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Suppose the matrix $B$ is holomorphic at zero and $0$ is an eigenvalue of $R$. Let $m$ be the largest nonnegative integer for which $-m$ is an eigenvalue of $R$. Seek $F=z^m{\sum }_{n\ge 0}{h}_n{z}^n$, with $h_0\in \ker (mI+R)$ nonzero. The Frobenius method gives $((m+n)I+R)h_n=-{\sum }_{j=0}^{n-1}{B}_j{h}_{n-1-j}$ for $n\ge 1$, where scalar terms multiply the identity. Every left-hand matrix is invertible by the choice of $m$. Its inverse is $O(1/n)$, so a scalar convergent majorant proves convergence of the series. This yields a nonzero analytic solution; it is necessarily nonconstant when $m>0$, but can be constant when $m=0$. For the displayed sign convention, indicial exponents are eigenvalues of $-R$.

For $x>0$, the Bessel differential equation of order $1/2$ has independent solutions $x^{-1/2}\sin x$ and $x^{-1/2}\cos x$. Substituting $y=x^{-1/2}u$ reduces it to $u^{\prime \prime }+u=0$. The normalization of the Bessel function of the first kind gives the displayed formulas for orders $1/2$ and $-1/2$. The Frobenius method obtains the same pair from exponents $1/2$ and $-1/2$: their even-coefficient series reduce to the sine and cosine power series. In the negative-exponent recurrence, the free coefficient at the next exponent can add the positive-exponent solution, an example of an undetermined coefficient at Frobenius resonance.

For an ellipse with semiaxes $1,x$, the perimeter integral has integrand $\sqrt{1-t{\cos }^2\theta }$. The uniformly convergent binomial series near $t=0$ permits termwise integration. Integrals of ${\cos }^2\theta$ and ${\cos }^4\theta$ over a full period give the displayed first corrections. The same coefficients follow from the Frobenius method for the transformed perimeter equation.

A mathematician associated with the Frobenius method and representation theory.

For $x^2{y}^{\prime \prime }+2x{y}^{\prime }+[x^2-\ell (\ell +1)]y=0$ with integer $\ell \ge 0$, the Frobenius method gives $\sigma =\ell$ or $-\ell -1$. The general coefficient equation is $[(j+\sigma )(j+\sigma +1)-\ell (\ell +1)]a_j=-a_{j-2}$, taking negative-index coefficients as zero. The regular branch has $a_1=0$ and $a_j=-a_{j-2}/[j(j+2\ell +1)]$. The singular branch has $a_j=-a_{j-2}/[j(j-2\ell -1)]$ except at $j=2\ell +1$. That resonant equation is compatible because its earlier odd coefficient vanishes; the newly free coefficient adds a multiple of the regular solution. Setting it to zero gives a pure even series times $x^{-\ell -1}$, independent of the regular branch, with no logarithmic term. For $\ell =0$ the resonance occurs already at $j=1$.

The normalization $a_0=2^{\ell }\ell !/(2\ell +1)!$ on the regular branch gives the Spherical Bessel functionIn particular $j_0(x)=\sin x/x$, with removable value one at zero.

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.

For $J(\epsilon )={\int }_0^{\infty }{e}^{-x-\epsilon /x}dx/x$, reciprocity $x\mapsto \epsilon /x$ and the small-argument expansion of the exponential integral give $J=-\log \epsilon -2\gamma +o(1)$. Differentiation and integration by parts give $\epsilon J^{\prime \prime }+J^{\prime }=J$. Its Frobenius method yieldsThe logarithm is a singular contribution that a naive termwise expansion at $x=0$ misses.

Seek a Frobenius series $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$, on the positive side of zero. The lowest coefficient gives the indicial equation from part (a); the next gives $(r+1)(2r-1)a_1=0$. For either permitted indicial exponent, $a_1=0$. Matching the subsequent powers givesThe two indicial exponents differ by a noninteger, so there are two independent Frobenius solutions without a logarithmic companion. Normalizing each leading coefficient to one, their forms areThus the general local solution is $C_0{f}_0+C_1{f}_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne 0$, so $y(x)\sim C_1{x}^{3/2}$ and $\alpha =3/2$, with the arbitrary nonzero amplitude understood.

