--- title: Euler Equations, Regular Singular Points, and Frobenius module: Series Solutions and Special Functions moduleNumber: 4 lessonNumber: 2 order: 402 summary: > The Euler equation x^2 y'' + a x y' + b y = 0 is solved outright by y = x^r, and its three root cases fix the behavior at any regular singular point. The Frobenius method multiplies x^r by a power series; the indicial equation chooses the exponents, and equal or integer-separated roots force a logarithm in the second solution. Gauss's hypergeometric equation is the archetype containing most classical functions as special cases. topics: [Series Solutions and Special Functions] sources: - book: Boyce ref: "Ch. 5 §5.4 Euler Equations; Regular Singular Points; §5.5 Series Solutions Near a Regular Singular Point, Part I; §5.6 Part II" - book: Simmons ref: "Ch. 5 §29–30 Regular Singular Points; §31 Gauss's Hypergeometric Equation; §32 The Point at Infinity" draft: false --- The [power-series method](/differential-equations/series-solutions/power-series-ordinary-points) solves $P(x)y'' + Q(x)y' + R(x)y = 0$ near any point where $P \neq 0$. At a singular point the analyticity of the solutions fails: fractional powers, negative powers, and logarithms appear, and no Taylor series can represent them. Three objects replace the ordinary-point machinery. The **Euler equation** is the one singular equation solvable in closed form, and its solutions exhibit every behavior a mild singularity produces. The classification of singular points into **regular** (Euler-like, tractable) and **irregular** (deferred to advanced texts) separates the cases the elementary theory reaches. The **method of Frobenius** handles the regular case: the solution is an Euler solution $x^r$ times a convergent power series, and the algebra reduces to a quadratic (the indicial equation) plus a recurrence. ## The Euler equation > **Definition (Euler equation).** The **Euler equation** (also equidimensional or Cauchy–Euler equation) is > $$ > L[y] = x^2 y'' + \alpha x y' + \beta y = 0, > $$ > with $\alpha, \beta$ real constants. Its only singular point is $x = 0$. Each term balances: $x^2$ against two derivatives, $x$ against one. A power $y = x^r$ is therefore reproduced up to a constant, $$ L[x^r] = x^2 \cdot r(r-1)x^{r-2} + \alpha x \cdot r x^{r-1} + \beta x^r = x^r \big( r(r-1) + \alpha r + \beta \big), $$ so $y = x^r$ solves the equation exactly when $r$ is a root of the quadratic $$ F(r) = r(r-1) + \alpha r + \beta = 0, \qquad r_{1,2} = \frac{-(\alpha - 1) \pm \sqrt{(\alpha-1)^2 - 4\beta}}{2}. $$ The parallel with the [characteristic equation](/differential-equations/second-order-linear/homogeneous-constant-coefficients) is exact — the substitution $x = e^t$ converts one problem into the other — and the same three cases appear, with $e^{rt}$ replaced by $x^r$.[^boyce-euler] - **Real distinct roots $r_1 \neq r_2$.** Then $y_1 = x^{r_1}$ and $y_2 = x^{r_2}$ have Wronskian $(r_2 - r_1)x^{r_1 + r_2 - 1} \neq 0$, so for $x > 0$ $$ y = c_1 x^{r_1} + c_2 x^{r_2}. $$ - **Repeated root $r_1 = r_2$.