5.2 — Frobenius Method (Regular Singular Points)

Learning Objectives
  • Classify singular points as regular or irregular
  • Apply the Frobenius method to find series solutions
  • Determine and solve the indicial equation
  • Handle all three cases of indicial root differences

Motivation

The power series method we studied in Section 5.1 works beautifully at ordinary points, but fails when we approach a singular point. What happens when we need a solution valid near a singularity?

The Frobenius method extends the power series approach by allowing an initial fractional power. Instead of:

$$y = \sum_{n=0}^{\infty} a_n x^n$$

we use:

$$y = (x - x_0)^r \sum_{n=0}^{\infty} a_n (x - x_0)^n = \sum_{n=0}^{\infty} a_n (x - x_0)^{n+r}$$

where $r$ is determined by an indicial equation. This method works at regular singular points and will give us access to solutions involving logarithms and all the classical special functions of physics and engineering.

Theory: Regular Singular Points and the Frobenius Method

Definition: Regular vs Irregular Singular Points

For the differential equation:

$$y'' + p(x)y' + q(x)y = 0$$

A point $x = x_0$ is a regular singular point if both of the following limits exist and are finite:

Regular Singular Point
$$(x - x_0)p(x) \quad \text{and} \quad (x - x_0)^2 q(x)$$

are both analytic at $x = x_0$ (or equivalently, can be expressed as convergent power series centered at $x_0$).

If at least one of these conditions fails, $x_0$ is an irregular singular point, where Frobenius typically does not apply.

The Frobenius Theorem

Theorem (Frobenius, 1873)

If $x_0$ is a regular singular point of $y'' + p(x)y' + q(x)y = 0$, then there exists at least one solution of the form:

$$y = (x - x_0)^r \sum_{n=0}^{\infty} a_n (x - x_0)^n$$

valid in some interval $0 < |x - x_0| < R$, where the exponent $r$ is determined by the indicial equation.

The Indicial Equation

Let:

$$p_0 = \lim_{x \to x_0} (x - x_0)p(x) \quad \text{and} \quad q_0 = \lim_{x \to x_0} (x - x_0)^2 q(x)$$

The indicial equation is:

Indicial Equation
$$r(r-1) + p_0 r + q_0 = 0$$

or equivalently: $r^2 + (p_0 - 1)r + q_0 = 0$

This is a quadratic equation in $r$. Its roots $r_1$ and $r_2$ (with $r_1 \geq r_2$) determine which solutions exist and whether logarithmic terms appear.

Three Cases for the Indicial Roots

Case 1: Distinct Roots Not Differing by an Integer

If $r_1 - r_2 \notin \mathbb{Z}$ (the difference is not an integer), then two independent Frobenius series solutions are:

$$y_1 = (x - x_0)^{r_1} \sum_{n=0}^{\infty} a_n^{(1)} (x - x_0)^n$$ $$y_2 = (x - x_0)^{r_2} \sum_{n=0}^{\infty} a_n^{(2)} (x - x_0)^n$$

General solution: $y = c_1 y_1 + c_2 y_2$

Case 2: Repeated Root

If $r_1 = r_2 = r$ (a double root), the second solution contains a logarithmic term:

$$y_1 = (x - x_0)^r \sum_{n=0}^{\infty} a_n (x - x_0)^n$$ $$y_2 = y_1 \ln(x - x_0) + (x - x_0)^r \sum_{n=1}^{\infty} b_n (x - x_0)^n$$

General solution: $y = c_1 y_1 + c_2 y_2$

Case 3: Roots Differing by a Positive Integer

If $r_1 - r_2 = m \in \mathbb{Z}^+$ (a positive integer), a logarithmic term may appear in the second solution:

$$y_1 = (x - x_0)^{r_1} \sum_{n=0}^{\infty} a_n (x - x_0)^n$$ $$y_2 = A y_1 \ln(x - x_0) + (x - x_0)^{r_2} \sum_{n=0}^{\infty} b_n (x - x_0)^n$$

where $A$ may be zero (no logarithm) or nonzero. The form must be determined by substitution.

