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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.

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 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 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.

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.

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.

For the second-order Fuchsian equation represented by the Papperitz symbol, the coefficient of $w^{\prime }$ isThis follows from the sums of local exponents at its two finite regular singular points. Abel's Wronskian identity, obtained by differentiating $w_1{w}_2^{\prime }-w_1^{\prime }{w}_2$ and using the differential equation, gives $W^{\prime }=-pW$. Thus, on every branch domain avoiding zero and one,The normalized expression is locally constant. Its analytic continuation around either singularity is unchanged: the power factors cancel the determinant of the local monodromy, equivalently the integral of $-p$ around that singularity. It therefore defines a single-valued holomorphic function on the punctured plane, locally constant near both punctures. Those punctures are removable, so the normalized expression extends to an entire function. This argument also covers resonant exponents and avoids assuming two pure power solutions in that case.

Write a power series $y={\sum }_{n\ge 0}{c}_n{x}^n$. Substitution givesThe lowest coefficients give $c_1=0$, $c_3=0$, with $c_0,c_2$ free. For $n\ge 4$, matching the coefficient of $x^{n-1}$ gives the recurrenceAll odd coefficients consequently vanish. For the first solution, $c_0=a$, $c_2=y_1^{\prime \prime }(0)/2=0$, so the first three nonzero terms, when $a\ne 0$, areFor the second, $c_0=0$, $c_2=b/2$, givingIf its prescribed constant is zero, the corresponding solution is identically zero and has no nonzero terms.

For the change of variable, put $y(x)=Y(u)$, $u=x^{\alpha }$, with $\alpha \ne 0$. The chain rule givesAfter division by $x^{2\alpha -1}$, the coefficient of $Y_u$ is $\alpha (\alpha -2)x^{-\alpha }$ and that of $Y$ is $4x^{4-2\alpha }$. A genuine constant-coefficient equation requires $\alpha =2$, which both removes $Y_u$ and makes the final coefficient constant. The transformed equation is $4Y_{uu}+4Y=0$.

This quadratic-variable reduction of a singular oscillator yieldsIt holds on each interval away from zero and both solutions extend smoothly through zero. Here $y(0)=A$ and $y^{\prime \prime }(0)=2B$, so the specified solutions are exactly $a\cos (x^2)$ and $(b/2)\sin (x^2)$. Their Taylor series agree with the recurrences above. Although the original equation has a regular singular point at zero, these two linearly independent smooth solutions supply the required data there.

For a first-order system, it is important to specify the frame in which its coefficients are written. In the simple-pole convention intended here, a Fuchsian linear differential system has, at a finite singular point $a$,The residue is the matrix $R_a={\lim }_{z\to a}(z-a)A(z)$, computed entry by entry. Such a point is a regular singular point: on a sector, the coefficient bound $\Vert A(z)\Vert \le C/|z-a|$ and integration along rays give at most power growth of solutions. Intrinsically, a regular singular point means moderate power growth of all solutions on sectors, allowing logarithmic factors. This intrinsic definition is more general than requiring a simple coefficient pole in an arbitrary meromorphic frame.

For $A(z)=R/z$, a fundamental matrix on a chosen complex logarithm branch isDifferentiating the matrix exponential verifies ${\Phi }^{\prime }=R\Phi /z$. A counterclockwise circuit replaces $\log z$ by $\log z+2\pi i$, soThis is the residue and monodromy of a constant Fuchsian system. A change of solution basis conjugates $M$. Its eigenvalues are $e^{2\pi i\lambda }$ for the residue eigenvalues $\lambda$; a nontrivial Jordan block supplies the corresponding logarithmic terms in $z^R$. The monodromy does not uniquely recover $R$, since integer shifts of residue eigenvalues can leave their exponentials unchanged.

The coordinate-independent coefficient is the matrix-valued one-form $A(z)dz$. With $t=1/z$, its coefficient isBeing Fuchsian at infinity means $A(z)=O(z^{-1})$, while being ordinary there means $A(z)=O(z^{-2})$. If the finite simple poles are $a_1,\dots ,a_s$, subtract their principal parts:Each entry of $H$ is entire and tends to zero at infinity. Liouville's theorem therefore gives $H=0$. Expanding this partial fraction decomposition at infinity yieldsThus the sum of all residues, including the residue at infinity of the one-form, is zero. If infinity is ordinary, $R_{\infty }=0$ and the finite residues alone sum to zero. Treating $A$ merely as a scalar function of $z$, without transforming $dz$, would give the wrong infinity convention.

