The Frobenius method · Differential Equations

Intuition

At a singular point a power series can fail, but at the mildest kind — a regular singular point, such as the origin for Bessel’s equation — a small change rescues the method. Multiply the series by a power of x whose exponent r is to be found: the lowest power in the equation then gives a quadratic for r, the indicial equation, and each root starts a series. The exponent need not be a whole number, so a solution can behave like the square root of x near the singular point, or like a negative power.

A key that almost fits a lock. The power series is the key; a regular singular point is a lock with one extra ward, and the factor x to the r is the notch filed into the key so that it turns.

$2xy''+y'+y=0$ has a regular singular point at $0$ and the exponents $0$ and $\tfrac12$ . The two Frobenius solutions are $\cos\sqrt{2x}$ and $\tfrac{1}{\sqrt2}\sin\sqrt{2x}$ ; the second starts like $\sqrt{x}$ , with a vertical tangent that no power series could have.

A power times a power series

A singular point $x_{0}$ is regular if $(x-x_{0})\,p(x)$ and are analytic there. At a regular singular point , try with . The lowest power gives the indicial equation below, where and are the values at of and . Its larger root always gives a solution; the smaller gives a second when the roots do not differ by a whole number — otherwise the second solution may need a logarithm, which is left to a later course.

y=x^{r}\sum_{n=0}^{\infty}a_{n}x^{n},\qquad r(r-1)+p_{0}r+q_{0}=0

Exponents at a singular point

$2xy''+y'+y=0$ : and , so , , and gives and .

The indicial equation

\begin{aligned}&y=x^{r}\textstyle\sum a_{n}x^{n},\ a_{0}\ne0\\&\Longrightarrow\ r(r-1)+p_{0}r+q_{0}=0\end{aligned}

Multiply the equation by x squared, so that its coefficients become x p and x squared q, both power series at a regular singular point. Substituting the trial solution, each term starts at the power x to the r: the second derivative with r times r minus 1, the first derivative with p0 times r, and y itself with q0. The coefficient of that lowest power must vanish, and since a0 is not zero, r must solve the indicial equation.

Proof steps

x^{2}y''+x\,(xp)\,y'+(x^{2}q)\,y=0

Multiply the standard form by x squared.

x^{2}y''=r(r-1)a_{0}x^{r}+\cdots,\quad xy'=ra_{0}x^{r}+\cdots

The lowest power in each term is x to the r.

\bigl[r(r-1)+p_{0}r+q_{0}\bigr]a_{0}x^{r}+\cdots=0

Near zero, x p starts with p0 and x squared q with q0.

a_{0}\ne0\ \Longrightarrow\ r(r-1)+p_{0}r+q_{0}=0

The coefficient of the lowest power must vanish.

Applications

Practice

Regular Singular Points

A singular point is regular when x times p and x squared times q are still power series there. Then a power times a power series works.

xp(x),\ x^{2}q(x)\ \text{analytic at}\ 0

Try it

Which equation has a regular singular point at $0$ ?