Section 7.3 Singular 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.
Subsection 7.3.1 Examples
Example 7.3.1.
Let us first look at a simple first order equation
Note that \(x=0\) is a singular point. If we try to plug in
we obtain
First, \(a_0 = 0\text{.}\) Next, the only way to solve \(0 = 2 k a_k  a_k = (2k1) \, a_k\) for \(k = 1,2,3,\dots\) is for \(a_k = 0\) for all \(k\text{.}\) Therefore we only get the trivial solution \(y=0\text{.}\) We need a nonzero solution to get the general solution.
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 \(y' = r x^{r1}\text{.}\) So
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
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
where \(f(x)\) is an analytic function.
Example 7.3.2.
Consider the equation
and again note that \(x=0\) is a singular point.
Let us try
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
Plugging into our equation we obtain
To have a solution we must first have \(\bigl( 4r(r1) + 1 \bigr) \, a_0 = 0\text{.}\) Supposing that \(a_0 \not= 0\) we obtain
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
If we plug in \(r=\nicefrac{1}{2}\) and solve for \(a_k\text{,}\) we get
Let us set \(a_0 = 1\text{.}\) Then
Extrapolating, we notice that
In other words,
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
which in our case is
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
We then fix \(b_0\) and obtain a solution \(y_2\text{.}\) Then we write the general solution as \(y = A y_1 + B y_2\text{.}\)
Subsection 7.3.2 The method of Frobenius
Before giving the general method, let us clarify when the method applies. Let
be an ODE. As before, if \(p(x_0) = 0\text{,}\) then \(x_0\) is a singular point. If, furthermore, the limits
both exist and are finite, then we say that \(x_0\) is a regular singular point.
Example 7.3.3.
Often, and for the rest of this section, \(x_0 = 0\text{.}\) Consider
Write
So \(x = 0\) is a regular singular point.
On the other hand if we make the slight change
then
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.
Theorem 7.3.1. Method of Frobenius.
Suppose that
has a regular singular point at \(x=0\text{,}\) then there exists at least one solution of the form
A solution of this form is called a Frobeniustype solution.
The method usually breaks down like this:

We seek a Frobeniustype solution of the form
\begin{equation*} y = \sum_{k=0}^\infty a_k x^{k+r} . \end{equation*}We plug this \(y\) into equation (7.3). We collect terms and write everything as a single series.
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 \(r\text{.}\)

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 Frobeniustype 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.

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 (7.3) and solving for the constants \(b_k\text{.}\)

If the indicial equation has two real roots such that \(r_1r_2\) 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 (7.3) and solve for the constants \(b_k\) and \(C\text{.}\)

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 complexvalued function—all the \(a_k\) are complex numbers. We obtain our two linearly independent solutions^{ 2 } by taking the real and imaginary parts of \(y\text{.}\)
The main idea is to find at least one Frobeniustype 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.
Subsection 7.3.3 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 consider Bessel's equation of order \(p\text{:}\)
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
into Bessel's equation of order \(p\text{,}\) we obtain the indicial 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
Exercise 7.3.1.
Verify that the indicial equation of Bessel's equation of order \(p\) is \((rp)(r+p)=0\text{.}\)
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
Notice that \(\Gamma(1) = 1\text{.}\) The gamma function also has a wonderful property
From this property, it follows that \(\Gamma(n) = (n1)!\) when \(n\) is an integer. So the gamma function is a continuous version of the factorial. We compute:
Exercise 7.3.2.
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
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
In this case
and so \(J_{n}\) is not a second linearly independent solution. The other solution is the socalled 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:
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 \(n\text{:}\)
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 \(J_{n1}(x)\) and \(J_{n+1}(x)\text{.}\) Furthermore, these functions oscillate, although they are not periodic. See Figure 7.4 for graphs of Bessel functions.
Example 7.3.4.
Other equations can sometimes be solved in terms of the Bessel functions. For example, given a positive constant \(\lambda\text{,}\)
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 \(t\text{:}\)
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 \(x\text{:}\)
This equation comes up, for example, when finding the fundamental modes of vibration of a circular drum, but we digress.
Subsection 7.3.4 Exercises
Exercise 7.3.3.
Find a particular (Frobeniustype) solution of \(x^2 y'' + x y' + (1+x) y = 0\text{.}\)
Exercise 7.3.4.
Find a particular (Frobeniustype) solution of \(x y''  y = 0\text{.}\)
Exercise 7.3.5.
Find a particular (Frobeniustype) solution of \(y'' +\frac{1}{x}y'  xy = 0\text{.}\)
Exercise 7.3.6.
Find the general solution of \(2 x y'' + y'  x^2 y = 0\text{.}\)
Exercise 7.3.7.
Find the general solution of \(x^2 y''  x y' y = 0\text{.}\)
Exercise 7.3.8.
In the following equations classify the point \(x=0\) as ordinary, regular singular, or singular but not regular singular.
\(x^2(1+x^2)y''+xy=0\)
\(x^2y''+y'+y=0\)
\(xy''+x^3y'+y=0\)
\(xy''+xy'e^xy=0\)
\(x^2y''+x^2y'+x^2y=0\)
Exercise 7.3.101.
In the following equations classify the point \(x=0\) as ordinary, regular singular, or singular but not regular singular.
\(y''+y=0\)
\(x^3y''+(1+x)y=0\)
\(xy''+x^5y'+y=0\)
\(\sin(x)y''y=0\)
\(\cos(x)y''\sin(x)y=0\)
a) ordinary, b) singular but not regular singular, c) regular singular, d) regular singular, e) ordinary.
Exercise 7.3.102.
Find the general solution of \(x^2 y'' y = 0\text{.}\)
\(y = A x^{\frac{1+\sqrt{5}}{2}} + B x^{\frac{1\sqrt{5}}{2}}\)
Exercise 7.3.103.
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 \(a_0 = 1\))
Exercise 7.3.104.
(tricky) Find the general solution of \(x^2 y''  x y' +y = 0\text{.}\)
\(y = Ax + B x \ln(x)\)