Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

Taking traces of the given matrices gives the character on the five classes,For a complex finite-group representation, the multiplicity of an irrep $\rho$ is its character inner product with ${\chi }_D$:Every linear character has value one at both $1$ and $-1$, so its multiplicity is $(4-4)/8=0$. For the two-dimensional character,ThereforeThe dimensions check: $2\times 2=4$.

The choice of scalar field matters here, even though the displayed matrices are real. Over $\mathbb{R}$, this particular four-dimensional representation is irreducible: for any nonzero real vector $v$, the vectors $v,D(i)v,D(j)v,D(k)v$ are mutually orthogonal with the same nonzero norm. This follows from the skew-symmetry and quaternion multiplication identities of the three matrices. A nonzero invariant real subspace contains all four vectors and is therefore the whole space. Its complexification is the two-copy decomposition above.

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.

The elements $1,-1$ commute with the whole quaternion group, so they form singleton classes. Each of the three remaining pairs is a conjugacy class: for example conjugation by $j$ sends $i$ to $-i$, and likewise cyclically. No such pair can mix with a different pair, because the quotient by $\{\pm 1\}$ is abelian and its cosets $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. HenceFor a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations equals the number of conjugacy classes, so there are five. The sum of squares of irreducible representation dimensions is the group order, eight. With five positive integer dimensions, the only possibility isIndeed five one-dimensional contributions give five, and replacing one by a two-dimensional contribution adds exactly three; any further enlargement would exceed eight. These are the dimensions used in the character table of the quaternion group.

The homomorphism law and $g{g}^{-1}=g^{-1}g=e$ giveThus $D(g^{-1})$ is both a left and right inverse of $D(g)$, provingIf only multiplicativity is initially stated, $D(e{)}^2=D(e)$ together with invertibility of representation matrices first implies $D(e)=I$. This explains why a group representation takes values in the general linear group rather than arbitrary square matrices.

Work over the complex numbers, the usual scalar field for the finite-group character theory used below. A group representation on a nonzero vector space $V$ is a homomorphism $D:G\to GL(V)$, so $D(gh)=D(g)D(h)$ and $D(e)=I$.

A faithful representation is injective: distinct group elements have distinct representing maps. Equivalently, its kernel is just $\{e\}$.

An irreducible representation has no nonzero proper invariant subspace: if $W\subseteq V$ and $D(g)W\subseteq W$ for every $g\in G$, then $W=\{0\}$ or $W=V$. Irreducibility concerns invariant subspaces, whereas faithfulness concerns whether any group element is represented by the identity.

If $f$ has no explicit $x$ dependence, the chain rule givesConsequentlyon an Euler-Lagrange equation solution. Thus the Beltrami identity supplies the first integralThe opposite sign is an equally valid convention. The conserved value can be zero for particular integrands or paths; absence of explicit $x$ alone does not guarantee a nonzero numerical value. The useful statement is that this expression is constant along an extremal.

For fixed endpoint values, a stationary differentiable path obeys the Euler-Lagrange equationIt follows by taking the first variation of the functional and integrating the term in $\delta y^{\prime }$ by parts; $\delta y$ vanishes at the endpoints.

At the smaller indicial root $\sigma =1/2$, the $k=1$ equation is $0=0$: it leaves $a_1$ free. For $k\ge 2$, the valid recurrence isChoosing $a_1=0$ makes every later coefficient zero. The resulting terminating Frobenius series is thereforeSubstitution gives $y^{\prime }=a_0/(2\sqrt{x})$ and $y^{\prime \prime }=-a_0/(4x^{3/2})$; the three terms in the equation sum to $a_0\sqrt{x}[-(1+x)-2+(x+3)]=0$.

Termination is a permitted choice, not a condition forced on every smaller-root solution. Retaining the free $a_1$ adds the solution in part (e), so the general real solution on $x>0$ is $\boxed{y=\sqrt{x}[C_1+C_2\log (1+x)]}$. This is exactly the freedom explained by undetermined coefficient at Frobenius resonance. On either negative interval avoiding $-1$ and $0$, analogous real formulas use $\sqrt{|x|}$ and $\log |1+x|$.

The Taylor series of the logarithm givesTherefore the closed form of the larger-root Frobenius solution isThe series derivation holds initially for $0<x<1$; the closed form extends as a real solution to all $x>0$. To check it without series, put $y=\sqrt{x}v(x)$ in the original equation. All undifferentiated $v$ terms cancel and the remaining expression is $4x^{5/2}[(1+x)v^{\prime \prime }+v^{\prime }]$. Hence $[(1+x)v^{\prime }{]}^{\prime }=0$, and $v=\log (1+x)$ indeed solves it. Its logarithm is analytic near zero, so this is not a forced $\log x$ companion term at the Frobenius point.

For the larger indicial root $\sigma =3/2$, the recurrence in part (b) becomesNo factor being cancelled is zero in this range, soThus the resulting Frobenius solution isThere is exactly one free parameter, $a_0$, setting the overall scale. The analytic series factor has radius of convergence one. For a real-valued expression take $x>0$; a complex version requires a local square-root branch.

Because $a_0\ne 0$, the lowest-power equation from part (b) requires the indicial equationThe two candidate indicial roots are thereforeTheir difference is the integer one, so the smaller-root coefficient equation at $k=1$ needs examination rather than automatic division.