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For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

Resolve the angular surface charge density into Legendre polynomials:For a single degree $\ell$, write the common surface potential as $V_{\ell }{P}_{\ell }$. Regularity and decay require radial factors $(r/R{)}^{\ell }$ inside and $(R/r{)}^{\ell +1}$ outside. Their exterior-minus-interior radial derivative at $R$ is $-(2\ell +1)V_{\ell }{P}_{\ell }/R$. Thus spherical harmonic matching with a derivative jump gives $V_{\ell }=R{\sigma }_{\ell }/(2\ell +1)$.

Only degrees zero and two occur, yieldingBoth expressions have surface value $2AR/3-2AR{P}_2/15$, and their derivative difference is $-2A/3+(2A/3)P_2=-A{\sin }^2\theta$, checking both interface conditions. The interior quadratic is regular at the origin, and the exterior potential tends to zero.

The leading far-field term is the monopole term. Indeed the total charge is $Q=R^2\int \sigma d\Omega =8\pi A{R}^2/3$, and the normalization of the specified jump gives $Q/(4\pi r)=2A{R}^2/(3r)$. No extra permittivity factor is inserted into the supplied normalization. Regularity here means finite piecewise smooth potential; its normal derivative necessarily jumps at the charged surface.

Use separation of variables, writing $\Psi =F(r)\Theta (\theta )$. After multiplication by $r^2/(F\Theta )$, the radial and angular terms are functions of independent variables, so setWith $x=\cos \theta$, the angular equation isRegularity at both poles selects $\lambda =\ell (\ell +1)$, $\ell =0,1,\dots$, and the Legendre polynomials $P_{\ell }(x)$; the other angular branch is singular at a pole. The radial Euler-Cauchy equation is $r^2{F}^{\prime \prime }+2r{F}^{\prime }-\ell (\ell +1)F=0$. A power $r^s$ gives $(s-\ell )(s+\ell +1)=0$, hence $F=a_{\ell }{r}^{\ell }+b_{\ell }{r}^{-\ell -1}$.

Superposing these axisymmetric spherical harmonics givesThe coefficients must give convergence appropriate to the domain. In a region containing the origin, regularity removes the negative powers; in an exterior region with zero potential at infinity, it removes the growing powers and the constant. No such radial restriction is needed for a general annular region.

Write the lower block of $T(a)$ as $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$. ThenA representation of $C_4$ must satisfy its defining relation $T(a{)}^4=I_3$, so $(bc{)}^2=1$. For real $b,c$, this means $bc=1$ or $bc=-1$; it also ensures invertibility.

If $bc=-1$, then $T(a{)}^2\ne I_3$ while $T(a{)}^4=I_3$. The image of the generator has order four, so its four powers are distinct and the representation is faithful. If $bc=1$, then $T(a{)}^2=I_3$ but $T(a)\ne I_3$, since the lower block has zero diagonal. The image has order two and the kernel is $\{I,a^2\}$. HenceFor other values there is no representation of $C_4$, rather than an additional unfaithful case; in particular $bc=0$ would make the assigned matrix singular. These conclusions are the faithfulness criterion for a block representation of the cyclic group of order four. There is no requirement that the proposed representation be orthogonal, so no separate conditions $|b|=|c|=1$ should be imposed.

The matrices are listed in the order $I,a,a^2,a^3$. In the first representation, $D(a)$ is the clockwise quarter-turn matrix and $D(a{)}^2=-I_2$. In the second, $E(a)=i{I}_2$, whose successive powers give the listed scalar matrices. Both therefore obey the group table and distinguish all four elements.

Taking traces gives the characters of a representation:In particular ${\chi }_D(a)=0$ but ${\chi }_E(a)=2i$. Since trace is invariant under similarity, the representations are inequivalent, even after allowing complex changes of basis for the real matrices of $D$.

One can also see the obstruction directly: conjugating the scalar matrix $i{I}_2$ leaves it unchanged, whereas $D(a)$ is not scalar. Over the complex numbers, $D(a)$ has eigenvalues $i,-i$, while $E(a)$ has $i,i$; those different multiplicities cannot be removed by changing basis.

Since $a^4=I$, multiplication adds exponents modulo four: $a^j{a}^k=a^{j+k\mathrm{mod}4}$. Hence the Cayley table of the cyclic group of order four, with the left factor indexing rows, isThe table is symmetric because the group is Abelian. Its row for each element permutes all four members, as cancellation requires.

A group representation assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:Choosing a basis turns those maps into matrices. A faithful representation is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.

Two group representations on spaces of the same dimension are called equivalent representations if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.

The character of a representation isThe cyclic property of the trace makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.

