Chapter 6 · Section 6.2

6.2 Solutions About Singular Points: The Frobenius Method

Section 6.1 gave us power series solutions about ordinary points. But some of the most important places in a physical problem are exactly the singular points: the center of a circular drum, the axis of a cylinder, the origin of a spherical wave. There the coefficients of the equation blow up, and the solutions often do too. An ordinary power series cannot describe a solution that behaves like $1/ x$ or $ln x$ near the point, so we need a more flexible form.

A familiar example shows what to expect. The Cauchy–Euler equation (Section 4.5)

2 x^{2} y^{''} + 3 x y^{'} - y = 0

has $x = 0$ as a singular point. Substituting $y = x^{m}$ gives $2 m (m - 1) + 3 m - 1 = (2 m - 1) (m + 1) = 0$ , so the solutions are $x^{1/2}$ and $x^{- 1}$ . Neither is a power series in $x$ : one has a fractional power, the other a negative one. The key observation of this section, due to Frobenius, is that for a large and important class of singular points the solutions look like a power of $x$ times a power series:

y = x^{r} (a_{0} + a_{1} x + a_{2} x^{2} + \dots),

where the exponent $r$ need not be a nonnegative integer. Sometimes a logarithm appears as well, and we will see exactly when.

Regular and irregular singular points

As before, write the equation in normalized form $y^{''} + p (x) y^{'} + q (x) y = 0$ .

Definition 6.2.1 (Regular and irregular singular points)

A singular point $x_{0}$ of $y^{''} + p (x) y^{'} + q (x) y = 0$ is a regular singular point if both

(x - x_{0}) p (x) and (x - x_{0})^{2} q (x)

are analytic at $x_{0}$ . A singular point that is not regular is called irregular.

In words: at a regular singular point, $p$ may blow up no worse than $1/ (x - x_{0})$ and $q$ no worse than $1/ (x - x_{0})^{2}$ . This is exactly how the coefficients of a Cauchy–Euler equation behave at $0$ , and that is why the Cauchy–Euler idea $y = x^{r}$ is the right starting point. At an irregular singular point the solutions can behave in far wilder ways, and the method of this section does not apply.

Multiplying the normalized equation by $(x - x_{0})^{2}$ gives a convenient standard form. Taking $x_{0} = 0$ (we can always shift to it with $t = x - x_{0}$ ):

x^{2} y^{''} + x P (x) y^{'} + Q (x) y = 0, P (x) = x p (x), Q (x) = x^{2} q (x) .

So $x = 0$ is a regular singular point exactly when the equation can be written this way with $P$ and $Q$ analytic at $0$ . When $P$ and $Q$ are constants, this is a Cauchy–Euler equation.

Example 6.2.1

Find and classify the singular points of

x^{2} (x - 2)^{2} y^{''} + 3 x y^{'} + (x - 2) y = 0.

Solution

Normalize. $p (x) = \frac{3 x}{x ^{2} ( x - 2 ) ^{2}} = \frac{3}{x ( x - 2 ) ^{2}}, q (x) = \frac{x - 2}{x ^{2} ( x - 2 ) ^{2}} = \frac{1}{x ^{2} ( x - 2 )} .$ The singular points are $x = 0$ and $x = 2$ .
At $x = 0$ . $x p (x) = \frac{3}{( x - 2 ) ^{2}}$ and $x^{2} q (x) = \frac{1}{x - 2}$ . Both are rational functions whose denominators do not vanish at $0$ , so both are analytic there: $x = 0$ is a regular singular point.
At $x = 2$ . $(x - 2) p (x) = \frac{3}{x ( x - 2 )}$ still has $x - 2$ in the denominator, so it is not analytic at $2$ . That is enough: $x = 2$ is an irregular singular point. (We need not even check $q$ .)

The Frobenius series and the indicial equation

Let $x = 0$ be a regular singular point, with the equation in standard form and

P (x) = p_{0} + p_{1} x + p_{2} x^{2} + \dots, Q (x) = q_{0} + q_{1} x + q_{2} x^{2} + \dots .

We look for a solution on $x > 0$ of the form

y = x^{r} n = 0 \sum \infty a_{n} x^{n} = n = 0 \sum \infty a_{n} x^{n + r}, a_{0} \neq = 0.

The condition $a_{0} \neq = 0$ is not a restriction: it simply says that $x^{r}$ is the lowest power that actually appears. We call such a series a Frobenius series.

