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For the normal-form coefficient $p(x)=-1/(1-x)$, Abel's identity gives the WronskianUse reduction of order: if $y_2=y_{1}v$, then $W=y_1^2{v}^{\prime }$. Absorbing a nonzero Wronskian constant into the second integration constant, the general solution isThis local representation applies on intervals where $y_1\ne 0$, in particular sufficiently near $x=1$ because $y_1(1)=a_0\ne 0$.

Normalize the free leading coefficient to $a_0=1$. The linear recurrence relation from part (b) givesConsequently the first four nonzero coefficients areand the single pure Frobenius solution isFor clarity about the second independent solution, a choice with no added multiple of $y_1$ isThese companion coefficients follow by reduction of order with the normalized Wronskian $W=z^{-3}(1+z^2{)}^{3/2}\exp [3z^2/(2(1+z^2))]$. The logarithm requires a local branch; it does not supply a second ordinary power series solution.

For an exponential trial $y=e^{rx}$, the characteristic equation isIts distinct complex roots are $r=-a\pm ib$. Thus the general real solution of the homogeneous second-order linear differential equation isThe two displayed solutions have nonzero Wronskian $b{e}^{-2ax}$ and hence linear independence. The same formula permits complex solutions when $C_1,C_2$ are complex.

Define the Wronskian by $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. Differentiating and using the normal-form differential equation givesIntegrating this first-order linear differential equation proves Abel's identity:Thus, on any interval avoiding zero and $m$,Absolute-value signs from real logarithms can be absorbed into the constant on each such interval. For linearly independent solutions, $C\ne 0$.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

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 $x^{n+1}$.

Multiply the equation for $y_i$ by $y_j$ and the equation for $y_j$ by $y_i$, then subtract. The $q$ terms cancel and the derivative terms combine into a Wronskian boundary derivative:Integrating over the interval eliminates the boundary term because both eigenfunctions vanish at both endpoints. Since their eigenvalues differ,This is orthogonality in a weighted inner product in the regular Sturm-Liouville setting; the endpoint values must justify the integration by parts.

The integral construction requires differentiation under the integral and vanishing boundary terms. With those assumptions, integration by parts givesThe standard construction therefore sets $g^{\prime }={\zeta }^{2}g$, giving $g=C{e}^{{\zeta }^3/3}$. Convergent contours run between rays where $\Re {\zeta }^3<0$: convenient central rays have arguments $-\pi /3$, $\pi /3$ and $\pi$. Two choices are the oriented pathsThe cubic exponential dominates the linear one at both ends. To verify independence directly, put $A={\int }_0^{\infty }{e}^{-r^3/3}dr>0$ and $B={\int }_0^{\infty }r{e}^{-r^3/3}dr>0$. At $z=0$ the value and derivative of the integral along rays ${\theta }_1,{\theta }_2$ are $A(e^{i{\theta }_2}-e^{i{\theta }_1})$ and $B(e^{2i{\theta }_2}-e^{2i{\theta }_1})$. For the two contours above their derivative-to-value ratios are respectively $B/A$ and $(B/A)(e^{i\pi /3}-1)$, which differ. Hence their Wronskian is nonzero.

For the second equation multiply by $z$ before integrating by parts:Setting the residual weight to zero gives $g=C(1+{\zeta }^2{)}^{-1/2}$. One contour is the segment from $-i$ to $i$, with the square root positive at $\zeta =0$. Its integral is a nonzero multiple of $J_0(z)$: substituting $\zeta =it$ gives $i{\int }_{-1}^1{e}^{izt}(1-t^2{)}^{-1/2}dt$, an entire function nonzero at zero.

A second contour starts at $i$ and follows $\zeta =i-t$, $0\le t<\infty$, choosing a continuous branch along the path. For $\Re z>0$ it converges at infinity; at the branch endpoint its integrable singularity is $t^{-1/2}$. In both cases the boundary term contains $\sqrt{1+{\zeta }^2}$ and vanishes at branch endpoints. The second integral has a nonzero logarithmic divergence as $z\to 0$ through positive real values, because its large-$t$ integrand is a nonzero multiple of $e^{-zt}/t$. It is therefore independent of the entire first solution; analytic continuation supplies solutions elsewhere away from the equation's singular point.

