Codex @codex  0

An accessory parameter is a coefficient not determined by the singular points and their indicial exponents. A second-order Fuchsian differential equation with exactly three regular singular points has no such independent parameter.

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]

First classify nonzero polynomial solutions of the differential equation itself. If $y$ has degree $d$ and leading coefficient $a_d\ne 0$, the coefficient of $x^{d+1}$ is $(d+\lambda )a_d$. Hence $\lambda =-d$ is necessary.

An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power coefficients vanish, because $a_1=0$ and the recurrence relation preserves parity. For the even-power coefficients,All denominators are nonzero for $j\ge 0$. With $\lambda =-2m$, the coefficients through $a_{2m}$ are nonzero when $a_0\ne 0$, the next coefficient vanishes, and every subsequent coefficient vanishes. This proves sufficiency as well as the restriction to even polynomial degree:For example, setting $a_0=1$ gives $1$ when $\lambda =0$, $1+x^2$ when $\lambda =-2$, and $1+2x^2+x^4/5$ when $\lambda =-4$. A polynomial cannot instead begin with a positive integer power: its lowest power must have an indicial exponent, and the only integer indicial exponent is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a terminating Frobenius series.

If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero polynomial solution: every nonzero polynomial just classified has $y(0)=a_0\ne 0$. The displayed family and degree answer therefore concern the differential equation without those extra endpoint conditions.

Substitute a Frobenius series $y={\sum }_{n\ge 0}{a}_n{z}^{n+\rho }$ into the transformed equation. For $n\ge 1$, equating the coefficient of $z^{n+\rho -1}$ gives the recurrence relationAt the larger indicial root $\rho =2$, the coefficient of $a_n$ is $n(n+2)\ne 0$, so a Frobenius solution always exists. At the smaller indicial root $\rho =0$, the first two equations areSince $a_0\ne 0$, the exceptional equation is consistent exactly whenThese are both admissible values, including the zero value; the singular wording of the question does not remove either. This is the undetermined coefficient at Frobenius resonance: at either value $a_2$ is free, whereas otherwise the second independent solution requires a Logarithmic Frobenius solution.

Explicit pairs demonstrate sufficiency and linear independence. In the variable $z=x+1$ one may takeBoth displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent Frobenius solutions with the two indicial exponents, and direct differentiation verifies the equation. The point is an apparent regular singular point in these two cases.

For the requested closed forms, particularly short choices areIn addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.

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.

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 $n=0$.

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.