** Then $F(r) = (r - r_1)^2$, and differentiating the identity $L[x^r] = x^r F(r)$ with respect to $r$ produces the second solution: $$ \frac{\partial}{\partial r} L[x^r] = L[x^r \ln x] = x^r\big( F'(r) + F(r)\ln x \big) \Big|_{r = r_1} = 0 \quad\Longrightarrow\quad y = (c_1 + c_2 \ln x)\,x^{r_1}. $$ - **Complex roots $r = \lambda \pm i\mu$.** Define $x^r = e^{r \ln x}$; Euler's formula turns $x^{i\mu}$ into $\cos(\mu \ln x) + i \sin(\mu \ln x)$, and the real and imaginary parts give $$ y = x^{\lambda}\big( c_1 \cos(\mu \ln x) + c_2 \sin(\mu \ln x) \big). $$ > **Worked example.** The three root cases on concrete equations, each solved by $y = x^r$.[^boyce-euler] > > **Distinct real roots.** $2x^2 y'' + 3x y' - y = 0$ gives $x^r(2r^2 + r - 1) = 0$, so $2r^2 + r - 1 = (2r - 1)(r + 1) = 0$ with roots $r = \tfrac12, -1$: > > $$ > y = c_1 x^{1/2} + c_2 x^{-1}, \qquad x > 0. > $$ > > **Repeated root.** $x^2 y'' + 5x y' + 4y = 0$ gives $r^2 + 4r + 4 = (r + 2)^2 = 0$, a double root $r = -2$: > > $$ > y = x^{-2}(c_1 + c_2 \ln x), \qquad x > 0. > $$ > > **Complex roots.** $x^2 y'' + x y' + y = 0$ gives $r^2 + 1 = 0$, roots $r = \pm i$ (so $\lambda = 0$, $\mu = 1$): > > $$ > y = c_1 \cos(\ln x) + c_2 \sin(\ln x), \qquad x > 0, > $$ > > bounded but oscillating faster and faster as $x \to 0^+$. For $x < 0$ the same formulas hold with $x$ replaced by $|x|$ (substitute $x = -\xi$ and check that the equation is unchanged); they are then valid on any interval not containing the origin.[^boyce-euler] The behavior near the singular point is read directly off the exponents: $x^r \to 0$ as $x \to 0^+$ when $r > 0$, blows up when $r < 0$, and the logarithmic and oscillatory factors modify this at the margins. $$ % caption: Solutions $x^r$ of an Euler equation near the singular point: positive % exponents vanish at the origin, negative exponents blow up, $r = 0$ is constant, % and the repeated-root companion $x \ln x$ leaves the origin with unbounded slope. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (4.9,0) node[below] {$x$}; \draw[->, black] (0,-0.7) -- (0,2.8) node[left] {$y$}; \draw[black] (2.2,0.05) -- (2.2,-0.05) node[below] {$1$}; \draw[acc, thick] plot[smooth] coordinates {(0.00,0.00) (0.11,0.01) (0.33,0.06) (0.66,0.16) (1.10,0.35) (1.54,0.59) (1.98,0.85) (2.42,1.15) (3.08,1.66) (3.74,2.22)}; \draw[acc, thick] plot[smooth] coordinates {(0.00,0.00) (0.11,0.22) (0.33,0.39) (0.66,0.55) (1.10,0.71) (1.54,0.84) (1.98,0.95) (2.42,1.05) (3.08,1.18) (3.74,1.30) (4.40,1.41)}; \draw[black, thick] (0,1.0) -- (4.4,1.0); \draw[acc, thick, dashed] plot[smooth] coordinates {(0.33,2.58) (0.44,2.24) (0.66,1.83) (0.88,1.58) (1.10,1.41) (1.54,1.20) (1.98,1.05) (2.42,0.95) (3.08,0.85) (3.74,0.77) (4.40,0.71)}; \draw[acc, thick, dashed] plot[smooth] coordinates {(1.54,1.71) (1.98,1.17) (2.42,0.87) (3.08,0.60) (3.74,0.45) (4.40,0.35)}; \draw[black, thick] plot[smooth] coordinates {(0.02,-0.08) (0.11,-0.15) (0.22,-0.23) (0.44,-0.32) (0.66,-0.36) (1.10,-0.35) (1.54,-0.25) (2.20,0.00) (2.86,0.34) (3.52,0.75)}; \node[acc, anchor=west] at (3.8,2.25) {$x^{3/2}$}; \node[acc, anchor=west] at (4.45,1.41) {$x^{1/2}$}; \node[anchor=west, text=black] at (4.45,1.0) {$x^{0}$}; \node[acc, anchor=west] at (4.45,0.71) {$x^{-1/2}$}; \node[acc, anchor=west] at (4.45,0.35) {$x^{-3/2}$}; \node[anchor=north, text=black] at (0.9,-0.46) {$x \ln x$}; \end{tikzpicture} $$ ## Regular versus irregular singular points Return to the general equation $P(x)y'' + Q(x)y' + R(x)y = 0$ with a singular point at $x_0$. Whether Euler-style methods extend depends on how fast $Q/P$ and $R/P$ blow up there. The usable notion of a "weak" singularity is calibrated against the Euler equation itself, where $Q/P \sim \alpha/x$ and $R/P \sim \beta/x^2$. > **Definition (Regular singular point).** A singular point $x_0$ is a **regular singular point** of $P y'' + Q y' + R y = 0$ if both limits > $$ > \lim_{x \to x_0}\,(x - x_0)\,\frac{Q(x)}{P(x)} = p_0, > \qquad > \lim_{x \to x_0}\,(x - x_0)^2\,\frac{R(x)}{P(x)} = q_0 > $$ > exist (equivalently, if $(x - x_0)\,Q/P$ and $(x - x_0)^2\,R/P$ are analytic at $x_0$). A singular point that is not regular is an **irregular singular point**. The conditions say the singularity of $Q/P$ is no worse than $(x - x_0)^{-1}$ and that of $R/P$ no worse than $(x - x_0)^{-2}$ — the profile of an Euler equation. Singular points cannot be ignored in practice: solutions concentrate their interesting behavior there, and geometric features of a physical problem (corners, edges, the axis of a cylinder) land on them. The classification is a two-limit computation:[^boyce-rsp] $$ % caption: Classifying a point of $Py'' + Qy' + Ry = 0$: check $P(x_0)$, then the % two weighted limits. Regular singular points admit Frobenius series; irregular % ones are outside the scope of these methods. \begin{tikzpicture}[>=stealth, font=\footnotesize, q/.style={draw, thick, minimum width=34mm, minimum height=11mm, align=center}, leaf/.style={draw, black, minimum width=30mm, minimum height=11mm, align=center}] \definecolor{acc}{HTML}{4A6FA5} \node[q, draw=acc, text=acc] (root) at (0,0) {$P(x_0) = 0$ ?}; \node[leaf] (ord) at (-4.1,-2.0) {ordinary point:\\power series}; \node[q, draw=acc, text=acc] (lim) at (2.6,-2.0) {$(x - x_0)Q/P$ and\\$(x - x_0)^2 R/P$\\both bounded?}; \node[leaf] (reg) at (-0.6,-4.3) {regular singular:\\Frobenius series}; \node[leaf] (irr) at (5.4,-4.3) {irregular singular:\\advanced methods}; \draw[->, acc, thick] (root) -- (ord) node[midway, above left=-2pt] {no}; \draw[->, acc, thick] (root) -- (lim) node[midway, above right=-2pt] {yes}; \draw[->, acc, thick] (lim) -- (reg) node[midway, above left=-2pt] {yes}; \draw[->, acc, thick] (lim) -- (irr) node[midway, above right=-2pt] {no}; \end{tikzpicture} $$ Worked classifications: - **Legendre's equation** $(1 - x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0$: singular points $x = \pm 1$. At $x = 1$, $$ \lim_{x \to 1}\,(x-1)\frac{-2x}{1 - x^2} = \lim_{x\to 1}\frac{2x}{1+x} = 1, \qquad \lim_{x \to 1}\,(x-1)^2\frac{\alpha(\alpha+1)}{1 - x^2} = 0, $$ both finite, so $x = 1$ is regular (and $x = -1$ likewise, by symmetry). - **$2x(x-2)^2 y'' + 3xy' + (x-2)y = 0$**: singular points $x = 0$ and $x = 2$. At $x = 0$ both limits are $0$: regular. At $x = 2$, $$ \lim_{x \to 2}\,(x-2)\,\frac{3x}{2x(x-2)^2} = \lim_{x \to 2}\frac{3}{2(x-2)}, $$ which does not exist: irregular. - **Bessel's equation** $x^2y'' + xy' + (x^2 - \nu^2)y = 0$: at $x = 0$, $p_0 = 1$ and $q_0 = -\nu^2$, both finite, so the origin is a regular singular point, treated in full for [Bessel's equation](/differential-equations/series-solutions/bessel-and-special-functions). Non-polynomial coefficients are handled by the analyticity form of the definition: for $(x - \pi/2)^2 y'' + (\cos x)y' + (\sin x)y = 0$, expanding $\cos x$ about $\pi/2$ shows $(x - \pi/2)\,Q/P$ is analytic there, so $x = \pi/2$ is a regular singular point. The point at infinity is classified the same way after the substitution $\xi = 1/x$; equations whose every singular point, including $\infty$, is regular are the Fuchsian equations, and the hypergeometric equation below is the fundamental example.[^simmons-infinity] ## The method of Frobenius Let $x = 0$ be a regular singular point (translate $x_0$ to the origin if necessary). By definition, $x p(x) = xQ/P$ and $x^2 q(x) = x^2 R/P$ are analytic: $$ x\,p(x) = \sum_{n=0}^{\infty} p_n x^n, \qquad x^2 q(x) = \sum_{n=0}^{\infty} q_n x^n, \qquad |x| < \rho . $$ Multiplying the equation by $x^2/P$ puts it in the standard form $$ x^2 y'' + x \big( x p(x) \big) y' + \big( x^2 q(x) \big) y = 0 . $$ If every $p_n, q_n$ with $n \ge 1$ were zero this would be the Euler equation $x^2y'' + p_0 x y' + q_0 y = 0$; in general the higher coefficients perturb the Euler equation without changing the character of its solutions near $0$. That suggests the ansatz — an Euler solution times a power series:[^boyce-frob1] > **Definition (Frobenius series).** A **Frobenius series** about the regular singular point $x = 0$ is > $$ > y = x^r \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{r+n}, > \qquad a_0 \neq 0, > $$ > where the exponent $r$ need not be an integer. The normalization $a_0 \neq 0$ pins $r$ down as the exponent of the leading term. $$ % caption: Structure of a Frobenius solution: the factor $x^r$ carries all the singular % behavior at the origin, and the power-series factor is analytic with $a_0 = 1$; % the indicial equation selects the admissible exponents $r$. \begin{tikzpicture}[>=stealth, font=\footnotesize, b/.style={draw, thick, minimum height=13mm, align=center}] \definecolor{acc}{HTML}{4A6FA5} \node[b, draw=acc, text=acc, minimum width=20mm] (euler) at (0,0) {$x^{r}$}; \node at (1.7,0) {$\times$}; \node[b, draw=black, minimum width=46mm] (series) at (4.7,0) {$1 + a_1 x + a_2 x^2 + \cdots$}; \node[anchor=south, acc, align=center] at (0,1.0) {Euler part:\\singular behavior}; \node[anchor=south, text=black, align=center] at (4.7,1.0) {analytic part:\\recurrence relation}; \draw[->, acc] (0,-0.85) -- (0,-1.55); \node[anchor=north, align=center] at (-0.4,-1.6) {$r$ from the indicial equation\\$F(r) = r(r-1) + p_0 r + q_0 = 0$}; \draw[->, black] (4.7,-0.85) -- (4.7,-1.55); \node[anchor=north, text=black, align=center] at (5.1,-1.6) {$a_n$ from the recurrence,\\each step dividing by $F(r + n)$}; \end{tikzpicture} $$ Substituting the series and collecting powers of $x$ gives $$ L[y] = a_0 F(r)\,x^r + \sum_{n=1}^{\infty} \left[ F(r+n)\,a_n + \sum_{k=0}^{n-1} a_k \big( (r+k)\,p_{n-k} + q_{n-k} \big) \right] x^{r+n} = 0, $$ where $F(r) = r(r-1) + p_0 r + q_0$. Setting each coefficient to zero yields the two moving parts of the method. - **The indicial equation.