📂Odinary Differential Equations

Solution of Differential Equations Using Series Solutions

Description

Differential equations with constant coefficients can be relatively easily solved using methods such as separation of variables or integrating factor method. However, differential equations with coefficients that include the independent variable, as shown below, cannot be easily solved.

$$\begin{equation} P(x)\dfrac{d^2 y}{dx^2} + Q(x)\dfrac{dy}{dx}+R(x)y=0 \label{1}\end{equation}$$

Here, $P$, $Q$, and $R$ are assumed to be polynomials without common factors. Equations of the above form include

Bessel’s equation

$$x^2 y^{\prime \prime} +xy^{\prime}+(x^2-\nu ^2)y=0,\quad \nu \text{ is constant}$$

Legendre’s equation

$$(1-x^2)y^{\prime \prime}-2xy^{\prime}+l(l+1)y=0,\quad l \text{ is constant}$$

among others. The goal in solving such differential equations is to find solutions in the form of power series.

Definition¹

In $\eqref{1}$, $x_{0}$ is called an ordinary point if $P(x_{0}) \ne 0$. Since $P$ is continuous, there exists an open interval that includes $x_{0}$, denoted as $P(x) \ne 0$. Our objective here is to find a power series solution to $\eqref{1}$ in the vicinity of an ordinary point $x_{0}$. In other words, the solution to $\eqref{1}$ is assumed to be a power series of the form

$$y=a_{0}+a_{1}(x-x_{0})+a_2(x-x_{0})^2+\cdots = \sum \limits _{n=0}^\infty a_{n}(x-x_{0})^n$$

and converges within the radius of convergence $|x-x_{0}| < \rho$. This method allows us to solve difficult differential equations with the independent variable $x$ in the coefficients.

On the other hand, $x_{0}$ is called a singular point if $P(x_{0})=0$. Among singular points,

$$\lim \limits_{x\rightarrow x_{0}}(x-x_{0})\frac{Q(x)}{P(x)} < \infty\quad \mathrm{and}\quad \lim \limits_{x\rightarrow x_{0}} (x-x_{0})^{2}\frac{R(x) }{P(x)}<\infty$$

those satisfying this condition are called regular singular points. If it is not a regular singular point, it is called an irregular singular point. When $x_{0}$ is a regular singular point, the solution is assumed to start as follows.

$$y=\sum \limits_{n=0}^{\infty}a_{n}x^{n+s}$$

This method of solving is known as the Frobenius method.

Example

Solve for the series solution of $y^{\prime \prime}+y=0$ and $-\infty < x < \infty$.

Although the content is lengthy, it is not difficult, so take your time reading it. The given equation for the example may be sufficiently easily solved without resorting to a series solution, but the importance lies in practicing the method of solving through series solutions. Firstly, $y^{\prime \prime}+y=0$ is when $P(x)=1$, $Q(x)=0$, and $R(x)=1$. Thus, every point is an ordinary point, but for simplification, let’s choose $x_{0}=0$. Assume the given differential equation’s solution converges for the power series below at $|x| < \rho$.

$$y=a_{0}+a_{1}x+a_2x^2+\cdots = \sum \limits_{n=0}^\infty a_{n}x^n$$

To insert into the differential equation, calculate $y^{\prime \prime}$ to get

$$y^{\prime \prime}=2a_2+3\cdot 2 a_{3} x + \cdots +n(n-1)a_{n}x^{n-2}+\cdots = \sum \limits_{n=2} ^\infty n(n-1)a_{n}x^{n-2}$$

Inserting $y$ and $y^{\prime \prime}$ into the given differential equation yields

$$\sum \limits_{n=2} ^\infty n(n-1)a_{n}x^{n-2} + \sum \limits_{n=0}^\infty a_{n}x^n=0$$

An important point in the series solution method is to align the order of $x$. When substituting $n+2$ instead of $n$ for the first term of the series, it becomes as follows.

