4.4.1: Introduction

IT MIGHT BE POSSIBLE TO USE POWER SERIES to obtain solutions to differential equations in terms of series involving noninteger powers. For example, we found in Example \(4.6\) that \(y_{1}(x)=\dfrac{\sin x}{x}\) and \(y_{2}(x)=\dfrac{\cos x}{x}\) are solutions of the differential equation \(x y^{\prime \prime}+2 y^{\prime}+x y=0 .\) Series expansions about \(x=0\) are given by

\[ \begin{aligned} \dfrac{\sin x}{x} &=\dfrac{1}{x}\left(x-\dfrac{x^{3}}{3 !}+\dfrac{x^{5}}{5 !}-\dfrac{x^{7}}{7 !}+\ldots\right) \\ &=1-\dfrac{x^{2}}{3 !}+\dfrac{x^{4}}{5 !}-\dfrac{x^{6}}{7 !}+\ldots \end{aligned} \nonumber \]

\[ \begin{aligned} \dfrac{\cos x}{x} &=\dfrac{1}{x}\left(1-\dfrac{x^{2}}{2 !}+\dfrac{x^{4}}{4 !}-\dfrac{x^{6}}{6 !}+\ldots\right) \\ &=\dfrac{1}{x}-\dfrac{x}{2 !}+\dfrac{x^{3}}{4 !}-\dfrac{x^{5}}{6 !}+\ldots \end{aligned} \nonumber \]

While the first series is a Taylor series, the second one is not due to the presence of the first term, \(x^{-1}\). We would like to be able to capture such expansions. So, we seek solutions of the form

\[y(x)=x^{r} \sum_{n=0}^{\infty} c_{n} x^{n}=\sum_{n=0}^{\infty} c_{n} x^{n+r} \nonumber \]

for some real number \(r\). This is the basis of the Frobenius Method.

Consider the differential equation,

\[y^{\prime \prime}(x)+a(x) y^{\prime}(x)+b(x) y(x)=0 \nonumber \]

If \(x a(x)\) and \(x^{2} b(x)\) are real analytic, i.e., have convergent Taylor series expansions about \(x=0\), then we can find a solution of the form

\[y(x)=\sum_{n=0}^{\infty} c_{n} x^{n+r} \nonumber \]

for some constant \(r\). Furthermore, \(r\) is determined from the solution of an indicial equation.

Example \(\PageIndex{1}\)

Show that \(x=0\) is a regular singular point of the equation

\[x^{2} y^{\prime \prime}+x(3+x) y^{\prime}+(1+x) y=0\nonumber \]

If \(x=0\) is a regular singular point, then we can apply the Frobenius Method.

and then find a solution in the form \(y(x)=\sum_{n=}^{\infty}\)
Rewriting the equation as

\[y^{\prime}+\dfrac{3+x}{x} y^{\prime \prime}+\dfrac{(1+x)}{x^{2}} y=0\nonumber \]

\[\begin{aligned} &\text { Rewriting the equation as } \\ &\qquad \begin{aligned} y^{\prime}+\dfrac{3+x}{x} y^{\prime \prime} &+\dfrac{(1+x)}{x^{2}} y=0 \\ a(x) &=\dfrac{3+x}{x} \\ b(x)=& \dfrac{(1+x)}{x^{2}} \end{aligned} \end{aligned} \nonumber \]

So, \(x a(x)=3+x\) and \(x^{2} b(x)=1+x\) are polynomials in \(x\) and are therefore real analytic. Thus, \(x=0\) is a regular singular point.

Example \(\PageIndex{2}\)

Solve

\[x^{2} y^{\prime \prime}+x(3+x) y^{\prime}+(1+x) y=0 \nonumber \]

using the Frobenius Method.

In order to find a solution to the differential equation using the Frobenius Method, we assume \(y(x)\) and its derivatives are of the form

\[\begin{aligned} y(x) &=\sum_{n=0}^{\infty} c_{n} x^{n+r} \\ y^{\prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ y^{\prime \prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2} \end{aligned} \label{4.34} \]

