# Introduction to Differential Equations

## Section4.3Singular points and the method of Frobenius

Note: 1 or 1.5 lectures, §8.4 and §8.5 in [EP], §5.4–§5.7 in [BD]
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.

### Subsection4.3.1Examples

#### Example4.3.1.

Let us first look at a simple first order equation
\begin{equation*} 2 x y' - y = 0 . \end{equation*}
Note that $$x=0$$ is a singular point. If we try to plug in
\begin{equation*} y = \sum_{k=0}^\infty a_k x^k , \end{equation*}
we obtain
\begin{equation*} \begin{split} 0 = 2 xy'-y &= 2x \, \Biggl( \sum_{k=1}^\infty k a_k x^{k-1} \Biggr) - \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ & = a_0 + \sum_{k=1}^\infty (2 k a_k - a_k) \, x^{k} . \end{split} \end{equation*}
First, $$a_0 = 0\text{.}$$ 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 $$k\text{.}$$ Therefore, in this manner we only get the trivial solution $$y=0\text{.}$$ We need a nonzero solution to get the general solution to the equation.
Let us try $$y=x^r$$ for some real number $$r\text{.}$$ Consequently our solution—if we can find one—may only make sense for positive $$x\text{.}$$ Then $$y' = r x^{r-1}\text{.}$$ So
\begin{equation*} 0 = 2 x y' - y = 2 x r x^{r-1} - x^r = (2r-1) x^r . \end{equation*}
Therefore $$r= \nicefrac{1}{2}\text{,}$$ or in other words $$y = x^{1/2}\text{.}$$ Multiplying by a constant, the general solution for positive $$x$$ is
\begin{equation*} y = C x^{1/2} . \end{equation*}
If $$C \not= 0\text{,}$$ 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 $$y=0\text{.}$$
Not every problem with a singular point has a solution of the form $$y=x^r\text{,}$$ of course. But perhaps we can combine the methods. What we will do is to try a solution of the form
\begin{equation*} y = x^r f(x) , \end{equation*}
where $$f(x)$$ is an analytic function.

