For a nonzero separated solution $v=X(x)T(t)$, separation of variables gives $X^{\prime \prime }+{\kappa }^{2}X=0$ and $T^{\prime }=-\nu {\kappa }^{2}T$. The endpoint conditions require $X=\sin \kappa x$ and $\cos \kappa L=0$. Thus the mixed Dirichlet-Neumann modes on an interval have ${\kappa }_n=(n+1/2)\pi /L$, $n\ge 0$. Their squared L2 norms are $L/2$, so the initial Fourier coefficients areThe integration uses $\cos {\kappa }_{n}L=0$ and $\sin {\kappa }_{n}L=(-1{)}^n$. ThereforeAt positive time the series can be differentiated termwise and satisfies both homogeneous endpoint conditions for $v$. At time zero it recovers $-Hx/k$, so $u$ recovers the initial zero temperature. If $H\ne 0$, the prescribed derivative at $(L,0)$ is incompatible with differentiating the initial zero profile; the temperature is still continuous there, while its spatial derivative has a corner layer. At long times the decaying modes leave $u\to Hx/k$.

Choose the stationary solution $w^{\prime \prime }=0$ satisfying $w(0)=0$ and $k{w}^{\prime }(L)=H$. Hence $\boxed{w(x)=Hx/k}$. Subtracting it gives the homogeneous problemThis boundary lifting leaves a Dirichlet boundary condition at one end and a Neumann boundary condition at the other, as in the fixed-temperature and fixed-flux heat equation on an interval.

With temperature measured relative to the fixed-temperature end, the heat equation and its data areHere $\nu >0$ is thermal diffusivity and $k>0$ is thermal conductivity. The sign of the derivative condition describes incoming heat flux at $L$: the outward conductive heat flux is $-k{u}_x=-H$.

For the normal modes of an equilateral spring triangle, the complement of the three rigid motions has dimension three. One convenient mutually orthogonal basis isThese vectors are also orthogonal to $t_x,t_y,r$, and each has squared length $12$. Their geometry comes from applying symmetric linear deformations to the equilibrium triangle: $b$ is uniform expansion, $d$ stretches one axis while compressing the other, and $s$ is the second independent shear pattern. Their extensions areInsert each row into the force equations. For example, $b$ gives $m{\ddot{x}}_1=3\sqrt{3}k=-3k{b}_{x_1}$ and $m{\ddot{y}}_3=-6k=-3k{b}_{y_3}$, with the same proportionality in every component. The other two rows similarly give force $-3k/2$ times their displacement vector. Thusand the remaining normal modes and normal frequencies areAny orthogonal pair spanning the same two-dimensional shear subspace is equally valid, because its eigenvalue is repeated. The six displayed vectors are nonzero and mutually orthogonal, so they supply all six independent normal modes without solving a characteristic polynomial.

For explicit normal coordinates, normalize $t_x,t_y$ by $\sqrt{3}$ and $r,b,d,s$ by $\sqrt{12}$, and write $q={\sum }_j{Q}_j{\hat{v}}_j$. Since all masses are equal,Positive-frequency coordinates have $Q_j=A_j\cos ({\omega }_{j}t)+B_j\sin ({\omega }_{j}t)$; the three rigid coordinates have the linear form found above.

Order coordinates as $q=(x_1,x_2,x_3;y_1,y_2,y_3)$. For either uniform displacement, $t_x=(1,1,1;0,0,0)$ or $t_y=(0,0,0;1,1,1)$, every relative displacement and every extension vanishes. These normal modes have zero normal frequency and represent two independent planar translations.

For $r=(-1,-1,2;\sqrt{3},-\sqrt{3},0)$, substitution givesThus $Kr=0$ too. A small rigid rotation through angle $\alpha$ moves equilibrium position $(X_i,Y_i)$ by $\alpha (-Y_i,X_i)$. With the specified positions this displacement is $-\alpha r/2$, identifying the third vector as infinitesimal rigid rotation. HenceA zero-frequency normal coordinate obeys $\ddot{Q}=0$, so $Q=A+Bt$. This permits a constant displacement or a uniform velocity at the level of the linearized equations. In particular, the rotational pattern preserves lengths to first order; finite-angle rotation follows a curved path outside a single linear displacement pattern.