Differentiate term by term. Because $r$ is in general not an integer, no terms drop out, and all sums still start at $n = 0$ :

y^{'} = n = 0 \sum \infty (n + r) a_{n} x^{n + r - 1}, y^{''} = n = 0 \sum \infty (n + r) (n + r - 1) a_{n} x^{n + r - 2} .

x^{2} y^{''} = n = 0 \sum \infty (n + r) (n + r - 1) a_{n} x^{n + r}, x y^{'} = n = 0 \sum \infty (n + r) a_{n} x^{n + r} .

The lowest power. Substitute into $x^{2} y^{''} + x P y^{'} + Q y = 0$ . The lowest power of $x$ that can appear is $x^{r}$ , and it comes only from the $n = 0$ terms multiplied by the constant terms $p_{0}$ , $q_{0}$ :

[r (r - 1) + p_{0} r + q_{0}] a_{0} x^{r} .

Its coefficient must vanish, and $a_{0} \neq = 0$ , so $r$ must satisfy

F (r) := r (r - 1) + p_{0} r + q_{0} = 0.

Definition 6.2.2 (Indicial equation)

For $x^{2} y^{''} + x P (x) y^{'} + Q (x) y = 0$ with $p_{0} = P (0)$ , $q_{0} = Q (0)$ , the quadratic equation

r (r - 1) + p_{0} r + q_{0} = 0

is the indicial equation; its roots $r_{1}, r_{2}$ are the exponents (or indicial roots) at the singular point. When both are real we label them so that $r_{1} \geq r_{2}$ .

Notice that this is exactly the Cauchy–Euler indicial equation of Section 4.5 for the "frozen" equation $x^{2} y^{''} + p_{0} x y^{'} + q_{0} y = 0$ . Near $x = 0$ the equation behaves like this Cauchy–Euler equation, and the exponents tell us how the solutions behave there: like $x^{r_{1}}$ and $x^{r_{2}}$ , possibly with a logarithm.

The higher powers. Collecting the coefficient of $x^{n + r}$ for $n \geq 1$ gives

F (n + r) a_{n} = - k = 1 \sum n [(n - k + r) p_{k} + q_{k}] a_{n - k}, n \geq 1.

You do not need to memorize this; in practice we substitute directly, as in the examples. But it shows the crucial point: each $a_{n}$ is found by dividing by $F (n + r)$ . Everything is fine as long as $F (n + r) \neq = 0$ .

For the larger root $r = r_{1}$ : $F (r_{1} + n) = 0$ would mean that $r_{1} + n$ is also a root. But the only roots are $r_{1}$ and $r_{2} \leq r_{1}$ , and $r_{1} + n > r_{1}$ . So $F (r_{1} + n) \neq = 0$ for all $n \geq 1$ , and the larger root always gives a Frobenius solution.
For the smaller root $r = r_{2}$ : $F (r_{2} + n) = 0$ exactly when $r_{2} + n = r_{1}$ , that is, when $n = r_{1} - r_{2}$ . So trouble can arise only if $r_{1} - r_{2}$ is a positive integer, at the step $n = N = r_{1} - r_{2}$ . And if $r_{1} = r_{2}$ , the two "solutions" coincide, so the second one must be found another way.

This is the whole story behind the famous three cases.

Figure 6.2.1. The three possibilities for real indicial roots. What matters is the difference

r_{1} - r_{2}

: not an integer (Case 1), zero (Case 2), or a positive integer (Case 3).

Frobenius' theorem: the three cases

Theorem 6.2.1 (Frobenius)

Let $x = 0$ be a regular singular point of $y^{''} + p (x) y^{'} + q (x) y = 0$ , and suppose the Taylor series of $P (x) = x p (x)$ and $Q (x) = x^{2} q (x)$ about $0$ both converge for $∣ x ∣ < R$ , where $R > 0$ . Let $r_{1}, r_{2}$ be the roots of the indicial equation, real, with $r_{1} \geq r_{2}$ . Then on $0 < x < R$ there are two linearly independent solutions $y_{1}, y_{2}$ , where always

y_{1} = x^{r_{1}} n = 0 \sum \infty a_{n} x^{n}, a_{0} = 1,

and $y_{2}$ has the following form.

Case 1: $r_{1} - r_{2}$ is not an integer. $y_{2} = x^{r_{2}} n = 0 \sum \infty b_{n} x^{n}, b_{0} = 1.$
Case 2: $r_{1} = r_{2}$ . $y_{2} = y_{1} (x) ln x + x^{r_{1}} n = 1 \sum \infty b_{n} x^{n} .$
Case 3: $r_{1} - r_{2} = N$ , a positive integer. $y_{2} = C y_{1} (x) ln x + x^{r_{2}} n = 0 \sum \infty b_{n} x^{n}, b_{0} = 1,$ where the constant $C$ may turn out to be zero.

