Continue with $\mu =2$. Decompose the initial displacement into the normal modes:There is no translational component. Because all initial velocities vanish, the oscillatory coordinates contain only cosines and the zero-frequency velocity is zero. ThereforeAt $t=0$ these give the prescribed displacements and zero velocities. Each term separately satisfies $M\ddot{x}+Kx=0$. Also $x_1+2x_2+x_3=0$ for all $t$, so the centre of mass remains at its original equilibrium position, consistent with the absence of external forces.

Substitution of $\mu =2$ into the normal modes of three masses connected by two equal springs givesThe corresponding shapes are $v_0=(1,1,1{)}^T$, $v_1=(1,0,-1{)}^T$, $v_2=(1,-1,1{)}^T$. For example, $K{v}_2=k(2,-4,2{)}^T=2(k/m)M{v}_2$, verifying the third eigenvalue directly.

Their relevant orthogonality uses the kinetic matrix $M=m\mathrm{diag}(1,2,1)$:This is mass-matrix orthogonality of normal modes. It also follows by subtracting $v_i^{T}K{v}_j={\omega }_j^2{v}_i^{T}M{v}_j$ from its transpose and using the symmetry of $K,M$. Ordinary Euclidean inner products need not vanish: in fact $v_0^T{v}_2=1$.

Use displacements $x_i$ from an equilibrium, so the two spring extensions are $x_2-x_1$ and $x_3-x_2$. The kinetic energy and elastic potential energy areThus $T={\dot{x}}^{T}M\dot{x}/2$ and $V=x^{T}Kx/2$, with the small-oscillation mass and stiffness matricesIf $x_i$ instead denote absolute positions, the appropriate natural spring lengths must first be subtracted from the differences in $V$. They do not affect the equations for displacements.

Euler-Lagrange equations give $M\ddot{x}+Kx=0$, so a normal mode $x=v{e}^{i\omega t}$ obeys the generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$. Reflection symmetry separates antisymmetric shapes $(A,0,-A{)}^T$ from symmetric shapes $(A,B,A{)}^T$. In the antisymmetric sector the end equation is $kA=m{\omega }^{2}A$, giving ${\omega }^2=k/m$. In the symmetric sector the two independent equations areTwice the first plus the second gives $m{\omega }^2(2A+\mu B)=0$. If $\omega =0$, then $B=A$. Otherwise $B=-2A/\mu$, and substitution into the first equation gives ${\omega }^2=(k/m)(1+2/\mu )$. Thus the three independent shapes are:For $v_0$ both springs retain their lengths, so this is uniform translation. For $v_1$ the end masses move oppositely and their forces on the middle mass cancel. For $v_2$ the end masses move together, while the middle mass moves oppositely so the centre of mass remains fixed. Direct substitution verifies all three eigenpairs; for positive $m,k,\mu$ the three eigenvalues are distinct and exhaust the three degrees of freedom. These are the normal modes of three masses connected by two equal springs.

A useful normalization is the weighted inner product induced by $M$. Since the squared mass norms are $m(\mu +2)$, $2m$ and $m(2+4/\mu )$, respectively, takeIf ordinary Euclidean unit eigenvectors are intended instead, divide $v_0,v_1,v_2$ by $\sqrt{3},\sqrt{2},\sqrt{2+4/{\mu }^2}$, respectively. The frequencies and physical motions are independent of this normalization choice. The zero-frequency coordinate has general motion $A+Bt$ rather than an oscillation.

Use the vector triple product expression from part (a). Put $s=\Omega \cdot x$ and $r^2=x\cdot x$. Since $u=\Omega +\beta x$,Expanding $W=(u\cdot u)\Omega -(u\cdot \Omega )u$ and canceling the zeroth-order terms givesThe first bracket is odd in $x$ and integrates to zero over the centred ball. With $\alpha =4\pi a^5/15$, the Cartesian moments of a three-dimensional ball giveThereforeThe resulting vector is parallel to $\Omega$, as the rotational symmetry of the centred domain requires.

Take the centre of the ball as the origin. An orthogonal change of variables leaves both this domain and its volume measure invariant. Consequently its Cartesian moment integrals are isotropic tensors. Equivalently, inversion $x\mapsto -x$ immediately makes every odd moment zero.

There is no nonzero invariant rank-one vector, soAn isotropic tensor of rank two is a scalar multiple of the Kronecker delta, giving ${\int }_V{x}_i{x}_{j}dV=\alpha {\delta }_{ij}$. Contract the indices and use spherical coordinates:Thus the Cartesian moments of a three-dimensional ball giveFor the last identity, one may also use the general rotation-invariant rank-three tensor $c{ϵ}_{ijk}$: the moment is fully symmetric in its indices, whereas the Levi-Civita symbol is antisymmetric, forcing $c=0$.