Strictly, nonzero $w$ and an arbitrary contour alone do not prove the printed necessity for $g$. Necessity also needs injectivity of the chosen Laplace representation, while the vanishing-integrand argument above is the usual sufficient construction. For a concrete counterexample to unconditional necessity in the Bessel case, take a positively oriented circle $|\zeta |=R>1$, choose the branch $(1+{\zeta }^2{)}^{-1/2}\sim 1/\zeta$ outside the cut $[-i,i]$, and add one to that weight. The added entire integrand integrates to zero. Expanding the original weight on this circle gives integral $2\pi i{J}_0(z)$, which is nonzero and solves the equation, although the modified weight is not a constant multiple of the printed one. The standard weights and contours above still give the requested independent solutions.

For an admissible Laplace integral solution, differentiation under the integral sign and integration by parts giveChoose endpoints for which the boundary term vanishes. Setting the density in braces to zero givesTwo suitable contours are the rays ${\Gamma }_{+}:0\to +i\infty$ and ${\Gamma }_{-}:0\to -i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O({\zeta }^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.

To check independence rather than merely exhibit different paths, defineWith unit normalization, the two solutions satisfy $w_{\pm }(0)=e^{\pm i\pi /4}{J}_0$ and $w_{\pm }^{\prime }(0)=\pm i{e}^{\pm i\pi /4}{J}_1$. Their Wronskian at zero is $-2i{J}_0{J}_1\ne 0$, so they form an independent pair.

Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a null weight in a contour Laplace representation, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.

Assume a real, time-independent potential with the same zero asymptote on both sides, so $k=\sqrt{2mE}/\hslash$ is the same incident and transmitted wave number. For unit incoming amplitudes,Multiplying the two Schrödinger equations crosswise and subtracting shows that the Wronskian $W={\psi }_L{\psi }_R^{\prime }-{\psi }_L^{\prime }{\psi }_R$ is constant. Its two asymptotic values are $-2ik{t}_R$ and $-2ik{t}_L$, hence $t_L=t_R$.

For a real potential the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is conserved, giving $|r_L{|}^2+|t_L{|}^2=|r_R{|}^2+|t_R{|}^2=1$. ThereforeThe reflection amplitudes can have different phases. Reality or self-adjointness is essential for the reflection-probability conclusion: a complex absorbing potential need not have equal left and right reflectances. Equal asymptotic speeds are also used in identifying transmission with the squared amplitude.

Let $x_{<}=\min (x,x^{\prime })$ and $x_{>}=\max (x,x^{\prime })$. A homogeneous solution vanishing at the left wall is $u(x)=\sin [k(x+a/2)]$; one vanishing at the right wall is $v(x)=\sin [k(a/2-x)]$. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-k\sin ka$. Continuity and the source derivative jump $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$ therefore giveThis is exactly the printed form because $x_{<}=(x+x^{\prime }-|x-x^{\prime }|)/2$ and $x_{>}=(x+x^{\prime }+|x-x^{\prime }|)/2$, with a minus sign from reversing the argument of the second sine.

The true poles occur at $k_n=n\pi /a$, $n=1,2,\dots$, corresponding toThe apparent zero at $k=0$ is removable: the numerator and $k\sin ka$ both vanish quadratically. Negative values of $k_n$ duplicate the same energies, rather than describing extra levels. The boundary conditions and source jump uniquely determine the inverse whenever $E$ is not an eigenvalue.

For $\gamma \ne 0$ and $x>0$, the change of variables $u=x^{\gamma }$ convertsinto $Y_{uu}+Y=0$, where $y(x)=Y(u)$. The chain rule cancels the first-derivative terms. Thus $\cos (x^{\gamma })$ and $\sin (x^{\gamma })$ form a pair of linearly independent solutions, with Wronskian $\gamma x^{\gamma -1}$. This is a variable-speed reparametrization of the harmonic oscillator equation.

For $Ly=-x^{-2}(x^2{y}^{\prime }{)}^{\prime }+y$ on $x>0$, require boundedness at zero and at infinity. The substitution $z=xy$ reduces the homogeneous equation to $-z^{\prime \prime }+z=0$. The adapted solutions are $\sinh x/x$ at zero and $e^{-x}/x$ at infinity, with Wronskian $-x^{-2}$. The displayed Green function is continuous and has first-derivative jump $-1$, so $L_{x}G=\delta (x-\xi )$ with ordinary source measure $d\xi$. It satisfies weighted reciprocity $x^{2}G(x,\xi )={\xi }^{2}G(\xi ,x)$. For forcing equal to one on $[0,R]$ and zero outside, integration gives $y=1-(R+1)e^{-R}\sinh x/x$ inside and $y=(R\cosh R-\sinh R)e^{-x}/x$ outside. Limits define the regular value at zero.

