7.3: Singular Points and the Method of Frobenius

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Examples

While behavior of ODEs at singular points is more complicated, certain singular points are not especially difficult to solve. Let us look at some examples before giving a general method. We may be lucky and obtain a power series solution using the method of the previous section, but in general we may have to try other things.

Example $$\PageIndex{1}$$

Let us first look at a simple first order equation

$2 x y' - y = 0 . \label{ex1eq1}$

Note that $$x=0$$ is a singular point. If we only try to plug in

$y = \sum_{k=0}^\infty a_k x^k ,\label{ex1eq2}$

we obtain

\begin{align}\begin{aligned} 0 = 2 xy'-y &=2x \, \left( \sum_{k=1}^\infty k a_k x^{k-1} \right)-\left( \sum_{k=0}^\infty a_k x^k \right) \\ &=a_0 +\sum_{k=1}^\infty (2 k a_k - a_k) \, x^{k} . \end{aligned}\end{align} \label{ex1eq3}

First, $$a_0 = 0$$. Next, the only way to solve $$0 = 2 k a_k - a_k = (2k-1) \, a_k$$ for $$k = 1,2,3,\dots$$ is for $$a_k = 0$$ for all $$k$$. Therefore we only get the trivial solution $$y=0$$. We need a nonzero solution to get the general solution.

Let us try $$y=x^r$$ for some real number $$r$$. Consequently our solution---if we can find one---may only make sense for positive $$x$$. Then $$y' = r x^{r-1}$$. So

$0 = 2 x y' - y = 2 x r x^{r-1} - x^r= (2r-1) x^r . \nonumber$

Therefore $$r= \dfrac{1}{2}$$, or in other words $$y = x^{1/2}$$. Multiplying by a constant, the general solution for positive $$x$$ is

$y = C x^{1/2} . \nonumber$

If $$C \not= 0$$ then the derivative of the solution "blows up" at $$x=0$$ (the singular point). There is only one solution that is differentiable at $$x=0$$ and that's the trivial solution $$y=0$$.

Not every problem with a singular point has a solution of the form $$y=x^r$$, of course. But perhaps we can combine the methods. What we will do is to try a solution of the form

$y = x^r f(x) \nonumber$

where $$f(x)$$ is an analytic function.

Example $$\PageIndex{2}$$

Suppose that we have the equation

$4 x^2 y'' - 4 x^2 y' + (1-2x)y = 0, \label{ex2eq1}$

and again note that $$x=0$$ is a singular point. Let us try

$y = x^r \sum_{k=0}^\infty a_k x^k = \sum_{k=0}^\infty a_k x^{k+r} , \label{ex2eq2}$

where $$r$$ is a real number, not necessarily an integer. Again if such a solution exists, it may only exist for positive $$x$$. First let us find the derivatives

\begin{align} y' &= \sum_{k=0}^\infty (k+r)\, a_k x^{k+r-1} ,\nonumber \\ y'' &= \sum_{k=0}^\infty (k+r)\,(k+r-1)\, a_k x^{k+r-2} . \label{ex2eq3} \end{align}

Plugging Equations \ref{ex2eq2} - \ref{ex2eq3} into our original differential equation (Equation \ref{ex2eq1}) we obtain