Inserting these series into the differential equation, we have

\[\begin{aligned} 0=& x^{2} y^{\prime \prime}+x(3+x) y^{\prime}+(1+x) y=0 \\ =& x^{2} \sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2}+x(3+x) \sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ &+(1+x) \sum_{n=0}^{\infty} c_{n} x^{n+r} \\ =& \sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r}+3 \sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r} \\ &+\sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r+1}+\sum_{n=0}^{\infty} c_{n} x^{n+r}+\sum_{n=0}^{\infty} c_{n} x^{n+r+1} \\ =& \sum_{n=0}^{\infty} c_{n}[(n+r)(n+r-1)+3(n+r)+1] x^{n+r}+\sum_{n=0}^{\infty} c_{n}[n+r+1] x^{n+r+1} \\ =& \sum_{n=0}^{\infty} c_{n}[(n+r)(n+r+2)+1] x^{n+r}+\sum_{n=0}^{\infty} c_{n}(n+r+1) x^{n+r+1} \end{aligned} \nonumber \]

Next, we re-index the last sum using \(k=n+1\) so that both sums involve the powers \(x^{k+r}\). Therefore, we have

\[\sum_{k=0}^{\infty} c_{k}[(k+r)(k+r+2)+1] x^{k+r}+\sum_{k=1}^{\infty} c_{k-1}(k+r) x^{k+r}=0 \nonumber \]

We can combine both sums for \(k=1,2, \ldots\) if we set the coefficients in the \(k=0\) term to zero. Namely,

\[c_{0}[r(r+2)+1]=0 \nonumber \]

If we assume that \(c_{0} \neq 0\), then

\[r(r+2)+1=0 \nonumber \]

This is the indicial equation. Expanding, we have

\[0=r^{2}+2 r+1=(r+1)^{2}\nonumber \]

So, this gives \(r=-1\). Inserting \(r=-1\) into Equation \(\PageIndex{5}\) and combining the remaining sums, we have

\[\sum_{k=1}^{\infty}\left[k^{2} c_{k}+(k-1) c_{k-1}\right] x^{k-1}=0 \nonumber \]

We find the indicial equation from the terms with lowest powers of \(x\). Setting \(n=0\) at the bottom of the previous page, the lowest powers are \(x\) and the coefficient yields \(c_{0}[r(r+2)+1]=0 .\) The indicial equation is then \(r^{2}+2 r+1=0\).

Setting the coefficients equal to zero, we have found that

\[c_{k}=\dfrac{1-k}{k^{2}} c_{k-1}, \quad k=1,2,3, \ldots \nonumber \]

So, each coefficient is a multiple of the previous one. In fact, for \(k=1\), we have that

\[c_{1}=(0) c_{0}=0\nonumber \]

Therefore, all of the coefficients are zero except for \(c_{0}\). This gives a solution as

\[y_{0}(x)=\dfrac{c_{0}}{x}\nonumber \]

We had assumed that \(c_{0} \neq 0 .\) What if \(c_{0}=0 ?\) Then, Equation \(\PageIndex{5}\) becomes

\[\sum_{k=1}^{\infty}\left[((k+r)(k+r+2)+1) c_{k}+(k+r) c_{k-1}\right] x^{k+r}=0\nonumber \]

Setting the coefficients equal to zero, we have

\[((k+r)(k+r+2)+1) c_{k}=-(k+r) c_{k-1}, \quad k=1,2,3, \ldots\nonumber \]

When \(k=1\), (the lowest power of \(x\) )

\[((1+r)(r+3)+1) c_{1}=-(1+r) c_{0}=0\nonumber \]

So, \(c_{1}=0\), or \(0=(1+r)(r+3)+1=r^{2}+4 r+4=(r+2)^{2} .\) If \(c_{1} \neq 0\) this gives \(r=-2\) and

\[c_{k}=-\dfrac{(k-2)}{(k-1)^{2}} c_{k-1}, \quad k=2,3,4, \ldots\nonumber \]

Then, we have \(c_{2}=0\) and all other coefficient vanish, leaving the solution as

\[y(x)=c_{1} x^{1-2}=\dfrac{c_{1}}{x}\nonumber \]

We only found one solution. We need a second linearly independent solution in order to find the general solution to the differential equation. This can be found using the Method of Reduction of