Method: Step-by-Step Solution Procedure

  1. Verify $x_0$ is a regular singular point
    Write the equation in the form $y'' + p(x)y' + q(x)y = 0$ (divide by the leading coefficient if necessary). Check that $(x - x_0)p(x)$ and $(x - x_0)^2 q(x)$ are both analytic at $x_0$.
  2. Compute the limit values $p_0$ and $q_0$
    $$p_0 = \lim_{x \to x_0} (x - x_0)p(x), \quad q_0 = \lim_{x \to x_0} (x - x_0)^2 q(x)$$
  3. Write and solve the indicial equation
    $$r(r-1) + p_0 r + q_0 = 0$$ Find the roots $r_1 \geq r_2$.
  4. Determine which case applies
    Check if $r_1 - r_2$ is an integer. This determines the form of the second solution.
  5. For the larger root $r_1$: assume the Frobenius series
    $$y = \sum_{n=0}^{\infty} a_n (x - x_0)^{n+r_1}$$ Substitute into the differential equation.
  6. Find the recurrence relation
    Collect coefficients of like powers of $(x - x_0)$ to obtain a recurrence formula relating $a_n$ to previous coefficients.
  7. Solve for coefficients
    Set $a_0 = 1$ (or any nonzero constant) and use the recurrence to find $a_1, a_2, a_3, \ldots$. This yields the first solution $y_1$.
  8. Find the second solution
    Depending on the case:
    • Case 1: Repeat steps 5–7 using $r_2$
    • Case 2: Use the formula with the logarithmic term
    • Case 3: Attempt the logarithmic form or use reduction of order

Worked Examples

Solution

Step 1: Verify $x = 0$ is regular singular

Divide by $2x^2$:

$$y'' + \frac{1}{2x}y' - \frac{x+1}{2x^2}y = 0$$

Here $p(x) = \frac{1}{2x}$ and $q(x) = -\frac{x+1}{2x^2}$.

Check: $(x - 0)p(x) = x \cdot \frac{1}{2x} = \frac{1}{2}$ ✓ and $(x - 0)^2 q(x) = x^2 \cdot \left(-\frac{x+1}{2x^2}\right) = -\frac{x+1}{2}$ ✓

Both analytic at $x = 0$, so $x = 0$ is regular singular.

Step 2: Compute $p_0$ and $q_0$

$$p_0 = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2}$$ $$q_0 = \lim_{x \to 0} \left(-\frac{x+1}{2}\right) = -\frac{1}{2}$$

Step 3: Indicial equation

$$r(r-1) + \frac{1}{2}r - \frac{1}{2} = 0$$ $$r^2 - r + \frac{1}{2}r - \frac{1}{2} = 0$$ $$r^2 - \frac{1}{2}r - \frac{1}{2} = 0$$ $$2r^2 - r - 1 = 0$$ $$(2r + 1)(r - 1) = 0$$

Roots: $r_1 = 1$ and $r_2 = -\frac{1}{2}$, so $r_1 - r_2 = \frac{3}{2} \notin \mathbb{Z}$ (Case 1).

Step 4: First solution with $r_1 = 1$

Assume $y = x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1}$.

Then: $y' = \sum_{n=0}^{\infty} (n+1)a_n x^n$ and $y'' = \sum_{n=0}^{\infty} (n+1)n a_n x^{n-1}$.

Substituting into $2x^2 y'' + xy' - (x+1)y = 0$ and collecting coefficients of $x^n$ yields a recurrence relation. Setting $a_0 = 1$ and solving gives the first Frobenius series solution.

Note: The recurrence relation and explicit coefficients follow standard Frobenius methods. For brevity, we focus on the procedure and classification.

