Put $x=\sqrt{s}R>0$. Orthogonality of the Legendre polynomials lets us match each degree separately. Continuity of the field and its radial derivative givesEliminate $A_{\ell }$ by multiplying the first equation by $(\ell +1)/R$ and adding it to the second. A nonzero mode therefore requiresThis argument avoids dividing by a potentially zero boundary value. Apply the supplied recurrence with $\nu =\ell -1$:Consequently spherical Helmholtz matching to a decaying harmonic exterior giveswhere $j_{-1}(x)=\cos x/x$ for the $\ell =0$ sector.

The original PDF has an index error here: it prints $f_{\ell }$ rather than $f_{\ell -1}$ in the resulting eigenvalue condition. This cannot be repaired by silently choosing another normalization. If $j_{\ell }(x)$ itself were zero at a positive matching root, the boundary equation would also give $j_{\ell }^{\prime }(x)=0$, contradicting uniqueness for this nonzero solution of a regular differential equation at that point.

Separate the interior Helmholtz equation in the same angular basis. Each radial coefficient $F_{\ell }(r)$ satisfiesFor $s>0$, set $x=\sqrt{s}r$. The equation becomes $x^2{f}_{\ell }^{\prime \prime }+2x{f}_{\ell }^{\prime }+[x^2-\ell (\ell +1)]f_{\ell }=0$. Its regular solution is a Spherical Bessel function $j_{\ell }(x)$, up to a multiplicative constant. The second independent solution is singular as $r^{-\ell -1}$ at the origin and is excluded. HenceOne may denote $j_{\ell }$ by $f_{\ell }$ as in the source. Its leading behaviour $j_{\ell }(x)\sim x^{\ell }/(2\ell +1)!!$ ensures regularity. Primes on $j_{\ell }$ below refer to its dimensionless argument; differentiating with respect to $r$ introduces the factor $\sqrt{s}$.

For separation of variables, write $B=F(r)\Theta (\theta )$. The exterior Laplace equation separates intoWith $\mu =\cos \theta$, angular regularity at the poles selects $\lambda =\ell (\ell +1)$ and the Legendre polynomial $P_{\ell }(\mu )$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }-\ell (\ell +1)F=0$, an Euler-Cauchy equation with solutions $r^{\ell },r^{-\ell -1}$. Decay at infinity excludes the first branch for every $\ell$, including its constant $\ell =0$ case. ThereforeThe expansion uses axisymmetric spherical harmonics. The written scalar boundary-value problem permits the $\ell =0$ term; any additional physical exclusion of that sector must be stated separately.

Write the finite cyclic group as $C_n=⟨g:g^n=e⟩$, the source's $Z_n$. It is Abelian, so $hg{h}^{-1}=g$ and every conjugacy class is a singleton: there are $n$ such classes.

Over the complex numbers, all its irreducible representations are one-dimensional. One direct proof is that the matrix representing $g$ has an eigenvector over $\mathbb{C}$; its line is preserved by every power of that matrix and hence by the whole group. Irreducibility forces the line to be the whole representation space. Its eigenvalue must be an $n$th root of unity.

The complex irreducible representations of a finite cyclic group are thereforeTheir generator values are distinct, so no two are equivalent. There are $n$ of them, and $n$ squared dimensions equal one, exhausting the sum of squares of irreducible representation dimensions: ${\sum }_j{d}_j^2=n=|G|$. Thus the list is complete.

For $n=3$, let $\omega =e^{2\pi i/3}$, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$. The character table isThe complex-field convention is essential. The preceding part used real matrices, but that does not make the one-dimensional classification valid over $\mathbb{R}$: conjugate nonreal characters combine into the two-dimensional irreducible real representations of a finite cyclic group. The character table and dimension-sum result used here have their usual complex meaning.

