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A regular singular point with a full set of analytic local solutions is an apparent singularity of the differential equation. Singular normalized coefficients do not by themselves force singular solutions. For example, $x{y}^{\prime \prime }-y^{\prime }=0$ has normalized coefficient $-1/x$, but its fundamental solutions $1,x^2$ are both analytic through zero. Their Wronskian is $2x$, which vanishes at the singular point while being nonzero on each adjacent ordinary interval. The nonvanishing-Wronskian criterion for a fundamental set applies on intervals where the normalized equation has regular coefficients.

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.

For a real pair of indicial roots $r_1\ge r_2$ at a regular singular point of a second-order linear differential equation, the second local solution is classified by the gap $\Delta$ and by the recurrence relation for the smaller indicial root. Case I has $\Delta \notin \mathbb{Z}$ and two pure Frobenius solutions. Case II has $\Delta =0$ and an independent Logarithmic Frobenius solution. With a positive integer gap, Case IIIa means the exceptional coefficient equation is inconsistent, so a nonzero multiple of the first Frobenius solution times a logarithm is necessary. Case IIIb means that equation is consistent and leaves a free coefficient, so there are two pure Frobenius solutions. This IIIa/IIIb convention is the one used in the Oxford course notes.^[ref]

The free coefficient in Case IIIb adds a multiple of the first Frobenius solution, as explained by the undetermined coefficient at Frobenius resonance. Stating whether the logarithm occurs is useful alongside the case name because other courses may choose different labels.

In a fixed frame, a first-order linear differential equation is Fuchsian at a finite point when its coefficient matrix has at most a simple pole there. Its residue is $R$. The global object is the matrix-valued one-form $A(z)dz$: in the coordinate $t=1/z$ its coefficient is $-t^{-2}A(1/t)$. A Fuchsian system has intrinsic regular singular points, but an intrinsically regular singular system may have higher-order coefficient poles in a different meromorphic frame.

The Gauss hypergeometric equation isIts only possible singular points on the Riemann sphere are $0,1,\infty$, and they are regular singular points when singular. Their characteristic exponent pairs are $(0,1-c)$, $(0,c-a-b)$ and $(a,b)$ respectively, with exponents at infinity defined by behavior $z^{-\rho }$. The sum is one, agreeing with the Fuchs relation.^[ref]

When $c\notin \{0,-1,-2,\dots \}$, the solution analytic at zero with value one is the Gauss hypergeometric function ${}_2{F}_1(a,b;c;z)$. Its coefficient recurrence comes directly from matching powers. Resonant parameter values require compatibility checks or logarithmic companions; the normalized exponent-zero series cannot simply be asserted for every $c$.

For $a=0$ the constant function one is a solution. For a nonpositive-integer numerator parameter, the normalized series can terminate to a polynomial. The equation provides one common framework for many elementary and special functions.

The equation $x^2{y}^{\prime \prime }-2x{y}^{\prime }+(2-x^2)y=0$ has a regular singular point at zero. Factoring $y=xu$ cancels the Euler derivative terms and leaves $u^{\prime \prime }-u=0$ away from zero. Its analytic solutions $x\cosh x$ and $x\sinh x$ extend across zero. Their coefficient recurrence $(n-1)(n-2)c_n=c_{n-2}$ has independent free coefficients $c_1,c_2$, giving two entire solutions without a logarithmic branch despite integer-separated indicial roots.

An indicial equation determines candidate leading powers in a Frobenius solution near a regular singular point. Fornear $x=0$, suppose $p(x)=xP(x)$ and $q(x)=x^{2}Q(x)$ extend analytically to zero. Substituting $y=x^r(a_0+a_{1}x+\cdots )$ with $a_0\ne 0$ and collecting the lowest power givesIts roots are the indicial exponents. Subsequent powers determine the coefficient recurrence; the leading polynomial alone does not decide whether the second solution contains a logarithm.

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }+x{y}^{\prime }+{\mu }^{2}y=0$, $I(r)=r^2+{\mu }^2$. If $\mu >0$, its complex roots $\pm i\mu$ give real solutions $\cos (\mu \log x)$ and $\sin (\mu \log x)$ on $x>0$, exhibiting log-periodic oscillation.^[ref]

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.

An indicial root is a root of the indicial equation for a regular singular point. For a leading term $x^r$ as $x\downarrow 0$, its magnitude is $x^{\mathrm{Re}r}$, so positive real part gives decay and negative real part gives divergence. At zero real part, logarithmic factors and oscillation must also be checked.

For $x^2{y}^{\prime \prime }+x{y}^{\prime }=0$, the repeated root is zero. Its independent solutions $1$ and $\log x$ have the same indicial root but different boundedness. For roots $r=\pm i\mu$, nonzero real solutions oscillate in $\log x$ with bounded magnitude and usually no limit at zero. Thus the sign of a real root is useful away from the borderline, but is not a complete boundedness rule for all Frobenius solutions.

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.

For $x{y}^{\prime \prime }+a(x)y=0$ with analytic coefficient and a continuous solution having $a(0)y(0)\ne 0$, the equation gives $y^{\prime \prime }\sim -a(0)y(0)/x$. Integrating towards the endpoint gives the displayed logarithmic divergence of the derivative. A bounded solution value therefore does not imply a finite derivative at a regular singular point.