\begin{align}\begin{aligned} 0 &= 4x^2y''-4x^2y'+(1-2x)y \\ &= 4x^2 \, \left( \sum_{k=0}^\infty (k+r)\,(k+r-1) \, a_k x^{k+r-2} \right)-4x^2 \, \left( \sum_{k=0}^\infty (k+r) \, a_k x^{k+r-1} \right)+(1-2x) \left( \sum_{k=0}^\infty a_k x^{k+r} \right) \\ &=\left( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \right)-\left( \sum_{k=0}^\infty 4 (k+r) \, a_k x^{k+r+1} \right)+\left( \sum_{k=0}^\infty a_k x^{k+r} \right)-\left( \sum_{k=0}^\infty 2a_k x^{k+r+1} \right) \\ &=\left( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \right)-\left( \sum_{k=1}^\infty 4 (k+r-1) \, a_{k-1} x^{k+r} \right)+\left( \sum_{k=0}^\infty a_k x^{k+r} \right)-\left( \sum_{k=1}^\infty 2a_{k-1} x^{k+r} \right)\\ &=4r(r-1) \, a_0 x^r + a_0 x^r +\sum_{k=1}^\infty \left( 4 (k+r)\,(k+r-1) \, a_k - 4 (k+r-1) \, a_{k-1}+a_k -2a_{k-1} \right) \, x^{k+r}\\ &=\left( 4r(r-1) + 1 \right) \, a_0 x^r +\sum_{k=1}^\infty\left( \left( 4 (k+r)\,(k+r-1) + 1 \right) \, a_k-\left( 4 (k+r-1) + 2 \right) \, a_{k-1} \right) \, x^{k+r} .\end{aligned}\end{align} \nonumber

To have a solution we must first have $$\left( 4r(r-1) + 1 \right) \, a_0 = 0$$. Supposing that $$a_0 \not= 0$$ we obtain

$4r(r-1) + 1 = 0 . \nonumber$

This equation is called the indicial equation. This particular indicial equation has a double root at $$r = \dfrac{1}{2}$$.

OK, so we know what $$r$$ has to be. That knowledge we obtained simply by looking at the coefficient of $$x^r$$. All other coefficients of $$x^{k+r}$$ also have to be zero so

$\left( 4 (k+r)\,(k+r-1) + 1 \right) \, a_k- \left( 4 (k+r-1) + 2 \right) \, a_{k-1} = 0 . \nonumber$

If we plug in $$r=\dfrac{1}{2}$$ and solve for $$a_k$$ we get

$a_k=\dfrac{4 (k+\dfrac{1}{2}-1) + 2}{4 (k+\dfrac{1}{2})\,(k+\dfrac{1}{2}-1) + 1} \, a_{k-1}=\dfrac{1}{k} \, a_{k-1} . \nonumber$

Let us set $$a_0 = 1$$. Then

$a_1 = \dfrac{1}{1} a_0 = 1 , \qquad a_2 = \dfrac{1}{2} a_1 = \dfrac{1}{2} ,\qquad a_3 = \dfrac{1}{3} a_2 = \dfrac{1}{3 \cdot 2} ,\qquad a_4 = \dfrac{1}{4} a_3 = \dfrac{1}{4 \cdot 3 \cdot 2} , \qquad \dots \nonumber$

Extrapolating, we notice that

$a_k = \dfrac{1}{k(k-1)(k-2) \cdots 3 \cdot 2} = \dfrac{1}{k!} . \nonumber$

In other words,

$y =\sum_{k=0}^\infty a_k x^{k+r}=\sum_{k=0}^\infty \dfrac{1}{k!} x^{k+1/2}=x^{1/2} \sum_{k=0}^\infty \dfrac{1}{k!} x^{k}=x^{1/2}e^x . \nonumber$

That was lucky! In general, we will not be able to write the series in terms of elementary functions. We have one solution, let us call it $$y_1 = x^{1/2} e^x$$. But what about a second solution? If we want a general solution, we need two linearly independent solutions. Picking $$a_0$$ to be a different constant only gets us a constant multiple of $$y_1$$, and we do not have any other $$r$$ to try; we only have one solution to the indicial equation. Well, there are powers of $$x$$ floating around and we are taking derivatives, perhaps the logarithm (the antiderivative of $$x^{-1}$$) is around as well. It turns out we want to try for another solution of the form

$y_2 = \sum_{k=0}^\infty b_k x^{k+r} + (\ln x) y_1 , \nonumber$

which in our case is

$y_2 = \sum_{k=0}^\infty b_k x^{k+1/2} + (\ln x) x^{1/2} e^x . \nonumber$

We now differentiate this equation, substitute into the differential equation and solve for $$b_k$$. A long computation ensues and we obtain some recursion relation for $$b_k$$. The reader can (and should) try this to obtain for example the first three terms

$b_1 = b_0 -1 , \qquad b_2 = \dfrac{2b_1-1}{4} , \qquad b_3 = \dfrac{6b_2-1}{18} , \qquad \ldots \nonumber$

We then fix $$b_0$$ and obtain a solution $$y_2$$. Then we write the general solution as $$y = A y_1 + B y_2$$.

