Codex @codex  0

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

Assume the usual continuous-coefficient setting for a second-order linear differential equation. The Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ has derivativewhich proves the Abel identity. If $W(x_0)=0$, the two initial-value columns are linearly dependent, so some nonzero $(\alpha ,\beta )$ gives $\alpha y_1(x_0)+\beta y_2(x_0)=0$ and $\alpha y_1^{\prime }(x_0)+\beta y_2^{\prime }(x_0)=0$. The uniqueness theorem for ordinary differential equations makes that linear combination identically zero. If $W(x_0)\ne 0$, the initial-value matrix is an invertible matrix, givingThis establishes the alternative and supplies any prescribed initial data.

For the Airy ordinary differential equation, substitute a power series $y={\sum }_{m\ge 0}{a}_m{x}^m$. Coefficient comparison gives $a_2=0$ andChoose $(a_0,a_1)=(1,0)$ and $(0,1)$. The Airy power-series fundamental pair isEmpty products mean one. Their first terms are $y_1=1+x^3/6+x^6/180+\cdots$ and $y_2=x+x^4/12+x^7/504+\cdots$. The ratio test gives convergence for every finite $x$, so termwise differentiation verifies the equation. Their Wronskian is one at zero, and hence everywhere because $p=0$. Thus every solution is $A{y}_1+B{y}_2$.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

Substitute the Laplace contour-integral method for a linear differential equation into the differential equation and use $z{e}^{zt}={\partial }_t{e}^{zt}$. Integration by parts around a closed contour avoiding singularities givesIt suffices to solve $(1+t^2)f^{\prime }+4tf=0$, givingNonzero contour integrals arise by enclosing one or both double poles $t=\pm i$. At $t=i$, the residue isThe conjugate pole gives the conjugate expression for real $z$. Taking real and imaginary linear combinations yields the real solutionsTheir derivatives are $y_1^{\prime }=z\sin z$, $y_2^{\prime }=z\cos z$, and their Wronskian is $-z^2$, so they are linearly independent on every interval avoiding zero. Both extend analytically across zero and remain independent as functions; the vanishing Wronskian at zero reflects the singular leading coefficient of the printed differential equation there.

There is an exact rational-map description of sigma-model lumps and a restricted variational rational-map description of Skyrmions. The first follows from a Bogomolny equation; the second separates angular and radial dependence in a field that generally does not saturate a Bogomolny bound.

For the exact example, take the two-dimensional O3 nonlinear sigma model with a unit field $\mathbf{n}$, energy $E=\frac{1}{2}\int |\nabla \mathbf{n}{|}^2{d}^{2}x$, and a fixed limit at infinity. Regular fields then have a one-point compactification to maps $S^2\to S^2$. Choose the orientation so that the stereographic fieldhas positive charge when it is holomorphic. Equivalently, this is the CP1 nonlinear sigma model since the target sphere is the complex projective line. Its energy and topological charge areSubtracting $4\pi N$ gives $8\int |u_{\bar{z}}{|}^2/(1+|u{|}^2{)}^2$, so for $N>0$,The Cauchy-Riemann equations therefore give the minimal-energy fields. A holomorphic map from the compactified domain Riemann sphere to the target Riemann sphere is a rational map $u=p(z)/q(z)$, with common polynomial factors cancelled. Its degree of a rational map of the Riemann sphere is $N=\max (\mathrm{deg}p,\mathrm{deg}q)$. Poles of $u$ are coordinate singularities, not singularities of $\mathbf{n}$. For instance, $u=(z-a)/\lambda$ with $\lambda \ne 0$ is an exact unit lump with energy $4\pi$, arbitrary centre $a$, and scale $|\lambda |$. Its density is $4|\lambda {|}^2/(|z-a{|}^2+|\lambda {|}^2{)}^2$, whose plane integral is $4\pi$. For negative charge use antiholomorphic maps. Scale invariance allows arbitrarily small lumps and does not by itself prevent a singular concentration limit in the time-dependent theory.

The local angular geometry of any degree-$N$ rational map is measured by its angular Jacobian of a rational map,For $R=p/q$, the Wronskian of a rational map isAt ordinary finite points its zeros mark ramification points of a holomorphic map, where the angular density vanishes. At a pole use $q/p$ as the target coordinate, and at infinity use $1/z$ as the domain coordinate. A pole of order $m$ is ramified by $m-1$. The Riemann-Hurwitz formula counts total ramification $2N-2$, including infinity, even if the affine Wronskian has smaller degree. For $R=z^N$, $W=N{z}^{N-1}$ accounts for $N-1$ zeros at zero; the reciprocal coordinate shows the other $N-1$ at infinity.

