Second-order linear differential equation

A second-order linear equation has the form $y\prime \prime +p(x)y\prime +q(x)y=f(x)$.

A point interaction concentrates a coefficient of a linear differential equation at a single interior point. On an interval $0<x<\ell$, with $0<c<\ell$, consider $L_{\alpha }y=y^{\prime \prime }+\alpha y(c){\delta }_c$. The domain consists of continuous functions that are twice differentiable away from $c$, with the required one-sided derivatives. Multiplication of the Dirac delta by $y$ means the unambiguous point value $y(c){\delta }_c$. It does not mean an undefined product of a delta with an arbitrary discontinuous function.

The equation is interpreted using distributional derivatives. Smooth forcing gives an ordinary equation on each side, with transmission conditions for a delta coefficient at the interaction. A delta potential in the one-dimensional Schrödinger equation is a physical example after choosing the corresponding sign and constants.

The boundary value problem $y^{\prime \prime }+\alpha y(c){\delta }_c=f$ on $(0,\ell )$ with Dirichlet boundary conditions has a unique solution for every continuous forcing and prescribed endpoint values exactly when $\alpha \ne {\alpha }_{\ast }:=\ell /(c(\ell -c))$. The left-normalized homogeneous solution iswhere $(t{)}_{+}=\max (t,0)$. Continuity at $c$ and the transmission conditions for a delta coefficient give this formula. The final boundary condition fixes the initial slope unless the displayed endpoint value is zero. This is a one-dimensional kernel obstruction, analogous to the Fredholm alternative.

At resonance the normalized null function isFor $y(0)=a$ and $y(\ell )=b$, the compatibility condition isIndeed, integrate $\varphi L_{\alpha }y-y{L}_{\alpha }\varphi$ on both sides of $c$. The internal terms cancel by the matching derivative jumps, leaving $[\varphi y^{\prime }-{\varphi }^{\prime }y{]}_0^{\ell }$. This equals the right-hand side. The initial-slope construction also proves sufficiency: when compatible there is a one-parameter family $y_0+t\varphi$; otherwise no solution exists.

For a point interaction in a second-order differential equation, the equation $y^{\prime \prime }+\alpha y(c){\delta }_c=f+J{\delta }_c$, with nonsingular continuous $f$, requiresThe distributional second derivative across a jump explains both conditions: there is no derivative-of-delta source to balance a value jump, and the coefficients of the delta source must agree. In particular, without a point source, $[y^{\prime }{]}_c=-\alpha y(c)$. With a unit point source at the same location, the condition is $[y^{\prime }{]}_c=1-\alpha y(c)$; one must combine the two contributions rather than impose two incompatible jump conditions.

This equation has a regular singular point at zero and an irregular singular point at infinity. For a nonnegative integer $\nu =n$, the solution analytic at zero with value one is the Laguerre polynomial $L_n$. Its weighted self-adjoint form is $(x{e}^{-x}{y}^{\prime }{)}^{\prime }+\nu e^{-x}y=0$, connecting it to the Laguerre differential operator on polynomials.

A Frobenius solution $y={\sum }_{k\ge 0}{a}_k{x}^{k+\sigma }$ has the repeated indicial root $\sigma =0$. Equating coefficients gives the displayed recurrence relation. The resulting analytic solution is an entire function: its consecutive term ratio has magnitude asymptotic to $|x|/k$. For $\nu =n\ge 0$ an integer, the recurrence terminates and $L_n(x)={\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})x^k/k!$. The second local solution is independent and has a logarithmic term at zero; repeated indicial roots do not supply two independent analytic series.

The chain rule with $y(x)=Y(x^2)$ gives $x{y}^{\prime \prime }-y^{\prime }=4x^3{Y}^{\prime \prime }$ and transforms the equation, away from zero, into $Y^{\prime \prime }+Y=0$. Thus its solution space is spanned by $\cos (x^2)$ and $\sin (x^2)$. Both extend smoothly through the regular singular point at zero. Their even power series allow the independent data $y(0)$ and $y^{\prime \prime }(0)$ to select the two solutions; every smooth solution has $y^{\prime }(0)=0$.