Use the equilibrium coordinates specified in the paper; their distances from the origin are one and their mutual separation is $\sqrt{3}$. Let $u_i=(x_i,y_i)$ be the small displacements. Each initially unstretched spring has, to first order, extension equal to the relative displacement projected along its original bond. Choose the three oriented extensionsThe kinetic energy is exact in these displacement coordinates, while the potential energy needed for small oscillations is its quadratic part:The factor $\sqrt{3}$ in the equilibrium bond length cancels when forming the unit bond directions. Applying the Euler-Lagrange equation, or differentiating $V$ for the force on each mass, givesThe total force vanishes componentwise, as expected for three masses coupled only by internal spring forces.

Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite $p_{ij}=p{\delta }_{ij}+s_{ij}+{ϵ}_{ijk}{a}_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p{\delta }_{ij}+s_{ij}-{ϵ}_{ijk}{a}_k$. Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, $a_k=0$, hence $A_k=0$ and $\boxed{P_{ij}=P_{ji}}$. No equality between $\mu$ and $\nu$ is needed for that conclusion.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

Now the inverse integrand can be writtenFor $0<\lambda <1/2$, the exponent has a positive real time parameter. Close the Bromwich contour to the left; the residue at $s=-1$ is $\frac{1}{2}(2\lambda )=\lambda$. For $1/2<\lambda <1$, close to the right, where there are no poles. The reversed orientation there still gives zero. ConsequentlyThe same result follows from the Laplace transform time-shift rule: $H(s)=\frac{1}{2}{e}^{-s\log 2}/(s+1)$ corresponds to $f(t)=e^{-t}\Theta (t-\log 2)$, where $\Theta$ is the Heaviside step function.

The original transform cannot fix a value at the single point $\lambda =1/2$. The Bromwich inversion formula chooses the average of the limiting values $1/2$ and zero, giving $\boxed{h(1/2)=1/4}$ under that convention. Indeed,regardless of the assigned value at the jump.

Choose $\gamma >1/2$. The inverse integrand is ${\lambda }^{-s}/[2(s-1/2)]$. Because $-\log \lambda >0$, the exponential factor decays towards the left. Closing the Bromwich contour to the left and using the residue theorem picks up just the simple pole at $s=1/2$, with residue ${\lambda }^{-1/2}/2$. HenceIts defining integral checks both the answer and its convergence condition:

Assume the regular real Sturm-Liouville problem, with $p,w>0$, ordered eigenvalues ${\lambda }_0<{\lambda }_1<\cdots$, and real orthonormal eigenfunctions in the weighted inner productThe Rayleigh quotient for an admissible trial function isThe second expression follows by integration by parts. It is the quantity used in the Rayleigh-Ritz method.

For $\tilde{y}=y_0+{\sum }_{n\ge 1}{a}_n{y}_n$, orthonormality eliminates every cross-term in the denominator:Applying $\mathcal{L}$ to the eigenfunction expansion and using the same orthogonality in the numerator givesHence, whenever the trial function has finite energy,Put $S=\sum a_n^2$ and $\Delta =\sum a_n^2({\lambda }_n-{\lambda }_0)$. The exact error and its expansion areIn particular, under a common small scaling $a_n=\epsilon b_n$ with fixed finite sums $\sum b_n^2$ and $\sum b_n^2({\lambda }_n-{\lambda }_0)$,The fourth-order remainder includes mixed products $a_n^2{a}_j^2$, not just individual fourth powers.