Method of Reduction of Order. Order from Section 2.2.1 For completeness we will seek a solution \(y_{2}(x)=v(x) y_{1}(x)\), where \(y_{1}(x)=x^{-1} .\) Then,

\[ \begin{aligned} 0=& x^{2} y_{2}^{\prime \prime}+x(3+x) y_{2}^{\prime}+(1+x) y_{2} \\ =& x^{2}\left(v y_{1}\right)^{\prime \prime}+x(3+x)\left(v y_{1}\right)^{\prime}+(1+x) v y_{1} \\ =& {\left[x^{2} y_{1}^{\prime \prime}+x(3+x) y_{1}^{\prime}+(1+x) y_{1}\right] v } \\ &+\left[x^{2} v^{\prime \prime}+x(3+x) v^{\prime}\right] y_{1}+2 x^{2} v^{\prime} y_{1}^{\prime} \\ =& {\left[x^{2} v^{\prime \prime}+x(3+x) v^{\prime}\right] y_{1}+2 x^{2} v^{\prime} y_{1}^{\prime} } \\ =& {\left[x^{2} v^{\prime \prime}+x(3+x) v^{\prime}\right] x^{-1}-2 x^{2} v^{\prime} x^{-2^{\prime}} } \\ =& x v^{\prime \prime}+(3+x) v^{\prime}-2 v^{\prime} \\ =& x v^{\prime \prime}+(1+x) v^{\prime} \end{aligned} \label{4.36} \]

Letting \(z=v^{\prime}\), the last equation can be written as

\[x \dfrac{d z}{d x}+(1+x) z=0 \nonumber \]

This is a separable first order equation. Separating variables and integrating, we have

\[\int \dfrac{d z}{z}=-\int \dfrac{1+x}{x} d x\nonumber \]

or

\[\ln |z|=-\ln |x|-x+C\nonumber \]

Exponentiating,

\[z=\dfrac{d v}{d x}=A \dfrac{e^{-x}}{x}\nonumber \]

Further integration yields

\[v(x)=A \int \dfrac{e^{-x}}{x} d x+B\nonumber \]

Thus,

\[y_{2}(x)=\dfrac{1}{x} \int \dfrac{e^{-x}}{x} d x\nonumber \]

Note that the integral does not have a simple antiderivative and defines the exponential function

\[E_{1}(x)=\int_{x}^{\infty} \dfrac{e^{-t}}{t} d t=-\gamma-\ln x-\sum_{n=1}^{\infty} \dfrac{(-1)^{2} x^{n}}{n ! n}\nonumber \]

where \(\gamma=0.5772 \ldots\) is the Euler-Mascheroni constant.

Thus, we have found the general solution

\[y(x)=\dfrac{c_{1}}{x}+\dfrac{c_{2}}{x} E_{1}(x)\nonumber \]

Another example is that of Bessel’s equation. This is a famous equation which occurs in the solution of problems involving cylindrical symmetry. We discuss the solutions more generally in Section 4.6. Here we apply the Frobenius method to obtain the series solution.

Example \(\PageIndex{3}\)

Solve Bessel’s equation using the Frobenius method,

\[x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0 \nonumber \]

We first note that \(x=0\) is a regular singular point. (Bessel’s equation). We assume \(y(x)\) and its derivatives are of the form

\[ \begin{aligned} y(x) &=\sum_{n=0}^{\infty} c_{n} x^{n+r} \\ y^{\prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ y^{\prime \prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2} \end{aligned} \label{4.37} \]

Inserting these series into the differential equation, we have

\[ \begin{aligned} 0 &=x^{2} y^{\prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y \\ &=x^{2} \sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2}+x \sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ &+\left(x^{2}-v^{2}\right) \sum_{n=0}^{\infty} c_{n} x^{n+r} \\ &=\sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r}+\sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r} \\ & \sum_{n=0}^{\infty} c_{n} x^{n+r+2}-\sum_{n=0}^{\infty} v^{2} c_{n} x^{n+r} \\ &=\sum_{n=0}^{\infty}\left[(n+r)(n+r-1)+(n+r)-v^{2}\right] c_{n} x^{n+r}+\sum_{n=0}^{\infty} c_{n} x^{n+r+2} \\ &=\sum_{n=0}^{\infty}\left[(n+r)^{2}-v^{2}\right] c_{n} x^{n+r}+\sum_{n=0}^{\infty} c_{n} x^{n+r+2} \end{aligned} \label{4.38} \]

We re-index the last sum with \(k=n+2\), or \(n=k-2\), to obtain

\[\begin{aligned} 0=& \sum_{n=2}^{\infty}\left(\left[(k+r)^{2}-v^{2}\right] c_{k}+c_{k-2}\right) x^{k+r} \\ &+\left(r^{2}-v^{2}\right) c_{0} x^{r}+\left[(1+r)^{2}-v^{2}\right] c_{1} x^{r+1} \end{aligned} \label{4.39} \]