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 a normalized second-order equation, a point is singular when its coefficient functions are not both analytic there. If multiplying them by the first and second powers of distance to the point makes them analytic, it is a regular singular point. This classification concerns the equation; some individual solutions may nevertheless extend analytically through the point.

For $t>0$, the change of variables $\tau =1/t$ and $u(t)=v(\tau )/\tau$ converts $t^4{u}^{\prime \prime }+{\lambda }^{2}u=0$ into $v^{\prime \prime }+{\lambda }^{2}v=0$. The chain rule gives $u^{\prime }=v-\tau v^{\prime }$ and $u^{\prime \prime }={\tau }^3{v}^{\prime \prime }$. Thus $u=t[A\cos (\lambda /t)+B\sin (\lambda /t)]$. The prefactor cancels the derivative term introduced by inverting the independent variable.

A smooth invertible change of variables in $u_{xx}-f(x)u=0$ introduces a first derivative in time. Rescaling by ${\dot{x}}^{-1/2}$ removes it. Choosing $\dot{x}=f^{-1/2}$ locally for $f>0$ produces the constant leading coefficient used in the WKB approximation; negative $f$ requires a complex branch or the corresponding oscillatory real formulation.

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.

The Mathieu equation is the displayed linear differential equation with periodic coefficients. Floquet theory describes its bounded and exponentially growing solutions, while its periodic Mathieu characteristic values mark the edges of parametric resonance gaps. The forcing normalization matters: replacing $t$ by $t/2$ rescales both the constant and periodic coefficients.

The Mathieu characteristic values are the parameter values $a$ for which the Mathieu equation has the even periodic mode traditionally denoted ${\mathrm{ce}}_n$, or the odd periodic mode ${\mathrm{se}}_n$. They are eigenvalues of $-d^2/d{t}^2+2q\cos 2t$ with the corresponding periodic or antiperiodic boundary conditions. Their small-$q$ expansions can be derived by projecting onto a Fourier series and perturbing the resulting tridiagonal matrix.

This second-order ordinary differential equation has the independent entire Airy functions $\mathrm{Ai}$ and $\mathrm{Bi}$ as a standard basis. It is distinct from the dispersive PDE in Airy equation. Its general solution is $a_1\mathrm{Ai}(Z)+a_2\mathrm{Bi}(Z)$.

On the unit interval, the weak form is $a(u,v)={\int }_0^{1}fv$, with $a(u,v)={\int }_0^1(u^{\prime }{v}^{\prime }+xuv)$ on the zero-boundary Sobolev space. The nonnegative potential makes this form coercive in the derivative norm, and the Poincaré inequality makes that norm equivalent to the Sobolev norm. The Lax-Milgram theorem gives a unique solution. A conforming Ritz method minimizes the corresponding strictly convex quadratic energy and produces a symmetric positive-definite tridiagonal matrix for piecewise-linear hats.

For the Airy ordinary differential equation $y^{\prime \prime }=xy$, a power series obeys $a_2=0$ and $a_{m+3}=a_m/[(m+3)(m+2)]$. Setting its first two coefficients to $(1,0)$ or $(0,1)$ givesEmpty products are one. The ratio test proves convergence on the whole complex plane. Termwise differentiation verifies the equation, and the initial normalization gives Wronskian one. By the Abel identity, the Wronskian remains one everywhere, so these solutions span the whole solution space.

Named after George Biddell Airy, the standard Airy functions solve the Airy ordinary differential equation. Their Wronskian is $\mathrm{Ai}(Z){\mathrm{Bi}}^{\prime }(Z)-{\mathrm{Ai}}^{\prime }(Z)\mathrm{Bi}(Z)=1/\pi$, so they form a fundamental solution pair, including for complex $Z$.

For $q=e^{i\pi /6}$ define $I_0(C)={\int }_0^{\infty }\mathrm{Ai}(q(s-C))ds$ and $I_1(C)={\int }_0^{\infty }s\mathrm{Ai}(q(s-C))ds$. Both converge by Airy function decay. Integrating its differential equation gives $I_1-C{I}_0=iq\mathrm{Ai}\prime (-qC)$. These moments convert a decaying vorticity profile into its velocity and displacement for three-layer long-wave shear-flow matching.