Denote the four listed matrices, in order, by $I,P,Q,W$. As permutation matrices, $P,Q,W$ act on the four basis vectors by $(12)(34)$, $(13)(24)$ and $(14)(23)$, respectively. Direct multiplication givesThese relations prove closure and show that each element is its own inverse. The identity is included and matrix multiplication is associative, so the matrices form a group.

The relations coincide with those of the Klein four-group $D_2$: take its half-turn generator to $P$, a reflection generator to $Q$, and their product to $W$. All four matrices are distinct, so this assignment is a faithful representation of $D_2$. It is the four-point regular representation of the Klein four-group.

Every matrix fixes $v=(1,1,1,1{)}^T$, since it permutes its coordinates. Thereforeis a one-dimensional invariant subspace, nonzero and strictly smaller than the four-dimensional space. This proves the representation is reducible. In fact the four mutually orthogonal sign vectors $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$ and $(1,-1,-1,1)$ give invariant lines and completely diagonalize the representation, as in the normalized Hadamard matrix construction.

For a group representation $D:G\to \mathrm{GL}(V)$, a faithful representation is injective, equivalently $D(g)=I$ only for $g=e$. Thus no distinct group elements are represented by the same linear map.

Two equivalent representations are related by one fixed invertible change of basis: $E(g)=S^{-1}D(g)S$ for every $g$. The same $S$ must work simultaneously for all group elements.

An irreducible representation acts on a nonzero vector space and has no invariant subspace other than $\{0\}$ and the whole space. An invariant subspace is one carried into itself by every representing matrix.

The character of a representation isThe cyclic property of the trace makes it invariant under equivalence and constant on conjugacy classes. Its value at the identity is the dimension of the representation.

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$, with $0\le \theta \le \pi$. The chain rule converts $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ to $Y^{\prime \prime }$, so $Y=\cos (n\theta )$ solves $Y^{\prime \prime }+n^{2}Y=0$. Hence $y_n(x)=\cos (n\arccos x)$ is a solution in the interior.

For nonnegative integer $n$, De Moivre's theorem makes its polynomial nature explicit:Every term is a polynomial of degree at most $n$. For $n\ge 1$, the coefficient of $x^n$ is ${\sum }_j\left(\genfrac{}{}{0ex}{}{n}{2j}\right)=2^{n-1}$, so its polynomial degree is exactly $n$. These are the Chebyshev polynomials $T_n$. Their polynomial form also proves the required boundedness of the function and derivative at both endpoints. Negative labels reproduce $T_{|n|}$ and have degree $|n|$.

The first three areUnder $wdx=-d\theta$, their cross inner products become integrals of $1$, $\cos \theta$ and $\cos 2\theta$ on $(0,\pi )$. The constant-cosine integrals vanish, and $\cos \theta \cos 2\theta =(\cos \theta +\cos 3\theta )/2$ also integrates to zero. Thus all three pairs are orthogonal. Their squared weighted norms are $\pi ,\pi /2,\pi /2$, respectively; more generally

Use the real eigenfunctions, or insert a complex conjugate on the first factor for complex functions. Multiply $-(p{y}_n^{\prime }{)}^{\prime }=n^{2}w{y}_n$ by $y_m$, do the same with $m,n$ interchanged, and subtract. Integration by parts givesThe boundary term vanishes because both functions and derivatives are bounded while $p\to 0$. For distinct squared indices, orthogonality of Sturm-Liouville eigenfunctions therefore yieldsA second integration by parts, using the equation for $y_n$, givesThus derivative orthogonality when the Sturm-Liouville potential vanishes givesThe index qualification matters. The source initially allows integer indices, so $n\ne m$ alone does not imply distinct eigenvalues: $m=-n$ gives the same $n^2$. Taking the conventional nonnegative labels $n,m\ge 0$ makes the stated conclusion valid. For the polynomial solutions below, $y_{-n}=y_n$, so opposite indices are an explicit counterexample to an unqualified orthogonality claim.