The useful coordinate for this scale-dependent differential operator is $t=\ln x$. With $Y(t)=y(e^t)$, the chain rule givesThe interval becomes $0\le t\le 1$, and the endpoint values become $Y(0)=Y(1)=0$. This logarithmic change of variable in a Sturm-Liouville problem reduces the calculation to a constant-coefficient Dirichlet problem. Also $dx/x=dt$, so the transformation preserves the appropriate weighted inner product.

For the weighted inner product, the factor $w$ cancels the $w^{-1}$ in the Sturm-Liouville operator. Assuming the coefficients are real, two applications of integration by parts giveThis is the boundary form in the Lagrange identity for a Sturm-Liouville operator. The potential terms cancel; if $q$ were complex, an additional integral of $(q-q^{\ast })f^{\ast }g$ would remain, so suitable real coefficients are part of formal self-adjointness. The endpoint restrictions in part (a) eliminate the displayed form for every pair of functions in the domain.

Use the angle orientation in the PDF: $M$ has coordinates $(R\sin \theta ,-R\cos \theta )$ relative to the hoop centre. The wrapped string segment has length $R(\pi /2-\theta )$. As $\theta$ increases, it releases length $R\theta$, so the suspended mass moves down by $R\theta$. Both speeds have magnitude $R|\dot{\theta }|$.

Thus the hoop with a suspended counterweight hasand $L=T-V$. The Euler-Lagrange equation givesAt equilibrium,The positive curvature $V^{\prime \prime }({\theta }_0)=MgR\cos {\theta }_0$ establishes stability. Put $\theta ={\theta }_0+\delta$ and linearize. The equilibrium force cancels, leavingHenceFor $M\gg m$, ${\theta }_0\simeq m/M$ and $\omega \to \sqrt{g/R}$, the small-oscillation frequency of a simple pendulum of length $R$. The linearization assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.

For a variation $q\mapsto q+ϵ\eta$ with $\eta (t_0)=\eta (t_1)=0$, the first variation isThe endpoint term vanishes. Since the interior variation is arbitrary, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationAlong such a trajectory,If $L$ has no explicit time dependence, the first integral is thereforeIts negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total mechanical energy $T+V$.

For a circle in the right half-plane use $0<\lambda <1$. Its centre and radius from part (ii) satisfy $d^2-R^2=a^2$. Hence choosePut ${\tau }_0=\mathrm{arcosh}(d/R)$. Then $\sinh {\tau }_0=a/R$ andIndeed the Apollonius circle has centre $a\coth {\tau }_0=d$ and radius $a/\sinh {\tau }_0=R$. Its logarithmic modulus is therefore $-{\tau }_0$, while on $x=0$ the modulus is one. The logarithmic Dirichlet potential between a circle and a line isThe logarithmic singularity at $a$ lies inside the excluded disk, since $d-a<R$. The other singularity, at $-a$, lies outside the right half-plane. Thus the real logarithmic modulus is harmonic throughout the required region and has exactly the two specified boundary values.

Local branches of the complex logarithm justify harmonicity; its real part is globally single-valued even though the complex logarithm need not be single-valued around the excluded disk. The displayed solution is bounded and tends to zero at infinity. A boundedness or corresponding far-field condition supplies uniqueness in this unbounded domain; finite boundary values alone do not rule out unbounded additional harmonic solutions. The same geometry appears in cylinder-plane image charges.

Squaring the ratio of moduli givesFor $\lambda \ne 1$, complete the square to obtain the Apollonius circleIts centre is $(a(1+{\lambda }^2)/(1-{\lambda }^2),0)$ and its radius is $2a\lambda /|1-{\lambda }^2|$. The centres are positive for $\lambda <1$ and negative for $\lambda >1$.

At the allowed exceptional value $\lambda =1$, the locus is the line $x=0$, the perpendicular bisector of the two fixed points. It is a generalized circle, but not an ordinary finite-radius circle. The point $z=-a$ is excluded from the original ratio.

For real-direction increments, complex differentiability gives $f^{\prime }=u_x+i{v}_x$. For imaginary-direction increments it gives $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating these expressions gives the Cauchy-Riemann equationsA holomorphic function is smooth, so mixed partial derivatives commute. Differentiating these equations givesThus both real and imaginary parts are harmonic functions:

The matrix is $M=2I-e{e}^T$, where $e=(1,1,1{)}^T$. Along $\hat{e}=e/\sqrt{3}$, its eigenvalue is $2-3=-1$. Every vector perpendicular to $e$ has eigenvalue two. For example, use the orthonormal basisThe principal-coordinate equation isIt depends only on distance from the $u_3$ axis and position along it, so every rotation about that axis preserves the surface. ThereforeThis is a one-sheet hyperboloid with waist radius $\sqrt{3/2}$. The repeated positive eigenvalue makes the two transverse principal axes nonunique.