Expanding the derivative and dividing by $1-x$ givesThe only finite singular point is $\boxed{x=1}$. There $(x-1)p(x)=1$ and $(x-1{)}^{2}q(x)=c^2(x-1{)}^2-\lambda (x-1)$ are analytic, so it is a regular singular point. At infinity the coefficient of $y$ tends to $c^2\ne 0$, giving an irregular singular point after reciprocal substitution; the finite classification is independent of $\lambda$.

At every $z=n\pi$, the inner function has a simple pole, with $\cot z=1/(z-n\pi )+h(z)$ for analytic $h$. ThusThe first factor is analytic and nonzero, while the second has infinitely many negative powers in its Laurent series. Therefore, by the exponential of a pole is an essential singularity result,At all other finite points the composition is analytic, so there are no additional finite singularities.

For $z\ne 0$, singularities occur when $1/z=n\pi$, with $n\ne 0$. Each is a simple pole, since the substitution has nonzero derivative there:More explicitly, $1/z-n\pi \sim -(z-z_n)/z_n^2$, so $\cot (1/z)\sim -z_n^2/(z-z_n)$ and its residue is $-1/(n^2{\pi }^2)$.

These poles accumulate at zero. Consequently $\boxed{z=0\text{ is a non-isolated singularity}}$. The removable/pole/essential classification for isolated singularities cannot be applied at that accumulation point; it is not an isolated essential singularity.

The zeros of $\sin z$ are $z=n\pi$ for $n\in \mathbb{Z}$, all simple, and $\cos (n\pi )=(-1{)}^n\ne 0$. Hence the finite singularities of $\cot z$ are exactlyLocally $\cot z\sim 1/(z-n\pi )$, so each pole has residue $1$.

For distinct nonzero recurrence roots, the determinant formula givesThe ratio one is excluded: at that repeated root, part (iii) gives $(N+1)(\pm k{)}^N\ne 0$. Thus the naive extra root corresponding to equal recurrence roots is spurious.

To count the physical solutions without ambiguity from exchanging ${\beta }_1$ and ${\beta }_2$, write ${\beta }_{1,2}=k{e}^{\pm iq}$ and $d=2k\cos q$ with $0<q<\pi$. ThenIts $N$ distinct zeros occur at $q=j\pi /(N+1)$, $j=1,\dots ,N$. They give exactly $N$ distinct positive normal frequencies, or equivalently $N$ distinct roots in ${\omega }^2$:If signed $\omega$ is counted, the characteristic equation has $2N$ roots $\pm {\omega }_j$. The sine quotient is the Chebyshev polynomial of the second kind representation of the fixed-end spring-chain determinant and its repeated-root limit.

Substitute $D_N={\beta }^N$ into the recurrence. Since $k>0$, its roots are nonzero and satisfyFor distinct roots, the general recurrence solution is $D_N=c_1{\beta }_1^N+c_2{\beta }_2^N$. The initial values require $c_1+c_2=1$ and $c_1{\beta }_1+c_2{\beta }_2=d={\beta }_1+{\beta }_2$, givingAt a repeated root, $d=\pm 2k$ and ${\beta }_1={\beta }_2=\beta =\pm k$, this quotient must be replaced by its limit:It also follows by solving the repeated-root recurrence as $(c_1+c_{2}N){\beta }^N$. This exceptional case is necessary to avoid introducing a false root in the next part; it is included in the fixed-end spring-chain determinant and its repeated-root limit.

Expand the tridiagonal determinant along its first row. The diagonal entry contributes $d{D}_{N-1}$. In the off-diagonal cofactor, expansion along its first column supplies the other factor $-k$, and the cofactor sign makes the net contribution $-k^2{D}_{N-2}$. ThereforeThe initial values are $D_1=d$ and $D_2=d^2-k^2$. Writing the same formula at $N=2$ forces $\boxed{D_0=1}$, the empty determinant convention. This is the tridiagonal determinant recurrence.

Introduce the fixed endpoint values $x_0=x_{N+1}=0$. The Lagrangian isThus $T=m{I}_N$, $V_{ij}=k(2{\delta }_{ij}-{\delta }_{i,j+1}-{\delta }_{i,j-1})$, andThis is a symmetric tridiagonal matrix, reflecting that only nearest-neighbour masses are coupled.

Using the three real eigenvectors and positive frequencies in part (ii), the general motion isThere are six independent real constants, $A_1,A_2,A_3,B_1,B_2,B_3$, corresponding to the three initial displacements and three initial velocities. With the displayed orthonormal eigenvectors, $A_j={\mathbf{u}}_j^T\mathbf{x}(0)$ and $B_j={\mathbf{u}}_j^T\dot{\mathbf{x}}(0)/{\omega }_j$. This spans all solutions because the normal-mode eigenvectors form a basis.

Write $d=2k-m{\omega }^2$. The characteristic determinant of the tridiagonal matrix isThe three values of $d$ are $\sqrt{2}k$, $0$ and $-\sqrt{2}k$. Therefore the normal mode frequencies and convenient Euclidean-normalized generalized eigenvectors areEach pair satisfies $V\mathbf{u}={\omega }^{2}T\mathbf{u}$. The first mode moves all masses in phase, the middle mode leaves the central mass fixed, and the last has the central mass moving opposite to the outer masses. All squared frequencies are positive, consistent with the positive spring potential and fixed endpoints.