** Since $a_0 \neq 0$, the lowest power forces $$ F(r) = r(r-1) + p_0 r + q_0 = 0, $$ the same quadratic the associated Euler equation would produce. Its roots $r_1 \ge r_2$ (real case) are the **exponents at the singularity**; they control the qualitative behavior of every solution near $0$ and require only two limits to compute: $p_0 = \lim_{x\to 0} x\,Q/P$ and $q_0 = \lim_{x \to 0} x^2 R/P$. - **The recurrence.** For $n \ge 1$, $$ F(r + n)\,a_n = -\sum_{k=0}^{n-1} a_k \big( (r + k)\,p_{n-k} + q_{n-k} \big), $$ determining each $a_n$ from its predecessors, provided the divisor $F(r + n)$ is never zero. For $r = r_1$ (the larger root) this is automatic: $r_1 + n$ lies to the right of both roots of $F$. For $r = r_2$ the divisor $F(r_2 + n)$ vanishes exactly when $r_2 + n = r_1$, i.e. when the roots differ by the positive integer $n$ — the source of all the case analysis below. ```algorithm caption: $\textsc{Frobenius}(P, Q, R)$ — series solution at a regular singular point $x = 0$ compute $p_0 = \lim_{x \to 0} x\,Q(x)/P(x)$ and $q_0 = \lim_{x \to 0} x^2 R(x)/P(x)$ solve the indicial equation $F(r) = r(r-1) + p_0 r + q_0 = 0$ for $r_1 \ge r_2$ substitute $y = \sum_{n \ge 0} a_n x^{r+n}$ with $a_0 = 1$ into the equation for each $n \ge 1$ do read the recurrence $F(r+n)\,a_n = -\sum_{k=0}^{n-1} a_k ((r+k)\,p_{n-k} + q_{n-k})$ set $r = r_1$ and solve the recurrence to obtain $y_1(x) = x^{r_1}(1 + \sum_{n \ge 1} a_n(r_1)\,x^n)$ if $r_1 - r_2$ is not an integer then set $r = r_2$ and repeat to obtain $y_2$ else obtain $y_2$ in the logarithmic form of the three-case theorem return $y = c_1 y_1 + c_2 y_2$ ``` A complete run of the method, with two independent series: > **Worked example.** Solve $2x^2 y'' - xy' + (1 + x)y = 0$ near $x = 0$.[^boyce-frob1] Here $x p(x) = -1/2$ and $x^2 q(x) = (1 + x)/2$, so $p_0 = -\tfrac12$, $q_0 = \tfrac12$, $q_1 = \tfrac12$, and all other coefficients vanish. Substituting the Frobenius series and collecting, > > $$ > a_0\big[ 2r(r-1) - r + 1 \big] x^r > + \sum_{n=1}^{\infty} \Big[ \big( 2(r+n)(r+n-1) - (r+n) + 1 \big) a_n + a_{n-1} \Big] x^{r+n} = 0 . > $$ > > The indicial equation is $2r^2 - 3r + 1 = (r - 1)(2r - 1) = 0$, with exponents $r_1 = 1$, $r_2 = \tfrac12$: distinct and not differing by an integer, so two Frobenius series exist. The recurrence is > > $$ > a_n = -\frac{a_{n-1}}{\big( (r+n) - 1 \big)\big( 2(r+n) - 1 \big)}, \qquad n \ge 1 . > $$ > > For $r = 1$ it reads $a_n = -a_{n-1}/\big( (2n+1)n \big)$, so > > $$ > a_1 = -\frac{a_0}{3 \cdot 1}, \qquad > a_2 = \frac{a_0}{(3 \cdot 5)(1 \cdot 2)}, \qquad > a_3 = -\frac{a_0}{(3 \cdot 5 \cdot 7)(1 \cdot 2 \cdot 3)}, > $$ > > which telescopes to $a_n = (-1)^n 2^n a_0/(2n+1)!$. For $r = \tfrac12$ the recurrence $a_n = -a_{n-1}/\big( n(2n-1) \big)$ gives $a_n = (-1)^n 2^n a_0/(2n)!$. With $a_0 = 1$, > > $$ > y_1(x) = x \left( 1 + \sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{(2n+1)!}\,x^n \right), > \qquad > y_2(x) = x^{1/2} \left( 1 + \sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{(2n)!