Multiply the differential equation by $(1-x^2{)}^{-1/2}$. Sincethe Sturm-Liouville form isThe weighted inner product is consequently $⟨u,v{⟩}_w={\int }_{-1}^1{u}^{\ast }vdx/\sqrt{1-x^2}$. These endpoints are singular: $p$ vanishes and $w$ diverges, though $w$ is integrable. The required boundedness of $y,y^{\prime }$ makes the boundary factors $p{y}^{\prime }$ vanish, which is the fact needed in the integration identities below.

Rearrange the first integral from part (iv) and use the half-angle identities:Eliminating $\lambda$ givesSquaring removes the sign of the slope; the rising and falling sides of the same profile have opposite tangent angles but obey the same height relation. Heights and angles must remain on the regular branch with $\cos \psi >0$.

Choose the tangent angle $\psi =\arctan y^{\prime }$ in $(-\pi /2,\pi /2)$, so $1/\sqrt{1+(y^{\prime }{)}^2}=\cos \psi$. At the maximum, the first integral has value $C=y_0(1-\lambda )$. HenceA nonconstant regular branch below its positive maximum has $y>0$ and $\psi \ne 0$, giving $\lambda >1$. For completeness, strictness follows from the differential equation rather than assuming it at every point: the Euler-Lagrange equation simplifies toAt the maximum, $y_0{y}^{\prime \prime }=1-\lambda$ implies $\lambda \ge 1$. If $\lambda =1$, the positive-height initial data $y=y_0$, $y^{\prime }=0$ give the constant solution uniquely, contradicting the existence of the lower-height branch. Thus $\lambda =1$ is only the degenerate horizontal solution. The first integral with $\lambda >1$ also excludes $y=0$ on this regular branch.

Here $F$ has no explicit $x$ dependence, so use the Beltrami identity. With $p=y^{\prime }$, one has $F_{y^{\prime }}=yp/\sqrt{1+p^2}$ and thereforeThis first integral reduces the second-order Euler-Lagrange equation to a first-order relation.

A missing explicit dependence produces a first integral of the Euler-Lagrange equation. The two cases follow by differentiating the candidate conserved quantity, as shown in the child solutions.

For a variation $y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$, the first variation of the functional isThe boundary term vanishes by the fixed endpoints. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

Write $P_n(x)\sim c_n{x}^n$ with $c_n\ne 0$. For $n\ge 1$, this branch grows. Choosing the other reduction of order solution by an integral from infinity gives, on $x>1$,Thus for $n\ge 1$, every bounded exterior solution decays as $|x{|}^{-n-1}$, up to a constant factor, on each exterior interval. Equivalently the dominant-balance exponents in the Legendre differential equation are $n$ and $-n-1$.

There is an important exception at degree zero. Since $P_0=1$,These solutions are bounded even when $C_1\ne 0$; boundedness alone does not then imply decay to zero. The constants on $x>1$ and $x<-1$ are independent unless continuation across the singular endpoints is specified.

For normalized Legendre polynomials, $P_n(1)=1$ and $P_n(-1)=(-1{)}^n$. Near the endpoints the integral factor has the asymptoticsandThus an independent solution, for which $C_2\ne 0$, has the logarithmic endpoint singularitiesThe constants are understood for a solution continued consistently across the ordinary interior points. Adding any multiple of $P_n$ cannot cancel these logarithms. This explains why boundedness at an endpoint selects the Legendre polynomial and excludes the independent Legendre function of the second kind.

Two integrations give the general reduction of order formulaThe integral is local on intervals avoiding zeros of $P_n$. A pole of the integral at an interior polynomial zero need not be a singularity of $y$: the product extends by reduction of order across a zero of the known solution. After a standard normalization the independent second solution is the Legendre function of the second kind $Q_n$.

Substitute $y=v{P}_n$ into the Legendre differential equation and expand using the product rule. The terms proportional to $v$ cancel because $P_n$ is itself a solution. What remains isMultiplying by $P_n$ identifies an exact derivative:The latter form is convenient for reduction of order on intervals where $P_n\ne 0$ and $x\ne \pm 1$.