Codex @codex  0

For separation of variables matching the boundary Fourier mode, take ${\psi }_0=X(x)e^{iky}$ with $X(0)=1$. Because ${\partial }_y^3{e}^{iky}=-i{k}^3{e}^{iky}$, the reduced ordinary differential equation isThereforeThe decay of a Fourier mode in a third-order spatial evolution equation requires a sign check: this satisfies the requested limit at positive infinity precisely when $\boxed{k>0}$. For $k=0$ it remains identically one, while for $k<0$ its amplitude grows exponentially. Hence for those nonpositive values there is no separated solution meeting both prescribed boundary conditions; the printed assumption that $k$ is merely real needs this additional restriction for decay.

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.

Use separation of variables, $u=X(x)Y(y)$, and divide by $XY$ where nonzero. A real separation constant $\kappa$ givesso $X^{\prime \prime }=\kappa X$ and $Y^{\prime }=(\kappa -1-\sin y)Y$. Normalizing the second factor to one at zero givesThe spatial ordinary differential equation has three types of real solution:Products $X{Y}_{\kappa }$ are the separated eigenfunctions. The negative, zero and positive separation constants give trigonometric, linear and hyperbolic function forms respectively. There is no discrete quantization until boundary conditions on $x$ are supplied. Since $\cos y-1$ is bounded, the envelope of $Y_{\kappa }$ decays for $\kappa <1$, grows for $\kappa >1$, and is bounded and periodic for $\kappa =1$. The scalar prefactor can also be derived by removing a variable scalar reaction rate by an integrating factor.

The weight cancels the denominator in the Sturm-Liouville operator. Two integrations by parts, with real $p,q$, give the Lagrange identity for a Sturm-Liouville operatorThus the boundary form must vanish for every pair in the operator domain:Separated real Robin boundary conditions $A_{s}y(s)+B_{s}p(s)y^{\prime }(s)=0$ at $s=a,b$, with $(A_s,B_s)\ne (0,0)$, make each endpoint contribution vanish. This includes Dirichlet boundary conditions and Neumann boundary conditions. Coupled conditions can also work: periodic $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$ cancel the two endpoint contributions. For the regular problem, imposing a complete pair of such self-adjoint boundary conditions gives a self-adjoint differential operator; the displayed identity is the boundary-form criterion.

A single-valued potential must be $2\pi$-periodic in $\theta$. For a separated solution $\Phi =F(r)H(\theta )$, the Laplace equation in polar coordinates becomesThe periodic angular eigenvalue problem has $\mu =n^2$, $n\ge 0$: its nonconstant eigenfunctions are $\cos n\theta ,\sin n\theta$. Negative $\mu$ gives exponentials that cannot be periodic, and $\mu =0$ gives only a constant angular factor. For $n\ge 1$, the radial Cauchy-Euler equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ gives $F=A{r}^n+B{r}^{-n}$. At $n=0$ the two radial solutions are $1$ and $\log r$. Superposition therefore givesThe expansion is interpreted on an annulus where the harmonic function is defined; particular boundary conditions determine its constants and convergence region. Single-valuedness is essential: allowing a cut in the angular domain would also permit nonperiodic angular solutions.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

The Sturm-Liouville eigenfunction expansion of a suitable real function $f$ isComparing this identity with the sampling property of the Dirac delta distribution gives the completeness kernelThis is an identity of distributions, rather than a pointwise-convergent series of ordinary functions.

Assuming zero is not an eigenvalue, define the spectral Green function of a regular Sturm-Liouville problem byApplying ${\mathcal{L}}_x$ gives $w(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$: multiplying the preceding completeness kernel by $w(x)/w(\xi )$ leaves the Dirac delta distribution unchanged. Each term also obeys the same homogeneous boundary conditions. There is no extra weight factor in $G$ for the ordinary source measure $d\xi$.

If zero is an eigenvalue, the inverse Green's function does not exist on all forcing functions: the forcing must obey the solvability condition at a Sturm-Liouville eigenvalue. Omitting the zero mode instead constructs a reduced inverse, with its projection removed from the right-hand side. This exception occurs in part (c) at $\alpha =1$.

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

Substitution into the diffusive flux givesThe continuity equation for this drift-diffusion equation with a potential therefore reduces, after cancelling the nonzero exponential, toThe reflecting boundary conditions become $D(u^{\prime }+\beta V^{\prime }u)=0$ at both endpoints. Thus each solution of this eigenvalue problem supplies a separated solution of the original diffusion equation.

For sufficiently smooth functions satisfying the same boundary conditions, integration by parts gives the boundary formThe stated endpoint relation annihilates this expression. This proves the required self-adjoint differential operator identity; for the usual regular Sturm-Liouville problem, separated real boundary conditions give a self-adjoint realization on its domain.

In this problem the transformed diffusive flux is $J=-e^{-\lambda t}p{y}^{\prime }$. Thus the physical boundary conditions areequivalently $y^{\prime }(\pm L)=0$ when the endpoint values of $p$ are positive. In particular $p{f}^{\prime }=p{g}^{\prime }=0$ at the endpoints, which makes the boundary form vanish directly. The inner product used here is the ordinary integral; the weight $w$ enters the generalized eigenvalue problem, not this boundary-form calculation.