}\,x^n \right), > \qquad x > 0, > $$ > > both convergent for all $x$ by the ratio test. Near the origin $y_1 \sim x$ and $y_2 \sim x^{1/2}$, so they are independent, and $y = c_1 y_1 + c_2 y_2$ is the general solution. The exponents were available after two limits and a quadratic, before any series work; they alone often settle whether a bounded solution exists and how fast the unbounded one grows. ## The three cases The gap $r_1 - r_2$ determines how the second solution is found. The full statement:[^boyce-thm] > **Theorem (Frobenius, three cases).** Let $x = 0$ be a regular singular point of $x^2y'' + x(xp(x))y' + (x^2q(x))y = 0$, with $\rho$ the smaller radius of convergence of $xp$ and $x^2q$, and let $r_1 \ge r_2$ be the (real) indicial roots. Then on $0 < x < \rho$ (and on $-\rho < x < 0$ with $|x|$ in place of $x$) there is always a solution > $$ > y_1(x) = |x|^{r_1} \left( 1 + \sum_{n=1}^{\infty} a_n(r_1)\,x^n \right). > $$ > The second solution takes one of three forms: > > 1. If $r_1 - r_2$ is neither zero nor a positive integer: $\displaystyle y_2 = |x|^{r_2} \Big( 1 + \sum_{n \ge 1} a_n(r_2)\,x^n \Big)$. > 2. If $r_1 = r_2$: $\displaystyle y_2 = y_1(x) \ln|x| + |x|^{r_1} \sum_{n \ge 1} b_n(r_1)\,x^n$. > 3. If $r_1 - r_2 = N$, a positive integer: $\displaystyle y_2 = a\,y_1(x)\ln|x| + |x|^{r_2} \Big( 1 + \sum_{n \ge 1} c_n(r_2)\,x^n \Big)$, where the constant $a$ may be zero. > > Each series converges at least for $|x| < \rho$, and $y_1, y_2$ form a fundamental set. $$ % caption: The gap between the indicial roots picks the case. Complex roots never % collide, equal roots always force a logarithm, and an integer gap may or may not, % depending on whether the recurrence survives the division by $F(r_2 + N) = 0$. \begin{tikzpicture}[>=stealth, font=\footnotesize, b/.style={draw, thick, minimum width=36mm, minimum height=11mm, align=center}, leaf/.style={draw, black, minimum width=40mm, minimum height=12mm, align=center}] \definecolor{acc}{HTML}{4A6FA5} \node[b, draw=acc, text=acc] (root) at (0,0) {indicial roots $r_1 \ge r_2$}; \node[leaf] (c1) at (-5.0,-2.3) {$r_1 - r_2$ not an integer:\\two Frobenius series}; \node[leaf] (c2) at (0,-2.3) {$r_1 = r_2$:\\log term always}; \node[leaf] (c3) at (5.0,-2.3) {$r_1 - r_2 = N \ge 1$:\\log term sometimes ($a$ may be $0$)}; \draw[->, acc, thick] (root) -- (c1); \draw[->, acc, thick] (root) -- (c2); \draw[->, acc, thick] (root) -- (c3); \node[anchor=north, text=black, align=center] at (-5.0,-3.05) {ex: $r = 1,\ \tfrac12$}; \node[anchor=north, text=black, align=center] at (0,-3.05) {ex: Bessel, order $0$}; \node[anchor=north, text=black, align=center] at (5.0,-3.05) {ex: Bessel, orders $\tfrac12$ and $1$}; \end{tikzpicture} $$ Working notes on the cases: | Situation | Second solution | How it is computed | | --- | --- | --- | | $r_1 - r_2 \notin \mathbb{Z}$ | second Frobenius series at $r_2$ | run the same recurrence with $r = r_2$ | | $r_1 = r_2$ | logarithm, always | $y_2 = \partial \phi(r, x)/\partial r \big\rvert_{r=r_1}$, i.e. coefficients $a_n'(r_1)$; or substitute the log form directly; or reduction of order | | $r_1 - r_2 = N$ | logarithm iff the recurrence at $r_2$ fails at $n = N$ | try the plain series at $r_2$ first: if $a_N$ is determinable, $a = 0$ and no log appears | | $r_{1,2}$ complex | two (complex) Frobenius series | roots can never be equal or differ by an integer; take real and imaginary parts | > **Worked example.