Method of Frobenius

Before giving the general method, let us clarify when the method applies. Let

$p(x) y'' + q(x) y' + r(x) y = 0 \nonumber$

be an ODE. As before, if $$p(x_0) = 0$$, then $$x_0$$ is a singular point. If, furthermore, the limits

$\lim_{x \to x_0} ~ (x-x_0) \dfrac{q(x)}{p(x)} \qquad \text{and} \qquad \lim_{x \to x_0} ~ (x-x_0)^2 \dfrac{r(x)}{p(x)} \nonumber$

both exist and are finite, then we say that $$x_0$$ is a regular singular point.

Example $$\PageIndex{3}$$: Expansion around a regular singular point

Often, and for the rest of this section, $$x_0 = 0$$. Consider

$x^2y'' + x(1+x)y' + (\pi+x^2)y = 0 . \nonumber$

Write

\begin{align}\begin{aligned} \lim_{x \to 0} ~x \dfrac{q(x)}{p(x)} &=\lim_{x \to 0} ~x \dfrac{x(1+x)}{x^2} = \lim_{x \to 0} ~(1+x) = 1 , \\ \lim_{x \to 0} ~x^2 \dfrac{r(x)}{p(x)} &=\lim_{x \to 0} ~x^2 \frac{(\pi+x^2)}{x^2} = \lim_{x \to 0} ~(\pi+x^2) = \pi \end{aligned}\end{align}. \nonumber

So $$x = 0$$ is a regular singular point.

On the other hand if we make the slight change

$x^2y'' + (1+x)y' + (\pi+x^2)y = 0 , \nonumber$

then

$\lim_{x \to 0} ~x \dfrac{q(x)}{p(x)} =\lim_{x \to 0} ~x \dfrac{(1+x)}{x^2} = \lim_{x \to 0} ~\dfrac{1+x}{x} =\text{DNE}. \nonumber$

Here DNE stands for does not exist. The point $$0$$ is a singular point, but not a regular singular point.

Let us now discuss the general Method of Frobenius$$^{1}$$. Let us only consider the method at the point $$x=0$$ for simplicity. The main idea is the following theorem.

Theorem $$\PageIndex{1}$$

Method of Frobenius

Suppose that

$\label{eq:26} p(x) y'' + q(x) y' + r(x) y = 0$

has a regular singular point at $$x=0$$, then there exists at least one solution of the form

$y = x^r \sum_{k=0}^\infty a_k x^k . \nonumber$

A solution of this form is called a Frobenius-type solution.

The method usually breaks down like this.

1. We seek a Frobenius-type solution of the form $y = \sum_{k=0}^\infty a_k x^{k+r} . \nonumber$ We plug this $$y$$ into equation $$\eqref{eq:26}$$. We collect terms and write everything as a single series.
2. The obtained series must be zero. Setting the first coefficient (usually the coefficient of $$x^r$$) in the series to zero we obtain the indicial equation, which is a quadratic polynomial in $$r$$.
3. If the indicial equation has two real roots $$r_1$$ and $$r_2$$ such that $$r_1 - r_2$$ is not an integer, then we have two linearly independent Frobenius-type solutions. Using the first root, we plug in $y_1 = x^{r_1} \sum_{k=0}^\infty a_k x^{k} , \nonumber$ and we solve for all $$a_k$$ to obtain the first solution. Then using the second root, we plug in $y_2 = x^{r_2} \sum_{k=0}^\infty b_k x^{k} , \nonumber$ and solve for all $$b_k$$ to obtain the second solution.
4. If the indicial equation has a doubled root $$r$$, then there we find one solution $y_1 = x^{r} \sum_{k=0}^\infty a_k x^{k} , \nonumber$ and then we obtain a new solution by plugging $y_2 = x^{r} \sum_{k=0}^\infty b_k x^{k} + (\ln x) y_1 , \nonumber$ into Equation $$\eqref{eq:26}$$ and solving for the constants $$b_k$$.
5. If the indicial equation has two real roots such that $$r_1-r_2$$ is an integer, then one solution is $y_1 = x^{r_1} \sum_{k=0}^\infty a_k x^{k} , \nonumber$ and the second linearly independent solution is of the form $y_2 = x^{r_2} \sum_{k=0}^\infty b_k x^{k} + C (\ln x) y_1 , \nonumber$ where we plug $$y_2$$ into $$\eqref{eq:26}$$ and solve for the constants $$b_k$$ and $$C$$.
6. Finally, if the indicial equation has complex roots, then solving for $$a_k$$ in the solution $y = x^{r_1} \sum_{k=0}^\infty a_k x^{k} \nonumber$ results in a complex-valued function---all the $$a_k$$ are complex numbers. We obtain our two linearly independent solutions$$^{2}$$ by taking the real and imaginary parts of $$y$$.