All the power series that appear converge at least for $∣ x ∣ < R$ . The general solution on $0 < x < R$ is $y = c_{1} y_{1} + c_{2} y_{2}$ .

Some remarks that make the theorem easier to use:

Finding $R$ . For polynomial coefficients, $R$ is at least the distance from $0$ to the nearest other singular point (complex ones included), just as in Section 6.1. If there is no other singular point, $R = \infty$ .
Negative $x$ . On $- R < x < 0$ the same statements hold with $x^{r}$ replaced by $∣ x ∣^{r}$ and $ln x$ by $ln ∣ x ∣$ .
Complex exponents. If $r_{1, 2} = α \pm i β$ with $β \neq = 0$ , the difference is not an integer, so Case 1 applies with complex coefficients; the real and imaginary parts of $y_{1}$ give two real solutions (compare the complex case of Cauchy–Euler, Section 4.5).
Case 3 with $C = 0$ . In Case 3 the step $n = N$ of the recurrence for $r_{2}$ reads $0 \cdot b_{N} = (something)$ . If the "something" is not zero, the equation is impossible, there is no second Frobenius series, and the logarithm is needed ( $C \neq = 0$ ). If the "something" happens to be zero, $b_{N}$ can be chosen freely and $C = 0$ : we get two Frobenius series after all.
Behaviour at $0$ . The exponents tell you which solutions stay bounded as $x \to 0^{+}$ . In applications (temperature at the center of a disc, displacement at the center of a drum) we usually keep only the bounded solution.

How to find the logarithmic solution

In Cases 2 and 3 there are two practical routes.

Substitute the form. Put $y_{2} = C y_{1} ln x + x^{r_{2}} \sum b_{n} x^{n}$ into the equation and solve for $C$ and the $b_{n}$ . The logarithms cancel because $y_{1}$ is a solution. The following identity, which you can check with the product rule, does most of the work: for $L [y] = x^{2} y^{''} + x P y^{'} + Q y$ , $L [y_{1} ln x] = ln x L [y_{1}] + 2 x y_{1}^{'} + (P - 1) y_{1} = 2 x y_{1}^{'} + (P - 1) y_{1} .$ Example 6.2.4 uses this route.
Reduction of order (Section 4.1): $y_{2} = y_{1} \int \frac{e ^{- \int p d x}}{y _{1}^{2}} d x$ , expanding the integrand in a series before integrating. Example 6.2.3 uses this route; it is convenient when $y_{1}$ has a closed form.

(A third route for Case 2, elegant but more abstract, is to keep $r$ as a variable in the series and differentiate with respect to $r$ at $r = r_{1}$ .)

Method in a nutshell

Check that $x = 0$ is a regular singular point; find $R$ .
Read off $p_{0} = P (0)$ and $q_{0} = Q (0)$ from the standard form and solve the indicial equation $r (r - 1) + p_{0} r + q_{0} = 0$ . Decide the case from $r_{1} - r_{2}$ .
Substitute $y = \sum a_{n} x^{n + r}$ (all sums from $n = 0$ ), shift so every sum contains the same power, and get the recurrence with $r$ still a letter. The lowest power reproduces the indicial equation, which is a useful check.
Put $r = r_{1}$ in the recurrence and compute $a_{1}, a_{2}, \dots$ (with $a_{0} = 1$ ). This gives $y_{1}$ .
Case 1: put $r = r_{2}$ and repeat to get $y_{2}$ . Case 3: try $r = r_{2}$ ; if the step $n = N$ is consistent, you get $y_{2}$ with $C = 0$ ; otherwise use the logarithmic form. Case 2: use the logarithmic form.
Write $y = c_{1} y_{1} + c_{2} y_{2}$ on $0 < x < R$ .

The solver below accepts the standard form $x^{2} y^{''} + x P (x) y^{'} + Q (x) y = 0$ with polynomial $P$ and $Q$ . It forms the indicial equation, classifies the case, derives the recurrence and computes the first terms. The first preset is Example 6.2.2 (multiplied by $x /3$ to reach standard form); the second is Example 6.2.4, where the logarithm is unavoidable. Enter your own $P$ and $Q$ to explore the three cases.

Worked examples

Example 6.2.2 (Case 1)

Find two linearly independent Frobenius solutions of $3 x y^{''} + 2 y^{'} + y = 0$ near $x = 0$ .