** $xy'' + y = 0$ has a regular singular point at $x = 0$: from $xQ/P = 0$ and $x^2 R/P = x$, the coefficients are $p_0 = q_0 = 0$, so the indicial equation is $r(r-1) = 0$ with exponents $r_1 = 1$ and $r_2 = 0$ differing by the integer $N = 1$.[^boyce-int] Substituting $y = \sum_{n \ge 0} a_n x^{r+n}$ gives the recurrence > > $$ > a_{n+1} = -\frac{a_n}{(r + n + 1)(r + n)}, \qquad n \ge 0. > $$ > > For the larger root $r = 1$ the denominators never vanish: > > $$ > a_1 = -\frac{a_0}{2}, \qquad a_2 = \frac{a_0}{12}, \qquad a_3 = -\frac{a_0}{144}, > $$ > > so $y_1 = x - \tfrac12 x^2 + \tfrac{1}{12}x^3 - \tfrac{1}{144}x^4 + \cdots$. For the smaller root $r = 0$ the step to $a_1$ divides by $(r+1)(r) = 0$, so the recurrence breaks at $n = 0$: no second Frobenius series exists, and the companion solution takes the Case 3 logarithmic form with $a \neq 0$. The equal-root mechanism is the Euler one again: with $F(r) = (r - r_1)^2$, the operator identity $L[\phi(r, x)] = a_0(r - r_1)^2 x^r$ still holds after one differentiation in $r$, so $\partial\phi/\partial r|_{r_1}$ is a second solution, and differentiating $x^r = e^{r \ln x}$ is what produces the $\ln x$.[^boyce-thm] [Bessel's equation](/differential-equations/series-solutions/bessel-and-special-functions) realizes all three exceptional outcomes concretely: order $0$ gives equal roots, order $\tfrac12$ an integer gap with no logarithm, and order $1$ an integer gap with a logarithm. A practical point: when $P, Q, R$ are polynomials it is usually easier to substitute the Frobenius series into the original equation $Py'' + Qy' + Ry = 0$ directly than to first expand $xQ/P$ and $x^2R/P$ as power series. ## The hypergeometric equation One equation is both the standard exercise in these techniques and the source of most of the special functions in this module. > **Definition (Hypergeometric equation).** **Gauss's hypergeometric equation** with parameters $a, b, c$ is > $$ > x(1 - x)\,y'' + \big[ c - (a + b + 1)x \big]\,y' - ab\,y = 0 . > $$ The singular points are $x = 0$, $x = 1$, and (after the substitution $\xi = 1/x$) the point at infinity; all three are regular, and this three-regular-singular-point profile characterizes the equation up to changes of variable.[^simmons-hyper] At $x = 0$: $x p(x) = c + \cdots$ and $x^2 q(x) = -abx - \cdots$, so $p_0 = c$, $q_0 = 0$, and the indicial equation $r(r - 1) + cr = 0$ has exponents $r = 0$ and $r = 1 - c$. At $x = 1$ the exponents are $0$ and $c - a - b$; at $\infty$ they are $a$ and $b$ — the six exponents are pure parameter arithmetic. For the exponent $r = 0$ (assume $c$ is not zero or a negative integer), the recurrence collapses to $$ a_{n+1} = \frac{(a + n)(b + n)}{(n+1)(c + n)}\,a_n , $$ and with $a_0 = 1$ the solution is the **hypergeometric series** $$ F(a, b, c, x) = 1 + \sum_{n=1}^{\infty} \frac{a(a+1)\cdots(a+n-1)\;b(b+1)\cdots(b+n-1)}{n!