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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 Schrödinger 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 Schrödinger 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

Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrödinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.

The spectral equation $(S)$ isThis is the KdV Schrodinger spectral problem. Differentiating it with respect to $t$ givesFora direct differentiation, followed by substitution of the two preceding identities, yieldsThereforeComparison with $(\ast )$ givesThis is the reusable KdV Schrodinger spectral problem Wronskian identity.

Now take $u$ to satisfy the Korteweg-De Vries equation and $\varphi ={\phi }_n$, where ${\lambda }_n=-{\kappa }_n^2$. Both ${\phi }_n$ and its derivative decay at infinity, as does $u$, so integration of $(\ast )$ over the real line givesThe normalization was used in the last equality. Hencewhich is the Isospectrality of the KdV discrete spectrum.

With the KdV equation and ${\dot{\lambda }}_n=0$, the Wronskian identity saysDecay at infinity makes the constant zero. Thus $(Q/{\phi }_n{)}_x=0$ between zeros, and continuation across the isolated zeros givesMultiplying by ${\phi }_n$ and integrating,The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x{\phi }_n^{2}dx$, soTo evaluate this, differentiate $(S)$ in $x$ and use it to writeIts integral is $-2\int {\phi }_{n,x}{\phi }_{n,xx}dx=0$ by decay. Hence

As $x\to +\infty$, rapid decay of $u$ and $u_x$, together withreduces $Q=0$ toSince ${\kappa }_n$ is constant,as stated by the Evolution of a KdV discrete norming constant.

Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore

The Wronskian is $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. Abel's identity givesIf $W\ne 0$, the two column vectors in the displayed representation form a basis of ${\mathbb{R}}^2$ at each $x$. Differentiating the first component and comparing it with the second givesSubstitution in the differential equation givesSolving this two-by-two system yields

The determinant in the hint is independent of $Y$ and equals the WronskianIt is nonzero because $P,Q$ are linearly independent. Also $\mathrm{deg}W<2n$, and coefficient convolution givesfor a constant $C_0$ depending only on $C$.

Fix $y$, and suppose $F(X,y)$ has multiplicity $r$ at $a=p/q$. Inthe two terms vanish to orders at least $r$ and $r-1$, so $W$ has multiplicity at least $r-1$ at $a$. The primitive polynomial $(qX-p{)}^{r-1}$ therefore divides $W$ in $\mathbb{Z}[X]$ by Gauss lemma for polynomials. Comparing leading coefficients givesIf $r>\delta n+1$, this implies $q^{\delta n}<C_0^n$ and hence $q<C_0^{1/\delta }$. Choosing $q$ larger than this bound proves that $r\le \delta n+1$, uniformly in $y$.

The function $y_1=x$ solves the linear ordinary differential equation. Apply reduction of order by writing $y=xv(x)$. Substituting $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$ and cancelling the terms in $v$ givesFor $w=v^{\prime }$, this becomes a first-order linear ordinary differential equation:A partial fraction decomposition provides the required antiderivative:ThereforeThe logarithm has positive argument on the required interval. The Wronskian of $x$ and $1+x\log ((x-1)/x)$ is $1/(x-1)\ne 0$, proving their linear independence and hence completeness of this general solution.

For $y=x$, the derivatives are $y^{\prime }=1$ and $y^{\prime \prime }=0$, and substitution gives $2(x-1)-2(x-1)=0$.

Use reduction of order: set $y=xv$, so $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$. Substitution and cancellation giveSince $x>0$, this reduces to $v^{\prime \prime }+2v^{\prime }=0$, hence $v^{\prime }=A{e}^{-2x}$ and $v=C_1+C_2{e}^{-2x}$. The general solution isThe two solutions are linearly independent, since their ratio $e^{-2x}$ is nonconstant; equivalently, their Wronskian is $-2x^2{e}^{-2x}\ne 0$.

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.