For a second-order linear differential equationthe substitution $y=u\exp [-\frac{1}{2}\int p(x)dx]$ eliminates the first-derivative term and gives

For two real self-adjoint equations $(p_i{y}_i\prime )\prime +q_i{y}_i=0$ with continuous coefficients, $0<p_2\le p_1$ and $q_1\le q_2$, between consecutive zeros of a nontrivial first solution a nontrivial second solution either has a zero or is proportional to the first there. This comparison controls oscillation through the coefficients.

If $q_1\le q_2$, then between consecutive zeros of a nontrivial solution ofthere is a zero of every nontrivial solution of $y^{\prime \prime }+q_2(x)y=0$, unless the two coefficients agree throughout that interval.

For solutions ${\phi }_i^{\prime \prime }+q_i{\phi }_i=0$, the mixed WronskiansatisfiesIts endpoint signs between consecutive zeros of a positive ${\phi }_1$ force a zero of ${\phi }_2$ unless $q_1=q_2$.

Kummer's equation isIts solution analytic at zero is the confluent hypergeometric function of the first kindWhen $b$ is not an integer, a second local solution is

Putting $Y=e^{z}V$ in the displayed Kummer differential equation gives $z{V}^{\prime \prime }+(z+b)V^{\prime }=0$. Thus $V^{\prime }=C{z}^{-b}{e}^{-z}$ on a chosen branch, yielding $Y=e^z[A+B\int z^{-b}{e}^{-z}dz]$. For $b=1/2$ and $z=x/2$, one solution is $e^{x/2}$ and a normalized second Frobenius solution is $\sqrt{x}{\sum }_{n\ge 0}{x}^n/(2n+1)!!$. The analytic factors in both solutions have infinite radius of convergence, since their coefficient ratios tend to zero. The second solution still has a square-root branch at zero; an entire series factor does not remove a nonintegral Frobenius exponent.

Confluent hypergeometric functions solve the Kummer differential equation $z{y}^{\prime \prime }+(b-z)y^{\prime }-ay=0$. The regular solution is the confluent hypergeometric function of the first kind; a second standard solution is conventionally denoted $U(a,b,z)$. The family arises by coalescing singular points of the Gauss hypergeometric equation.

For $b$ not a nonpositive integer, the regular solution normalized by $M(a,b,0)=1$ has the power seriesHere $(a{)}_n$ is the rising factorial. Substitution into the Kummer differential equation proves the coefficient recurrence. This normalization agrees with NIST's definitions.

For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.

Two differentiable functions can be linearly independent on a connected interval while their Wronskian vanishes everywhere. For example, let $f(x)=(\max \{-x,0\}{)}^2$ and $g(x)=(\max \{x,0\}{)}^2$. Both have a continuous first derivative, and their values and first derivatives vanish at zero. On either side of zero one of the functions is identically zero, so $f{g}^{\prime }-f^{\prime }g=0$ everywhere. Evaluating a constant linear combination $Af+Bg=0$ at $-1$ and $1$ forces $A=B=0$.

The obstruction is that the two halves supply independent information without simultaneous nonzero value-and-derivative data at any point. For solutions of a common regular second-order linear differential equation, the uniqueness theorem for ordinary differential equations removes this possibility; Abel's identity gives the zero-or-nowhere-zero alternative for their Wronskian.

For $y\prime \prime +py\prime +qy=0$, the Wronskian satisfies $W\prime =-pW$.

More precisely, let $y_1,y_2$ be solutions of a differential equation for the same homogeneous linear differential equationon an interval $I$ where $p,q$ are continuous functions. For any $x_0\in I$, Abel's identity givesOnly the coefficient of $y^{\prime }$ enters the evolution of the Wronskian. In an unnormalized equation $A(x)y^{\prime \prime }+B(x)y^{\prime }+C(x)y=0$, first use $p=B/A$ on an interval where $A\ne 0$ and the coefficient ratios are continuous functions.^[ref]

The proof uses cancellation. Differentiating the Wronskian by the product rule and substituting the common equation givesMultiplying this first-order linear differential equation by the integrating factor $\mu (x)=\exp ({\int }_{x_0}^{x}p(s)ds)$ gives $(\mu W{)}^{\prime }=0$. Since $\mu (x_0)=1$, the constant is $W(x_0)$, proving Abel's identity. This argument also covers $W(x_0)=0$ without dividing by $W$.

