Let $Q$ be the real orthogonal matrix. If $Qv=\lambda v$ for a nonzero complex eigenvector, then preservation of the complex Euclidean norm givesso $|\lambda |=1$. Its characteristic polynomial has real coefficients, hence nonreal eigenvalues occur in conjugate pairs. Such a pair is $e^{i\alpha },e^{-i\alpha }$, and the remaining real eigenvalue is $+1$ or $-1$.

If all three eigenvalues are real, each is $\pm 1$; two have the same sign and can be paired as $e^{\pm i\alpha }$ with $\alpha =0$ or $\pi$. Thus in every case the spectrum has the formTheir product shows that $\epsilon =\det Q$.

The inverse is assumed to exist. First,By matrix commutation with an inverse, multiplying on both sides by $(I-M{)}^{-1}$ givesAn equivalent calculation writes $I+M=2I-(I-M)$, so both products equal $2(I-M{)}^{-1}-I$. No convergent infinite matrix series is needed.

At $\Omega =3/4$, the two continuous triangles overlap near zero. Their amplitudes add, giving a central plateau. Explicitly, write $C(\omega )$ for the continuous part of $\tilde{\varphi }$:There are also the two carrier Dirac delta lines $\pi \delta (\omega \mp 3/4)$. The last panel of the preceding figure shows this spectrum and the retained central portion. The prescribed strict cutoff removes the carriers and keepsThe values of the ordinary function at individual endpoints do not affect its inverse transform. To handle the carrier lines, one must additionally assign zero filter gain to those lines, as the carrier-rejecting prescription indicates. A discontinuous mask has no canonical product with a Dirac delta at its discontinuity; a different physical boundary gain would leave a corresponding multiple of $\cos (3t/4)$.

The retained spectrum is real and even. Its inverse Fourier transform isIntegrating the linear factors yieldsThe removable value is $\boxed{\rho (0)=1/4}$, obtained either by expansion or by integrating $\tilde{\rho }/(2\pi )$. Because the carrier frequency is below the message bandwidth, this is Fourier filtering of overlapping amplitude-modulation sidebands; the usual picture of two isolated bands in single-sideband modulation does not apply without that qualification.

A direct way to avoid the singular transform of $1/t^2$ is to invert a triangular function. Evenness givesAt $t=0$ the limit is $1/2$, so there is no singularity in the original signal. ThusAlternatively, the hint uses the Hadamard finite-part reciprocal-power distribution, whose distributional Fourier transform is $\mathcal{F}[\mathrm{fp}(1/t^2)]=-\pi |\omega |$; combining its three translates gives $\pi (|\omega -1|+|\omega +1|-2|\omega |)/2$, the same triangle. Treating $1/t^2$ as an ordinary integrable function would be invalid.

The message spectrum has height $\pi$ at zero, vanishing at $\omega =\pm 1$. For $\Omega >1$, the modulated continuous spectrum consists of disjoint triangles of height $\pi /2$, centred at $\pm \Omega$, each with half-width one. In addition there is a Dirac delta carrier line of weight $\pi$ at each centre. These lines have specified area, not a finite height; arrows mark them schematically.

Use the Fourier transform conventionSince $\mathcal{F}[e^{i\Omega t}]=2\pi \delta (\omega -\Omega )$ in distribution theory, the carrier has transformThe modulation property of the Fourier transform translates the message spectrum to the two carrier frequencies. ThereforeThe first term represents the unmodulated carrier wave; the other terms are the amplitude modulation sidebands.

The left-hand value at the junction isThe right-hand value isThus the piecewise Green function solution is continuous, withIts derivative is continuous as well: both one-sided values equal $-\frac{1}{2}(1-e^{-\mu }{)}^2$. Only the second derivative jumps, consistently with the step in the forcing and the absence of a Dirac delta source.

Away from $x=\xi$, the homogeneous equation has solutions $e^x,e^{-x}$. A basis adapted to the two boundary conditions is $u(x)=\sinh x$ at the origin and $v(x)=e^{-x}$ at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-1$. The Green function must be continuous at $\xi$, while integration of $-G_{xx}+G=\delta (x-\xi )$ across the source givesThese conditions give the Dirichlet half-line Green function for minus d2 plus 1:where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The second term is the image that enforces $G(0,\xi )=0$; both terms decay as $x\to \infty$.

For the finite forcing interval, $y(x)={\int }_0^{\mu }G(x,\xi )d\xi$. When $x\le \mu$, split the integral at $\xi =x$:When $x\ge \mu$, the entire source interval lies to the left, giving the Dirichlet half-line response to uniform interval forcingThis solution is unique: a homogeneous difference satisfying the origin condition is proportional to $\sinh x$, which cannot decay at infinity unless its coefficient is zero.

With the specified ratios, the two factors in the modal amplitude become $\sin (n\pi /5)$ and $\sin (n\pi /14)$. Hence a normal mode is unexcited precisely when one of these factors is zero:That includes all multiples of five or fourteen, with multiples of seventy counted once. The mode $n=0$ is not a fixed-end string mode.