Solution

Classify. $p = \frac{2}{3 x}$ and $q = \frac{1}{3 x}$ , so $x = 0$ is singular. But $x p = \frac{2}{3}$ and $x^{2} q = \frac{x}{3}$ are analytic: regular singular point. There is no other singular point, so $R = \infty$ .
Indicial equation. $p_{0} = \frac{2}{3}$ , $q_{0} = 0$ : $r (r - 1) + \frac{2}{3} r = r (r - \frac{1}{3}) = 0 ⟹ r_{1} = \frac{1}{3}, r_{2} = 0.$ The difference $\frac{1}{3}$ is not an integer: Case 1.
Substitute $y = \sum_{n \geq 0} a_{n} x^{n + r}$ into the original equation (this avoids fractions): $n = 0 \sum \infty 3 (n + r) (n + r - 1) a_{n} x^{n + r - 1} + n = 0 \sum \infty 2 (n + r) a_{n} x^{n + r - 1} + n = 0 \sum \infty a_{n} x^{n + r} = 0.$
Combine the first two sums. $3 (n + r) (n + r - 1) + 2 (n + r) = (n + r) (3 n + 3 r - 1)$ .
Shift the third sum to the power $x^{n + r - 1}$ : replace $n$ by $n - 1$ , so it becomes $\sum_{n = 1}^{\infty} a_{n - 1} x^{n + r - 1}$ .
Peel off and combine. $r (3 r - 1) a_{0} x^{r - 1} + n = 1 \sum \infty [(n + r) (3 n + 3 r - 1) a_{n} + a_{n - 1}] x^{n + r - 1} = 0.$ The first term gives $r (3 r - 1) = 0$ , the same indicial roots (good check).
Recurrence: $(n + r) (3 n + 3 r - 1) a_{n} = - a_{n - 1}$ for $n \geq 1$ .
$r = r_{1} = \frac{1}{3}$ . Then $(n + \frac{1}{3}) (3 n) = n (3 n + 1)$ , so $a_{n} = - \frac{a _{n - 1}}{n ( 3 n + 1 )} : a_{1} = - \frac{1}{4}, a_{2} = \frac{1}{4} \cdot \frac{1}{2 \cdot 7} = \frac{1}{56}, a_{3} = - \frac{1}{56} \cdot \frac{1}{3 \cdot 10} = - \frac{1}{1680} .$
$r = r_{2} = 0$ . Then $n (3 n - 1) a_{n} = - a_{n - 1}$ : $a_{1} = - \frac{1}{2}, a_{2} = \frac{1}{2} \cdot \frac{1}{2 \cdot 5} = \frac{1}{20}, a_{3} = - \frac{1}{20} \cdot \frac{1}{3 \cdot 8} = - \frac{1}{480} .$
General solution for $x > 0$ : $y = c_{1} x^{1/3} (1 - \frac{x}{4} + \frac{x ^{2}}{56} - \frac{x ^{3}}{1680} + \dots) + c_{2} (1 - \frac{x}{2} + \frac{x ^{2}}{20} - \frac{x ^{3}}{480} + \dots) .$ Both series converge for all $x$ (the ratio of consecutive coefficients tends to $0$ ).

Notice that $y_{2}$ is an ordinary power series (its exponent is $0$ ) while $y_{1}$ is not: $y_{1} \approx x^{1/3}$ has a vertical tangent at the origin. Both behaviours are visible in Figure 6.2.2.

Figure 6.2.2. The two Frobenius solutions of Example 6.2.2.

y_{1}

starts like

x^{1/3}

, with infinite slope at

0

;

y_{2}

starts at

1

with slope

- \frac{1}{2}

. Both are bounded near the origin.

Example 6.2.3 (Case 2: equal roots)

Solve $x y^{''} + (1 - x) y^{'} - y = 0$ near $x = 0$ .

