Frobenius method

Near a regular singular point $x_0$, the Frobenius method seeksThe lowest power gives the indicial equation for $r$, and the remaining powers give a coefficient recurrence.

For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta }$ is nonzero, and the exceptional equation $0=-b{d}_{N-\beta }$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the Frobenius case classification, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.

The coefficient of the necessary logarithm can also be calculated. Let $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis proves that the logarithm cannot disappear for a special nonzero value of $b$. For $a=0$, reduction of order instead produces the repeated-root Logarithmic Frobenius solution directly.

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.

When two indicial roots differ by a positive integer, the recurrence for the smaller-root Frobenius solution can have a zero coefficient at the index of the larger root. If the corresponding right-hand side also vanishes, that coefficient is free and adds the larger-root solution; if it does not vanish, a pure Frobenius series is obstructed and a logarithmic term may be required. For $x^2{y}^{\prime \prime }-2x{y}^{\prime }-x^{2}y=0$, the recurrence $n(n-3)a_n=a_{n-2}$ has $a_1=0$ and leaves $a_3$ free. Choosing $a_3=0$ selects the even solution, not the only solution with exponent zero.

With $s=\sqrt{z}$ and $Y(s)=y(s^2)$, the displayed linear differential equation becomes $Y_{ss}+Y=0$ away from zero. Its Frobenius solutions have exponents $0$ and $1/2$: $\cos \sqrt{z}={\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n)!$ and $\sin \sqrt{z}=z^{1/2}{\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n+1)!$. The first is an entire function of $z$; the second needs a local square-root branch. This illustrates how a regular singular point can arise from a degenerate change of independent variable.

A local series solution at a regular singular point, with exponent $r$ satisfying the indicial equation. Resonant exponents can require a logarithmic companion solution.

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.

A terminating Frobenius series is a Frobenius method solution whose analytic power series factor has only finitely many nonzero coefficients. Its coefficient linear recurrence relation must allow the first omitted coefficient and all following coefficients to be zero. At an integer separation of characteristic exponents at a regular singular point, a resonant coefficient equation can become $0=0$, permitting such a choice rather than forcing a logarithmic term. With exponent zero, termination produces a polynomial solution.