Each reflection has order two, giving exactly the three order-two subgroupsConjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. None of these subgroups is therefore normal.

The two nonidentity rotations generate the same order-three subgroup:It is invariant under conjugation by $r$, and $srs=r^{-1}$ shows invariance under $s$ as well; since $r,s$ generate the group, this proves normality. The complete partition into conjugacy classes isThe rotations are mutually conjugate by $s$, and the reflections by $r$; no rotation is conjugate to a reflection because conjugation preserves the order of a group element.

Label the triangle's vertices counterclockwise as $1,2,3$. Let $r$ rotate it by $2\pi /3$ and let $s$ reflect it in the axis through vertex 1 and the midpoint of the opposite side. ThenThe six dihedral group elements areThe first three are the identity and the rotations by $120^{\circ },240^{\circ }$; the last three are reflections fixing vertices $1,3,2$, respectively. These are all possible symmetries: the image of one vertex has three choices, and the order of the other two has two choices.

Use rightmost-first composition. From the generator relation,with exponents reduced modulo three and two. The row-times-column multiplication table isThus every symmetry is expressed in the two group generators $r,s$.

The group order of a finite group is its number of elements, $\boxed{|G|=\#G}$. A subgroup $H$ is normal when conjugation by any group element leaves it unchanged:Equivalently, its left and right cosets agree: $gH=Hg$ for every $g$.

The pinned third bar has zero displacement, so remove its row and column from the stiffness matrix while retaining the springs that connect the moving bars to it. The remaining unit-mass system hasIts eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $2k$ and $4k$. HenceFor $k=1$ these are $\sqrt{2}$ and $2$. The fixed value $X_3=0$ changes the static offsets and reaction force but not this linearized matrix.

For three unit masses, $M=I$ andThe uniform vector has $11^{T}1=31$, whereas every vector with coordinate sum zero is annihilated by $11^T$. One orthonormal choice of normal modes is thereforeIf $k=1$ is retained from (b), the angular frequencies are $1,2,2$. The equal upper frequencies follow from degenerate normal modes from permutation symmetry: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower frequency. The motion is an arbitrary sine-cosine superposition of these modes about the gravitational equilibrium.

The diagram has a connection between every pair of indices $0\le i<j\le n$, including the fixed beam $0$. Two parallel strings, each with spring constant $k/2$, together have energy $k(X_i-X_j{)}^2/2$. Thus the printed $E_{ij}$ is the energy of a connecting pair; calling it the energy of one string would introduce a factor-of-two discrepancy.

Take vertical coordinates positive upwards, with $X_0$ fixed. Each bar has gravitational potential energy $m_{i}g{X}_i$ and kinetic energy $m_i{\dot{X}}_i^2/2$. The Lagrangian isDifferentiating gives the Euler-Lagrange equationsEquivalently, with $S={\sum }_{j=1}^n{X}_j$, the spring term is $k[(n+1)X_i-S-X_0]$. Gravity changes the equilibrium but not the stiffness matrix. Subtracting a static equilibrium and writing ${\eta }_i=X_i-X_i^{\mathrm{eq}}$ gives the normal mode systemThis is the spring system underlying normal modes of an anchored complete spring graph.

The required contractions areSubstituting into the axis decomposition of a symmetric matrix givesIndeed $A^{T}B=-4+4=0$, $CA=0$ and $\mathrm{tr}C=-15+3+12=0$. Adding $25I+5A{A}^T+B{A}^T+A{B}^T+C$ reconstructs the given tensor.

Let $s=A_i{A}_i>0$, $q=A_i{T}_{ij}{A}_j$ and $\tau =T_{ii}$. Taking the trace of the proposed decomposition givessince $A_i{B}_i=0$ and $C_{ii}=0$. Contracting twice with $A$ gives $q=\alpha s+\beta s^2$. Solving these two scalar equations yieldsA single contraction gives $T_{ij}{A}_j=(\alpha +\beta s)A_i+s{B}_i$, and $\alpha +\beta s=q/s$. HenceFinally setThis is manifestly symmetric. The formulas give $A_i{B}_i=q/s-q/s=0$; substituting them into $C_{ij}{A}_j$ gives zero, and taking the trace gives $C_{ii}=0$. Since $C$ is symmetric, $A_i{C}_{ij}=0$ also holds.

A geometric form makes the axis decomposition of a symmetric matrix explicit. With the transverse orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/s$,Here $\alpha$ is half the transverse trace and $C$ is its symmetric traceless rank-two tensor part. Every term was forced by contractions, so the decomposition is unique.