The origin convention is necessary. For a ball centred at a nonzero vector $c$, its first moment would instead be $(4\pi a^3/3)c_i$, so the displayed claims implicitly refer to coordinates measured from its centre.

In orthonormal Cartesian coordinates, a rank-$n$ tensor has components transforming under a change of axes $R$ asRepeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product asBoth dot products are invariant under an orthogonal change of axes, while $u^{\prime }=Ru$ and ${\Omega }^{\prime }=R\Omega$. ThereforeHence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

Let $Y=\hat{y}$. Transform the three terms of the order-zero Bessel differential equation using part (a):Adding and dividing by $i$ givesOn $-1<\omega <1$, separation of variables gives $Y^{\prime }/Y=\omega /(1-{\omega }^2)$, hence $Y=C/\sqrt{1-{\omega }^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extensionThis is a genuine distributional weak solution: the differential equation can be written $[(1-{\omega }^2)Y{]}^{\prime }+\omega Y=0$, and $(1-{\omega }^2)Y=C\sqrt{1-{\omega }^2}{1}_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint Dirac delta terms.

Fourier inversion now givesThis integral is smooth at the origin and solves the original equation. Its value there is $C/2$. Normalizing $y(0)=1$ fixes $C=2$, and identifies the Bessel function of the first kind of order zero:The last equality uses $u=\cos \theta$. Without normalization, the regular solution is an arbitrary constant times this integral; this is the Fourier transform of the Bessel function J0 pair.

Regularity qualification. The source specifies the differential equation but no condition at its singular point $t=0$. The transformed equation separately has solutions $C_{\pm }/\sqrt{{\omega }^2-1}$ on the two exterior intervals. Thus it does not, by itself, determine the support or a single global multiplicative constant. The compactly supported choice above constructs the regular Bessel branch. The other independent branch, the Bessel function of the second kind, is singular at the origin. Also $J_0$ is not absolutely integrable on the real line, so its transform is properly taken as a tempered distribution, rather than relying on the stronger classical integrability assumptions of part (a).

For the Gaussian function, direct differentiation gives $f^{\prime }=-tf$. Applying the Fourier transform identities from part (a),Solving this first-order ordinary differential equation givesIn fact, the normalization can also be fixed: the Gaussian integral gives $F(0)={\int }_{\mathbb{R}}{e}^{-t^2/2}dt=\sqrt{2\pi }$, so $C=\sqrt{2\pi }$ with the convention used here. The transform has the same Gaussian shape.

Write $F(\omega )=\tilde{f}(\omega )$ to distinguish the Fourier transform from the original function. Integration by parts, using the stated decay, givesApplying the same argument to $f^{\prime }$ gives $\widehat{f^{\prime \prime }}=i\omega \widehat{f^{\prime }}=-{\omega }^{2}F$. Differentiation with respect to the frequency givesConsequentlyThe classical steps require existence of the relevant integrals and justification for differentiation under the integral, for example sufficient integrability of $f$, its derivatives and $tf$. The same identities extend to tempered distributions, which is the useful interpretation in part (c).

First use $\sin \theta /(a-\sin \theta )=a/(a-\sin \theta )-1$. If $J={\int }_{-\pi }^{\pi }(a-\sin \theta {)}^{-1}d\theta$, the required integral is $aJ-2\pi$.

Set $z=e^{i\theta }$. The unit circle is traversed counterclockwise once, $d\theta =dz/(iz)$, and $\sin \theta =(z-z^{-1})/(2i)$. ThusThe function $h(z)/z=-2i/(z^2-2iaz-1)$ is regular at zero. Its only enclosed pole is $z_{-}$ from part (b), and its residue isBy the residue theorem, $J=(1/i)2\pi i/\Delta =2\pi /\Delta$. ThereforeAs a sign and scale check, this is positive and behaves as $\pi /a^2$ for large $a$, in agreement with expanding the integrand and integrating its first nonzero, ${\sin }^2\theta /a^2$, term.

Clear the apparent negative power of $z$ to obtain the rational functionLet $\Delta =\sqrt{a^2-1}>0$. Its denominator factors as $(z-z_{-})(z-z_{+})$, withThese are distinct simple poles. Since $(a-\Delta )(a+\Delta )=1$ and $a+\Delta >1$, we have $0<a-\Delta <1$: only $z_{-}$ lies inside the unit circle. The apparent singularity at $z=0$ in the original expression is a removable singularity, with the continued value $h(0)=0$; it is not another pole.

Using the residue formula from part (a), with denominator derivative $2z-2ia$, givesThere are no other finite singularities of the continued rational function.

A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.

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.