Assume the struck interval lies within the string, so $0\le \ell -a/2<\ell +a/2\le L$. The initially stationary straight string has $A_n=0$. Write its initial velocity as a Fourier sine series $g(x)={\sum }_n{g}_n\sin (k_{n}x)$. ThenThus $y(x,t)={\sum }_n(g_n/{\omega }_n)\sin ({\omega }_{n}t)\sin (k_{n}x)$. Immediately after the strike all the energy is kinetic energy:The orthogonality integral ${\int }_0^L\sin (k_{n}x)\sin (k_{m}x)dx=(L/2){\delta }_{nm}$ assigns energy $E_n=\rho L{g}_n^2/4$ to mode $n$. Since the ideal modes evolve independently, their total kinetic-plus-elastic energies remain constant. Therefore the modal energy of a localized velocity impulse on a string givesThe centre factor selects modes having significant motion at the impact location; the width factor accounts for cancellation across the finite hammer footprint. Parseval's identity ensures ${\sum }_{n\ge 1}{E}_n/E=1$. The velocity discontinuities are understood in the mean-square sense and do not alter the energy integrals.

Set $y(x,t)=X(x)T(t)$ in the wave equation. A nonzero product obtained by separation of variables obeysThe fixed-end Dirichlet boundary conditions require $X(0)=X(L)=0$. The zero and negative spatial eigenvalue cases produce only the trivial solution; the nontrivial eigenfunctions areFor each normal mode, $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$. By superposition and completeness of the Fourier sine basis, the general finite-energy solution has the expansionIf $f(x)=y(x,0)$ and $g(x)=y_t(x,0)$, orthogonality gives $A_n=(2/L){\int }_0^{L}f(x)\sin (k_{n}x)dx$ and $B_n=2{\int }_0^{L}g(x)\sin (k_{n}x)dx/(L{\omega }_n)$. Individual products are separable; their sum is the general solution.

Use the intended infinite square well interpretation: the kinetic-energy operator on $(-a,a)$ with zero Dirichlet boundary conditions. The variational energy is most safely evaluated through its quadratic form,The boundary term vanishes because the trial function vanishes at both walls. Its zero extension has derivative jumps there, so the form interpretation avoids assigning an unrestricted full-line second derivative to it.

Set $u=x/a$ and $b=\lambda a^2$. Apart from an irrelevant factor $a^2$, the trial function is $q_b(u)=(1-u^2)(1+b{u}^2)$. Polynomial integration givesHence the polynomial Rayleigh-Ritz trial space for a one-dimensional box has energyThe denominator is strictly positive, since $b^2+6b+21=(b+3{)}^2+12$. Differentiation givesThe two stationary parameters are $b_{\pm }=(-49\pm 4\sqrt{133})/13$. The smaller is a local maximum; the larger is a local minimum. Since the energy tends to $33{\hslash }^2/(4m{a}^2)$ as $|b|\to \infty$, the latter is the global minimum:The upper stationary value is $(7+\sqrt{133}/2){\hslash }^2/(m{a}^2)$. Exact roots give a slightly tighter bound than substituting the rounded hints.

There are two qualifications to the printed operator and comparison energy. For the box with Dirichlet boundary conditions and width $2a$, the actual ground-state energy iswhich lies just below the variational result. The coefficient $6/5$ printed as an exact eigenvalue is inconsistent with this box problem. Also, merely declaring the differential expression to be zero outside the interval does not impose hard-wall confinement or specify a self-adjoint domain. Read literally on a space admitting exterior-supported states, it has zero-energy states outside and does not have the claimed positive ground energy. The boxed variational result is the well-defined answer for the intended confined particle; a different full-line operator would need its domain and matching conditions supplied.

The polynomial vector field is smooth throughout the solid, so the divergence theorem applies. Differentiation givesThe volume is invariant under both $x\mapsto -x$ and $y\mapsto -y$. Hence the integrals of the two odd terms vanish, and the flux is simply the volume. Using the circular cross-sections,The seams between the surface pieces have zero surface measure and contribute no extra term.

The upper piece is a downward-opening circular elliptic paraboloid cap, running from the circle $r=1,z=1$ to the apex $r=0,z=2$. The middle piece is the side of a circular cylinder of radius one and height one. The lower piece is the unit disk in the horizontal plane. Their union is the closed boundary ofThe four panels show the three separate pieces and their union; there is no additional fourth component. In a meridional section, the boundary consists of the base segment, two vertical sides, and the parabolic cap. The outward normals point downward on the base, radially outward on the cylindrical side, and upward/outward on the cap.

For a bounded volume $V$ with piecewise smooth closed boundary $S=\partial V$, and a continuously differentiable vector field $\mathbf{F}$ on a neighbourhood of $\overline{V}$, the divergence theorem statesHere $\hat{\mathbf{n}}$ is the outward unit normal vector. The left side is the outward flux; the right side integrates the divergence over the enclosed volume.