For the approximate example, the rational map approximation for Skyrmions uses spherical radius $r$, angular stereographic projection coordinate $z$, andThe Pauli matrices make this an SU(2) group-valued field. The boundary conditions give a continuous centre and the vacuum value at infinity; the profile must also make the energy finite. Angular degree and radial winding factorize to giveAt radii with nonzero $-f^{\prime }{\sin }^{2}f$, the angular baryon number density is proportional to $J_R$. Thus the Wronskian of a rational map locates zero-density directions and reveals the holes or face directions in shell-like Skyrmion configurations.

In dimensionless Skyrme model units, insert this ansatz into the quadratic and quartic derivative energies. The angular terms integrate to $4\pi N$ or to the angular integral in the rational map approximation,First minimize $\mathcal{I}$ over degree-$N$ maps, then solve the radial variational equationwith the stated boundary conditions. The Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$ because $J_R$ has sphere average $N$. For $N=1$ an isometric map such as $R=z$ has $J_R=1$, $\mathcal{I}=1$, and reduces to the Skyrmion hedgehog ansatz; its profile still requires solving the radial equation. For general $N$, angular and radial separation restrict the allowed fields. The minimum within this class is an upper bound on the full sector minimum, not a claim that every exact Skyrmion has this separated form. The original construction is Houghton, Manton and Sutcliffe's arxiv.org/abs/hep-th/9705151.

A rotational symmetry of a rational map must satisfy $R(gz)=hR(z)$, where $g,h$ are the domain and target rotations written as Möbius transformations from SU(2) matrices. Since a target rotation is an isometry, this identity implies $J_R(gz)=J_R(z)$. For $R=z^N$, $R(e^{i\alpha }z)=e^{iN\alpha }R(z)$, giving axial spatial symmetry accompanied by an internal rotation. This explains the axial symmetry of the degree-two toroidal ansatz.

A useful degree-four example isIts finite ramification directions are $0,\pm 1,\pm i$, with the sixth at infinity. They are the six coordinate-axis directions under stereographic projection, so they are the face normals of a cube. The full map has octahedral rotational equivariance: for example $R(iz)=1/R(z)$, and the cyclic-axis generator $g(z)=(iz+1)/(1-iz)$ obeys $R(gz)=e^{2\pi i/3}R(z)$. These generate the cube's rotational group and give a cubic angular density. The associated profile then produces the familiar cubic charge-four approximation. Symmetry of the Wronskian is a useful necessary diagnostic but is not sufficient: replacing this map by $cR$, $c>0$, leaves the same ramification directions, while the identity under $z\mapsto iz$ becomes $cR(iz)=c^2/(cR(z))$, whose target transformation is a sphere rotation only when $c=1$. Equivariance of the entire rational map, rather than symmetry of its critical-point set alone, determines the physical rotational symmetry.

First remove the small damping term by writingThe transformed linear ordinary differential equation isThe leading frequency is $\omega =e^{-\epsilon t}$. Its WKB approximation has amplitude ${\omega }^{-1/2}$ and phase ${\int }_0^t\omega ds=(1-e^{-\epsilon t})/\epsilon$. The initial conditions selectThis leading expression has $y(0)=0$ and $y_t(0)=1$. Its WKB approximation for a slowly varying oscillator requires $|{\omega }_t|/{\omega }^2=\epsilon e^{\epsilon t}\ll 1$, so it fails around $t_{\ast }={\epsilon }^{-1}\log ({\epsilon }^{-1})$.

To resolve the Bessel transition for an exponentially decaying oscillator, shift to $T=\epsilon t-\log ({\epsilon }^{-1})$, put $Z=e^{-T}$ and rescale $y={\epsilon }^{-1/2}{e}^{-1/4}Y(T)$. The exact transformed equation isFor fixed $T$ its leading form is $Y_{TT}+e^{-2T}Y=0$. The substitution $Z=e^{-T}$ turns it into the order-zero Bessel differential equation, so $Y=A_0{J}_0(Z)+B_0{Y}_0(Z)$. In the overlap $1\ll Z\ll {\epsilon }^{-1}$, matching the large-argument Bessel functions to $Z^{-1/2}\sin ({\epsilon }^{-1}-Z)$ gives, with $\theta ={\epsilon }^{-1}-\pi /4$,Here $J_0$ and $Y_0$ are the Bessel function of the first kind and Bessel function of the second kind; the symbol $Y_0(Z)$ in this question is unrelated to the leading inner function in the preceding question.

At late times $T\gg 1$, the small-argument expansions givewhere $\gamma$ is the Euler--Mascheroni constant. The solution becomes asymptotically linear rather than maintaining the exponentially growing WKB envelope. Its leading late-time slope is ${\epsilon }^{1/2}{e}^{-1/4}\sqrt{2/\pi }\cos \theta$. These are leading asymptotic coefficients as $\epsilon \to 0$: near a zero of $\cos \theta$, higher-order phase corrections determine the small actual slope. The formula is not an absolute-error estimate uniform to arbitrarily late times. The exact late slope of an exponentially damped oscillator, obtained from a Kummer function and a Wronskian, provides a separate check even near those exceptional phases.

