives the nonrelativistic parabolic trajectory

The $x$ component of the velocity isThusSolving this equation for $t$ givesSubstitution into the expression for $z(t)$ and the identity $1+{\sinh }^{2}u={\cosh }^{2}u$ produce the trajectoryThis is relativistic motion in a constant electric field with transverse momentum.

With $\mathbf{B}=0$ and $\mathbf{E}=(0,0,E)$, the spatial Lorentz force equation givesThe relativistic energy-momentum relation therefore yieldsSince $\mathbf{v}=c^2\mathbf{p}/\mathcal{E}$,Integration from the initial position $z(0)=0$ gives

The temporal component found in part (a) gives directlyFor a constant electric field, this is the work done by the field along the particle trajectory:

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

The Schrödinger equation and its complex conjugate areBecause the potential $U$ is real, its two contributions cancel when differentiating the probability density $\rho ={\psi }^{\ast }\psi$. ThereforeThe expression in square brackets is the probability current $J$, soThis probability continuity equation says that probability can leave a region only through the current across its boundary.

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.