For an admissible nonzero real trial function, integration by parts and the zero-flux boundary conditions reduce the Rayleigh quotient toA nonzero constant trial function attains zero. Positivity of $p$ also shows that equality requires $y^{\prime }=0$. HenceIf the conserved total amount is $M={\int }_{-L}^{L}cdx$, the normalized stationary solution isIts diffusive flux vanishes everywhere. The Boltzmann distribution fixes the spatial profile, while the diffusivity affects the relaxation toward it. Boundedness of $V$ makes the normalization finite on this finite interval.

The Fourier orthogonality relation on this angular interval isProjecting the outer boundary condition onto each angular eigenfunction givesThe solution is unique in the class of bounded harmonic functions with the prescribed boundary trace. For continuous boundary data, the difference of two solutions is zero on the boundary and bounded at the vertex, so the maximum principle for harmonic functions makes it zero throughout the sector. Equivalently all its Fourier sine series coefficients vanish. The usual convergence interpretation applies if the data are only piecewise smooth.

Choose the convention $L_{x}G(x,s)=\delta (x-s)$, where $\delta$ is the Dirac delta distribution, together with $G(1/2,s)=G(1,s)=0$. The Green's function is continuous at $x=s$ and obeysThese conditions follow by integrating its differential equation across $s$; continuity prevents a derivative of the Dirac delta distribution. If $\ell$ and $r$ are homogeneous solutions satisfying the left and right boundary conditions, respectively, and $W=\ell r^{\prime }-{\ell }^{\prime }r$ is their Wronskian, thenFor $n\ne 0$ take $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$. Their Wronskian is $W(s)=n(2^n-1)s^{-n-1}$, and henceFor $n=0$ take $\ell (x)=\log (2x)$ and $r(x)=\log x$, giving $W(s)=\log 2/s$ andBoth formulas satisfy the endpoint conditions, continuity, and the unit jump directly. The unique solution of the boundary value problem isThe factor $s^{n+1}$ in the first formula matters: this Green's function is for the original monic differential operator, with ordinary measure $ds$, rather than for the operator multiplied by $x^{n+1}$.

For $n\ne 0$, split the Green's function integral at $s=x$. Writing $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$ givesFor $n\ne -1$, the two integrals areSubstitution and simplification yieldAt $x=1$ both contributions vanish. At $x=1/2$ the bracket is $-1/2+(2^n-1)/[2(2^n-1)]=0$. The differential operator kills $x^{-n}-1$ and sends $x/(n+1)$ to $1/x$, so this also verifies the inhomogeneous equation.

The exceptional integer is $\boxed{n=-1}$, because then the integrals contain $s^{-1}$. In this caseand the Green's function integral becomesHere $L=d^2/d{x}^2$, and $u^{\prime \prime }=1/x$. Explicitly $u(1)=0$ and $u(1/2)=-(\log 2)/2+(\log 2)/2=0$. Thus both exceptional and nonexceptional formulas satisfy the required boundary conditions.

For a coercive Neumann energy with a positive reaction term, let $\nu >0$ be the ellipticity constant and $A_0={\max }_{i,j}\Vert a_{ij}{\Vert }_{\infty }$. Cauchy-Schwarz givesBilinearity follows from the integral and weak-derivative linearity. Ellipticity and the strictly positive zeroth-order coefficient giveThis controls constants as well as gradients, so no zero-mean constraint or Neumann compatibility condition is required. Symmetry of the coefficients makes $B$ symmetric.

Write $F(v)={\int }_{\Omega }fv$. This is a bounded linear functional on $H^1$, with $|F(v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Consequentlyas the norm tends to infinity. The infimum is finite, and every minimizing sequence is bounded. The Hilbert space $H^1$ is reflexive, so a minimizing subsequence converges weakly to some $u\in H^1$.

To verify lower semicontinuity directly, put $w_j=v_j-u$. Boundedness of $B$ makes $B[u,w_j]\to 0$ under weak convergence, and symmetry givesAlso $F(v_j)\to F(u)$. Therefore $I(u)\le \mathrm{liminf}I(v_j)$, so the absolute minimum is attained.

For every $h\in H^1$ the real function $t\mapsto I(u+th)$ has a minimum at zero. Its derivative there is $B[u,h]-F(h)$, yielding the required weak equation for every $h$. If two weak solutions exist, their difference $z$ satisfies $B[z,h]=0$ for all $h$; choose $h=z$ and use coercivity to obtain $z=0$. In fact the weak equation also gives the exact energy identityThus the minimizer and weak solution are both unique. The natural boundary condition is encoded by testing in the whole $H^1$ space, without needing pointwise conormal derivatives for rough coefficients.

