Series Solution of ODEs

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Let us consider the linear homogenious second order ODE with $z \in \mathbb{C}$

\begin{align} y'' + p(z) y' +q(z)y=0 \tag{1} \end{align} If (1) the fucntions $p(z)$ and $q(z)$ are at some point $z=z_0$ finite, then they can be expressed as a power series \begin{align} p(z)=\sum_{n=0}^{\infty} p_n(z-z_0)^n \\ q(z)=\sum_{n=0}^{\infty} q_n(z-z_0)^n \\ \end{align} Functions for which a power series exists are called analytic at $z=z_0$ and $z_0$ is called an ordinary point. If a function $p(z)$ or $q(z)$ diverges at $z=z_0$, then $z_0$ is called a singular point.

Even if a function diverges at $z=z_0$ it may possess a nonsingular (finite) solution at $z_0$. The nessessary and sufficient condition of for existence is that the functions \begin{align} p(z)(z-z_0)\\ q(z)(z-z_0)^2 \end{align} are analytic at $z-z_0$. Such a point is called a regular singular point. Problems of this kind can be solved by the Method of Frobenius.

Series solution about an ordinary point

In the following discussion we make the transformiation $Z=z-z_0$, if $z_0$ is not zero already. Since any solution $y$ of equation (1) can be represented by a power series \begin{align} y(z)=\sum_{n=0}^{\infty} a_n \, z^n \tag{2} \end{align} we take (2) as an ansatz for (1). The derivatives of (2) is \begin{align} y'&=\sum_{n=0}^{\infty} n \cdot a_n \, z^{n-1}=\sum_{n=0}^{\infty} (n+1) \cdot a_{n+1} \, z^{n} \end{align} \begin{align} y''&=\sum_{n=0}^{\infty} n(n-1) \cdot a_n \, z^{n-2}=\sum_{n=0}^{\infty} (n+2)(n+1) \cdot \, a_{n+2} z^{n} \end{align}

The solution to an ODE (1) is obtained by pluggin in the derivatives of $y(z)$ with the same power of $z$. This enables us to use only one summation sign and we can pull out $z^n$. From there we can go on and find a recurrence relation for the coefficients of the power series. This is best illustrated at an example.


Solve the ODE \begin{align} y'' + y=0 \end{align} We simply plug in the derivatives of $y$ \begin{align} \sum_{n=0}^{\infty} (n+2)(n+1) \cdot a_{n+1} \, z^{n} + \sum_{n=0}^{\infty} a_{n} \, z^{n} =0\\ \sum_{n=0}^{\infty} \left[ (n+2)(n+1) \cdot a_{n+1} + a_{n} \right] \, z^{n} =0 \end{align} This equation must be zero for every $n$. Since we assume $z^n$ not to be zero its coefficient must vanish. \begin{align} (n+2)(n+1) \cdot a_{n+1} + a_{n}=0 \end{align} So we find the recurrence relation \begin{align} a_{n+2}=\frac{-a_n}{(n+2)(n+1)} \end{align} If we choose one initial value zero and the other one e.g. $a_0=0$ and $a_1=1$ we get $a_{2k}=0$ for $k=0,1,2, \dots$ and \begin{align} a_{3}&=\frac{-a_1}{(1+2)(1+1)}=- \frac{1}{6}\\ a_{5}&=\frac{-a_3}{(3+2)(3+1)}=\frac{-a_3}{15}=\frac{a_1}{15 \cdot 6}=\frac{1}{90}\\ \vdots\\ a_{n+2}&=(-1)^n \frac{1}{(2n+1)!} \end{align} Plugging in the values of $a_n$ into the ansatz (2) we have \begin{align} y_1(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}- \dots =\sum_{n=0}^{\infty} \frac{(-1)^n \, z^{2n+1}}{(2n+1)!} \end{align} Similarly for $a_1=0$ and $a_0=1$ we get \begin{align} y_2(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}- \dots =\sum_{n=0}^{\infty} \frac{(-1)^n \, z^{2n}}{(2n)!} \end{align} We may be familiar with the solution from an exponential ansatz or recognize the series $y_1$ and $y_2$ as sine and cosine, respectively. The general solution to the problem is \begin{align} y(z) = c_1 \, cos (z) + c_2 \, sin(z) \end{align}

Series solution about a regular singular point

If the ODE (1) has functions $p(y)$ and $q(x)$ have associated functions \begin{align} z \cdot p(z)=s(z)=\sum_{n=0}^{\infty} s_n(z-z_0)^n \\ z^2 \cdot q(z)=t(z)=\sum_{n=0}^{\infty} t_n(z-z_0)^n \\ \end{align} which are analytic, than we can make the Frobenius ansatz

\begin{align} y(z)=z^{\sigma} \sum_0^{\infty} a_n z^n \end{align} with derivatives \begin{align} y'=\sum_{n=0}^{\infty} (n+ \sigma) a_n \, z^{n + \sigma -1} \end{align} and \begin{align} y''=\sum_{n=0}^{\infty} (n+ \sigma)(n + \sigma -1) a_n \, z^{n + \sigma -2} \end{align} The original equation (1) in terms of $s(z)$ and $t(z)$ is \begin{align} y''+ \frac{s(z)}{z}y'+ \frac{t(z)}{z^2}y=0 \tag{3} \end{align}

If we now substitute the Frobenius series into (3) we get \begin{align} \sum_{n=0}^{\infty} (n+ \sigma)(n + \sigma -1) a_n \, z^{n + \sigma -2} + \frac{s(z)}{z} \sum_{n=0}^{\infty} (n+ \sigma) a_n \, z^{n + \sigma -1} + \frac{t(z)}{z^2} \sum_0^{\infty} a_n z^{\sigma +n} =0 \\ \sum_{n=0}^{\infty} (n+ \sigma)(n + \sigma -1) a_n \, z^{n + \sigma -2} + s(z) \sum_{n=0}^{\infty} (n+ \sigma) a_n \, z^{n + \sigma -2} + t(z) \sum_0^{\infty} a_n z^{n + \sigma -2} =0 \\ \end{align} deviding by $z^{\sigma - 2}$ we find \begin{align} \sum_{n=0}^{\infty} \left[ (n+ \sigma)(n + \sigma -1) + s(z) (n+ \sigma)+ t(z) \right] \, a_n \, z^{n} =0 \\ \end{align} For a nonzero $z$ and $n=0$ this equation must satisfy

\begin{align} (\sigma)(\sigma -1) + s(z) \sigma + t(z) =0 \\ \end{align} The solutions to this quadratic equation determine the properties of the solution to the ODE (in the same way as the charakteristik equation in the an exponential ansatz). Its roots can be repeated ($\sigma_1=\sigma_2$) or distinct($\sigma_1 \not =\sigma_2$). If they are distinct and differ by an integer this can still be a problem since the solutions may be linearly dependent. If we find only one solution the second solution can be obtained by the Wronskian Method.

The solution of (1) is given by $y(z)=c_1 y_1(z) + c_2 y_2(z)$ with \begin{align} y_1(z)=z^{\sigma_1} \sum_0^{\infty} a_n z^n \hspace{2cm} y_2(z)=z^{\sigma_2} \sum_0^{\infty} a_n z^n \end{align}

Examples of equations that can be solved by series solution about a regular singular point are the celebrated

that appear in many Problems of Partial Differential Equations, where a PDE is reduced to several ODEs.

For more practice get the books:

  • Riley, Hobson, Bence - Mathematical Methods for Physics and Engineering
  • Schaums outlines - Differential Equations