Error comparison and sign. Since $\Vert y_0{\Vert }_w=1$, the relative weighted norm error of the supplied trial function isFor ${\lambda }_0\ne 0$, the relative eigenvalue error is $\Delta /(|{\lambda }_0|(1+S))=O({\epsilon }^2)$. Thus the Rayleigh quotient has a second-order error for a first-order error in the trial eigenfunction, provided the energy-weighted coefficient sum is controlled. Small weighted norm error alone does not control arbitrarily high-frequency contamination, because ${\lambda }_n$ can grow without bound. If ${\lambda }_0=0$, an absolute error must replace the undefined relative one.

Finally every summand of $\Delta$ is nonnegative. Thuswith strict inequality whenever some $a_n\ne 0$. This is the upper-bound property of the Rayleigh-Ritz variational principle for the lowest eigenvalue.

For the real Sturm-Liouville operator $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:Set $u=y_n,v=\tilde{y}$. Both vanish at the endpoints, so the boundary term is zero. Since $\mathcal{L}{y}_n={\lambda }_{n}w{y}_n$,ConsequentlyThis is self-adjointness with the homogeneous Dirichlet boundary conditions, and expresses the Fredholm solvability condition at the eigenvalue ${\lambda }_n$. In this unweighted integral $F_n$ already contains its factor of $w$; adding another weight would be incorrect.

By superposition, $y(x)={\int }_1^{e}G(x,\xi )f(\xi )d\xi$. Split the integral at $\xi =x$: the portion $\xi >x$ uses the first branch of the Green's function, and the portion $\xi <x$ uses the second. This yieldsFor $f(\xi )=1/\xi$, put $t=\ln x$. The two integrals areSubstitution and cancellation giveIndeed $y(1)=y(e)=0$. For a direct differential check,and thereforeThe homogeneous uniqueness shown in part (a) makes this the unique solution.

Choose the homogeneous solutions $u(x)=x\ln x$, which vanishes at $1$, and $v(x)=x(\ln x-1)$, which vanishes at $e$. Their Wronskian in the order $u{v}^{\prime }-u^{\prime }v$ is $x$. For a source point $1<\xi <e$, seek the Green's function asIt is continuous at $x=\xi$; otherwise its second distributional derivative would contain a derivative of the Dirac delta function, which the source does not contain. Integrating the differential equation across the source, using continuity and the leading coefficient one, gives the jump conditionContinuity and this derivative jump requireSolving with the Wronskian $\xi$ gives $A=v(\xi )/\xi =\ln \xi -1$ and $B=u(\xi )/\xi =\ln \xi$. ThereforeThe two values at $x=\xi$ are identical. The derivative jump is explicitlyBoth endpoint values vanish and the differential expression is zero off the diagonal, completing the verification of the Green's function.

Use reduction of order by setting $y=xv$. Then $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$, so the differential expression simplifies toIn the homogeneous case $x{v}^{\prime }$ is constant. Hence $v=c_1+c_2\ln x$, and a basis of the solution space isTheir Wronskian is $x\ne 0$ on $[1,e]$, proving linear independence. No nonzero homogeneous solution satisfies both endpoint conditions: $y(1)=0$ sets $c_1=0$, and $y(e)=0$ then sets $c_2=0$.

Mixed endpoint spectrum. For $y^{\prime \prime }+\lambda y=0$, integration against $y$ and integration by parts, using $y(0)=0,y^{\prime }(\pi /2)=0$, gives $\lambda \int y^2=\int (y^{\prime }{)}^2>0$ for every nonzero real eigenfunction. Thus $\lambda =k^2>0$ and the first condition makes $y=A\sin (kx)$. The second requires $\cos (k\pi /2)=0$, so $k=2n+1$, $n\ge 0$. Sincethe normalized eigenfunctions and eigenvalues areThe orthogonality also follows by applying integration by parts to two different eigenfunctions of the same self-adjoint differential operator.

