By the definition of convolution and Fourier inversion,The second line uses $\nu =\omega -\eta$. The factor $2\pi$ comes from the inverse-transform normalization. This is the product form of the convolution theorem, initially justified for sufficiently decaying functions and extended when the indicated products and convolutions exist.

With the stated Fourier transform convention, the exponential transform is a shifted Dirac delta distribution:Writing the sine as $(e^{i{\omega }_{0}t}-e^{-i{\omega }_{0}t})/(2i)$ then givesThese are tempered distributions, since the time-domain functions are not integrable on the full real line. The second formula also correctly vanishes when ${\omega }_0=0$.

Choose the separated cosine solutions instead:Its derivatives are ${\Phi }_x=-m\sin (mx)\cos (py)$ and ${\Phi }_y=-p\cos (mx)\sin (py)$. They vanish on the appropriate opposite sides of the square for integer $m,p$, so the outward normal derivative is zero everywhere on the boundary. Also $\Delta \Phi =-(m^2+p^2)\Phi$.

This is nonzero for every pair of integer indices, including a zero index. For $m=p=0$ it is the constant solution, illustrating the zero mode allowed by Neumann boundary conditions. Negative indices give the same cosine modes. The construction is the Neumann part of the Dirichlet and Neumann spectra of a square.

For $\Phi =X(x)Y(y)$, separation of variables gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-(m^2+p^2)$. With the standard separation constants $m^2$ and $p^2$, the zero-endpoint solutions are sines. Thus, provided $mp\ne 0$,is nonzero, has zero value on all four sides, and satisfies $\Delta \Phi =-(m^2+p^2)\Phi$.

There is a missing qualification in the printed request for arbitrary integer $m,p$. The Dirichlet and Neumann spectra of a square show that a nonzero Dirichlet solution exists precisely whenWhen one of $m,p$ is zero, the displayed sine product vanishes identically, and a nonzero solution need not exist: for example $(m,p)=(0,0)$ or $(0,1)$ admits none. On the other hand $(0,5)$ does admit $\sin (3x)\sin (4y)$. Thus the natural intended case is two nonzero integer indices, with the stated spectral criterion covering the exceptional cases.

Set $z=x^3/3$ and write $y(x)=Y(z)$. Thenso the two first-derivative contributions cancel exactly, leavingThe Dirichlet spectrum of $-d^2/d{z}^2$ on this interval is $(3n\pi {)}^2$, $n\ge 1$. If $\lambda +a\le 0$, the linear or hyperbolic solutions cannot satisfy both zero endpoints nontrivially, so no additional modes have been omitted. ThusTo verify normalization, use $dz=x^{2}dx$:This proves weighted orthonormality and also transfers completeness from the ordinary Fourier sine series. The construction is the cubic-coordinate reduction of a singular Sturm-Liouville operator. The positive constant $a$ shifts the entire spectrum down; a zero eigenvalue occurs if $a=9n^2{\pi }^2$, so the no-zero-eigenvalue proviso in part (a)(iii) must be imposed separately if that inverse formula is used here.

Expanding the general Sturm-Liouville form givesMatching the second-derivative coefficients requires $p/w=x^{-4}$. Matching the first-derivative coefficients then gives $p^{\prime }/p=-2/x$. Up to a common positive normalization of $p,q,w$, one may therefore takeIndeed $-x^{-2}(x^{-2}{y}^{\prime }{)}^{\prime }-ay=-x^{-4}{y}^{\prime \prime }+2x^{-5}{y}^{\prime }-ay$. The left endpoint is singular in these coordinates. Its self-adjoint interpretation is supplied by the regular transformed problem in part (ii), rather than by assuming that $p(0)$ is finite.

Choose a complete orthonormal system of eigenfunctions ${\varphi }_n$ for the self-adjoint problem. Expand $f={\sum }_n{f}_n{\varphi }_n$ and $y={\sum }_n{c}_n{\varphi }_n$, withSubstitution into $Ly=f$ gives ${\lambda }_n{c}_n=f_n$. Since no eigenvalue is zero, the Sturm-Liouville eigenfunction expansion isFor eigenfunctions not normalized to one, divide each term additionally by $⟨{\varphi }_n,{\varphi }_n⟩$. The expansion is understood in the weighted function space and, with sufficient data regularity, in the operator domain. A zero eigenvalue would instead require its corresponding source coefficient to vanish and would leave an arbitrary homogeneous component.

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

The downward drop from the starting rim to angle $\theta$ is $R\cos \theta$. With no dissipative forces, conservation of energy gives $m{v}^2/2=mgR\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. Dividing the line element by this speed yieldsThe endpoint singularities are integrable for the minimizing path. The integrand is independent of $z$; its Euler-Lagrange equation gives a constant $K$, after absorbing the overall prefactor:Solving for $z^{\prime }$ and taking the positive branch required by $z_2>z_1$ givesThe constant is fixed implicitly byThis integral is strictly increasing from zero to infinity as $K$ increases from zero to one, so every finite positive separation selects a unique $K$. The time integrand has second derivative $R^2/[\sqrt{\cos \theta }(R^2+z^{\prime 2}{)}^{3/2}]>0$ with respect to $z^{\prime }$. Its supporting-line inequality, integrated between fixed endpoints with constant first derivative with respect to $z^{\prime }$, proves this stationary path is the unique global minimum within the stated single-valued class.