$$\begin{align*} && \sum \limits_{n=0} ^\infty (n+2)(n+1)a_{n+2}x^{n} + \sum \limits_{n=0}^\infty a_{n}x^n&=0 \\ \implies && \sum \limits_{n=0} ^\infty \left[ (n+2)(n+1)a_{n+2}+ a_{n} \right] x^{n}&=0 \end{align*}$$

From the properties of power series, for the above equation to hold, all coefficients must be $0$. Thus, we obtain the following equation.

$$\begin{align*} && (n+2)(n+1)a_{n+2}+ a_{n}&=0 \\ \implies && (n+2)(n+1)a_{n+2}&=-a_{n} \\ \implies&& a_{n+2}&=\dfrac{-1}{(n+2)(n+1)} a_{n} \end{align*}$$

The formula that describes the relationship between the preceding and succeeding coefficients is called a recurrence relation. From the recurrence relation, each term’s coefficient can be determined. Since the $n+2$th coefficient can be found from the $n$th coefficient, knowing the first two coefficients, $a_{0}, a_{1}$, allows us to know all coefficients. Therefore, the series can be divided into two parts, grouped as $a_{0}$ and $a_{1}$. For even $n$ in general, it can be expressed as follows for $n=2k(k=1,2,\dots)$.

$$\begin{align*} a_2 &=\dfrac{-1}{2\cdot 1}a_{0}=\dfrac{-1}{2!}a_{0} \\ a_{4} &=\dfrac{-1}{4\cdot 3}{a_2}=\dfrac{-1}{4\cdot 3}\dfrac{-1}{2!}a_{0}=\dfrac{1}{4!}a_{0} \\ a_{6}&=\dfrac{-1}{6!}a_{0} \\ &\vdots \\ a_{n} &=a_{2k}=\dfrac{(-1)^k}{(2k)!}a_{0} \end{align*}$$

For odd $n$ in general, it can be expressed as follows for $n=2k+1(k=1,2,\dots)$.

$$\begin{align*} a_{3} &=\dfrac{-1}{3\cdot 2}a_{1}=\dfrac{-1}{3!}a_{1} \\ a_{5}&=\dfrac{-1}{5 \cdot 4}a_{3}=\dfrac{-1}{5\cdot 4}\dfrac{-1}{3!}a_{0}=\dfrac{1}{5!}a_{1} \\ a_{7}&=\dfrac{-1}{7!}a_{1} \\ &\vdots \\ a_{n} &=a_{2k+1}=\dfrac{(-1)^k}{(2k+1)!}a_{1} \end{align*}$$

Inserting and organizing the above results into $y$ gives the following.

$$\begin{align*} y&= a_{0}+a_{1}x-\dfrac{a_{0}}{2!}x^2-\dfrac{a_{1}}{3!}x^3+\cdots +\dfrac{(-1)^na_{0}}{(2n)!}x^{2n}+\dfrac{ (-1)^{n} a_{1}}{(2n+1)!} x^{2n+1} +\cdots \\ &=a_{0}\left[ 1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4+\cdots + \dfrac{(-1)^n}{(2n)!}x^{2n} + \cdots \right] + a_{1} \left[ x-\dfrac{1}{3!}x^3 +\dfrac{1}{5!}x^5+\cdots +\dfrac{ (-1)^n}{(2n+1)!}x^{2n+1} +\cdots \right] \\ &= a_{0} \sum \limits_{n=0}^\infty \dfrac{(-1)^n } {(2n)!} x^{2n}+a_{1}\sum \limits_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \end{align*}$$

We have found the two solutions and the general solution for the given second-order differential equation. The general solution is represented as a linear combination of the below two independent solutions. $$y_{1}(x)=\sum \limits_{n=0}^\infty \dfrac{(-1)^n } {(2n)!} x^{2n},\quad y_{2}(x)=\sum \limits_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Using the ratio test, it can be seen that both series $y_{1}, y_{2}$ converge for all $x$. Moreover, these two series exactly match the Taylor series of $\cos$ and $\sin$. This means $y=a_{0}\cos x+a_{1}\sin x$, which is the same as the solution obtained through solving second-order differential equations with constant coefficients.

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