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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Section 4.3 Power Series Method

It is a method to find the solution of homogeneous linear differential equation of second order in the form of series. We know that the solution of differential equations
  1. \begin{equation*} y''-y=0 \end{equation*}
    \begin{equation*} C_{1}e^{x}+C_{2}e^{-x} \end{equation*}
    where,
    \begin{equation*} e^{x}=1+\frac{x}{1!}+ \frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \end{equation*}
    and
    \begin{equation*} e^{-x}= 1-\frac{x}{1!}+ \frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots \end{equation*}
    \begin{equation*} \therefore y = C_{1}\left\{1+\frac{x}{1!}+ \frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \right\} \end{equation*}
    \begin{equation*} +C_{2}\left\{1-\frac{x}{1!}+ \frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots \right\} \end{equation*}
    It is a power series solution of the given differential equation.
  2. If \(y''+y=0\) then \(y=A\cos x+B\sin x \) is the solution of a given equation. where,
    \begin{equation*} \cos x= 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots \end{equation*}
    and
    \begin{equation*} \sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots \end{equation*}
    \begin{equation*} \therefore y = A\left\{1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots \right\} \end{equation*}
    \begin{equation*} +B\left\{x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots \right\} \end{equation*}
    is the power series solution of the given equation.
Let us consider the second order homogeneous linear differential equation
\begin{equation*} f_{o}(x)\frac{\,d^{2}y}{\,dx^{2}}+f_{1}(x)\frac{\,dy}{\,dx}+f_{2}(x)y=0 \end{equation*}
\begin{equation} \frac{\,d^{2}y}{\,dx^{2}}+q_{1}(x)\frac{\,dy}{\,dx}+q_{2}(x)y=0\tag{4.3.1} \end{equation}
where,
\begin{equation*} q_{1}(x)=\frac{f_{1}(x)}{f_{o}(x)}, \quad \text{and} \quad q_{2}(x)=\frac{f_{2}(x)}{f_{o}(x)}. \end{equation*}
and suppose it has no solution expressible as a finite linear combination of elementary function. Algebric, trigonomertric, logerithmic, and exponential functions are elementary functions. Then, we can find that it has a solution which can be expressable in the form of infinite series. Let us assume such a solution of equation (4.3.1) is
\begin{equation} y=\sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda}\tag{4.3.2} \end{equation}
is called a power series in \(x\text{.}\) Where \(a_{o}{,}a_{1}{,}a_{2}{,} \cdots\) are constants. In the expression (4.3.2) at least one \(a's\) is not zero. But in case when \(a_{o}=a_{1}=a_{2}= \cdots =0\) then there is no series satisfying the given differential equation. Hence the power series solution method is not valid for such an equation. Some important terms are defining here helps us in establishing the validity of series solution method.
  1. Ordinary Point: \(x=0\) (or origin) is called an ordinary point of the differential equation (4.3.1) if \(q_{1}(x)\) and \(q_{2}(x)\) are finite or regular at \(x=0\) and these can be expressed as power series given by
    \begin{equation*} f(x) = \sum\limits_{k =0}^{\infty}a_{k}x^{k}. \end{equation*}
    e.g., in
    \begin{equation*} y'' -\frac{2x}{1-x^{2}}y'+\frac{2}{1-x^{2}}y =0 \end{equation*}
    where
    \begin{equation*} q_{1}(x)=-\frac{2x}{1-x^{2}}=-2x(1-x^{2})^{-1}=-2x(1+x^{2}+x^{4}+\cdots) \end{equation*}
    and
    \begin{equation*} q_{2}(x)=\frac{2}{1-x^{2}}=2(1-x^{2})^{-1} =2(1+x^{2}+x^{4}+\cdots) \end{equation*}
    and are finite at \(x=0\text{.}\)
    Thus \(x=0\) is an ordinary point of the given diffn. eqn. We assume the solution of this equation as
    \begin{equation*} y = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda}. \end{equation*}
  2. Singular Point: When \(q_{1}(x)\) and \(q_{2}(x)\) are not finite at \(x=0 \text{,}\) then \(x=0\) is called a singular point of the diffn. eqn. If \(xq_{1}(x)\) and \(x^{2}q_{2}(x)\) tend to finite values at \(x=0\) then the equation has a regular singularity at the origin. But if \(xq_{1}(x)\) and \(x^{2}q_{2}(x)\) one or both not finite at \(x=0\) then \(x=0\) is called an irregular singular points. For example in an equation
    \begin{equation*} y''-\frac{x}{2x^{2}}y' +\frac{(x^{2}+1)}{2x^{2}}y =0 \end{equation*}
    has rugular singularity at \(x=0\) and
    \begin{equation*} y''+\frac{4}{x}y' +\frac{5}{x^{3}}y=0 \end{equation*}
    has a singularity at \(x=0\) but not regular, i.e., it has an irregular singularity at \(x=0\text{.}\) Hence such type of equation has no series solution because on solving such equation we find all the constants \(a_{o}{,}a_{1}{,}a_{2}{,}\cdots\) are zero.
    The solution of a equation having \(x=0\) is a singular points is assumed in the form
    \begin{equation*} y = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k}. \end{equation*}
    where \(k\) may be positive or negative integer or fraction. Since many equations in physics fall to second category and equations under first category are just the special cases of second category, we concentrate here only on second category. This method of finding the solution of equation in power series is developed by Frobenius and Fuchs.
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