As $\theta \downarrow -\pi /2$, $z^{\prime }\to 0$. Therefore the skater should initially head directly down the curved ramp, perpendicular to the cylinder axis, and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the brachistochrone on a horizontal half-cylinder.

On the cylindrical surface $r=R$, the line element is $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. For a single-valued path parametrized by increasing $\theta$,The integrand has no explicit $z$, so its Euler-Lagrange equation gives the constant conjugate quantity $z^{\prime }/\sqrt{R^2+z^{\prime 2}}$. Its derivative with respect to $z^{\prime }$ is strictly positive, hence $z^{\prime }$ itself is constant. The two endpoints determineThis is a constant-pitch helix on the cylinder. To verify global minimality, unroll the cylinder with planar coordinates $(R\theta ,z)$. The path becomes a straight line between points separated by $\pi R$ and $z_2-z_1$, so the triangle inequality for arc length givesThis is the standard description of geodesics on a circular cylinder.

For $|z|<1$, multiply the absolutely convergent Taylor series of the sine by the geometric series:The product is odd, so all even coefficients vanish. Collecting the coefficient of $z^{2m+1}$ givesFor example, the first terms are $z+5z^3/6+101z^5/120+\cdots$. The denominator vanishes at $z=\pm 1$, and the numerator is nonzero there, so these are genuine simple poles. Therefore $\boxed{R=1}$ is the radius of convergence.

If $z_0\ne 0$, factor out $z_0$ and use the geometric series:ThusThe radius of convergence is the distance to the nearest singularity, the simple pole at $z=-z_0$. The case $z_0=0$ must be treated separately: the function is $1/z$, has no Taylor series about zero and is already its own Laurent series on $0<|z|<\infty$.

A real matrix is a Hermitian matrix exactly when it is a symmetric matrix. Since $R_1$ is orthogonal, this is equivalent to $R_1=R_1^{-1}$, or $R_1^2=I$. A generic rotation does not satisfy this; for example $\theta =0$, $\varphi =\pi /2$ gives a nonsymmetric quarter-turn. Thus $\boxed{R_1\text{ is not Hermitian in general.}}$

The exceptions are the identity and half-turns, as characterized by real rotation matrices that are Hermitian. For this product, a half-turn occurs when either $\theta$ or $\varphi$ is an odd multiple of $\pi$; the identity occurs when both are even multiples. For example, the half-turn condition follows from $\mathrm{tr}{R}_1=(1+\cos \theta )(1+\cos \varphi )-1=-1$.

Both factors and their product are real, so the Hermitian conjugate equals the transpose. Part (i) therefore givesHence $\boxed{R_1\text{ is unitary for every }\theta ,\varphi .}$ A real orthogonal matrix is always a unitary matrix.

A product of orthogonal matrices is orthogonal. In fact,Thus $\boxed{R_1\text{ is orthogonal for every }\theta ,\varphi .}$ Its determinant is also one, since each factor has determinant one, so it remains a proper rotation matrix.

Now $G$ is continuous, so $G_x$ has no delta contribution. Integrate its normalized differential equation over $(\xi -ϵ,\xi +ϵ)$. The bounded lower-order terms have integrals tending to zero, while the integral of $G_{xx}$ is the difference of the two one-sided first derivatives. Hence the jump condition isThe denominator is evaluated at the source point; it is not a variable $x$ left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of $L$.

Suppose the piecewise smooth Green function has jump $[G{]}_{\xi }=G({\xi }^{+};\xi )-G({\xi }^{-};\xi )$. Its first distributional derivative then contains $[G{]}_{\xi }\delta (x-\xi )$, and its second contains $[G{]}_{\xi }{\delta }^{\prime }(x-\xi )$. In the normalized equation from part (a), the first-derivative term can supply a delta but no delta derivative. The right side has only a delta. Thus its delta-derivative coefficient must vanish:Equivalently, the two homogeneous branches must join continuously. This is the continuity part of the jump condition for a second-order Green function.

The added sine vanishes at zero. Its derivative at one is $B\cos B$, so nontrivial admissible choices areNegative values give the same trial directions after changing the sign of $C$; $B=0$ adds nothing. For each choice, minimize the Rayleigh quotient of $x{e}^{-x}+C\sin (Bx)$ over $C$, or optimize over several coefficients simultaneously. The original trial is included, so the best bound is no larger than $5/4$. Enlarging the trial subspace successively is the Rayleigh-Ritz method; the boundary-compatible half-integer sines provide additional independent shapes.

For the added trial term $q_n=x^n{e}^{-Ax}$, the left Dirichlet boundary condition is automatic for a positive integer $n$. At the right endpoint,To preserve $y^{\prime }(1)=0$ for an arbitrary variational coefficient $C$, choose $\boxed{A=n}$. Then minimize the Rayleigh quotient of $x{e}^{-x}+C{x}^n{e}^{-nx}$ over $C$.

Since $C=0$ remains available, this optimization cannot worsen the original upper bound. For $n=1$, the added function is a multiple of the first trial, so the quotient does not change: an independent choice $n\ge 2$ is needed for extra freedom. More generally, several such functions form the exponential-polynomial trial space for a mixed Sturm-Liouville problem, and the Rayleigh-Ritz method reduces minimization to a small generalized eigenvalue problem.