Solution

Classify. Multiplying by $x$ gives the standard form $x^{2} y^{''} + x (1 - x) y^{'} - x y = 0$ , with $P = 1 - x$ , $Q = - x$ . So $x = 0$ is a regular singular point and $R = \infty$ .
Indicial equation. $p_{0} = 1$ , $q_{0} = 0$ : $r (r - 1) + r = r^{2} = 0$ , a double root $r_{1} = r_{2} = 0$ . Case 2.
Substitute. The terms $x y^{''} + y^{'}$ together give $\sum_{n \geq 0} (n + r)^{2} a_{n} x^{n + r - 1}$ . The terms $- x y^{'} - y$ give $- \sum_{n \geq 0} (n + r + 1) a_{n} x^{n + r}$ , which after the shift $n \to n - 1$ becomes $- \sum_{n \geq 1} (n + r) a_{n - 1} x^{n + r - 1}$ .
Recurrence. $(n + r)^{2} a_{n} = (n + r) a_{n - 1}$ , so $a_{n} = \frac{a _{n - 1}}{n + r}$ for $n \geq 1$ .
First solution. With $r = 0$ : $a_{n} = a_{n - 1} / n$ , hence $a_{n} = 1/ n!$ and $y_{1} = n = 0 \sum \infty \frac{x ^{n}}{n !} = e^{x} .$ Check: $x e^{x} + (1 - x) e^{x} - e^{x} = 0$ .
Second solution by reduction of order. In normalized form $p (x) = \frac{1 - x}{x} = \frac{1}{x} - 1$ , so $\int p d x = ln x - x$ and $e^{- \int p d x} = \frac{e ^{x}}{x}$ . Then $y_{2} = y_{1} \int \frac{e ^{- \int p d x}}{y _{1}^{2}} d x = e^{x} \int \frac{e ^{x}}{x e ^{2 x}} d x = e^{x} \int \frac{e ^{- x}}{x} d x .$
Expand and integrate. $\frac{e ^{- x}}{x} = \frac{1}{x} (1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots) = \frac{1}{x} - 1 + \frac{x}{2} - \frac{x ^{2}}{6} + \dots,$ so, term by term, $\int \frac{e ^{- x}}{x} d x = ln x - x + \frac{x ^{2}}{4} - \frac{x ^{3}}{18} + \dots = ln x + n = 1 \sum \infty \frac{( - 1 ) ^{n} x ^{n}}{n \cdot n !} .$
Multiply by $e^{x}$ . $y_{2} = e^{x} ln x + (1 + x + \frac{x ^{2}}{2} + \dots) (- x + \frac{x ^{2}}{4} - \frac{x ^{3}}{18} + \dots) = e^{x} ln x - x - \frac{3}{4} x^{2} - \frac{11}{36} x^{3} - \dots$ (for example, the $x^{2}$ coefficient is $\frac{1}{4} - 1 = - \frac{3}{4}$ ). This is exactly the form promised in Case 2: $y_{1} ln x$ plus a power series starting at $n = 1$ .
General solution for $x > 0$ : $y = c_{1} e^{x} + c_{2} (e^{x} ln x - x - \frac{3}{4} x^{2} - \frac{11}{36} x^{3} - \dots) .$

As $x \to 0^{+}$ , $y_{2} \to - \infty$ like $ln x$ , while $y_{1} \to 1$ . So the only solutions that stay bounded at the origin are the multiples of $e^{x}$ . (For the curious: $y_{2} = - e^{x} [E_{1} (x) + γ]$ , where $E_{1}$ is the exponential integral and $γ \approx 0.5772$ is Euler's constant.)

Figure 6.2.3. Example 6.2.3. The second solution inherits the logarithmic singularity of

ln x

(dashed) at the origin; the first solution

e^{x}

is perfectly smooth. Equal indicial roots always produce such a logarithm.

Example 6.2.4 (Case 3: the logarithm is needed)

Find two linearly independent solutions of $x y^{''} + 3 y^{'} - y = 0$ near $x = 0$ . (In Example 6.1.5 we expanded the same equation about the ordinary point $x = 1$ ; now we work at the singular point itself.)