The condition at one selects values of $\lambda$ for which this branch also vanishes there; it does not change the selected local exponent. The zero solution, available for every $\lambda$, has no nonzero leading coefficient and is excluded when asking for a leading exponent. This is a direct application of the Frobenius method and the noninteger-separated-root case.^[ref]

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.

Multiply the growth equation by $2{\eta }^2(1+\eta )$:Use the Frobenius method, writing $D={\sum }_{n\ge 0}{b}_n{\eta }^{\alpha +n}$ with $b_0\ne 0$. The lowest power gives the indicial equationThus the leading powers are $\alpha =1$ and $\alpha =-3/2$. For $n\ge 1$, equating the coefficient of ${\eta }^{\alpha +n}$ givesFor the growing branch $\alpha =1$, $b_n=-2(n+1)b_{n-1}/(2n+5)$, so $b_1=-4b_0/7$ and $b_2=8b_0/21$. Taking $b_0=A_k$,Since $\eta \propto a$, the leading behavior is precisely the usual growing density contrast ${\delta }_C\propto a\propto t^{2/3}$ during matter domination. The negative first correction shows the beginning of suppressed growth as the string fraction increases. The other branch starts as $a^{-3/2}$, the familiar decaying matter mode; its exact form is proportional to $\sqrt{1+\eta }/{\eta }^{3/2}$.

For $z\ne 0$, division by $4z$ gives $p(z)=1/(2z)$ and $q(z)=1/(4z)$. Thus zero is singular, but $zp=1/2$ and $z^{2}q=z/4$ are analytic. Zero is a regular singular point.

Apply the Frobenius method with $y=z^r{\sum }_{n=0}^{\infty }{a}_n{z}^n$, $a_0\ne 0$. The lowest power $z^{r-1}$ has coefficient $[4r(r-1)+2r]a_0$, so the indicial equation is $2r(2r-1)=0$. Its characteristic exponents at a regular singular point are $0$ and $1/2$. At the next powers,With $a_0=1$, the two Frobenius solutions are thereforeBoth analytic power-series factors converge everywhere by the ratio test. Recognizing the trigonometric series givesOn a punctured neighbourhood choose a consistent square-root branch for the second solution; the first is entire in $z$. These solutions are independent, since their leading orders are $1$ and $z^{1/2}$. A direct square-root reduction of a regular-singular differential equation also verifies the closed forms: for $s=\sqrt{z}$ and $Y(s)=y(s^2)$, $4z{y}^{\prime \prime }=Y_{ss}-Y_s/s$ and $2y^{\prime }=Y_s/s$, so the equation reduces to $Y_{ss}+Y=0$.

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.

Multiply the equation by $x$ to obtain $x{y}^{\prime \prime }-P(x)y^{\prime }-Q(x)y=0$. The leading terms after a Frobenius method trial $y=x^r$ arethe term involving $Q_0$ first appears at order $x^r$. The indicial equation is therefore

To obtain the first correction to a regular singular solution by reduction of order, start from the assumed solution $y_1=1+\beta x+O(x^2)$ and use the Abel identity:We work initially on $x>0$ near zero; since $P_0$ is a positive integer, the resulting expansion is an ordinary power series. Also $y_1^{-2}=1-2\beta x+O(x^2)$, so the reduction of order integral has integrandChoose its integration constant to be zero, eliminating an added multiple of $y_1$. Integrating and multiplying by $y_1$ givesWith $C=P_0+1$ the requested normalized answer isIts nonzero Wronskian makes it independent of $y_1$. The existence of the asserted $y_1$ also forces $P_0\beta +Q_0=0$; there is no need to introduce $Q_0$ into the normalized second coefficient.

With $t=1-x^2$ the chain rule gives $y^{\prime }=-2x{u}^{\prime }$ and $y^{\prime \prime }=-2u^{\prime }+4x^2{u}^{\prime \prime }$. Substitution, initially for $x>0$, givesThe transformed coefficients are $P(t)=1/t$ and $Q(t)=1/[4t(1-t)]$. At $t=0$, both $tP$ and $t^{2}Q$ are analytic; at $t=1$, both $(t-1)P$ and $(t-1{)}^{2}Q$ are analytic. Each point is singular because a normalized coefficient has a pole, but satisfies the regular singular point criterion for a second-order equation. Thus both points are regular singular points.