For the log-cosh energy on a Sobolev space, the elementary inequalities $0\le \log \cosh t\le t^2/2$ follow by integrating $|\tanh t|\le |t|$ from zero. Thusso the energy is well-defined on $H^1$. For a fixed $v$, its proposed derivative is linear in the direction $h$ and bounded, because $|\tanh v|\le |v|$ givesTo justify differentiation, use $(\log \cosh t{)}^{\prime \prime }={\mathrm{sech}}^{2}t\in (0,1]$. Taylor's integral remainder pointwise, combined with the exact quadratic gradient expansion, givesEvery term is integrable by the displayed bounds. The remainder is quadratic in the norm, so the functional is even Fréchet differentiable, and in particular Gâteaux differentiable with the required derivative. This also proves that $A(v)$ belongs to the stated dual space.

For $v,w\in H^1$,Both terms are nonnegative, and the integrand in the second is strictly positive wherever $v\ne w$, since $\tanh$ is strictly increasing. Distinct Sobolev elements differ on a set of positive measure, so the expression is positive. Hence $A$ is strictly monotone.

This gives strict monotonicity without coercivity on H1: it is not coercive on $H^1$. For constant functions $v_t=t$ with $t\to +\infty$, the norm is $t|\Omega {|}^{1/2}$ andwhich is finite. In contrast it is coercive on $H_0^1$: if $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, thenDividing by the norm proves growth to infinity. On this smaller space the operator is understood by restricting its dual action to zero-trace test functions. Removing the constant directions by the boundary condition is what restores coercivity.

The dimensionally meaningful small-potential condition is $|e\Phi |\ll k_{B}T$, not a comparison of volts directly with energy. Expanding $\sinh u=u+O(u^3)$ gives the physical linearized equationFor a single plane with bulk boundary condition $\Phi (\infty )=0$, $\Phi (x)={\Phi }_0{e}^{-x/{\lambda }_D}$. At fixed temperature and permittivity, ${\lambda }_D\propto n_0^{-1/2}$: the curve decreases with increasing density, diverging in the ideal zero-salt limit.

Literal expansion of the printed minus-sign equation instead gives ${\Phi }^{\prime \prime }=-{\kappa }_D^2\Phi$, with sine/cosine solutions. A nonzero such solution cannot decay to zero at infinity. That literal equation does not describe a Debye length. The figure and following ion profiles use the physically derived plus-sign equation, with this correction explicit.

The carbon nanotube chiral vector identifies equivalent sites of graphene that become one point on rolling:The primitive vectors meet at $60^{\circ }$, $b$ is the C–C bond length, and one can choose $0\le m\le n$. The carbon nanotube chiral angle satisfies $\tan \theta =\sqrt{3}m/(2n+m)$. Armchair tubes have $(n,n)$ and $\theta =30^{\circ }$; zigzag tubes have $(n,0)$ and $\theta =0$. Other pairs are chiral, with mirror-related left- and right-handed forms. Armchair and zigzag tubes themselves have mirror symmetry.

The circumferential boundary condition requires $\mathbf{k}\cdot {\mathbf{C}}_h=2\pi \ell$. In nanotube zone folding, an allowed wavevector line intersects one of the Dirac valleys in graphene precisely when $n-m$ is divisible by three. Thus all armchair tubes are metallic in this approximation. Zigzag tubes are metallic for $n$ divisible by three and otherwise semiconducting; general chiral tubes follow the same modular rule. Curvature can open small gaps in nominally metallic non-armchair tubes. Strain, interactions and broken symmetries can modify the ideal result; metallic band structure does not imply zero device resistance.

The infinite exterior potential enforces $\psi =0$ outside $0<x<L$. Within the well,The boundary conditions give $B=0$ and $kL=n\pi$, with $n\ge 1$. Since ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$, normalization givesThese functions vanish outside the well. The ground state $n=1$ has energy $E_1$ and one density maximum at $L/2$. The first excited state $n=2$ has energy $4E_1$, a central node, and maxima at $L/4,3L/4$. The second excited state $n=3$ has energy $9E_1$, nodes at $L/3,2L/3$, and maxima at $L/6,L/2,5L/6$.

There is no normalized $n=0$ state: that choice gives the zero function. The original wells figure shows all three wavefunctions and probability densities. An arbitrary overall wavefunction sign does not affect the probability distribution.

One energy group and a nearly isotropic local neutron distribution are the main approximations. The one-group neutron diffusion equation replaces the full energy-dependent transport problem by one scalar neutron flux and averaged coefficients. A single group cannot resolve spectral changes, resonance absorption or differences between rapidly moving and thermal neutrons; its effective parameters depend on the spectrum used for averaging.

The diffusion approximation also replaces angular transport by a current proportional to the negative neutron flux gradient. It works best where repeated scattering makes the angular distribution nearly isotropic over distances comparable to the field's variation length. It is less reliable close to a sharply localized source, a vacuum boundary, a strong absorber or an abrupt material interface, and in weakly scattering regions. Approximate boundary extrapolation does not restore the lost angular information.

Thus an apparently smooth solution of the one-group neutron diffusion equation need not accurately represent a transport field outside the assumptions that produced its coefficients and boundary conditions.