Divide by $2x$ away from zero to obtain $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda /2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^{2}Q(x)=\lambda x^2/2$ are analytic. The pole in $P$ makes zero singular, so $x=0$ is a regular singular point. Its indicial equation isAt $x=1$, both normalized coefficients are analytic functions, so $x=1$ is an ordinary point. These are the standard normalized-coefficient and indicial tests.^[ref]

For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The chain rule gives $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, so the transformed normalized equation isThe product of the first-derivative coefficient with $t$ still has a second-order pole at zero. Hence the criterion for a regular singular point fails and $x=\infty$ is an irregular singular point, including when $\lambda =0$. The primes on $Y$ here refer to differentiation with respect to $t$.

Move the expansion point to the origin with $z=x+1$. The transformed ordinary differential equation isAfter division by $z$, the functions $z(1-1/z)=z-1$ and $z^2(c/z)=cz$ are analytic functions at zero, so this is a regular singular point. For a Frobenius solution with leading term $a_0{z}^{\rho }$, $a_0\ne 0$, the lowest power is $z^{\rho -1}$ and its coefficient isTherefore the indicial equation and its indicial exponents areThe exponent $\rho$ denotes the quantity called $\alpha$ in the paper, and avoids confusion with other uses of that letter.

Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.

Normalize the Laguerre differential equation as $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, where $P(x)=1/x-1$ and $Q(x)=n/x$. At $x=0$, the products $xP=1-x$ and $x^{2}Q=nx$ are analytic. Thus zero is a regular singular point, with indicial equationAt $x=1$ the coefficients $P,Q$ are analytic, so it is an ordinary point. No singular-point indicial equation is needed there.

For infinity put $z=1/x$ and $Y(z)=y(1/z)$. Then the equation becomesAt $z=0$, the coefficient $z(z^{-1}+z^{-2})=1+1/z$ is not analytic. Therefore infinity is an irregular singular point, including when $n=0$. A terminating polynomial solution does not change the classification of the differential equation.

After division by $x$, the ordinary differential equation has coefficients $P(x)=(m+1)/x-1$ and $Q(x)=n/x$. At zero, $xP\to m+1$ and $x^{2}Q\to 0$, making zero a regular singular point. The indicial equation isso the indicial exponents are $\boxed{0,-m}$, coincident when $m=0$.

At infinity put $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isThe coefficient of $Y^{\prime }$ already fails the regular-singularity test at $t=0$. Thus $\boxed{x=\infty \text{ is an irregular singular point}}$, including when $n=0$.

Treat $F$ as a vector-valued solution; matrix solutions are handled columnwise. It is important to distinguish the coefficient convention from the indicial exponent. In a fixed Fuchsian frame, a finite singularity at $a$ meanswhere $B_a$ is holomorphic near $a$. The residue is $R_a$, equivalently the residue of the matrix-valued one-form $A(z)dz$. Such a Fuchsian linear differential system has a regular singular point: on sectors, its solutions have at most power and logarithmic growth. Intrinsically a regular singular system can be made Fuchsian by a meromorphic change of frame; a higher coefficient pole in some other frame does not itself prove irregularity. Residues here refer to the displayed fixed coefficient one-form.

At infinity use $w=1/z$ and $\tilde{F}(w)=F(1/w)$. The equation becomesA Fuchsian singularity at infinity means $\tilde{A}$ has at most a simple pole at $w=0$, and $R_{\infty }={\mathrm{Res}}_{w=0}\tilde{A}(w)$. The residue theorem applied entrywise on the Riemann sphere givesFor a direct calculation when all coefficient poles are simple, write $A(z)=P(z)+{\sum }_j{R}_j/(z-a_j)$ by rational partial fractions. Regularity at infinity in this frame requires $A(z)=O(1/z)$, so $P=0$. Transforming coordinates gives $R_{\infty }=-{\sum }_j{R}_j$. More generally the entrywise residue theorem remains valid even when a different meromorphic frame has higher-order poles.

The printed nonconstant-solution claim is false. A counterexample satisfying its spectral hypothesis isIts residue has eigenvalues $0,-1/3$, with zero of largest real part. The equations are $u^{\prime }=0$ and $v^{\prime }-v/(3z)=0$. If $v$ is analytic at zero, coefficient comparison gives $(n-1/3)v_n=0$ for every nonnegative integer $n$, so $v=0$. Hence every analytic solution is $(c,0{)}^T$, which is constant. This disproves nonconstancy, rather than silently replacing it with a weaker conclusion.

