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This is the Dirichlet principle. Put $v=w-\psi$, so $v$ vanishes on the boundary. For real, sufficiently regular functions of finite Dirichlet energy, Green's first identity givesThe surface term is zero because $v=0$, and the volume term is zero because $\psi$ is a harmonic function. Expanding the square of $\nabla w=\nabla \psi +\nabla v$ now gives the stronger identityThus the harmonic function minimizes Dirichlet energy among functions with the same boundary values. Equality requires $\nabla v=0$, hence $v$ is constant on each connected component; the zero boundary value makes that constant zero on every component having boundary. For an unbounded domain one additionally needs the boundary term at infinity to vanish; the usual bounded regular-domain setting needs no such extra condition.

Let $d=b-a$. In the mixed Dirichlet-Neumann problem in an elliptic annulus, the angular Fourier modes separate. The outer data excite the $\cos \nu$ mode, while the inner derivative excites $\cos 2\nu$.

For the first mode, a radial factor with zero derivative at $\mu =a$ is $\cosh (\mu -a)$. Normalizing its outer value gives $F_1(\mu )=\cosh (\mu -a)/\cosh d$. For the second mode, a radial factor vanishing at $\mu =b$ is $\sinh [2(b-\mu )]$. Its derivative at $a$ must be $-1$, fixing the coefficient to $1/[2\cosh (2d)]$. ThereforeEach term satisfies Laplace's equation; at $\mu =b$ the second term vanishes and the first is $\cos \nu$. At $\mu =a$, differentiating gives $0-\cos 2\nu$, as required. All other modes, including the constant mode, vanish by the homogeneous mixed conditions. Alternatively, integration by parts applied to the difference of two solutions gives a zero integral of $|\nabla \Phi {|}^2$, proving uniqueness.

Boundary convention. The calculation follows the explicitly printed coordinate derivative ${\partial }_{\mu }\Phi$. In elliptic coordinates, the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$ and the outward unit normal at the inner boundary points toward decreasing $\mu$. A physical normal derivative there would be $-h^{-1}{\partial }_{\mu }\Phi$. Thus the phrase “normal derivative” in the source must be understood in its displayed coordinate sense; imposing the same numerical data on a unit-normal derivative would be a different problem.

Use separation of variables with $\Phi =M(\mu )N(\nu )$. A nonzero product givesThe angular factor must be a periodic function of period $2\pi$. If $\lambda <0$, its exponential solutions cannot be periodic unless they vanish. If $\lambda =0$, $N=c+d\nu$ forces $d=0$, while $M=A_0+B_0\mu$. If $\lambda >0$, periodicity requires $\sqrt{\lambda }=n\in \mathbb{N}$, giving angular factors $\cos (n\nu ),\sin (n\nu )$ and radial factors $\cosh (n\mu ),\sinh (n\mu )$.

Superposing these separated harmonic functions yieldsTo see completeness, expand an arbitrary smooth harmonic function in a Fourier series in $\nu$ on each ellipse. Substitution into the equation forces every Fourier coefficient to obey the corresponding radial equation above. The series is interpreted with the convergence needed for a harmonic function, for example locally on compact subannuli. No boundedness condition at infinity or extension across $\mu =0$ has been imposed, so there is no reason to discard either radial branch in these elliptic coordinates.

Put $Y_p=D(Y)$. Direct matrix multiplication givesfor every real $p$. Since $D(X)=-I_2$ commutes with all matrices and $D(Z)=-Y_p$, the remaining relations also hold:All matrices are invertible, and the relations reproduce the multiplication table of the Klein four-group. ThereforeA representation need not be an orthogonal representation; imposing orthogonality on the displayed matrices would be an extra condition.

The characters are their traces, independent of $p$:In terms of part (b), $\chi ={\chi }_{-+}+{\chi }_{--}$, so the representation is reducible. For an explicit change of basis, takeThen $S^{-1}D(Y)S=\mathrm{diag}(-1,1)$ and $S^{-1}D(Z)S=\mathrm{diag}(1,-1)$, while $D(X)$ remains $-I_2$. These are the equivalent sheared representations of the Klein four-group; the parameter changes the basis, not the two irreducible constituents.