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.

Put $J_0=4\sigma D^4/3=M{D}^2/3$. The normal to the lamina is already an eigenvector, with moment $3J_0$. The planar block has characteristic polynomialso the principal moments of inertia areA corresponding orthonormal set of principal axes isThe sign of each axis vector is immaterial. The smaller-moment axis makes angle $\frac{1}{2}\arctan (2)$ with the positive $x_1$ direction, approximately $31.7^{\circ }$; the other planar axis is perpendicular. As a check, $I_{-}+I_{+}=3J_0=I_3$, the perpendicular axis theorem for a planar lamina.

Integrating the Dirac delta distribution in $x_3$ reduces the inertia integral to the planar lamina. Use $u=x_1-x_2$, $v=x_2$, so $x_1=u+v$ and the parallelogram becomes the square $-D\le u,v\le D$, with unit Jacobian determinant. The mass is $M=4\sigma D^2$, and symmetry gives a centre of mass at the origin.

Odd rectangular integrals vanish. The required second moments areThe mixed moments involving $x_3$ vanish. Hence the inertia tensor isThe off-diagonal entries are minus the product moment, not the product moment itself. This is the unit-shear case of the inertia tensor of a sheared rectangular lamina.

A pseudotensor has an additional determinant factor under an orthogonal transformation. Thus its rank-three transformation law isThe contracted expression $L_{ia}{L}_{jb}{L}_{kc}{ϵ}_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of ${ϵ}_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the determinant expansion of $L$, fixing that scalar to $\det L$. ThereforeThe Levi-Civita symbol consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three tensor transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of $L$. Thus a tensor of order $n$ transforms asRepeated old-basis indices are summed from one to three. Orthogonality of the new basis gives $L{L}^T=I$. Taking determinants yields $(\det L{)}^2=1$, hence $\boxed{\det L=+1\text{ or }-1}$. The positive case preserves orientation and the negative case reverses it.

Expand in the complete Fourier sine series $\tilde{y}(\omega ,x)={\sum }_{n\ge 1}{b}_n(\omega )\sin (k_{n}x)$ and apply Fourier inversion. Substitution of part (ii) givesTo check the source normalization, apply ${\partial }_t^2-{\partial }_x^2$ mode by mode. It cancels the denominator; the remaining inverse transform gives $2\pi F(t)$, and completeness givesThe endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.

The displayed real-axis integrals in the paper are not ordinary convergent integrals at their poles. One definite choice is the Cauchy principal value at each $\omega =\pm k_n$. If a causal solution is wanted, use $[k_n^2-(\omega -i0{)}^2{]}^{-1}$ instead. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, so the causal solution iswhen the past integral exists. The principal-value choice instead uses the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the causal Fourier resolvent of a point-driven fixed string.

Multiply the transformed equation by $\sin (k_{n}x)$ and integrate from zero to $L$. Two integrations by parts, using the zero endpoint values of both $\tilde{y}$ and the sine, giveThe delta source contributes $\tilde{F}(\omega )\sin (k_n{x}_0)$. Multiplying by $2/L$ and using the coefficient definition therefore yieldsAway from the resonant frequencies this givesAt $\omega =\pm k_n$ the quotient must be interpreted with a pole prescription if the source excites that mode. In distributional form, arbitrary delta terms at those frequencies may also be added; they are the free homogeneous oscillations. This is the frequency-space issue addressed explicitly in part (iii).

Transform only the time variable. Two integrations by parts give ${\mathcal{F}}_t[y_{tt}]=(i\omega {)}^2\tilde{y}=-{\omega }^2\tilde{y}$, when the ordinary transform boundary terms vanish; the same derivative rule holds for tempered distributions. Spatial derivatives and the fixed spatial delta commute with the transform. ThusorThe spatial Dirichlet boundary conditions transform to $\tilde{y}(\omega ,0)=\tilde{y}(\omega ,L)=0$.

Multiply $H$ by the exponential form of the sine and use the modulation property of the Fourier transform:Inserting part (iii) givesThe reciprocal terms are real principal-value distributions and the delta terms are imaginary. Equivalently, exponential damping $H(t)e^{-ϵt}$ gives ${\omega }_0/[(ϵ+i\omega {)}^2+{\omega }_0^2]$, whose distributional limit is the boxed expression. The shifted-difference formula remains valid at ${\omega }_0=0$, where all terms cancel.

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform isusing the given Dirichlet integral and the sign of $t$. Also ${\mathcal{F}}^{-1}[\delta ](t)=1/(2\pi )$. The inverse of the proposed expression is consequently $(iA/2)\mathrm{sgn}t+B/(2\pi )$. Comparison with $H(t)=[1+\mathrm{sgn}t]/2$ for $t\ne 0$ givesThe value assigned at the single point $t=0$ does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating $1/\omega$ as an ordinary integral would miss the required prescription.