Put $\omega =e^{2\pi i/3}$. The possible finite singular points of $(B+zC)/(1-z^3)$ are $1,\omega ,{\omega }^2$, andA point $\zeta$ is removable exactly when $B+\zeta C=0$ as a matrix. At infinity,which is holomorphic at $t=0$. Consequently the actual set is $S=\{\zeta :{\zeta }^3=1,B+\zeta C\ne 0\}$, with $\{1,\omega ,{\omega }^2\}$ the generic set. The finite residues sum to zero, as follows also from ${\sum }_{{\zeta }^3=1}{\zeta }^{-1}={\sum }_{{\zeta }^3=1}{\zeta }^{-2}=0$.

Conversely, suppose a Fuchsian linear differential system has its finite singular points among these three roots and has no singularity at infinity. Then $Q(z)=(1-z^3)A(z)$ extends to an entire matrix function, and ordinariness at infinity gives $Q(z)=O(z)$. By the Cauchy estimate, each entry has zero derivatives of order two or higher, so $Q(z)=B+zC$. The same argument applies to any subset $S$, with additional cancellations at the omitted roots. This proves the requested normal form.

There is a genuine convention limitation: the normal form is not true under intrinsic regular singularity alone in an unrestricted frame. For example, if $N\ne 0$ and $N^2=0$, then $A(z)=N/(z-1{)}^2$ has fundamental matrix $I-N/(z-1)$. Its solutions have only power growth, so $1$ is intrinsically a regular singular point, and infinity is ordinary. Nevertheless its coefficient has a double pole and cannot equal $(B+zC)/(1-z^3)$. The simple-pole convention makes the preceding residue and normal-form assertions correct.

The chain rule gives $z_x=\sin 2x$, $z_x^2=4z(1-z)$, and $z_{xx}=2(1-2z)$. Thereforewhich, after division, is the printed equation. It has only three regular singular points, $0,1,\infty$, and is thus a Papperitz equation. At zero and one the indicial roots are $0,1/2$. At infinity, inserting $w\sim z^{-\rho }$ gives $\rho =\pm n/2$. Its Papperitz symbol isThe exponent sum is one, as required for a three-singularity second-order equation.

Multiplying the equation by $z(1-z)$ puts it in Gauss hypergeometric equation form with $a=n/2$, $b=-n/2$, $c=1/2$. The solution analytic at zero and normalized to one is ${}_2{F}_1(a,b;c;z)$. Composing it with $z={\sin }^{2}x$ gives an even analytic solution of $w_{xx}+n^{2}w=0$ with $w(0)=1$, $w_x(0)=0$. Uniqueness of this ordinary differential equation impliesThis Gauss hypergeometric function identity initially holds near zero, and on $|x|<\pi /2$ with the usual real branch; it extends by matching analytic continuation. It is not an identity on every real $x$ if the left side is always reevaluated on its principal $z$ branch: for $n=1$ that principal value is $\sqrt{1-{\sin }^{2}x}=|\cos x|$. Continuing the branch consistently restores the stated cosine.

The origin is a regular singular point of this Kummer differential equation. Apply the Frobenius method, writing $y={\sum }_{n=0}^{\infty }{a}_n{x}^{n+r}$ with $a_0\ne 0$. The lowest power gives the indicial equationFor $n\ge 1$, comparison of the coefficient of $x^{n+r-1}$ givesSince $0<c<1$, both roots give valid independent branches. Normalize $a_0=1$:Here $(c{)}_n$ is the rising factorial, and the first series is the confluent hypergeometric function of the first kind $M(1,c,x)$. The two leading powers are distinct, proving linear independence. We describe real solutions on $x>0$, with their one-sided behavior at zero.

For the inhomogeneous equation a particular solution is $-(x+c)/2$, verified by direct substitution. Thus $y=A{y}_1+B{y}_2-(x+c)/2$. Since $y_2^{\prime }\sim (1-c)x^{-c}$, a finite derivative at zero forces $B=0$. Then $y(0)=0$ forces $A=c/2$, and the required solution isIn particular $y^{\prime }(0)=0$. The singular homogeneous branch cannot be added without violating the derivative condition.

For the normalized second-order equation, an ordinary point is one where both coefficient functions $p,q$ are analytic. A singular point of a second-order linear ODE is a point that is not ordinary. It is a regular singular point if $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; a singular point failing that condition is irregular.