Solution

Classify. Standard form: $x^{2} y^{''} + 3 x y^{'} - x y = 0$ , so $P = 3$ , $Q = - x$ : regular singular point, $R = \infty$ .
Indicial equation. $r (r - 1) + 3 r = r (r + 2) = 0$ : $r_{1} = 0$ , $r_{2} = - 2$ , and $N = r_{1} - r_{2} = 2$ . Case 3.
Recurrence. Substituting $\sum a_{n} x^{n + r}$ : the terms $x y^{''} + 3 y^{'}$ give $\sum (n + r) (n + r + 2) a_{n} x^{n + r - 1}$ and $- y$ gives (after shifting) $- \sum_{n \geq 1} a_{n - 1} x^{n + r - 1}$ . So $(n + r) (n + r + 2) a_{n} = a_{n - 1}, n \geq 1.$
First solution ( $r = 0$ ). $n (n + 2) a_{n} = a_{n - 1}$ : $a_{1} = \frac{1}{3}, a_{2} = \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{24}, a_{3} = \frac{1}{24} \cdot \frac{1}{15} = \frac{1}{360}, \dots, a_{n} = \frac{2}{n ! ( n + 2 )!} .$ So $y_{1} = 1 + \frac{x}{3} + \frac{x ^{2}}{24} + \frac{x ^{3}}{360} + \dots$ .
Try $r = - 2$ . The recurrence becomes $n (n - 2) a_{n} = a_{n - 1}$ . For $n = 1$ : $- a_{1} = a_{0}$ , so $a_{1} = - a_{0}$ . For $n = 2$ : $0 \cdot a_{2} = a_{1} = - a_{0}$ . Since $a_{0} \neq = 0$ , this is impossible. There is no second Frobenius series, so $C \neq = 0$ .
The logarithmic form. Write $y_{2} = C y_{1} ln x + w$ with $w = \sum_{n \geq 0} b_{n} x^{n - 2}$ , $b_{0} = 1$ . For our operator $L [y] = x y^{''} + 3 y^{'} - y$ , the product rule gives $L [y_{1} ln x] = ln x L [y_{1}] + x (\frac{2 y _{1}^{'}}{x} - \frac{y _{1}}{x ^{2}}) + \frac{3 y _{1}}{x} = 2 y_{1}^{'} + \frac{2 y _{1}}{x},$ since $L [y_{1}] = 0$ . So $L [y_{2}] = 0$ becomes $L [w] = - C (2 y_{1}^{'} + \frac{2 y _{1}}{x})$ .
Both sides as series in $x^{n - 3}$ . On the left, as in step 3 with $r = - 2$ , $L [w] = n = 0 \sum \infty [n (n - 2) b_{n} - b_{n - 1}] x^{n - 3} (b_{- 1} = 0) .$ On the right, $2 y_{1}^{'} + 2 y_{1} / x = \sum_{m \geq 0} 2 (m + 1) a_{m} x^{m - 1}$ ; with $m = n - 2$ this is $\sum_{n \geq 2} 2 (n - 1) a_{n - 2} x^{n - 3}$ .
Match coefficients. For each $n$ : $n (n - 2) b_{n} - b_{n - 1} = - 2 C (n - 1) a_{n - 2}$ (right side zero for $n < 2$ ). $n = 0 : n = 1 : n = 2 : n = 3 : n = 4 : 0 = 0 (b_{0} = 1 is free), - b_{1} - b_{0} = 0 \Rightarrow b_{1} = - 1, 0 \cdot b_{2} - b_{1} = - 2 C a_{0} \Rightarrow 1 = - 2 C \Rightarrow C = - \frac{1}{2}, 3 b_{3} - b_{2} = - 2 C \cdot 2 a_{1} = \frac{2}{3}, 8 b_{4} - b_{3} = - 2 C \cdot 3 a_{2} = \frac{1}{8} .$
The free coefficient. At $n = 2$ the coefficient $b_{2}$ is multiplied by $0$ , so it is arbitrary. Changing it only adds a multiple of $y_{1}$ to $y_{2}$ (look: $b_{2} x^{2 - 2} = b_{2}$ is the start of $b_{2} y_{1}$ ). We choose $b_{2} = 0$ . Then $b_{3} = \frac{2}{9}$ and $b_{4} = \frac{1}{8} (\frac{2}{9} + \frac{1}{8}) = \frac{25}{576}$ .
Result. $y_{2} = - \frac{1}{2} y_{1} (x) ln x + \frac{1}{x ^{2}} (1 - x + \frac{2}{9} x^{3} + \frac{25}{576} x^{4} + \dots),$ and the general solution is $y = c_{1} y_{1} + c_{2} y_{2}$ for $x > 0$ .

Near the origin $y_{2} \approx x^{- 2}$ : it is the exponent $r_{2} = - 2$ that sets the strength of the singularity; the logarithm is a milder correction.

Example 6.2.5 (Case 3 without a logarithm)

Solve $x^{2} y^{''} + 2 x y^{'} + (x^{2} - 2) y = 0$ near $x = 0$ , and identify the solutions in closed form.