\;c(c+1)\cdots(c+n-1)}\,x^n , $$ convergent for $|x| < 1$ (the distance to the singular point $x = 1$, as the general theory predicts), and a polynomial if $a$ or $b$ is zero or a negative integer. When $c$ is not a positive integer, the other exponent gives the second solution via the substitution $y = x^{1-c} z$, which reproduces the hypergeometric equation with shifted parameters; the general solution near $0$ is $$ y = c_1\,F(a, b, c, x) + c_2\,x^{1-c}\,F(a - c + 1,\; b - c + 1,\; 2 - c,\; x) . $$ The name records that $F(1, b, b, x) = 1 + x + x^2 + \cdots$ is the geometric series, and with suitable parameters $F$ specializes to much of elementary analysis: | Function | Hypergeometric form | | --- | --- | | $(1 + x)^p$ | $F(-p,\, b,\, b,\, -x)$ | | $\ln(1 + x)$ | $x\,F(1,\, 1,\, 2,\, -x)$ | | $\arcsin x$ | $x\,F\!\big(\tfrac12,\, \tfrac12,\, \tfrac32,\, x^2\big)$ | | $\arctan x$ | $x\,F\!\big(\tfrac12,\, 1,\, \tfrac32,\, -x^2\big)$ | | $\dfrac{1}{1 - x}$ | $F(1,\, b,\, b,\, x)$ | Legendre's equation is also hypergeometric in disguise: the substitution $t = \tfrac12(1 - x)$ moves its singular points $\pm 1$ to $1, 0$ and identifies the bounded solutions as $F\big({-n},\, n+1,\, 1,\, \tfrac12(1 - x)\big)$, which is how [the Legendre polynomials](/differential-equations/series-solutions/bessel-and-special-functions) are constructed. Gauss's 1812 study of this series was the first rigorous convergence analysis of an infinite series, and the equation stays central because so many of the special functions of analysis are hypergeometric specializations.[^simmons-hyper] [^boyce-euler]: **Boyce**, _Elementary Differential Equations and Boundary Value Problems_, §5.4 — the Euler equation: the three root cases, the $\partial/\partial r$ derivation of the $\ln x$ solution, the complex-power definition $x^r = e^{r \ln x}$, and the $|x|$ extension to $x < 0$. [^boyce-rsp]: **Boyce**, §5.4 — regular and irregular singular points: the limit conditions, why singular points cannot be ignored, and the Legendre/irregular worked classifications (Examples 4–6). [^simmons-infinity]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §32 — the point at infinity via $\xi = 1/x$; **Boyce**, §5.4, Problem 32. Fuchs's name for the all-regular class is noted in Boyce §5.3. [^boyce-frob1]: **Boyce**, §5.5 — the Frobenius ansatz as Euler solutions times power series, and the worked solution of $2x^2y'' - xy' + (1+x)y = 0$; **Simmons**, §29–30. [^boyce-thm]: **Boyce**, §5.6, Theorem 5.6.1 — the general recurrence, the indicial equation, the three cases for the second solution, and the computational routes ($a_n'(r_1)$ coefficients, direct substitution of the log form, or reduction of order). [^boyce-int]: **Boyce**, §5.6, Problem 11 — $xy'' + y = 0$: the regular singular point at $0$, exponents $0$ and $1$, and the larger-root series; the smaller root's recurrence breaks, forcing the logarithmic second solution. [^simmons-hyper]: **Simmons**, §31 — Gauss's hypergeometric equation: exponents at $0$, $1$, $\infty$, the hypergeometric series and its convergence, the $x^{1-c}$ second solution, and the special-case identities; **Boyce**, §5.6, Problem 14.