#### Example4.3.2.

Consider the equation
\begin{equation*} 4 x^2 y'' - 4 x^2 y' + (1-2x)y = 0, \end{equation*}
and again note that $$x=0$$ is a singular point.
Let us try
\begin{equation*} y = x^r \sum_{k=0}^\infty a_k x^k = \sum_{k=0}^\infty a_k x^{k+r} , \end{equation*}
where $$r$$ is a real number, not necessarily an integer. Again if such a solution exists, it may only exist for positive $$x\text{.}$$ First let us find the derivatives
\begin{equation*} \begin{aligned} y' & = \sum_{k=0}^\infty (k+r)\, a_k x^{k+r-1} , \\ y'' & = \sum_{k=0}^\infty (k+r)\,(k+r-1)\, a_k x^{k+r-2} . \end{aligned} \end{equation*}
Plugging into our equation we obtain
\begin{equation*} \begin{split} 0 & = 4x^2y''-4x^2y'+(1-2x)y \\ &= 4x^2 \, \Biggl( \sum_{k=0}^\infty (k+r)\,(k+r-1) \, a_k x^{k+r-2} \Biggr) - 4x^2 \, \Biggl( \sum_{k=0}^\infty (k+r) \, a_k x^{k+r-1} \Biggr) + (1-2x) \Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr) \\ &= \Biggl( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \Biggr) \\ & \phantom{mmm} - \Biggl( \sum_{k=0}^\infty 4 (k+r) \, a_k x^{k+r+1} \Biggr) + \Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr) - \Biggl( \sum_{k=0}^\infty 2a_k x^{k+r+1} \Biggr) \\ &= \Biggl( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \Biggr) \\ & \phantom{mmm} - \Biggl( \sum_{k=1}^\infty 4 (k+r-1) \, a_{k-1} x^{k+r} \Biggr) + \Biggl( \sum_{k=0}^\infty a_k x^{k+r} \Biggr) - \Biggl( \sum_{k=1}^\infty 2a_{k-1} x^{k+r} \Biggr) \\ &= 4r(r-1) \, a_0 x^r + a_0 x^r + \sum_{k=1}^\infty \Bigl( 4 (k+r)\,(k+r-1) \, a_k - 4 (k+r-1) \, a_{k-1} + a_k - 2a_{k-1} \Bigr) \, x^{k+r} \\ &= \bigl( 4r(r-1) + 1 \bigr) \, a_0 x^r + \sum_{k=1}^\infty \Bigl( \bigl( 4 (k+r)\,(k+r-1) + 1 \bigr) \, a_k - \bigl( 4 (k+r-1) + 2 \bigr) \, a_{k-1} \Bigr) \, x^{k+r} . \end{split} \end{equation*}
To have a solution we must first have $$\bigl( 4r(r-1) + 1 \bigr) \, a_0 = 0\text{.}$$ Supposing that $$a_0 \not= 0$$ we obtain
\begin{equation*} 4r(r-1) + 1 = 0 . \end{equation*}
This equation is called the indicial equation. This particular indicial equation has a double root at $$r = \nicefrac{1}{2}\text{.}$$
OK, so we know what $$r$$ has to be. That knowledge we obtained simply by looking at the coefficient of $$x^r\text{.}$$ All other coefficients of $$x^{k+r}$$ also have to be zero so
\begin{equation*} \bigl( 4 (k+r)\,(k+r-1) + 1 \bigr) \, a_k - \bigl( 4 (k+r-1) + 2 \bigr) \, a_{k-1} = 0 . \end{equation*}
If we plug in $$r=\nicefrac{1}{2}$$ and solve for $$a_k\text{,}$$ we get
\begin{equation*} a_k = \frac{4 (k+\nicefrac{1}{2}-1) + 2}{4 (k+\nicefrac{1}{2})\,(k+\nicefrac{1}{2}-1) + 1} \, a_{k-1} = \frac{1}{k} \, a_{k-1} . \end{equation*}
Let us set $$a_0 = 1\text{.}$$ Then
\begin{equation*} a_1 = \frac{1}{1} a_0 = 1 , \qquad a_2 = \frac{1}{2} a_1 = \frac{1}{2} , \qquad a_3 = \frac{1}{3} a_2 = \frac{1}{3 \cdot 2} , \qquad a_4 = \frac{1}{4} a_3 = \frac{1}{4 \cdot 3 \cdot 2} , \qquad \cdots \end{equation*}
Extrapolating, we notice that
\begin{equation*} a_k = \frac{1}{k(k-1)(k-2) \cdots 3 \cdot 2} = \frac{1}{k!} . \end{equation*}
In other words,
\begin{equation*} y = \sum_{k=0}^\infty a_k x^{k+r} = \sum_{k=0}^\infty \frac{1}{k!} x^{k+1/2} = x^{1/2} \sum_{k=0}^\infty \frac{1}{k!} x^{k} = x^{1/2} e^x . \end{equation*}
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 $$y_1 = x^{1/2} e^x\text{.}$$ 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 $$y_1\text{,}$$ 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
\begin{equation*} y_2 = \sum_{k=0}^\infty b_k x^{k+r} + (\ln x) y_1 , \end{equation*}
which in our case is
\begin{equation*} y_2 = \sum_{k=0}^\infty b_k x^{k+1/2} + (\ln x) x^{1/2} e^x . \end{equation*}
We now differentiate this equation, substitute into the differential equation and solve for $$b_k\text{.}$$ A long computation ensues and we obtain some recursion relation for $$b_k\text{.}$$ The reader can (and should) try this to obtain for example the first three terms
\begin{equation*} b_1 = b_0 -1 , \qquad b_2 = \frac{2b_1-1}{4} , \qquad b_3 = \frac{6b_2-1}{18} , \qquad \ldots \end{equation*}
We then fix $$b_0$$ and obtain a solution $$y_2\text{.}$$ Then we write the general solution as $$y = A y_1 + B y_2\text{.}$$

### Subsection4.3.2The method of Frobenius

Before giving the general method, let us clarify when the method applies. Let
\begin{equation*} p(x) y'' + q(x) y' + r(x) y = 0 \end{equation*}
be an ODE. As before, if $$p(x_0) = 0\text{,}$$ then $$x_0$$ is a singular point. If, furthermore, the limits
\begin{equation*} \lim_{x \to x_0} ~ (x-x_0) \frac{q(x)}{p(x)} \qquad \text{and} \qquad \lim_{x \to x_0} ~ (x-x_0)^2 \frac{r(x)}{p(x)} \end{equation*}
both exist and are finite, then we say that $$x_0$$ is a regular singular point.