The main idea is to find at least one Frobenius-type solution. If we are lucky and find two, we are done. If we only get one, we either use the ideas above or even a different method such as reduction of order (Exercise 2.1.8) to obtain a second solution.

Bessel Functions

An important class of functions that arises commonly in physics are the Bessel functions$$^{3}$$. For example, these functions appear when solving the wave equation in two and three dimensions. First we have Bessel's equation of order $$p$$:

$x^2 y'' + xy' + \left(x^2 - p^2\right)y = 0 . \nonumber$

We allow $$p$$ to be any number, not just an integer, although integers and multiples of $$\dfrac{1}{2}$$ are most important in applications. When we plug

$y = \sum_{k=0}^\infty a_k x^{k+r} \nonumber$

into Bessel's equation of order $$p$$ we obtain the indicial equation

$r(r-1)+r-p^2 = (r-p)(r+p) = 0 . \nonumber$

Therefore we obtain two roots $$r_1 = p$$ and $$r_2 = -p$$. If $$p$$ is not an integer following the method of Frobenius and setting $$a_0 = 1$$, we obtain linearly independent solutions of the form

\begin{align}\begin{aligned} y_1 &= x^p \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k}}{2^{2k}k!(k+p)(k-1+p) \cdots (2+p)(1+p)}, \\ y_2 &= x^{-p} \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k}}{2^{2k}k!(k-p)(k-1-p) \cdots (2-p)(1-p)}.\end{aligned}\end{align} \nonumber

Exercise $$\PageIndex{1}$$
1. Verify that the indicial equation of Bessel's equation of order $$p$$ is $$(r-p)(r+p)=0$$.
2. Suppose that $$p$$ is not an integer. Carry out the computation to obtain the solutions $$y_1$$ and $$y_2$$ above.

Bessel functions will be convenient constant multiples of $$y_1$$ and $$y_2$$. First we must define the gamma function

$\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt . \nonumber$

Notice that $$\Gamma(1) = 1$$. The gamma function also has a wonderful property

$\Gamma(x+1) = x \Gamma(x) . \nonumber$

From this property, one can show that $$\Gamma(n) = (n-1)!$$ when $$n$$ is an integer, so the gamma function is a continuous version of the factorial. We compute:

\begin{align}\begin{aligned} \Gamma(k+p+1)=(k+p)(k-1+p)\cdots (2+p)(1+p) \Gamma(1+p) ,\\ \Gamma(k-p+1)=(k-p)(k-1-p)\cdots (2-p)(1-p) \Gamma(1-p) .\end{aligned}\end{align} \nonumber

Exercise $$\PageIndex{2}$$

Verify the above identities using $$\Gamma(x+1) = x \Gamma(x)$$.