Use the identity relation as printed in the PDF, $XYZ=I$. Since $Z^2=I$, multiplying on the right by $Z$ gives $XY=Z$. Alsobecause $X,Y,Z$ are involutions. Thus $X$ and $Y$ commute, and so does their product $Z$ with either of them. Therefore the group is Abelian, and is the Klein four-group.

Each conjugacy class is a singleton. Over the complex numbers every irreducible representation of a finite abelian group is one-dimensional. Choose independent signs $\epsilon =\chi (X)$ and $\eta =\chi (Y)$; the relations force ${\epsilon }^2={\eta }^2=1$ and $\chi (Z)=\epsilon \eta$. Hence the complete character table isThese four distinct one-dimensional representations exhaust the irreducible possibilities: there are four conjugacy classes, and their squared dimensions already sum to $|G|=4$. The rows also satisfy character orthogonality. The local transcription damages the final symbol in the defining relation; the PDF makes the identity $I$ unambiguous.

A finite-dimensional representation of a group $G$ on a vector space $V$ is a group homomorphism $D:G\to \mathrm{GL}(V)$. Thus each $D(X)$ is invertible andHere the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to $X{X}^{-1}=X^{-1}X=I$ givesBy the defining property of a matrix inverse,If the definition only states the multiplication law together with invertibility, $D(I{)}^2=D(I)$ already forces $D(I)=I_V$ after multiplication by its inverse.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

The relevant weighted inner product is $⟨u,v{⟩}_w={\int }_0^{1}w(x)u(x)v(x)dx$. The normalized eigenfunctions of the Sturm-Liouville problem have the orthonormality propertyHere ${\delta }_{nm}$ is the Kronecker delta: the integral is one for $n=m$ and zero otherwise. The weight is $w$, rather than $p$ or the unweighted measure.

Define $h=f-y^{\prime }{f}_{y^{\prime }}$. For a density without explicit $x$ dependence, the chain rule givesThus the Euler-Lagrange equation implies the Beltrami identityWhere $y^{\prime }\ne 0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

Insert the Frobenius solution and its differentiated series into the equation. Collecting the terms before shifting an index givesThe coefficient of the lowest power $x^{\sigma -1}$ is $a_0\sigma (\sigma +\gamma -1)$. Because $a_0\ne 0$, the indicial equation givesFor the coefficient of $x^{n+\sigma -1}$ with $n\ge 1$, shift the index in the second sum. The hypergeometric Frobenius recurrence isWhenever the left coefficient is nonzero, division yieldsFor noninteger $\gamma$, the two independent local solutions can therefore be identified aswith a consistent local branch for the fractional power. The hypergeometric function coefficients follow by iterating the recurrence using rising factorials.

There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an undetermined coefficient at Frobenius resonance, use the undivided coefficient equation to test compatibility. For example, $\gamma =0,\sigma =0,n=1$ requires $0=\alpha \beta a_0$, so a pure exponent-zero series with $a_0\ne 0$ is impossible if $\alpha \beta \ne 0$. When $\gamma =1$ the two indicial roots coincide, and a second independent solution generally needs a logarithmic term. The indicial roots and undivided relation remain the valid coefficient conclusions in these exceptional cases.

The origin is an ordinary point of the Hermite differential equation, so put $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$. Substitution and matching powers givesThus the linear recurrence relation isEven and odd coefficients evolve independently from arbitrary $a_0,a_1$. The general power series solution iswith an empty product equal to one. Its first terms areBoth series converge at every finite $x$, since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree $N$, the highest coefficient in the differential equation requires $\lambda =2N$. Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. HenceHere $H_N$ is a physicists' Hermite polynomial. Two explicit choices areDirect differentiation verifies both. These are polynomial solutions up to an arbitrary overall factor. Without a polynomial requirement or additional boundary conditions, the equation also has nonpolynomial solutions for other real $\lambda$; discreteness here comes from termination.