The rational map approximation writes the angular dependence of a Skyrmion using a rational map $R:S^2\to S^2$ and determines a radial profile variationally. The degree of $R$ equals the baryon number, while zeros of its Wronskian mark directions of vanishing angular baryon density.

Let $F(X,Y)=P(X)+YQ(X)$ with $P,Q\in \mathbb{Z}[X]$ linearly independent, ${\mathrm{deg}}_{X}F\le n$, and $H(F)\le C^n$. For every $\delta >0$, if a reduced rational $p/q$ has sufficiently large denominator in terms of $C$ and $\delta$, then $F(X,y)$ has multiplicity at most $\delta n+1$ at $p/q$, for every fixed real $y$.

Indeed, the nonzero Wronskianhas degree below $2n$ and height at most $C_0^n$. A zero of multiplicity $r$ of $F(X,y)$ forces a zero of multiplicity at least $r-1$ of $W$. By Gauss lemma for polynomials, $(qX-p{)}^{r-1}$ then divides $W$ in $\mathbb{Z}[X]$, so $q^{r-1}$ divides the leading coefficient of $W$ and is at most $C_0^n$.

Given one nonzero solution of a homogeneous second-order linear equation, reduction of order constructs a second by writing it as a variable multiple of the first.

The integral representation $y_2=y_1\int e^{-\int p}/y_1^2$ is derived where $y_1\ne 0$. At a zero, its integral factor may have a pole while the product has a removable singularity. For regular coefficients, continue the actual second solution by its finite initial data and initial-value uniqueness; the Abel identity preserves a nonzero Wronskian. In the degree-one Legendre differential equation, $y_1=x$ and $y_2=x\mathrm{arctanh}x-1$ illustrate this cancellation at zero.

The homogeneous equation has linearly independent solutions $e^{rx}$ and $x{e}^{rx}$, with nonzero Wronskian $e^{2rx}$. Setting $y=e^{rx}u$ converts the forced equation to $u^{\prime \prime }=e^{-rx}f$. For resonant forcing $f=e^{rx}$, the particular response is $x^2{e}^{rx}/2$. This substitution explains the extra powers of $x$ required by repeated-root resonance.

For $L>0$, constant $\beta >0$ and $\mathrm{Re}p>0$, on $(0,L)$ put $b=\beta /2$, $\kappa =\sqrt{p+b^2}$ with the principal square root, andThese satisfy $(p-{\partial }_x^2-\beta {\partial }_x)u_p=(p-{\partial }_x^2-\beta {\partial }_x)v_p=0$ and the left and right homogeneous Neumann boundary conditions, respectively. The weighted Wronskian givesThe first derivative jumps by $-e^{-\beta y}$, so the resolvent integrates against $e^{\beta y}dy$. Its time Laplace transform interpretation is ${\mathcal{R}}_p={\int }_0^{\infty }{e}^{-pt}{K}_{\beta }(t)dt$. The factor $p$ records the zero eigenvalue and constant stationary eigenfunction.

For $0<x<1$, consider the self-adjoint differential operator $L=(pd/dx{)}^{\prime }+q$ with $p=e^{-x}/(1-x)$ and $q=-e^{-x}/(1-x{)}^2$. The homogeneous solutions $u=x$ and $v=e^x-ex$ satisfy the left and right zero Dirichlet boundary conditions. Their Wronskian is $(x-1)e^x$, and $pW=-1$. Thus the displayed Dirichlet Green function is continuous, zero at both endpoints, and has unit flux jump $p(s)[G_x(s^{+},s)-G_x(s^{-},s)]=1$.

The right solution has a double zero, $v(s)\sim e(1-s{)}^2/2$. Consequently $G(x,s)F(s)$ is integrable at $s=1$ for fixed interior $x$ when $F(s)=-e^{-s}/(1-s{)}^2$, even though $F$ itself is not integrable there. The endpoint zero of the Wronskian causes no loss of this construction because the weighted Wronskian $pW$ stays nonzero. This illustrates the need to retain the integrating factor when constructing a Green function at a singular endpoint.