The exponential function is positive for real $p$, so a real Wronskian either vanishes everywhere or has a fixed nonzero sign throughout $I$. In the preceding example, $p=-1$ and $W(0)=3$ give $W(x)=3e^x$ immediately. If $p=0$, every Wronskian of two solutions of a differential equation for that same equation is constant, regardless of $q$. Thus Abel's identity can certify linear independence and find the denominator used in variation of parameters without explicit formulas for both solutions of a differential equation.

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.

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.

Suppose $y^{\prime \prime }-P(x)y^{\prime }/x-Q(x)y/x=0$ has analytic $P,Q$ near zero, $P(x)=P_0+P_{1}x+\cdots$, and a solution $y_1=1+\beta x+\cdots$, with $P_0$ a positive integer. The indicial roots are $0,P_0+1$. The Abel identity gives $W=C{x}^{P_0}(1+P_{1}x+\cdots )$. Dividing by $y_1^2$ gives $C{x}^{P_0}[1+(P_1-2\beta )x+\cdots ]$. Integrating this reduction of order expression from zero, multiplying by $y_1$, and choosing $C=P_0+1$ proves the displayed normalized expansion. The integral's zero constant eliminates the added multiple of $y_1$. Its nonzero Wronskian proves independence. In this setting the assumed exponent-zero solution prevents a logarithm in the constructed exponent-$P_0+1$ solution.

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.

A Cauchy–Euler equation has powers of the independent variable matched to derivative order and is solved using power laws or a logarithmic change of variable.

For $x>0$, write $t=\log x$ and $y=xv(t)$. The chain rule gives $y^{\prime }=v+v_t$, $y^{\prime \prime }=x^{-1}(v_t+v_{tt})$, and thereforeOn $1<x<e^{\ell }$ with homogeneous Dirichlet boundary conditions, the eigenfunctions are $x\sin (n\pi \log x/\ell )$ and the eigenvalues are $-(n\pi /\ell {)}^2$. The weighted inner product with weight $x^{-3}$ becomes ${\int }_0^{\ell }\overline{v}wdt$, so their squared norms are $\ell /2$ and their orthogonality is exactly that of the Fourier sine series. The weight is the factor that makes this conjugacy preserve the inner product.

For a Cauchy-Euler equation, logarithmic coordinates turn scale changes in $x>0$ into translations in $t$. With $Y(t)=y(e^t)$, the chain rule gives $x{y}^{\prime }=Y^{\prime }$ and $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Consequently the Sturm-Liouville operator $-x^2{y}^{\prime \prime }-x{y}^{\prime }-cy$ becomes $-Y^{\prime \prime }-cY$. Its natural weighted inner product has measure $dx/x=dt$, so this change of variable also preserves the norm when the transformed problem uses ordinary integration in $t$. On $1<x<e^{\ell }$ with Dirichlet boundary conditions, the orthonormal eigenfunctions are $\sqrt{2/\ell }\sin (n\pi \log x/\ell )$ and the eigenvalues are $(n\pi /\ell {)}^2-c$, for $n\ge 1$.

For $-y^{\prime \prime }=\lambda x^{-2}y$ on $1<x<e^{\ell }$ with zero endpoint values, the natural weighted inner product has measure $dx/x^2$. Set $v=\ln x$ and $y(x)=\sqrt{x}Z(v)$. The chain rule gives $x^2{y}^{\prime \prime }=\sqrt{x}(Z^{\prime \prime }-Z/4)$, so $-Z^{\prime \prime }+Z/4=\lambda Z$ with zero endpoints on $(0,\ell )$. Moreover $|y{|}^{2}dx/x^2=|Z{|}^{2}dv$, so the transformation preserves the norm. The ordinary sine orthonormal eigenbasis therefore gives exactly the displayed spectrum and eigenfunctions for $n\ge 1$. The factor $\sqrt{x}$ and the additive $1/4$ distinguish this problem from the operator $-x^2{y}^{\prime \prime }-x{y}^{\prime }$ with measure $dx/x$.