Forced expansion. Let $k_n=2n+1$ and write $x=\sum f_n{u}_n$, $y=\sum c_n{u}_n$. The Fourier coefficients of the forcing arebecause the boundary cosine vanishes and $\sin (k_n\pi /2)=(-1{)}^n$. Applying $d^2/d{x}^2+\kappa$ to the eigenfunction expansion gives $(\kappa -k_n^2)c_n=f_n$. Since $\kappa$ is not an eigenvalue,The coefficients are $O(k_n^{-4})$, so the series and its first two differentiated series converge absolutely and uniformly on the interval. Each term satisfies the two endpoint conditions, and $y^{\prime \prime }+\kappa y$ equals the mixed-endpoint Fourier series of $x$. The forcing coefficients are $O(k_n^{-2})$, giving uniform convergence of that series as well. Uniqueness follows because the homogeneous problem has a nonzero solution only at one of the excluded eigenvalues.

Odd fourth-power sum. Set $\kappa =0$. Direct integration of $y^{\prime \prime }=x$, with the same endpoint conditions, givesAt $x=\pi /2$, the eigenfunction expansion instead gives $y(\pi /2)=-(4/\pi ){\sum }_{n\ge 0}{k}_n^{-4}$. Equating the two values yieldsAbsolute and uniform convergence justify evaluation at the endpoint.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.

Let $M=|m|$ for integer $m$. In the even or odd power series whose parity matches $M$, the numerator in the recurrence relation vanishes at the coefficient of $x^M$. All coefficients of lower powers in that parity are nonzero, and all higher ones vanish. Therefore a polynomial solution exists, of polynomial degree $M$, containingFor nonnegative $m$ these are the Chebyshev polynomials $T_m$. The normalized low-degree cases follow directly from the recurrence relation:Enforcing $T_m(1)=1$ givesTo see that normalization at one is always possible, set $x=\cos \theta$ in the Chebyshev differential equation. For $Y(\theta )=y(\cos \theta )$, the differential expression $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ becomes $Y^{\prime \prime }$. Thus $Y=\cos (M\theta )$ gives $T_M(x)=\cos (M\arccos x)$ on $[-1,1]$ and has value one at $x=1$. It is a polynomial because the cosine addition formula gives $T_{M+1}=2x{T}_M-T_{M-1}$. Negative integer $m$ produces the same solution as $M=|m|$.

Divide the Chebyshev differential equation by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite singular points of a differential equation are $x=\pm 1$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$, $a_0\ne 0$. The coefficient of the lowest power $x^{\sigma -2}$ is $\sigma (\sigma -1)a_0$, hence $\sigma =0$ or $1$. The next power gives $\sigma (\sigma +1)a_1=0$, and the remaining powers yieldFor $\sigma =0$, $a_1$ is free andChoosing $a_1=0$ selects the even power series, with $y(0)=a_0,y^{\prime }(0)=0$. For $\sigma =1$, the next-power equation forces $a_1=0$ andThis selects the odd power series, with $y(0)=0,y^{\prime }(0)=a_0$. Their different initial values prove linear independence. Although the Frobenius method produces two possible leading exponents, zero is an ordinary point, and a general $\sigma =0$ series already includes both initial constants.

The original PDF uses the conjugate transpose $N^{†}$, rather than the transpose found in the transcription. The printed diagonalisation criterion is correct when diagonalisation means unitary diagonalization of a normal matrix. Since $N^{†}=A-iB$,Consequently $N{N}^{†}=N^{†}N$ if and only if $AB=BA$. By part (ii), this gives a common unitary matrix $U$ andwhich is diagonal. Conversely, if $N=UD{U}^{†}$ for a diagonal $D$ and a unitary matrix $U$, thenThus the precise conclusion isIf arbitrary invertible changes of basis are allowed, the printed assertion is false. For example,satisfy $N=A+iB$. Both Hermitian matrices have distinct eigenvalues: $(3\pm \sqrt{2})/2$ for $A$ and $\pm 1/2$ for $B$. The distinct eigenvalues $1,2$ make $N$ diagonalizable by similarity, but $N{N}^{†}=\left(\begin{array}{cc}2& 2\\ 2& 4\end{array}\right)$ and $N^{†}N=\left(\begin{array}{cc}1& 1\\ 1& 5\end{array}\right)$ differ. This distinguishes ordinary diagonalization from the required unitary version.