#### Example4.3.3.

Often, and for the rest of this section, $$x_0 = 0\text{.}$$ Consider
\begin{equation*} x^2y'' + x(1+x)y' + (\pi+x^2)y = 0 . \end{equation*}
Write
\begin{equation*} \begin{aligned} & \lim_{x \to 0} ~x \frac{q(x)}{p(x)} = \lim_{x \to 0} ~x \frac{x(1+x)}{x^2} = \lim_{x \to 0} ~(1+x) = 1 , \\ & \lim_{x \to 0} ~x^2 \frac{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{equation*}
So $$x = 0$$ is a regular singular point.
On the other hand if we make the slight change
\begin{equation*} x^2y'' + (1+x)y' + (\pi+x^2)y = 0 , \end{equation*}
then
\begin{equation*} \lim_{x \to 0} ~x \frac{q(x)}{p(x)} = \lim_{x \to 0} ~x \frac{(1+x)}{x^2} = \lim_{x \to 0} ~\frac{1+x}{x} = \text{DNE}. \end{equation*}
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 . We only consider the method at the point $$x=0$$ for simplicity. The main idea is the following theorem.
The method usually breaks down like this:
1. We seek a Frobenius-type solution of the form
\begin{equation*} y = \sum_{k=0}^\infty a_k x^{k+r} . \end{equation*}
We plug this $$y$$ into equation (4.3). 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 $$r\text{.}$$
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
\begin{equation*} y_1 = x^{r_1} \sum_{k=0}^\infty a_k x^{k} , \end{equation*}
and we solve for all $$a_k$$ to obtain the first solution. Then using the second root, we plug in
\begin{equation*} y_2 = x^{r_2} \sum_{k=0}^\infty b_k x^{k} , \end{equation*}
and solve for all $$b_k$$ to obtain the second solution.
4. If the indicial equation has a doubled root $$r\text{,}$$ then there we find one solution
\begin{equation*} y_1 = x^{r} \sum_{k=0}^\infty a_k x^{k} , \end{equation*}
and then we obtain a new solution by plugging
\begin{equation*} y_2 = x^{r} \sum_{k=0}^\infty b_k x^{k} + (\ln x) y_1 , \end{equation*}
into equation (4.3) and solving for the constants $$b_k\text{.}$$
5. If the indicial equation has two real roots such that $$r_1-r_2$$ is an integer, then one solution is
\begin{equation*} y_1 = x^{r_1} \sum_{k=0}^\infty a_k x^{k} , \end{equation*}
and the second linearly independent solution is of the form
\begin{equation*} y_2 = x^{r_2} \sum_{k=0}^\infty b_k x^{k} + C (\ln x) y_1 , \end{equation*}
where we plug $$y_2$$ into (4.3) and solve for the constants $$b_k$$ and $$C\text{.}$$
6. Finally, if the indicial equation has complex roots, then solving for $$a_k$$ in the solution
\begin{equation*} y = x^{r_1} \sum_{k=0}^\infty a_k x^{k} \end{equation*}
results in a complex-valued function—all the $$a_k$$ are complex numbers. We obtain our two linearly independent solutions 3  by taking the real and imaginary parts of $$y\text{.}$$
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 (see Section 2.1) to obtain a second solution.

