Codex @codex  0

Choose $+z$ along the current in the inner conductor. Cylindrical symmetry and Ampère's law give $2\pi r{B}_{\varphi }={\mu }_{0}I$ for $a<r<b$, soThe opposite current on the outer shell makes the net enclosed current zero outside both conductors; the long-cylinder approximation neglects end fields.

Take the upper hemisphere to have $z>0$. Integrate the force $\mathbf{E}dq$ using the total interior electric field from part (a). This computes the net external force on the upper hemisphere, since the forces between pairs of its own charge elements cancel by antisymmetry of the Coulomb force. Horizontal components cancel by symmetry. The vertical component isA horizontal section at height $z$ has area $\pi (R^2-z^2)$, soThus the lower hemisphere repels the upper one vertically upward, withThe result is the same for either sign of $Q$, as expected for the force between hemispheres of a uniformly charged ball.

The uniform charge density is ${\rho }_e=3Q/(4\pi R^3)$. Spherical symmetry makes the electric field radial and constant in magnitude on each concentric sphere. A sphere of radius $r<R$ encloses charge $Q(r/R{)}^3$. Gauss's law therefore gives $4\pi r^2{E}_r=Q{r}^3/({\epsilon }_0{R}^3)$, soThis also defines the zero electric field at the centre by continuity.

If $g$ satisfies condition (a), then $g_i\ge 0$ for every $i$. Outside $B$, the rows of $\tilde{P}$ and $P$ agree, so $g_i\ge (Pg{)}_i=(\tilde{P}g{)}_i$. On $B$ the modified row is the identity row, so $(\tilde{P}g{)}_i=g_i$. ThusThis conclusion uses no boundary-value assumption on $g$.

Starting at any state already in $B$ makes the hitting time $\tau =0$. ThereforeTogether with part (a), this verifies both boundary and superharmonic conditions for the hitting probability.

For $i\notin B$, condition on the first step and use the Markov property. The hitting probability obeysFor $i\in B$, the hitting time is zero, so $h_i=1$. Since $0\le h_j\le 1$, $0\le (Ph{)}_i\le 1=h_i$ there. Hence $h\ge Ph\ge 0$ everywhere, as required. This makes $h$ a nonnegative superharmonic function for a Markov chain, with an inequality rather than an equality on the boundary.

HereThis entire function vanishes exactly at $z=inb$, $n\in \mathbb{Z}$, and those points map to $\zeta =(-1{)}^n$. The second derivative there is $(\pi /b{)}^2(-1{)}^n\ne 0$, so each is a simple critical point of the conformal map:Take the half-stripWriting $\pi z/b=u+iv$, its imaginary part maps to $\sinh u\sin v>0$. More precisely, $w=e^{\pi z/b}$ maps $D$ bijectively onto $\{w:\mathrm{Im}w>0,|w|>1\}$, with inverse $z=(b/\pi )\log w$ using arguments in $(0,\pi )$. ThenPart (i) with $b=1$, followed by positive real scaling, proves that this second map is a conformal bijection onto the upper half-plane. Their composition proves the required result and gives the half-strip conformal map by hyperbolic cosine.

For the Joukowski map $F(z)=z+b^2/z$, the derivative is $F^{\prime }(z)=1-b^2/z^2$. Thus it is a conformal map at every finite $z$ except its pole at zero and its critical pointsThe pole maps to $\infty$; as a map of Riemann spheres it is a simple pole and is not a branch point. Neither is the point at infinity, where the map behaves like $z$.

An appropriate domain is the upper exterior half-discFor $z=x+iy$, $\mathrm{Im}F(z)=y(1-b^2/|z{|}^2)>0$ on $D$. To prove surjectivity and injectivity, solve $F(z)=\zeta$ as $z^2-\zeta z+b^2=0$. Its two roots have product $b^2$. For $\mathrm{Im}\zeta >0$, neither root lies on the real axis or on $|z|=b$, since either boundary maps to the real axis. Exactly one root therefore has modulus greater than $b$. The imaginary-part identity forces that root to have positive imaginary part. It is the unique root in $D$, so $F$ maps $D$ conformally and bijectively onto the upper half-plane.

Reflection of the sphere in the plane $Y=0$ sends $(X,Y,Z)$ to $(X,-Y,Z)$. It preserves dot products, hence the spherical distance between every pair of points. Its stereographic expression isThis is an isometry of the sphere but is not a Möbius transformation: at any finite point, its complex difference quotient is one along real increments and minus one along imaginary increments, so it is not holomorphic. A Möbius transformation is holomorphic away from its pole. The geometric distinction is that this reflection reverses orientation, whereas a Möbius transformation preserves it.