Solution

Indicial equation. $P = 2$ , $Q = x^{2} - 2$ , so $p_{0} = 2$ , $q_{0} = - 2$ : $F (r) = r (r - 1) + 2 r - 2 = r^{2} + r - 2 = (r + 2) (r - 1) .$ $r_{1} = 1$ , $r_{2} = - 2$ , $N = 3$ : Case 3.
Recurrence. Substituting $\sum a_{n} x^{n + r}$ , the terms $x^{2} y^{''} + 2 x y^{'} - 2 y$ give $\sum F (n + r) a_{n} x^{n + r}$ and $x^{2} y$ gives $\sum_{n \geq 2} a_{n - 2} x^{n + r}$ . Hence $F (r) a_{0} = 0, F (r + 1) a_{1} = 0, F (n + r) a_{n} = - a_{n - 2} (n \geq 2) .$
$r = 1$ . $F (n + 1) = (n + 3) n$ . Since $F (2) = 4 \neq = 0$ , $a_{1} = 0$ , and then all odd coefficients vanish. For even $n$ : $a_{n} = - \frac{a _{n - 2}}{n ( n + 3 )}$ , giving $a_{2} = - \frac{1}{10}$ , $a_{4} = \frac{1}{280}$ , $a_{6} = - \frac{1}{15120}$ : $y_{1} = x (1 - \frac{x ^{2}}{10} + \frac{x ^{4}}{280} - \frac{x ^{6}}{15120} + \dots) .$
$r = - 2$ . Now $F (n - 2) = n (n - 3)$ : $n = 1 : n = 2 : n = 3 : - 2 a_{1} = 0 \Rightarrow a_{1} = 0, - 2 a_{2} = - a_{0} \Rightarrow a_{2} = \frac{1}{2}, 0 \cdot a_{3} = - a_{1} = 0.$ The dangerous step $n = N = 3$ reads $0 = 0$ . It is consistent, $a_{3}$ is arbitrary, and no logarithm is needed ( $C = 0$ ). Choose $a_{3} = 0$ (a nonzero choice would only add $a_{3} y_{1}$ ).
Continue. $4 a_{4} = - a_{2}$ gives $a_{4} = - \frac{1}{8}$ ; $18 a_{6} = - a_{4}$ gives $a_{6} = \frac{1}{144}$ : $y_{2} = x^{- 2} (1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{8} + \frac{x ^{6}}{144} - \dots) .$
Closed forms. One can check by direct substitution that $y_{1} = \frac{3 ( sin x - x cos x )}{x ^{2}}, y_{2} = \frac{cos x + x sin x}{x ^{2}}$ are solutions, and expanding $sin$ and $cos$ reproduces the two series above. For instance $cos x + x sin x = 1 + (1 - \frac{1}{2}) x^{2} + (\frac{1}{24} - \frac{1}{6}) x^{4} + \dots = 1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \dots$ .

Why did the logarithm disappear? The recurrence links $a_{n}$ to $a_{n - 2}$ , so odd and even coefficients never interact. The trouble spot $n = 3$ is odd, and all odd coefficients were already zero. Whenever the recurrence has step 2 and $N$ is odd, Case 3 comes without a logarithm. These two functions are, up to constant factors, the spherical Bessel functions of order 1; they and their combinations describe standing and outgoing waves around a sphere in acoustics and electromagnetics.

Figure 6.2.4. Example 6.2.5. The solution for the larger exponent

r_{1} = 1

vanishes at the origin; the one for

r_{2} = - 2

blows up like

x^{- 2}

. The dashed curve is the three-term Frobenius polynomial for

y_{1}

, accurate for small

x

Common mistakes

Applying the method at an irregular singular point. Always check that $x p$ and $x^{2} q$ are analytic first.
Forgetting that in a Frobenius series no terms drop out on differentiation: the sums for $y^{'}$ and $y^{''}$ start at $n = 0$ , unlike in Section 6.1.
Starting with the smaller root in Case 3. Always build $y_{1}$ from the larger root; it never fails.
Concluding "logarithm needed" in Case 3 without checking the step $n = N$ . As Example 6.2.5 shows, it may be consistent.
Reading $p_{0}, q_{0}$ from the wrong form. They are the constant terms of $P$ and $Q$ in $x^{2} y^{''} + x P y^{'} + Q y = 0$ , so first divide or multiply to reach this form.