We define the Bessel functions of the first kind of order $$p$$ and $$-p$$ as

\begin{align}\begin{aligned} J_p(x) &= \dfrac{1}{2^p\Gamma(1+p)} y_1=\sum_{k=0}^\infty \dfrac{{(-1)}^k}{k! \Gamma(k+p+1)}{\left(\dfrac{x}{2}\right)}^{2k+p} , \\ J_{-p}(x) &= \dfrac{1}{2^{-}\Gamma(1-p)} y_2=\sum_{k=0}^\infty \dfrac{{(-1)}^k}{k! \Gamma(k-p+1)}{\left(\dfrac{x}{2}\right)}^{2k-p} .\end{aligned}\end{align} \nonumber

As these are constant multiples of the solutions we found above, these are both solutions to Bessel's equation of order $$p$$. The constants are picked for convenience.

When $$p$$ is not an integer, $$J_p$$ and $$J_{-p}$$ are linearly independent. When $$n$$ is an integer we obtain

$J_n(x) =\sum_{k=0}^\infty \dfrac{{(-1)}^k}{k! (k+n)!}{\left(\dfrac{x}{2}\right)}^{2k+n} . \nonumber$

In this case it turns out that

$J_n(x) = {(-1)}^nJ_{-n}(x) , \nonumber$

and so we do not obtain a second linearly independent solution. The other solution is the so-called Bessel function of second kind. These make sense only for integer orders $$n$$ and are defined as limits of linear combinations of $$J_p(x)$$ and $$J_{-p}(x)$$ as $$p$$ approaches $$n$$ in the following way:

$Y_n(x) = \lim_{p\to n} \dfrac{\cos(p \pi) J_p(x) - J_{-p}(x)}{\sin(p \pi)} . \nonumber$

As each linear combination of $$J_p(x)$$ and $$J_{-p}(x)$$ is a solution to Bessel's equation of order $$p$$, then as we take the limit as $$p$$ goes to $$n$$, $$Y_n(x)$$ is a solution to Bessel's equation of order $$n$$. It also turns out that $$Y_n(x)$$ and $$J_n(x)$$ are linearly independent. Therefore when $$n$$ is an integer, we have the general solution to Bessel's equation of order $$n$$

$y = A J_n(x) + B Y_n(x) , \nonumber$

for arbitrary constants $$A$$ and $$B$$. Note that $$Y_n(x)$$ goes to negative infinity at $$x=0$$. Many mathematical software packages have these functions $$J_n(x)$$ and $$Y_n(x)$$ defined, so they can be used just like say $$\sin(x)$$ and $$\cos(x)$$. In fact, they have some similar properties. For example, $$-J_1(x)$$ is a derivative of $$J_0(x)$$, and in general the derivative of $$J_n(x)$$ can be written as a linear combination of $$J_{n-1}(x)$$ and $$J_{n+1}(x)$$. Furthermore, these functions oscillate, although they are not periodic. See Figure $$\PageIndex{1}$$ for graphs of Bessel functions.

Example $$\PageIndex{4}$$: Using Bessel functions to Solve a ODE

Other equations can sometimes be solved in terms of the Bessel functions. For example, given a positive constant $$\lambda$$,

$x y'' + y' + \lambda^2 x y = 0 , \nonumber$

can be changed to $$x^2 y'' + x y' + \lambda^2 x^2 y = 0$$. Then changing variables $$t = \lambda x$$ we obtain via chain rule the equation in $$y$$ and $$t$$:

$t^2 y'' + t y' + t^2 y = 0 , \nonumber$

which can be recognized as Bessel's equation of order 0. Therefore the general solution is $$y(t) = A J_0(t) + B Y_0(t)$$, or in terms of $$x$$:

$y = A J_0(\lambda x) + B Y_0(\lambda x) . \nonumber$

This equation comes up for example when finding fundamental modes of vibration of a circular drum, but we digress.

Footnotes

[1] Named after the German mathematician Ferdinand Georg Frobenius (1849 – 1917).

[2] See Joseph L. Neuringera, The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, Volume 9, Issue 1, 1978, 71–77.

[3] Named after the German astronomer and mathematician Friedrich Wilhelm Bessel (1784 – 1846).

This page titled 7.3: Singular Points and the Method of Frobenius is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.