Here $p=0$, $q=1/x$. Thus zero is singular but regular, because $xp=0$ and $x^{2}q=x$ are analytic. Write the analytic solution as $y={\sum }_{n\ge 0}{a}_n{x}^n$. Its constant equation gives $a_0=0$, and for $n\ge 1$,Using $a_1=1$ givesThe factorial denominators show convergence for every finite $x$.

For the other branch, integrating the first correction equation twice and imposing its two conditions gives $y_1^{\prime }=-\log x$ andThen $y_2^{\prime \prime }=\log x-1$, so $y_2^{\prime }=x\log x-2x+2$ andEach correction tends to zero at the origin and its derivative vanishes at one. Although the second solution has a finite limiting value one, it has a logarithmically divergent endpoint derivative. Indeed the differential equation gives $y^{\prime \prime }\sim -1/x$, and integration givesA finite initial derivative cannot be prescribed for this second branch. The logarithm is consistent with the integer separation of the two exponents in the Frobenius method; this is not the repeated-exponent case.

The Papperitz symbol describes a homogeneous second-order Fuchsian differential equation on the Riemann sphere, with three regular singular points and the two indicial exponents at each. At a finite point $z_j$, an exponent $r$ corresponds locally to a factor $(z-z_j{)}^r$; at infinity it corresponds to $z^{-r}$. Repeated or resonant exponents can require logarithms. The six exponents obey the Fuchs relation that their sum is one. With three singularities there is no independent accessory parameter, so these data determine the normalized equation.

A Möbius transformation of a Papperitz symbol moves the singular points without changing their exponents. Multiplication of the dependent variable by a local power shifts the two exponents by that power, using the reciprocal local coordinate at infinity.

Set $w=z/(z-1)$ and $v(w)=F(a,c-b,c;w)$. At $w=0,1,\infty$, the exponent pairs for $v$ are respectively $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. The Möbius transformation sends $z=0,1,\infty$ to $w=0,\infty ,1$. Before rescaling, $v(w(z))$ therefore has exponent pairs $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(1-z{)}^{-a}$ subtracts $a$ at one and adds $a$ at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with $(1-z{)}^{-a}=1$ at zero. The resulting solution is analytic there and has value one. Uniqueness of the normalized analytic solution yieldsThis is the Pfaff transformation. The normalization presupposes the usual admissible parameters, in particular $c\notin \{0,-1,-2,\dots \}$. Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.

Classification and characteristic exponents. For a normalized second-order linear differential equation, an ordinary point at zero means $p$ and $q$ are real analytic functions there. A regular singular point is a singular point where $xp$ and $x^{2}q$ nevertheless extend as real analytic functions to zero. Write $p_0={\lim }_{x\to 0}xp(x)$ and $q_0={\lim }_{x\to 0}{x}^{2}q(x)$. The lowest-power coefficient in the Frobenius method gives the indicial equationIts roots are the characteristic exponents at a regular singular point. If their difference is not an integer, there are two linearly independent solutions $x^{{\sigma }_j}{A}_j(x)$, with $A_j$ real analytic and $A_j(0)\ne 0$. For distinct real roots ${\sigma }_1>{\sigma }_2$ with integer difference, a solution for the larger root exists, and a second has the form $x^{{\sigma }_2}B(x)+c{y}_1\log x$; the logarithmic coefficient can vanish. For a repeated root, the second solution has the form $y_1\log x+x^{{\sigma }_1}B(x)$ after normalization. This is a Logarithmic Frobenius solution. Complex conjugate roots $a\pm ib$ give real solutions whose leading behaviour is $x^a\cos (b\log x)$ and $x^a\sin (b\log x)$, with analytic-series corrections. All these descriptions are on $x>0$, or on a fixed complex branch.

For the present equation, division by $x$ gives $p=1+(1-m)/x$ and $q=(1-m)/x$. Therefore $xp=x+1-m$ and $x^{2}q=(1-m)x$ are real analytic. Zero is regular singular when $m\ne 1$; when $m=1$, it is ordinary after removing the common factor $x$. In that exceptional case the normalized equation is $y^{\prime \prime }+y^{\prime }=0$.

For the Frobenius method substitution, the lowest coefficient and the general linear recurrence relation areThese equations come respectively from the coefficients of $x^{\sigma -1}$ and $x^{n+\sigma -1}$. At $m=1$, the same formal exponents $0,1$ describe the two possible leading powers at an ordinary point, rather than singular behaviour.