Exercises

Find and classify the singular points: (a) $x (x + 1)^{2} y^{''} + (x + 1) y^{'} - 2 y = 0$ ; (b) $x^{3} (x - 4) y^{''} + x y^{'} + 5 y = 0$ .
Answer
(a) $x = 0$ and $x = - 1$ , both regular. (b) $x = 0$ irregular ( $x p = \frac{1}{x ( x - 4 )}$ is not analytic at $0$ ); $x = 4$ regular.
Find the indicial roots and decide which case of Theorem 6.2.1 applies (do not solve): (a) $2 x^{2} y^{''} + 7 x y^{'} + (2 + x^{2}) y = 0$ ; (b) $x^{2} y^{''} + x (3 - 2 x) y^{'} + (1 - x^{2}) y = 0$ ; (c) $x^{2} y^{''} + x (x - 4) y^{'} + (6 + x) y = 0$ .
Answer
(a) $r = - \frac{1}{2}, - 2$ : Case 1. (b) $r = - 1$ (double): Case 2. (c) $r = 3, 2$ : Case 3 with $N = 1$ .
Find two Frobenius solutions of $2 x y^{''} + 3 y^{'} - y = 0$ (terms through $x^{3}$ inside the brackets).
Answer
$r = 0, - \frac{1}{2}$ ; $y_{1} = 1 + \frac{x}{3} + \frac{x ^{2}}{30} + \frac{x ^{3}}{630} + \dots$ , $y_{2} = x^{- 1/2} (1 + x + \frac{x ^{2}}{6} + \frac{x ^{3}}{90} + \dots)$ . (In fact $y_{1} = \frac{sinh 2 x}{2 x}$ , $y_{2} = \frac{cosh 2 x}{x}$ .)
Find two Frobenius solutions of $3 x^{2} y^{''} - x y^{'} + (1 - x) y = 0$ .
Answer
$r = 1, \frac{1}{3}$ ; $y_{1} = x (1 + \frac{x}{5} + \frac{x ^{2}}{80} + \frac{x ^{3}}{2640} + \dots)$ , $y_{2} = x^{1/3} (1 + x + \frac{x ^{2}}{8} + \frac{x ^{3}}{168} + \dots)$ .
Show that $y_{1} = e^{2 x}$ solves $x y^{''} + (1 - 2 x) y^{'} - 2 y = 0$ , and find the second solution in the form $y_{1} ln x + \sum_{n \geq 1} b_{n} x^{n}$ (terms through $x^{3}$ ).
Answer
$r = 0$ (double); $y_{2} = e^{2 x} ln x - 2 x - 3 x^{2} - \frac{22}{9} x^{3} - \dots$ $(= e^{2 x} \int e^{- 2 x} x^{- 1} d x)$ .
For $x^{2} y^{''} + 3 x y^{'} + (1 - 2 x) y = 0$ , find the Frobenius solution $y_{1}$ (general term) and write down the form of a second solution.
Answer
$r = - 1$ (double); $y_{1} = \frac{1}{x} n = 0 \sum \infty \frac{2 ^{n} x ^{n}}{( n ! ) ^{2}} = \frac{1}{x} (1 + 2 x + x^{2} + \frac{2}{9} x^{3} + \dots)$ ; $y_{2} = y_{1} ln x + x^{- 1} \sum_{n \geq 1} b_{n} x^{n}$ .
For $x y^{''} - y^{'} + y = 0$ : find $y_{1}$ , show that the smaller root does not give a Frobenius series, and find $C$ and the first terms of $y_{2}$ (take $b_{0} = 1$ , $b_{2} = 0$ ).
Answer
$r = 2, 0$ ; $y_{1} = x^{2} (1 - \frac{x}{3} + \frac{x ^{2}}{24} - \frac{x ^{3}}{360} + \dots)$ ; for $r = 0$ the step $n = 2$ gives $0 = - a_{0}$ , impossible. $C = - \frac{1}{2}$ , $y_{2} = - \frac{1}{2} y_{1} ln x + 1 + x - \frac{2}{9} x^{3} + \dots$
Solve $x^{2} y^{''} - 2 x y^{'} + 9 x^{2} y = 0$ . Show that Case 3 occurs but no logarithm is needed, and find both solutions in closed form.
Answer
$r = 3, 0$ ( $N = 3$ , recurrence with step 2); $y = c_{1} (cos 3 x + 3 x sin 3 x) + c_{2} (sin 3 x - 3 x cos 3 x)$ .
Explain why the Frobenius method cannot be used at $x = 0$ for $x^{2} y^{''} + y^{'} - y = 0$ .
Answer
$p = 1/ x^{2}$ , so $x p = 1/ x$ is not analytic at $0$ : the singular point is irregular.
Without solving, give an interval on which the Frobenius series about $0$ of $x^{2} (x^{2} + 16) y^{''} + x y^{'} + y = 0$ are guaranteed to converge.
Answer
$P = Q = \frac{1}{x ^{2} + 16}$ , analytic for $∣ x ∣ < 4$ (singular points $\pm 4 i$ ): the series converge on $0 < x < 4$ (and on $- 4 < x < 0$ with $∣ x ∣^{r}$ ).