For a Frobenius method substitution $u=t^{\rho }{\sum }_{n\ge 0}{a}_n{t}^n$, the lowest power gives the indicial equationThe repeated indicial root is zero. Setting $\rho =0$ and equating the coefficient of $t^n$ givesConsequently the analytic power-series solution of a differential equation isTaking $a_0=2\pi$ reproduces the ellipse perimeter expansion.

For a second independent solution, normalize $u_1(0)=1$. Reduction of order and $P=1/t$ giveSince $u_1(s{)}^{-2}=1+O(s)$, this has the Logarithmic Frobenius solution formThe logarithm is $\log |t|$ on the real negative side, or a chosen complex logarithm on a slit complex neighborhood. Its nonzero logarithmic coefficient establishes independence. The physical perimeter is finite at $t=0$, so it selects the analytic branch rather than this logarithmic one.

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.

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.

Here the indicial equation has the repeated root $0$. The first Frobenius method solution is $y_1=e^{-x}$. Since $p=1+1/x$, the Abel identity gives $W=C{e}^{-x}/x$. Reduction of order with $C=1$ produceswhere the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent real analytic solution does not exist.

The general local form isThis is the repeated-root Logarithmic Frobenius solution. On the negative real side one may use $\log |x|$, or use a fixed logarithm branch for a complex solution.

The expansion is singular near the moving lower limit, so expanding $e^{-\epsilon /x}$ throughout the original integral would lose its constant term. Introduce the symmetric exponential integralUnder $x=\epsilon /t$, the discarded piece becomesThis is a uniform Taylor expansion because $t\ge 1$. To derive the logarithmic expansion of a symmetric exponential integral without needing another special function, first exploit the same substitution to writeThe estimate follows from $0\le 1-e^{-\epsilon /x}\le \epsilon /x$. Here $\gamma$ is Euler's constant and the small-argument expansion of the exponential integral fixes the constant as well as the logarithm.

Differentiate under the integral and integrate the derivative of $e^{-x-\epsilon /x}/x$ over $(0,\infty )$; the exponentially small endpoint terms vanish. This gives the ordinary differential equationFor the Frobenius method at its regular singular point, write $J=A\log \epsilon +B+\epsilon (C_1\log \epsilon +D)+\cdots$. Substitution gives $C_1=A$ and $D=B-2A$. The preceding leading calculation sets $A=-1$, $B=-2\gamma$, henceSubtracting the moving-cutoff correction to a symmetric exponential integral yields the requested answer, including the logarithmically enhanced first-order term:Thus the $\epsilon \log \epsilon$ switchback term must be retained when asking for all contributions through order $\epsilon$.

After division by $x^2$, the equation is $y^{\prime \prime }+x^{-1}{y}^{\prime }-(1+{\alpha }^2{x}^{-2})y=0$. Every $x\ne 0$ is an ordinary point. The origin is singular, but it is a regular singular point because $xP(x)=1$ and $x^{2}Q(x)=-(x^2+{\alpha }^2)$ are analytic there. An ordinary point has analytic normalized coefficients $P,Q$; a singular point failing the displayed regularity test is irregular.

The Frobenius method ansatz $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$ gives the indicial equation $r^2-{\alpha }^2=0$, $a_1=0$, andFor nonintegral $\alpha$, two independent solutions are

For integral $\alpha$, put $m=|\alpha |$. In the $r=-m$ recurrence the denominator vanishes at $n=2m$, so that series fails or coincides in the exceptional $m=0$ case. Up to scale the single Frobenius series isFor $\alpha =1$, $y_1=x(1+x^2/8+\cdots )$, soIts integral contains both $s^{-2}$ and $\log s$; multiplying by $y_1$ leaves a pole and a logarithmic term. Thus the reduction-of-order solution is not a power series at zero.

Fora finite point $z_0$ is a regular singular point when $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.

For the displayed Bessel differential equation,At $z=0$, the functions $zP(z)=1$ and $z^{2}Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomesHere $w^{2}Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an irregular singular point. Thus

Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:Hence $f^{\prime \prime }+f=0$, and on any domain carrying a branch of $\sqrt{z}$ two linearly independent solutions areNear the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two Frobenius exponents. At infinity their combinations are $z^{-1/2}{e}^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.