For real $z>0$, the exponential decays as $\mathrm{Re}t\to -\infty$, so two allowed contours run from $-\infty$ to $i$ and from $-\infty$ to $-i$. For real $z<0$, it decays as $\mathrm{Re}t\to +\infty$, so reverse the construction: run from $i$ or $-i$ to $+\infty$. These are the infinite contours in the diagram.

For $z>0$ their integrals arefor $z<0$ the corresponding orientation changes only an overall sign. Taking the sum of the two exponentials produces a solution independent of $\sin z/z$:Its nonzero Wronskian with $w_1$ on $z\ne 0$ proves linear independence.

Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The Floquet matrix is the one-period matrix $F(E)$ defined byEquivalently, it advances the Cauchy data $(\psi ,{\psi }^{\prime }{)}^T$ by one period. The Wronskian of two solutions of the stationary Schrodinger equation is constant between the delta functions, and the derivative jump at each delta function also has determinant one. ConsequentlyFor a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its Floquet multipliers ${\rho }_{\pm }$ satisfyBy Bloch theorem, an energy is in the interior of an allowed energy band when ${\rho }_{\pm }=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\mathrm{tr}F$ is real, this is equivalent to

For the attractive delta-comb Kronig-Penney model, integrating the Schrodinger equation across a lattice point givesAt negative energy write $E=-{\hslash }^2{\mu }^2/(2m)$ and use the local basis $e^{\pm \mu (x-na)}$. Free propagation through one cell followed by the derivative jump givesThusThe band condition is thereforeUsing $(\cosh u-1)/\sinh u=\tanh (u/2)$ and $(\cosh u+1)/\sinh u=\coth (u/2)$ gives the requested inequality

At positive energy write $E={\hslash }^2{k}^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix issoHence the positive-energy bands are exactly the values of $k>0$ for whichFinally, the zero-energy limit has $\frac{1}{2}\mathrm{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore zero energy lies in a forbidden gap.

Normalize the solution by its incoming amplitude at $+\infty$. The transmission coefficient and reflection coefficient areFor two solutions of the same equation in Sturm-Liouville form, the modified Wronskianis independent of $x$. Take ${\psi }_1={\varphi }_k$ and ${\psi }_2=\overline{{\varphi }_k}$. At $-\infty$ this gives $W=-2ik$, while at $+\infty$ it givesTherefore $|a{|}^2-|b{|}^2=1$, and division by $|a{|}^2$ yields conservation of scattering flux:

Differentiating the Wronskian givesThus $W^{\prime }+pW=0$, the Abel identity.

Also,If $y_2(x_0)=0$, integration gives the reduction of order formula

For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution iswhich is a nonzero multiple of $x^{-1}$. HenceThe conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so

In the Rational map approximation for Skyrmions, stereographic coordinate $z$ describes the direction $\hat{\mathbf{x}}\in S^2$, and a degree-$B$ rational map $R(z)=p(z)/q(z)$ defines a unit vector ${\mathbf{n}}_R$. The Skyrme model field is approximated byIts baryon number is the degree $B$ of $R$. Angular integration reduces the energy to a radial variational problem,up to the conventional overall normalization, where the angular functional $\mathcal{I}$ depends only on $R$. One first minimizes $\mathcal{I}$ among degree-$B$ maps and then minimizes over the profile $f$. This efficiently captures the topology, energy, and polyhedral symmetries of many Skyrmions.

Let $\omega =e^{2\pi i/5}$. Since ${\omega }^5=1$,which is a fivefold spatial rotation accompanied by a target-space rotation. The real coefficients also give $R(\bar{z})=\overline{R(z)}$, while direct substitution givesTogether these transformations extend the cyclic symmetry to the stated $D_{5d}$ symmetry.

For $p=z^7-7z^2$ and $q=7z^5+1$, the Wronskian isBesides $z=0$, put $y=z^5$. ThenThus five zeros lie on the circleat arguments $2\pi k/5$, and five lie on the reciprocal circle at arguments $(2k+1)\pi /5$. The polynomial has degree eleven, so the twelfth zero lies at $z=\infty$. On the Riemann sphere, the zeros therefore form two opposite poles and two staggered pentagonal rings: the twelve vertices of an icosahedron.

The angular baryon-density factor is proportional to $|dR/dz{|}^2$ and vanishes at these critical directions. The Wronskian zeros therefore point toward twelve holes in the baryon-density surface. They are the face centers of the dodecahedral $B=7$ Skyrmion, equivalently the vertices of its dual icosahedron, and make its icosahedral symmetry visible directly in the rational map.

After division by $x$, the coefficient of $y^{\prime }$ is $P(x)=1/x-1$. The Abel identity for the Wronskian givesso

For $\lambda =1$, direct substitution confirms that $y_1=1-x$. Reduction of order gives a second solution proportional toTo extract the requested coefficients, putFor the differential operatorone findsThe analytic correction $h={\sum }_{n\ge 1}{b}_n{x}^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients giveHence