### Subsection4.3.3Bessel functions

An important class of functions that arises commonly in physics are the Bessel functions 4 . For example, these functions appear when solving the wave equation in two and three dimensions. First consider Bessel’s equation of order $$p\text{:}$$
\begin{equation*} x^2 y'' + xy' + \left(x^2 - p^2\right)y = 0 . \end{equation*}
We allow $$p$$ to be any number, not just an integer, although integers and multiples of $$\nicefrac{1}{2}$$ are most important in applications.
When we plug
\begin{equation*} y = \sum_{k=0}^\infty a_k x^{k+r} \end{equation*}
into Bessel’s equation of order $$p\text{,}$$ we obtain the indicial equation
\begin{equation*} r(r-1)+r-p^2 = (r-p)(r+p) = 0 . \end{equation*}
Therefore we obtain two roots $$r_1 = p$$ and $$r_2 = -p\text{.}$$ If $$p$$ is not an integer, then following the method of Frobenius and setting $$a_0 = 1\text{,}$$ we obtain linearly independent solutions of the form
\begin{equation*} \begin{aligned} & y_1 = x^p \sum_{k=0}^\infty \frac{{(-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 \frac{{(-1)}^k x^{2k}}{2^{2k} k! (k-p)(k-1-p)\cdots (2-p)(1-p)} . \end{aligned} \end{equation*}

#### Exercise4.3.1.

1. Verify that the indicial equation of Bessel’s equation of order $$p$$ is $$(r-p)(r+p)=0\text{.}$$
2. Suppose $$p$$ is not an integer. Carry out the computation to obtain the solutions $$y_1$$ and $$y_2$$ above.
Bessel functions are convenient constant multiples of $$y_1$$ and $$y_2\text{.}$$ First we must define the gamma function
\begin{equation*} \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt . \end{equation*}
Notice that $$\Gamma(1) = 1\text{.}$$ The gamma function also has a wonderful property
\begin{equation*} \Gamma(x+1) = x \Gamma(x) . \end{equation*}
From this property, it follows that $$\Gamma(n) = (n-1)!$$ when $$n$$ is an integer. So the gamma function is a continuous version of the factorial. We compute:
\begin{equation*} \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{equation*}

#### Exercise4.3.2.

Verify the identities above using $$\Gamma(x+1) = x \Gamma(x)\text{.}$$
We define the Bessel functions of the first kind of order $$p$$ and $$-p$$ as
\begin{equation*} \begin{aligned} & J_p(x) = \frac{1}{2^p\Gamma(1+p)} y_1 = \sum_{k=0}^\infty \frac{{(-1)}^k}{k! \, \Gamma(k+p+1)} {\left(\frac{x}{2}\right)}^{2k+p} , \\ & J_{-p}(x) = \frac{1}{2^{-p}\Gamma(1-p)} y_2 = \sum_{k=0}^\infty \frac{{(-1)}^k}{k! \,\Gamma(k-p+1)} {\left(\frac{x}{2}\right)}^{2k-p} . \end{aligned} \end{equation*}
As these are constant multiples of the solutions we found above, these are both solutions to Bessel’s equation of order $$p\text{.}$$ 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
\begin{equation*} J_n(x) = \sum_{k=0}^\infty \frac{{(-1)}^k}{k! \,(k+n)!} {\left(\frac{x}{2}\right)}^{2k+n} . \end{equation*}
In this case
\begin{equation*} J_n(x) = {(-1)}^nJ_{-n}(x) , \end{equation*}
and so $$J_{-n}$$ is not 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 $$J_{-p}(x)\text{,}$$ as $$p$$ approaches $$n$$ in the following way:
\begin{equation*} Y_n(x) = \lim_{p\to n} \frac{\cos(p \pi) J_p(x) - J_{-p}(x)}{\sin(p \pi)} . \end{equation*}
Each linear combination of $$J_p(x)$$ and $$J_{-p}(x)$$ is a solution to Bessel’s equation of order $$p\text{.}$$ Then as we take the limit as $$p$$ goes to $$n\text{,}$$ we see that $$Y_n(x)$$ is a solution to Bessel’s equation of order $$n\text{.}$$ 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 $$n\text{:}$$
\begin{equation*} y = A J_n(x) + B Y_n(x) , \end{equation*}
for arbitrary constants $$A$$ and $$B\text{.}$$ Note that $$Y_n(x)$$ goes to negative infinity at $$x=0\text{.}$$ 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 $$\cos(x)\text{.}$$ In fact, Bessel functions have some similar properties. For example, $$-J_1(x)$$ is a derivative of $$J_0(x)\text{,}$$ and in general the derivative of $$J_n(x)$$ can be written as a linear combination of $$J_{n-1}(x)$$ and $$J_{n+1}(x)\text{.}$$ Furthermore, these functions oscillate, although they are not periodic. See Figure 4.4 for graphs of Bessel functions.