For $\ell >0$, the logarithmic change of variable in a Sturm-Liouville problem sends this equation to $-Y^{\prime \prime }-\alpha Y=1$ on $(0,\ell )$. The normalized eigenfunctions are $\sqrt{2/\ell }\sin (n\pi t/\ell )$, with eigenvalues ${\lambda }_n=(n\pi /\ell {)}^2-\alpha$. If every ${\lambda }_n$ is nonzero, the Sturm-Liouville eigenfunction expansion givesIndeed the constant function has sine coefficients $2[1-(-1{)}^n]/(n\pi )$; division by each eigenvalue solves the equation in the operator domain. The series is uniformly convergent, with coefficients $O(n^{-3})$, while applying the differential operator is justified in the weighted square-integrable sense. At a zero eigenvalue the solvability condition at a Sturm-Liouville eigenvalue must instead be checked: an odd resonant index makes the constant forcing incompatible, whereas an even resonant index permits solutions with an arbitrary multiple of that zero mode.

Normalize the second-order equation as $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$. If every point other than zero and infinity is ordinary and both exceptions are at worst regular singular points, then $zP$ and $z^{2}Q$ extend holomorphically across zero and stay bounded at infinity. Liouville's theorem makes them constants, yielding the displayed Euler-Cauchy equation. If both exceptions must be genuinely singular, exclude $(A,B)=(0,0)$ and $(2,0)$, which make zero and infinity respectively ordinary.

Forthe power ansatz $w=z^{\lambda }$ gives the indicial equation $\lambda (\lambda -1)+a\lambda +b=0$. Distinct roots give $C_1{z}^{{\lambda }_1}+C_2{z}^{{\lambda }_2}$; a repeated root $\lambda$ gives $z^{\lambda }(C_1+C_2\log z)$ on a chosen logarithm branch.

A power-law ansatz assumes that an unknown function is a monomial $y=x^s$. Scale-invariant differential equations then reduce to an algebraic equation for the exponent $s$.

An indicial equation is the algebraic equation for the leading exponent of a power-law or Frobenius solution near a regular singular point.

An indicial root is a root of the indicial equation obtained from a power-law ansatz. For a Cauchy-Euler differential equation, such roots give the exponents of its power solutions. Their sign determines which modes remain bounded near a singular endpoint.

An indicial exponent is a root of the indicial equation at a regular singular point. It gives the leading local power in a Frobenius method; resonant roots can require logarithmic terms.

A log-periodic oscillation is periodic in the logarithm of its argument, for example $\cos (\mu \log x)$. Its successive zeros or extrema therefore occur at radii in a geometric progression.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has polynomial solutions for nonnegative integer $\ell$.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has the Legendre polynomial of degree $\ell$ as its normalized regular polynomial solution when $\ell$ is a nonnegative integer.

For the Legendre differential equation $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$, put $t=1+x$. The equation becomes $t(2-t)y_{tt}+2(1-t)y_t+n(n+1)y=0$. Its indicial equation is $2r^2=0$. The exponent-zero power series $\sum a_k{t}^k$ obeys the displayed linear recurrence relation. For a nonnegative integer $n$, it terminates at degree $n$, giving a Legendre polynomial with $a_0\ne 0$. Every analytic solution at the endpoint is proportional to this one, since the recurrence determines all coefficients from $a_0$.

The second independent solution has the local form $P_n(x)\log (1+x)+B(x)$, where $B$ is analytic near $-1$. Indeed, reduction of order gives $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$, whose integrand is $1/[2P_n(-1{)}^2(1+x)]$ plus an analytic function. Rescale the second solution to obtain the unit coefficient of $P_n\log (1+x)$. Reflection gives the corresponding expansion at $x=1$. This also proves the logarithmic divergence of the independent Legendre function of the second kind at each endpoint.

Substitute $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, into the Legendre differential equation. The lowest powers give $\sigma (\sigma -1)=0$ and $\sigma (\sigma +1)a_1=0$; subsequent powers give the recurrence in the title. Thus exponent zero has independent even and odd coefficients, while exponent one requires $a_1=0$ and represents the odd solution. Zero is an ordinary point of the equation, so the exponent-one choice simply selects a solution that vanishes there.