Frobenius method — regular singular points

A way to solve $y'' + P(x)y' + Q(x)y = 0$ near a point where the simple power-series method fails.

Why do we even need this?

So the WHAT: we look for $y = x^r\sum_{n\ge0} a_n x^n$ where $r$ is to be determined, not assumed integer.

Classifying the singular point (the gatekeeper)

Write the ODE in standard form: $y'' + P(x)\,y' + Q(x)\,y = 0.$

WHY these exact powers? Because if $P$ blows up no worse than $1/(x-x_0)$ and $Q$ no worse than $1/(x-x_0)^2$, then multiplying by $x^r$ exactly balances the worst singularity in the equation. Stronger blow-ups can't be balanced by a single power — that's why irregular points break the method.

Deriving the indicial equation FROM SCRATCH

Take $x_0=0$ and write the RSP equation in the cleanest form by multiplying by $x^2$: $x^2 y'' + x\,p(x)\,y' + q(x)\,y = 0,$ where $p(x)=\sum_{k\ge0}p_k x^k$ and $q(x)=\sum_{k\ge0}q_k x^k$ are analytic.

Step 1 — Assume the Frobenius ansatz. $y = \sum_{n=0}^{\infty} a_n x^{n+r}, \quad a_0\neq 0.$ Why $a_0\neq0$? We demand the lowest power present is $x^r$; if $a_0=0$ we just relabel $r$, so insist it's nonzero.

Step 2 — Differentiate term by term. $y' = \sum_n a_n (n+r)x^{n+r-1}, \qquad y'' = \sum_n a_n (n+r)(n+r-1)x^{n+r-2}.$

Step 3 — Plug in and look ONLY at the lowest power $x^r$ (i.e. $n=0$, $p\to p_0$, $q\to q_0$): $x^2 y'' \to a_0 r(r-1)x^r,\quad x p\, y' \to a_0 p_0 r\, x^r,\quad q\,y \to a_0 q_0 x^r.$ Adding and setting the coefficient of $x^r$ to zero (with $a_0\neq0$):

Step 4 — Recurrence for the rest. Setting the coefficient of $x^{n+r}$ to zero gives $\big[(n+r)(n+r-1)+p_0(n+r)+q_0\big]a_n = -\sum_{k=1}^{n}\big[p_k(n-k+r)+q_k\big]a_{n-k}.$ The bracket on the left is $F(n+r)$ where $F(s)=s(s-1)+p_0 s+q_0$ is the indicial polynomial. So $a_n$ is solvable as long as $F(n+r)\neq0$.

The three cases (WHY the second solution changes form)

Let $r_1 - r_2 =$ the root difference.

WHY a log in Cases 2 & 3? When $F(n+r_2)=0$ at some $n$, the recurrence demands division by zero. Reduction of order $y_2=v(x)y_1$ then produces a $\int dx/x = \ln x$ term. The log is the mathematics' way of supplying a second independent solution when the power series ran out of room.

Worked Example 1 — get $p_0,q_0$ and solve the indicial equation

ODE: $\;2x^2 y'' + 3x\,y' - (1+x)y = 0.$

Step 1: Standard form. Divide by $2x^2$: $y'' + \frac{3}{2x}y' - \frac{1+x}{2x^2}y=0.$ Why? We need $P,Q$ to read off the singularity.

Step 2: Check RSP. $xP = \tfrac32$ (analytic), $x^2Q = -\tfrac{1+x}{2}$ (analytic). ✔ So $x=0$ is a regular singular point. Why this step? Frobenius is only guaranteed at an RSP.

Step 3: Read $p_0,q_0$. $p_0 = \tfrac32$, $q_0 = x^2Q|_{x=0} = -\tfrac12$.

Step 4: Indicial equation. $r(r-1)+\tfrac32 r -\tfrac12 = r^2+\tfrac12 r -\tfrac12=0 \Rightarrow 2r^2+r-1=0 \Rightarrow (2r-1)(r+1)=0.$ Roots $r_1=\tfrac12,\; r_2=-1$. Difference $=\tfrac32\notin\mathbb Z$ → Case 1, two clean Frobenius series. Why this matters? We instantly know NO logarithm appears.

Worked Example 2 — full recurrence (Bessel-type, equal roots)

ODE: $\;x^2y'' + xy' + x^2 y = 0$ (Bessel order 0).

Step 1: $P=1/x,\ Q=1$. $xP=1$, $x^2Q=x^2$, both analytic → RSP. $p_0=1,\ q_0=0$.

Step 2: Indicial. $r(r-1)+r+0=r^2=0\Rightarrow r_1=r_2=0$ (Case 2 — expect a log in $y_2$). Why this step? Equal roots predict the structure before we compute anything.

Step 3: Plug $y=\sum a_n x^{n}$ (since $r=0$). Then $\sum a_n n(n-1)x^n+\sum a_n n x^n + \sum a_n x^{n+2}=0.$ Combine first two: $\sum a_n n^2 x^n + \sum a_n x^{n+2}=0$.

Step 4: Match powers $x^m$. $a_m m^2 + a_{m-2}=0\Rightarrow a_m = -\dfrac{a_{m-2}}{m^2}.$ Why this step? Shifting index $n\to m-2$ aligns powers so we can equate coefficients.

Step 5: Iterate. $a_1=0$ (since $1^2 a_1=0$), so all odd terms vanish. $a_2=-\frac{a_0}{4},\quad a_4=-\frac{a_2}{16}=\frac{a_0}{64},\dots,\quad a_{2k}=\frac{(-1)^k a_0}{2^{2k}(k!)^2}.$ With $a_0=1$ this is $J_0(x)=\sum_{k}\frac{(-1)^k}{(k!)^2}(x/2)^{2k}$ — the first Bessel function. The second solution $Y_0$ carries the predicted $\ln x$.

Common mistakes (Steel-manned)

Active Recall

Flashcards

Connections

• Power series solutions — ordinary points
• Bessel's equation and Bessel functions
• Legendre's equation and polynomials
• Reduction of order
• Wronskian and linear independence
• Analytic functions and radius of convergence
• Euler–Cauchy equation (the indicial equation is literally its characteristic equation)

Concept Map

Hinglish (regional understanding)