#### Example4.3.4.

Other equations can sometimes be solved in terms of the Bessel functions. For example, given a positive constant $$\lambda\text{,}$$
\begin{equation*} x y'' + y' + \lambda^2 x y = 0 , \end{equation*}
can be changed to $$x^2 y'' + x y' + \lambda^2 x^2 y = 0\text{.}$$ Then changing variables $$t = \lambda x\text{,}$$ we obtain via chain rule the equation in $$y$$ and $$t\text{:}$$
\begin{equation*} t^2 y'' + t y' + t^2 y = 0 , \end{equation*}
which we recognize as Bessel’s equation of order 0. Therefore the general solution is $$y(t) = A J_0(t) + B Y_0(t)\text{,}$$ or in terms of $$x\text{:}$$
\begin{equation*} y = A J_0(\lambda x) + B Y_0(\lambda x) . \end{equation*}
This equation comes up, for example, when finding the fundamental modes of vibration of a circular drum, but we digress.

### Subsection4.3.4Exercises

#### Exercise4.3.3.

Find a particular (Frobenius-type) solution of $$x^2 y'' + x y' + (1+x) y = 0\text{.}$$

#### Exercise4.3.4.

Find a particular (Frobenius-type) solution of $$x y'' - y = 0\text{.}$$

#### Exercise4.3.5.

Find a particular (Frobenius-type) solution of $$y'' +\frac{1}{x}y' - xy = 0\text{.}$$

#### Exercise4.3.6.

Find the general solution of $$2 x y'' + y' - x^2 y = 0\text{.}$$

#### Exercise4.3.7.

Find the general solution of $$x^2 y'' - x y' -y = 0\text{.}$$

#### Exercise4.3.8.

In the following equations classify the point $$x=0$$ as ordinary, regular singular, or singular but not regular singular.
1. $$\displaystyle x^2(1+x^2)y''+xy=0$$
2. $$\displaystyle x^2y''+y'+y=0$$
3. $$\displaystyle xy''+x^3y'+y=0$$
4. $$\displaystyle xy''+xy'-e^xy=0$$
5. $$\displaystyle x^2y''+x^2y'+x^2y=0$$

#### Exercise4.3.101.

In the following equations classify the point $$x=0$$ as ordinary, regular singular, or singular but not regular singular.
1. $$\displaystyle y''+y=0$$
2. $$\displaystyle x^3y''+(1+x)y=0$$
3. $$\displaystyle xy''+x^5y'+y=0$$
4. $$\displaystyle \sin(x)y''-y=0$$
5. $$\displaystyle \cos(x)y''-\sin(x)y=0$$
a) ordinary, b) singular but not regular singular, c) regular singular, d) regular singular, e) ordinary.

#### Exercise4.3.102.

Find the general solution of $$x^2 y'' -y = 0\text{.}$$
$$y = A x^{\frac{1+\sqrt{5}}{2}} + B x^{\frac{1-\sqrt{5}}{2}}$$

#### Exercise4.3.103.

Find a particular solution of $$x^2 y'' +(x-\nicefrac{3}{4})y = 0\text{.}$$
$$y = x^{3/2} \sum\limits_{k=0}^\infty \frac{{(-1)}^{-1}}{k!\,(k+2)!} x^k$$ (Note that for convenience we did not pick $$a_0 = 1\text{.}$$)
(tricky)   Find the general solution of $$x^2 y'' - x y' +y = 0\text{.}$$
$$y = Ax + B x \ln(x)$$
en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
en.wikipedia.org/wiki/Friedrich_Bessel