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Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:Using ${\sin }^{2}x+{\cos }^{2}x=1$, the answer isBoth endpoint values are zero. As a direct check, conjugating by $e^x$ sends $y^{\prime \prime }-2y^{\prime }+2y$ to $e^x(v^{\prime \prime }+v)$, and $v=1-\cos x-\sin x$ gives $v^{\prime \prime }+v=1$. Thus the differential equation and both boundary conditions are satisfied.

The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.

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 the two sine modes, orthogonality givesThe diagonal potential integrals vanish, while the mixed one isThus ${\int }_0^1\cos (\pi x){\tilde{\psi }}^{2}dx=\alpha /2$, giving the two-sine variational bound for a Dirichlet Mathieu equation:Its derivative vanishes when $1+6{\pi }^2\alpha -{\alpha }^2=0$. Hence the two stationary parameters are $3{\pi }^2\pm \sqrt{9{\pi }^4+1}$. To identify the minimum, use the normalized sine basis: the restricted self-adjoint operator has real symmetric matrixIts smaller eigenvalue is the minimum of the Rayleigh quotient, attained at the negative stationary parameter. ThereforeThe estimate improves on the single-sine value ${\pi }^2$. The Rayleigh-Ritz variational principle remains valid here even though $q(x)=\cos (\pi x)$ changes sign: on this finite interval the Sturm-Liouville operator is self-adjoint and bounded below, which is sufficient for the upper bound.

Choose the causal Green function convention $L_{x}G(x,\xi )=\delta (x-\xi )$ for each source point $\xi >0$, with $G(x,\xi )=0$ when $x<\xi$. This automatically gives the two zero-origin limits. The kernel is continuous at $x=\xi$, and integrating its equation across the source requires the unit derivative jump ${\partial }_{x}G({\xi }^{+},\xi )-{\partial }_{x}G({\xi }^{-},\xi )=1$.

For $x>\xi$, the homogeneous basis from part (a) gives a continuous kernel proportional to $x^r\log (x/\xi )$. Its derivative at $x=\xi$ is ${\xi }^{r-1}$, so the normalization is ${\xi }^{1-r}$. Hence the causal Green function for a repeated-root Euler operator isThe same formula follows from $[y_1(\xi )y_2(x)-y_2(\xi )y_1(x)]/W(\xi )$ on the right of the source, with $W(\xi )={\xi }^{-\alpha }$.

The singular endpoint matters for uniqueness. If $\alpha <-1$, then $r>1$, and both $x^r$ and $x^r\log x$, together with their first derivatives, vanish at zero. Adding any source-dependent linear combination of these homogeneous solutions to $G$ preserves the two stated origin conditions. Thus those conditions alone do not determine a unique Green's function in that parameter range. The causal requirement used here selects the displayed kernel. For $\alpha \ge -1$ no nonzero homogeneous solution satisfies both zero-origin limits, so this ambiguity is absent.

Using the causal Green's function,Put $\xi =xs$. Since $\alpha >-5$ is equivalent to $r<3$, the improper integral converges and givesThe elementary identity ${\int }_0^1{s}^{p-1}(-\log s)ds=1/p^2$ follows by differentiating ${\int }_0^1{s}^{p-1}ds=1/p$ for $p>0$. Direct substitution checks $L(x^3)=(\alpha +5{)}^{2}x/4$, and the displayed solution and its first derivative vanish at zero.

To give all solutions allowed by the printed boundary conditions, retain the singular-endpoint qualification from part (b):Thus the causal integral gives a particular member of a two-parameter family in the first range and the unique solution in the second.

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 Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

Multiply the Sturm-Liouville form for $y_n$ by $y_m$ and subtract the equation for $y_m$ multiplied by $y_n$. Integration givesThe boundary expression is zero because $p\to 0$ and the functions and first derivatives remain finite. Since $n,m$ are distinct positive integers, this provesFor the derivatives, apply integration by parts once to the same Sturm-Liouville form:Thus the new weight isThis is derivative orthogonality when the Sturm-Liouville potential vanishes. For $n=1$ the derivative is identically zero, which is consistent with the identity.

Use separation of variables in the form $u=X(x)T(t)$, with $T^{\prime }/T=X^{\prime \prime }/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial Sturm-Liouville problem isThese real separated Robin boundary conditions give a self-adjoint differential operator, so the eigenvalues are real. All three sign cases must be considered.

If $\lambda ={\mu }^2>0$, write $X=A\cos \mu x+B\sin \mu x$. The first condition gives $B=A/\mu$, while the second reduces toThus $\mu =n\pi$, and scaling the eigenfunction gives $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$.

If $\lambda =0$, then $X=A+Bx$. The first condition gives $B=A$, and the second gives $B=A+B$, so $A=B=0$; there is no zero eigenvalue.

If $\lambda =-{\kappa }^2<0$, write $X=A\cosh \kappa x+B\sinh \kappa x$. The first condition gives $B=A/\kappa$, and the second becomesHence $\kappa =1$, giving the single negative eigenvalue $-1$ and spatial eigenfunction $e^x$. Completeness of this regular Sturm-Liouville problem now givesThis is Robin diffusion with a growing mode. The growth is compatible with the stated conditions: they prescribe the same coordinate derivative at both ends, not identical dissipative outward-normal conditions.

The factor $z^{-1/4}$ has algebraic branch points at zero and infinity. Near zero the leading behaviour is $z^{-1/4}$; with $w=1/z$, the behaviour near infinity is $g(1/w)\sim w^{9/4}$. These are four-sheeted branch points.

The polynomial denominator vanishes at $z=\pm i$, with nonzero derivative $2z$, and the fractional factor is finite and nonzero near either point on any chosen local branch. HenceA single-valued holomorphic branch is obtained by choosing a branch cut joining zero to infinity. Points along the chosen cut mark its discontinuity, rather than additional isolated poles.

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.

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

Set $L=d^2/d{t}^2-4$ and choose $L_{t}G(t,t^{\prime })=\delta (t-t^{\prime })$, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are $e^{\pm 2t}$. Decay at each end and continuity at $t=t^{\prime }$ therefore give $G=C{e}^{-2|t-t^{\prime }|}$. Integrating its equation across the source gives a unit jump in its first derivative:ThusThe solution of the boundary value problem is the convolutionThere is no extra homogeneous term: no nonzero combination of $e^{2t},e^{-2t}$ decays at both infinities. The forcing must of course permit the imposed decay; integrable $f$, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.