Bessel's Equation: $x^2 y'' + xy' + x^2 y = 0$

Step 1: Verify regularity

Divide by $x^2$:

$$y'' + \frac{1}{x}y' + y = 0$$

Here $p(x) = \frac{1}{x}$ and $q(x) = 1$. Check: $xp(x) = 1$ ✓ and $x^2 q(x) = x^2$ ✓. Both analytic at $x = 0$. Regular singular point ✓

Step 2: Limits

$$p_0 = \lim_{x \to 0} 1 = 1, \quad q_0 = \lim_{x \to 0} x^2 = 0$$

Step 3: Indicial equation

$$r(r-1) + r + 0 = 0$$ $$r^2 = 0$$

Double root: $r = 0$ (Case 2: repeated root with logarithmic second solution).

Step 4: First solution $y_1$ with $r = 0$

Assume $y = \sum_{n=0}^{\infty} a_n x^n$ (ordinary power series). Substitution yields the recurrence:

$$a_n = \frac{-a_{n-2}}{n^2} \quad \text{for } n \geq 2$$

With $a_0 = 1$ and $a_1 = 0$, we get:

$$y_1(x) = J_0(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2} = 1 - \frac{x^2}{4} + \frac{x^4}{64} - \cdots$$

This is the Bessel function of the first kind, order 0.

Step 5: Second solution $y_2$ (Case 2)

Since we have a repeated root, the second solution is:

$$y_2(x) = y_1(x) \ln x + \sum_{n=1}^{\infty} b_n x^n$$

This involves the Bessel function of the second kind, $Y_0(x)$, which is far more complex to compute directly but is the standard companion solution to $J_0$.

Key Insight

Bessel's equation appears everywhere in physics: circular membranes, cylindrical heat flow, vibrations, electromagnetic waves, etc. The Frobenius method gives us direct access to these fundamental special functions.

Solution

Step 1: Verify regularity

Divide by $x^2$:

$$y'' + \frac{3}{x}y' + \frac{1}{x^2}y = 0$$

Here $p(x) = \frac{3}{x}$ and $q(x) = \frac{1}{x^2}$.

Check: $xp(x) = 3$ ✓ and $x^2 q(x) = 1$ ✓. Regular singular point ✓

Step 2: Limits

$$p_0 = 3, \quad q_0 = 1$$

Step 3: Indicial equation

$$r(r-1) + 3r + 1 = 0$$ $$r^2 + 2r + 1 = 0$$ $$(r + 1)^2 = 0$$

Double root: $r = -1$ (Case 2: repeated root with logarithmic second solution).

Step 4: Solution forms

For a Cauchy-Euler equation with repeated root $r$, the solutions are:

$$y_1 = x^{-1}, \quad y_2 = x^{-1} \ln x$$

General solution:

$$y(x) = c_1 x^{-1} + c_2 x^{-1} \ln x = \frac{c_1 + c_2 \ln x}{x}$$
Verification by Substitution

You can verify that both $y_1 = 1/x$ and $y_2 = (\ln x)/x$ satisfy the equation. Try differentiating and substituting!

Solution

Step 1: Verify regularity

Divide by $x^2$:

$$y'' - \frac{1+x}{x}y' + \frac{1}{x^2}y = 0$$

Here $p(x) = -\frac{1+x}{x}$ and $q(x) = \frac{1}{x^2}$.

Check: $xp(x) = -(1+x)$ ✓ and $x^2 q(x) = 1$ ✓. Regular singular point ✓

Step 2: Limits

$$p_0 = \lim_{x \to 0} (-(1+x)) = -1, \quad q_0 = \lim_{x \to 0} 1 = 1$$

Step 3: Indicial equation

$$r(r-1) - r + 1 = 0$$ $$r^2 - 2r + 1 = 0$$ $$(r - 1)^2 = 0$$

Wait—this is also a repeated root $r = 1$. But actually, re-examining the original problem: $-r + 1$ appears. The recurrence relation should be checked to see if the second solution truly requires logarithm (Case 2) or if we are in a subtle Case 3 scenario.

