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For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.

Put $\varphi =R(r)\Theta (\theta )Z(z)$ in the Laplace equation in cylindrical coordinates. Separation of variables first gives ${\Theta }^{\prime \prime }+m^2\Theta =0$; $2\pi$-periodicity forces $m=0,1,2,\dots$, with angular factors $\cos m\theta ,\sin m\theta$ (only the constant for $m=0$). Taking $Z^{\prime \prime }={\kappa }^{2}Z$ givesFor $\kappa >0$ the separated modes are thereforeHere $J_m$ and $Y_m$ are the Bessel function of the first kind and the Bessel function of the second kind. For the opposite sign $Z^{\prime \prime }=-{\kappa }^{2}Z$, the modes instead involve $I_m(\kappa r),K_m(\kappa r)$, the modified Bessel functions, paired with $\cos \kappa z,\sin \kappa z$. At zero separation constant use $Z=C+Dz$ and $R=A{r}^m+B{r}^{-m}$ for $m\ge 1$, or $R=A+B\log r$ for $m=0$. Superpositions over the integer angular orders and appropriate sums or integrals over the separation parameter give the general separated expansion; boundary conditions select its spectrum and coefficients. Regularity at the axis excludes $Y_m,K_m,r^{-m},\log r$ terms.

For the specified decaying side data, bounded separated modes use $e^{-\kappa z}{J}_m(\kappa r)$. A particular solution isAll three denominators are nonzero. Each term satisfies the radial Bessel differential equation, is regular at the axis and bounded for $0\le r\le 1,z\ge 0$, and substitution at $r=1$ gives the stated side values. The modes behave as $r^m$ at the axis, so their apparent angular dependence there causes no singularity.

There is an actual nonuniqueness in the printed problem: no values are specified at $z=0$. If $j_{0,1}$ is the first positive zero of $J_0$, then for every real $a$,is another bounded solution with the same side values. It remains smooth at the axis and even decays as $z\to \infty$. This explicitly proves that side boundary data do not determine a bounded harmonic function in a half-cylinder. Thus the boxed expression supplies a bounded solution; the phrase “the bounded solution” is not justified without an additional base boundary condition. The PDF has $z\ge 0$, whereas the TeX transcription displays $z>0$.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.

Separation of variables with $u=X(x)Y(y)$ and the three homogeneous boundary conditions gives $X_n=\sin (n\pi x)$ and $Y_n=\sinh (n\pi y)$. The boundary value at $y=1$ contains only the third Fourier sine series mode. ThereforeThis harmonic function has all four prescribed boundary values.

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.

For this charged scalar field, $\sqrt{-g}=r^3\sin \theta \cos \theta$. On $e^{-i\omega t}$, the gauge-covariant derivative of a charged scalar field supplies the shifted frequencyThe wave equation becomesSeparation of variables with constant $\lambda$ givesandThe angular equation is the axisymmetric spherical harmonic equation on the three-sphere; regular solutions have $\lambda =\ell (\ell +2)$ with $\ell =0,2,4,\dots$, the even values selected by the absence of $\varphi$ and $\psi$ dependence.

Expand $g$ in its Fourier sine seriesBy separation of variables, the decaying solution of Laplace equation in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin (k_{n}y)$. Continuity at $x=0$ and integration of the equation across the Dirac delta function require the normal-derivative jumpThe jump of $C_n{e}^{-k_n|x|}$ is $-2k_n{C}_n$, so

For $g(y)=\delta (y-c)$, its Fourier coefficients areThusTranslation in $x$ gives the Dirichlet Green function for an infinite strip with source $(b,c)$:

By separation of variables, the axisymmetric normal modes satisfying regularity at $r=0$ and the fixed-edge condition at $r=1$ arewhere $J_0({\gamma }_k)=0$. The zero initial displacement removes all cosine terms, soDifferentiating at $t=0$ gives the weighted Fourier-Bessel expansionMultiply by $r{J}_0({\gamma }_{\ell }r)$ and integrate from zero to one. The orthogonality and norm from part (i) giveUsingthe remaining integral isHenceand therefore

Linearization about rest replaces the material derivative by ${\partial }_t$, so the quasi-geostrophic potential-vorticity equation becomesInsert the separation of variables $\psi =\varphi (x,y,t)P(z)$ and imposeThe horizontal amplitude then satisfiesThe Linear Rossby-wave equation therefore governs each vertical normal mode, whose Rossby deformation radius is $c/f_0$.

For constant width, separation of variables with $\varphi =X(x)\sin (n\pi y/h_0)$ givesHence $X=A_n{e}^{i{k}_{n}x}+B_n{e}^{-i{k}_{n}x}$, with imaginary $k_n$ representing an evanescent mode.

For $X=ϵx$ and varying width,Projection of the next-order equation onto $\sin (n\pi y/h)$, whose squared norm is proportional to $h$, givesThus the WKB amplitude in a slowly varying duct isThis applies for $n\ge 1$ away from turning points $k_n=0$.

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.

Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

Open ends impose $C(0,t)=C(L,t)=0$. By separation of variables,Expanding the initial constant gives $b_n=2C_0(1-(-1{)}^n)/(n\pi )$, hence

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.