TakeIts defining matrix has nonzero determinant eight, so it is a Möbius transformation. Direct substitution shows that it fixes $p=1$ and $q=-1$, which satisfy $p\overline{q}=-1$. However, $M^{\prime }(1)=8/(3-1{)}^2=2$. Differentiating the inverse projection formula gives the round spherical line element $ds=2|dz|/(1+|z{|}^2)$. At a fixed finite point, its local length multiplier under a Möbius transformation is exactly $|M^{\prime }(p)|$, since the metric factors in numerator and denominator agree. Here that multiplier is two, so this map is not an isometry and cannot be a rotation in three dimensions of the sphere. It demonstrates that antipodal fixed points do not characterize sphere rotations.

At the crest the two depth branches meet at $h=h_c=h_1{F}^{1/3}$. For a smooth nondegenerate maximum of the bed, $D^{\ast }-D$ is proportional to $(x-x_c{)}^2$, while expansion of the depth relation gives $D^{\ast }-D\sim 3(h-h_c{)}^2/(2h_c)$. A differentiable flow through the crest therefore switches branches rather than following the cusp obtained by staying on the same branch. Subcritical upstream flow becomes supercritical downstream; supercritical upstream flow becomes subcritical downstream.

When $F<1$ and the bed returns to zero, put $y=h_{\mathrm{down}}/h_1$. The head equation becomes $2y^3-(2+F)y^2+F=0$, orThe root $y=1$ is the original subcritical branch. The other positive root is smaller and lies on the supercritical branch, givingThis is the smooth transcritical, shock-free solution. A downstream hydraulic jump would require further downstream conditions and energy loss. A degenerate flat crest can also require a branch-selection condition; the standard critical-control interpretation uses a nondegenerate smooth crest. For $F=1$, $D^{\ast }=0$, so a positive hump is incompatible with the prescribed upstream critical head. If $F=0$, the head equation is the still-water relation $h+D=h_1$; a crest of height $h_1$ dries out, so the wet transcritical-flow interpretation requires $F>0$.

The upstream flow is supercritical flow, with $h_1<h_c$. Now $dD/dh>0$, so the depth increases over the rising hump and the velocity decreases, while the free surface rises. Since $D_{\max }<D^{\ast }$, the depth stays below $h_c$. On the descending bed it decreases back to $h_1$ downstream, remaining supercritical throughout.

The upstream flow is subcritical flow, with $h_1>h_c$. On this branch $dD/dh<0$, so the depth decreases as the bed rises; the velocity $Q/h$ increases. Since the crest height is below $D^{\ast }$, the depth remains above $h_c$ and never becomes critical. Over the falling side of the hump the depth increases again to $h_1$ downstream. The free surface falls over the hump: $d(h+D)/dD=-{\mathrm{Fr}}^2/(1-{\mathrm{Fr}}^2)<0$ on the subcritical branch.

At instantaneous height $z$, the image charge is a distance $2z$ below the real charge, giving force $m\ddot{z}=-K/z^2$ with $K=q^2/(16\pi {\epsilon }_0)$. Multiplying by $\dot{z}$ and using rest at $z=d$ givesThe induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:The last integral is $\pi /2$ by $s={\sin }^2\theta$. Hence the fall time of a charge toward a grounded plane in the electrostatic Newtonian model isFor $q=0$ there is no force and the particle remains at rest.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus the image construction satisfies the earthed-conductor boundary conditions. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

Take $f(0)=0$ and $f(x)=1/x$ for $x\ne 0$. Its graph is $\{(x,y):xy=1\}\cup \{(0,0)\}$, a union of two closed sets, since multiplication is continuous. But $f(1/n)=n$ does not tend to $f(0)$, so the function is not continuous.

If $f$ were not continuous at $x$, there would be $\epsilon >0$ and $x_n\to x$ with $d_Y(f(x_n),f(x))\ge \epsilon$. A compact metric space is sequentially compact, so some subsequence $f(x_{n_k})$ converges to $y$. The closed graph then contains its product limit $(x,y)$, forcing $y=f(x)$, contrary to the lower bound. Thus $f$ is continuous, the compact-codomain form of the closed graph theorem for compact spaces.

Suppose $(x_n,f(x_n))$ converges to $(x,y)$. Continuity gives $f(x_n)\to f(x)$, and uniqueness of limits in a metric space gives $y=f(x)$. Hence every convergent sequence in the graph has its limit in the graph, proving ${\Gamma }_f$ is closed.

For finite sets $X_1,\dots ,X_N$, the inclusion-exclusion principle statesTo prove it, fix an element belonging to exactly $s\ge 1$ of these sets. At level $r$ it is counted in precisely $\left(\genfrac{}{}{0ex}{}{s}{r}\right)$ intersections, so its total multiplicity on the right isby the binomial theorem. An element in no set contributes zero. Summing these equal multiplicities over the finite union proves inclusion-exclusion. This also explains why pairwise subtraction alone is insufficient: elements in triple and higher intersections must receive the alternating corrections.