Note: In practice, once you have $r_1 = r_2 = 1$, you must substitute both the first Frobenius solution and the logarithmic form into the original equation to determine if the logarithmic term actually appears. This is a common pitfall in Frobenius method applications.

Solution

Step 1: Verify regularity

Divide by $x^2$:

$$y'' + \frac{1}{x}y' + \left(1 - \frac{1}{4x^2}\right)y = 0$$

Here $p(x) = \frac{1}{x}$ and $q(x) = 1 - \frac{1}{4x^2}$.

Check: $xp(x) = 1$ ✓ and $x^2 q(x) = x^2 - \frac{1}{4}$ ✓. Regular singular point ✓

Step 2: Limits

$$p_0 = 1, \quad q_0 = -\frac{1}{4}$$

Step 3: Indicial equation

$$r(r-1) + r - \frac{1}{4} = 0$$ $$r^2 - \frac{1}{4} = 0$$ $$r = \pm \frac{1}{2}$$

Distinct roots: $r_1 = \frac{1}{2}$ and $r_2 = -\frac{1}{2}$. Difference $r_1 - r_2 = 1 \in \mathbb{Z}$ (Case 3: difference is a positive integer).

Step 4: Solutions

For Bessel's equation of order $\nu$, the recurrence relations yield:

$$y_1 = x^{1/2} J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin x$$ $$y_2 = x^{-1/2} J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \cos x$$

Remarkable fact: Although roots differ by an integer, the second solution does not contain a logarithmic term! The general solution is:

$$y(x) = c_1 \sqrt{\frac{2}{\pi x}} \sin x + c_2 \sqrt{\frac{2}{\pi x}} \cos x$$
Warning

When $r_1 - r_2 \in \mathbb{Z}$, you cannot conclude whether a logarithm appears without checking the recurrence relation. The logarithmic term appears sometimes (Case 2 and most of Case 3) but not always (as here).

Practice: Self-Assessment Quiz

Test your understanding of the Frobenius method with these multiple-choice questions. After answering all questions, your score will be displayed.

0 / 6 correct
Question 1: Which of the following is a regular singular point?
Question 2: For the indicial equation $r^2 - 3r + 2 = 0$, what are the roots?
Question 3: If the indicial roots are $r_1 = 3$ and $r_2 = 1/2$, which case applies?
Question 4: For Bessel's equation of order 0, $x^2 y'' + xy' + x^2 y = 0$, the indicial equation is:
Question 5: For the Cauchy-Euler equation $x^2 y'' + 3xy' + y = 0$ with repeated root $r = -1$, the general solution is:
Question 6: If $r_1 - r_2 = 0$ (repeated root), which solution form is guaranteed?

Final Score: 0 / 6

Review any incorrect answers above and refer to the Theory and Examples sections for clarification.

Quick Reference Card

Regular Singular Point Test

For $y'' + p(x)y' + q(x)y = 0$ at $x = x_0$:

(x-x₀)p(x) and (x-x₀)²q(x) must be analytic at $x_0$.

Indicial Equation Form

$$r(r-1) + p_0 r + q_0 = 0$$

where $p_0 = \lim_{x \to x_0}(x-x_0)p(x)$ and $q_0 = \lim_{x \to x_0}(x-x_0)^2 q(x)$

Case 1: Distinct Roots

If $r_1 - r_2 \notin \mathbb{Z}$:

Two Frobenius series, both analytic, no logarithms.

Case 2: Repeated Root

If $r_1 = r_2 = r$:

$y_2 = y_1\ln(x-x_0) + \ldots$ (logarithm always present)

Case 3: Integer Difference

If $r_1 - r_2 = m \in \mathbb{Z}^+$:

Logarithm may or may not appear; use recurrence to check.

Frobenius Series Form

$$y = (x-x_0)^r \sum_{n=0}^{\infty} a_n(x-x_0)^n$$

Exponent $r$ determined by indicial equation roots.