For a Cartesian change of components $x_i^{\prime }=R_{ij}{x}_j$, a rank-$n$ tensor transforms according towith repeated indices summed. An isotropic tensor has unchanged components under every proper rotation matrix.

For rank two write $T^{\prime }=RT{R}^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turnsEach off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. ThereforeConversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the isotropic second-rank tensor classification without assuming that the original tensor was symmetric. Also $\lambda =T_{kk}/3$ is a scalar, since the trace is rotation-invariant.

At the origin, combine the rational coefficients before taking the limit:Equivalently $y(0)=L(3L+1)/[6(L+1)(2L+1)]$. The apparent poles at $L=1$ cancel, givingThe entire limiting profile is also finite:which independently gives the same central value.

Put $a=1/L\in (0,1)$. The Fourier transform of a derivative turns the differential operator into $k^4+5k^2+4=(k^2+1)(k^2+4)$, soUse the partial fraction decompositionThe inverse-transform pair ${\mathcal{F}}^{-1}[(k^2+b^2{)}^{-1}]=e^{-b|x|}/(2b)$ is the one-dimensional modified Helmholtz Green function. HenceIn terms of the original length scale,This is the convolution of $h$ with the fourth-order Green kernel with two decay scales, here $K(x)=e^{-|x|}/6-e^{-2|x|}/12$. It decays at both infinities. Although the individual exponential pieces have cusps, their derivative jumps through order three cancel, so the result has no unintended Dirac delta distribution at zero. The decaying solution is unique: the homogeneous solutions $e^{\pm x},e^{\pm 2x}$ have no nonzero globally decaying combination.

Use the Fourier transform convention $\tilde{h}(k)={\int }_{\mathbb{R}}{e}^{-ikx}h(x)dx$. Split at zero and integrate the decaying exponentials:The transform is real and even because $h(x)=e^{-|x|/L}$ is real and even.

Specify the branch before defining a signed jump. Choose $\varphi <\arg z<\varphi +2\pi$, and let $F_{+}$ be the value on the side with argument tending to $\varphi$ from above, while $F_{-}$ is the value on the other side, with argument tending to $\varphi +2\pi$ from below. The entire function $f$ has the same limiting value on both sides, soThus the jump of a fractional power across a branch cut isEquivalently the ratio of the limiting values is $F_{-}/F_{+}=e^{-2\pi i\alpha }$ wherever $f$ is nonzero. Reversing the definition of the jump reverses its sign. Choosing a different branch multiplies both limits by the corresponding common phase.

Reflect the physical source across $z=0$, writing $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$. Its radius is still $s=|{\mathbf{r}}^{\prime }|$, and its spherical image is $\overline{{\mathbf{r}}^{\ast }}$. The Dirichlet Green function inside a hemisphere isExplicitly this four-source method of images isBoth $G_0$ terms vanish on the spherical cap. On the flat equatorial disc, reflection equates the two physical-source distances and also the two image-source distances, so both pairs cancel. The reflected physical source lies in the lower half-ball and the two inverted images lie outside the sphere. Consequently the only source in the upper half-ball is ${\mathbf{r}}^{\prime }$, with the correct coefficient:The Dirichlet uniqueness for the Poisson equation then identifies this as the required Green function.

The three-dimensional fundamental solution of the Laplace equation, with $\Delta F=+\delta$, isIt is harmonic away from the source, and its gradient has total outward flux one through a small source-centred sphere.

For $s=|{\mathbf{r}}^{\prime }|\in (0,1)$, place the image at ${\mathbf{r}}^{\ast }={\mathbf{r}}^{\prime }/s^2$, outside the unit ball. On $|\mathbf{r}|=1$, expansion of squared distances gives $|\mathbf{r}-{\mathbf{r}}^{\ast }|=|\mathbf{r}-{\mathbf{r}}^{\prime }|/s$. The method of images therefore gives the Dirichlet Green function inside a sphereThe boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ there. At the central source, take the finite limit of the weighted image term:

Use the Dirichlet Green function with the sign conventionThe normal derivative is taken with respect to the integration variable and the outward normal of $V$. Green's second identity givesThe Dirac delta distribution evaluates the first volume term at ${\mathbf{r}}^{\prime }$. Substituting $\Delta u=\rho$, $u=f$ on $S$ and $G=0$ there yieldsHenceOne can justify applying the identity to the singular Green function by removing a small sphere around the source and then taking its radius to zero.