We again obtain the indicial equation from the \(k=0\) terms, \(r^{2}-v^{2}=0 .\) The solutions are \(r=\pm v .\)

We consider the case \(r=v .\) The \(k=1\) terms give

\[\begin{aligned} 0 &=\left[(1+r)^{2}-v^{2}\right] c_{1} \\ &=\left[(1+v)^{2}-v^{2}\right] c_{1} \\ &=[1+2 v] c_{1} \end{aligned} \nonumber \]

For \(1+2 v \neq 0, c_{1}=0\). [In the next section we consider the case \(\left.v=-\dfrac{1}{2} .\right]\)

For \(k=2,3, \ldots\), we have

\[\left[(k+v)^{2}-v^{2}\right] c_{k}+c_{k-2}=0 \nonumber \]

Or

\[c_{k}=-\dfrac{c_{k-2}}{k(k+2 v)} \nonumber \]

Noting that \(c_{k}=0, k=1,3,5, \ldots\), we evaluate a few of the nonzero coefficients:

\[\begin{aligned} k &=2: \quad c_{2}=-\dfrac{1}{2(2+2 v)} c_{0}=-\dfrac{1}{4(v+1)} c_{0} . \\ k &=4: \quad c_{4}=-\dfrac{1}{4(4+2 v)} c_{2}=-\dfrac{1}{8(v+2)} c_{2}=\dfrac{1}{2^{4}(2)(v+2)(v+1)} c_{0} . \\ k &=6: \quad c_{6}=-\dfrac{1}{6(6+2 v)} c_{4}=-\dfrac{1}{12(v+3)} c_{4} \\ &=-\dfrac{1}{\left.2^{6}(6)(v+3)\right)(v+2)(v+1)} c_{0} . \end{aligned} \nonumber \]

Continuing long enough, we see a pattern emerge,

\[c_{2 n}=\dfrac{(-1)^{n}}{2^{2 n} n !(v+1)(v+2) \cdots(v+n)}, \quad n=1,2, \ldots \nonumber \]

The solution is given by

\[y(x)=c_{0} \sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{2^{2 n} n !(v+1)(v+2) \cdots(v+n)} x^{2 n+v} \nonumber \]

As we will see later, picking the right value of \(c_{0}\), this gives the Bessel function of the first kind of order \(v\) provided \(v\) is not a negative integer.

The case \(r=-v\) is similar. The \(k=1\) terms give

\[\begin{aligned} 0 &=\left[(1+r)^{2}-v^{2}\right] c_{1} \\ &=\left[(1-v)^{2}-v^{2}\right] c_{1} \\ &=[1-2 v] c_{1} \end{aligned} \nonumber \]

For \(1-2 v \neq 0, c_{1}=0 .\) [In the next section we consider the case \(v=\dfrac{1}{2} .\) ] For \(k=2,3, \ldots\), we have

\[\left[(k-v)^{2}-v^{2}\right] c_{k}+c_{k-2}=0 \nonumber \]

Or

\[c_{k}=\dfrac{c_{k-2}}{k(2 v-k)} \nonumber \]

Noting that \(c_{k}=0, k=1,3,5, \ldots\), we evaluate a few of the nonzero coefficients:

\[k=2: \quad c_{2}=\dfrac{1}{2(2 v-2)} c_{0}=\dfrac{1}{4(v-1)} c_{0} \nonumber \]

\[\begin{aligned} k &=4: \quad c_{4}=\dfrac{1}{4(2 v-4)} c_{2}=\dfrac{1}{8(v-2)} c_{2}=\dfrac{1}{2^{4}(2)(v-2)(v-1)} c_{0} . \\ k &=6: \quad c_{6}=\dfrac{1}{6(2 v-6)} c_{4}=\dfrac{1}{12(v-3)} c_{4} \\ &=\dfrac{1}{\left.2^{6}(6)(v-3)\right)(v-2)(v-1)} c_{0} . \end{aligned} \nonumber \]

Continuing long enough, we see a pattern emerge,

\[c_{2 n}=\dfrac{1}{2^{2 n} n !(v-1)(v-2) \cdots(v-n)}, \quad n=1,2, \ldots\nonumber \]

The solution is given by

\[y(x)=c_{0} \sum_{n=0}^{\infty} \dfrac{1}{2^{2 n} n !(v-1)(v-2) \cdots(v-n)} x^{2 n+v}\nonumber \]

provided \(v\) is not a positive integer. The example \(v=1\) is investigated in the next section.