A useful correct conclusion is that a nonzero analytic solution exists under the stated hypotheses. Write $A=R/z+{\sum }_{j\ge 0}{B}_j{z}^j$, and let $m$ be the largest nonnegative integer with $-m\in \mathrm{spec}R$; such an $m$ exists because $0\in \mathrm{spec}R$. Choose a nonzero $h_0\in \ker (mI+R)$ and seekThe Frobenius method givesAll left-hand matrices are invertible by maximality of $m$. For large $n$ their inverse norm is $O(1/n)$ by a geometric-series inverse; enlarging the constant bounds every $n\ge 1$ by $C/n$. If $\Vert B_j\Vert \le M{r}^{-j}$ on a smaller disc, the scalar majorant with generating functionsatisfies $v^{\prime }=CM(1-t/r{)}^{-1}v$ and dominates the recurrence coefficientwise. It converges for $|t|<r$, proving convergence of $H$. Thus the analytic solution at a maximal integer Fuchsian exponent is nonzero, and is nonconstant if $m>0$. When $m=0$, nonconstancy requires additional information, as the counterexample shows.

There is also a sign problem in the printed exponent terminology. Substitution of $F=z^{\mu }H$ with $H(0)\ne 0$ into the actual equation givesThus the indicial exponents are eigenvalues of $-R$, not $R$. With the printed sign, an exponent of maximal real part admits a pure Frobenius solution, because the corresponding coefficient matrices are invertible at all subsequent positive integer steps. Lesser exponents can meet resonance.

The supplied example does not demonstrate such failure. Solving it directly givesEvery solution is analytic. The residue eigenvalue $-1$ corresponds to the true exponent $1$, and setting $c_1=0$ gives $F=z(-c_{2}z/2,c_2{)}^T$ with nonzero leading coefficient. Even with the literal printed factor $z^{-1}$, every displayed solution has the form $z^{-1}H$ with $H=zF$ analytic, though $H(0)=0$.

Nevertheless the last existence-of-a-counterexample request, read literally, can be fulfilled by the diagonal counterexample above. For its nonmaximal residue eigenvalue $\lambda =-1/3$, substitution of $F=z^{-1/3}H$ yieldsfor every integer $n\ge 0$. Neither diagonal entry, $n-1/3$ or $n-2/3$, vanishes, so $H=0$. Therefore there is no nonzero solution of that literal form, even without imposing $H(0)\ne 0$.

For the intended logarithmic resonance phenomenon with correctly signed exponents, take insteadIts equation gives $v=c_2/z$ and $u=c_1+c_2\mathrm{Log}z$ on a chosen branch. The exponent $-1$ would require $H(0)=(0,c_2{)}^T\ne 0$, but $z\mathrm{Log}z$ prevents $H=zF$ from being analytic when $c_2\ne 0$. The exponent zero has the constant solution $(1,0{)}^T$. This supplies an explicit resonance obstruction under the correct convention and completes the qualified resolution of the printed defects.

At zero an ordinary point requires $p,q$ to be analytic. A regular singular point requires $zp$ and $z^{2}q$ to extend analytically to zero, with at least one of $p,q$ failing to be analytic if “singular” excludes ordinary points.

Put $\zeta =1/z$ and $v(\zeta )=w(1/\zeta )$. Differentiation gives $w^{\prime }=-{\zeta }^2{v}^{\prime }$ and $w^{\prime \prime }={\zeta }^4{v}^{\prime \prime }+2{\zeta }^3{v}^{\prime }$, so the transformed equation isThe point at infinity is ordinary precisely when both transformed coefficients are analytic at $\zeta =0$. Solving these conditions for the original coefficients giveswith $P,Q$ analytic near zero; the sign can be absorbed into the arbitrary analytic $P$.

If the only singular point on the sphere is the regular singularity at zero, then $zp(z)$ and $z^{2}q(z)$ are entire. The condition at infinity makes the first tend to $2$ and the second to $0$. By Liouville theorem they are respectively the constants $2$ and $0$. Thus the unique equation iswith general solution $w=A+B/z$. Zero is genuinely regular singular and infinity is ordinary. If “no other singular points” were restricted to finite points only, the general coefficients would instead be $p=P(z)/z$ and $q=Q(z)/z^2$ with entire $P,Q$; the preceding question about infinity indicates the spherical interpretation.

In the normalized second-order linear differential equation, an ordinary point $z_0$ is one at which both $p$ and $q$ are holomorphic functions in a neighbourhood of $z_0$. A singular point of a differential equation is a point that is not ordinary. A singular point $z_0$ is a regular singular point if $(z-z_0)p(z)$ and $(z-z_0{)}^{2}q(z)$ extend analytically to $z_0$. Equivalently, $p$ has at most a simple pole and $q$ at most a double pole there. If a singular point fails this test, it is an irregular singular point. The normalized form matters: zeros of a leading coefficient must first be assessed after division by that coefficient.

Continuation around one positively oriented loop induces an invertible linear map $M$ on the two-dimensional solution space. Its inverse is continuation around the reversed loop. Over $\mathbb{C}$, choose an eigenvector with eigenvalue $\lambda \ne 0$, and let $w$ be the corresponding nonzero solution. After continuation it becomes $\lambda w$.

Choose $\sigma$ with $e^{2\pi i\sigma }=\lambda$. The product $h(z)=z^{-\sigma }w(z)$ has trivial monodromy, so it is single-valued and holomorphic on the punctured disc. Its Laurent series, convergent on compact subannuli, gives the Monodromy eigenfunction Laurent representationNo assumption of a regular singular point has been made, so the negative part need not terminate.