Series Solutions

Power series

Definition of Power series

Definition: Power Series

An expression of the form

$$A_0+A_1(x-x_0)+\cdots +A_n(x-x_0{)}^n+\cdots =\sum _{n=0}^{\mathrm{\infty }}{A}_n(x-x_0{)}^n.$$

is called a power series and is defined as the limit

$$\underset{N\to \mathrm{\infty }}{lim}\sum _{n=0}^N{A}_n(x-x_0{)}^n$$

for those values of $x$ for which the limit exists

Ratio Test for the Convergence of Power Series

To determine for what values of $x$ the series converges, we may make use of the ratio test, which states that if the absolute value of the ratio of the $(n+1{)}^{\text{th}}$ term to the $n^{\text{th}}$ term in any infinite series approaches a limit $\rho$ as $n$ goes to infinity, then the series converges when $\rho <1$ and diverges when $\rho >1$.

In the case of the power series

$$\underset{N\to \mathrm{\infty }}{lim}\sum _{n=0}^N{A}_n(x-x_0{)}^n$$

The ratio is

$$\rho =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{A_{n+1}}{A_n}\right||x-x_0|=L|x-x_0|$$

where

$$L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{A_{n+1}}{A_n}\right|.$$

In this case, the power series converge if

$$|x-x_0|<\frac{1}{L}$$

Or, the interval of convergence is $(x_0-\frac{1}{L},x_0+\frac{1}{L})$.

The distance $1/L$ is frequently called the radius of convergence.

Taylor Series Expansion

Suppose the function $f(x)$ can be expressed as a power series, which converges in a non-zero interval about $x=x_0$

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}{A}_n(x-x_0{)}^n.$$

Differentiating both sides of the above equation $k$ times and setting $x=x_0$

$$f^{(k)}(x)=k!A_k,(k=0,1,2,\cdots ).$$

Therefore, we obtain the infinite Taylor series expansion

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n.$$

If $f(x)$ and all its derivatives are continuous in an interval including $x=x_0$, then $f(x)$ can be expressed as a finite Taylor series plus a remainder,

$$f(x)=\sum _{n=0}^{N-1}\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+R_N(x).$$

Here, $R_N$, the remainder after $N$ terms, is given by

$$R_N(x)=\frac{f^{(N)}(\xi )}{N!}(x-x_0{)}^N$$

where $\xi$ is some point in the interval $(x_0,x)$.

Example1: Solving ODE with Power Series

To solve the differential equation

$$Ly\equiv \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}-y=0.$$

we assume a solution in the form

$$y=A_0+A_{1}x+A_2{x}^2+A_3{x}^3+A_4{x}^4+A_5{x}^5+\cdots .$$

Hence

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=2A_2+6A_{3}x+12A_4{x}^2+20A_5{x}^3+\cdots .$$

By substituting the above two expressions into the ODE, we can obtain

$$\begin{array}{rl}Ly\equiv & (2A_2-A_0)+(6A_3-A_1)x+(12A_4-A_2)x^2\\ & +(20A_5-A_3)x^3+\cdots =0.\end{array}$$

In order that this equation be valid over an interval, it is necessary that the coefficients of all powers of $x$ vanish independently to promise it works for all x in an interval, giving the equations

$$2A_2=A_0,6A_3=A_1,12A_4=A_2,20A_5=A_3,\cdots$$

from which there follows

$$\begin{array}{rl}{A}_2=\frac{1}{2}{A}_0,& A_3=\frac{1}{6}{A}_1,\\ A_4=\frac{1}{12}{A}_2=\frac{1}{24}{A}_0,& A_5=\frac{1}{20}{A}_3=\frac{1}{120}{A}_1,\\ \cdots \end{array}$$

Therefore, the solution becomes

$$\begin{array}{rl}y=& A_0+A_{1}x+A_2{x}^2+A_3{x}^3+A_4{x}^4+A_5{x}^5+\cdots \\ =& A_0+A_{1}x+\frac{1}{2}{A}_0{x}^2+\frac{1}{6}{A}_1{x}^3+\frac{1}{24}{A}_0{x}^4+\frac{1}{120}{A}_1{x}^5+\cdots \\ =& A_0(1+\frac{1}{2}{x}^2+\frac{1}{24}{x}^4+\cdots )+A_1(x+\frac{1}{6}{x}^3+\frac{1}{120}{x}^5+\cdots )\end{array}$$

It is seen that the coefficients $A_0$ and $A_1$ are undetermined, and hence arbitrary, but that succeeding coefficients are determined in terms of them. The general solution is thus of the form

$$y=A_0{u}_1(x)+A_1{u}_2(x)$$

$$y=A_0{u}_1(x)+A_1{u}_2(x)=A_1\cosh (x)+A_2\sinh (x)$$

Let $C_1=(A_0+A_1)/2,C_2=(A_0-A_1)/2$, then

$$y=C_1{e}^x+C_2{e}^{-x}.$$

where $u_1(x)$ and $u_2(x)$ are two linearly independent solutions

Example II: Solving ODE with Power Series

As a second example we consider the equation

$$Ly\equiv x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+(x^2+x)\frac{\mathrm{d}y}{\mathrm{d}x}-y=0.$$

Assuming a solution of the form

$$y=\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k$$

we obtain

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\sum _{k=0}^{\mathrm{\infty }}k{A}_k{x}^{k-1},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\sum _{k=0}^{\mathrm{\infty }}k(k-1)A_k{x}^{k-2}$$

hence

$$\begin{array}{rl}Ly\equiv & \sum _{k=0}^{\mathrm{\infty }}k(k-1)A_k{x}^k+\sum _{k=0}^{\mathrm{\infty }}k{A}_k{x}^{k+1}+\sum _{k=0}^{\mathrm{\infty }}k{A}_k{x}^k\\ & -\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k=0.\end{array}$$

The first, third, and fourth summations may be combined to give

$$\sum _{k=0}^{\mathrm{\infty }}[k(k-1)+k-1]A_k{x}^k=\sum _{k=0}^{\mathrm{\infty }}(k^2-1)A_k{x}^k$$

Hence there follows

$$Ly\equiv \sum _{k=0}^{\mathrm{\infty }}(k^2-1)A_k{x}^k+\sum _{k=0}^{\mathrm{\infty }}k{A}_k{x}^{k+1}=0$$

In order to combine these sums, we replace $k$ by $k-1$ in the second, to obtain

$$Ly\equiv \sum _{k=0}^{\mathrm{\infty }}(k^2-1)A_k{x}^k+\sum _{k=1}^{\mathrm{\infty }}(k-1)A_{k-1}{x}^k=0$$

In this way, we find

$$Ly\equiv -A_0+\sum _{k=1}^{\mathrm{\infty }}[(k^2-1)A_k+(k-1)A_{k-1}]x^k=0.$$

In order that $Ly$ may vanish identically, the constant term, as well as the coefficients of the successive powers of $x$, must vanish independently, giving the condition

$$\begin{array}{c}{A}_0=0\\ (k-1)[(k+1)A_k+A_{k-1}]=0(k=1,2,3,\cdots )\end{array}$$

The recurrence formula is automatically satisfied when $k=1$. When $k\ge 2$, it becomes

$$A_k=-\frac{A_{k-1}}{k+1}(k=2,3,4,\cdots )$$

Hence, we obtain

$$A_2=-\frac{A_1}{3},A_3=-\frac{A_2}{4}=\frac{A_1}{3\cdot 4},A_4=-\frac{A_3}{5}=-\frac{A_1}{3\cdot 4\cdot 5},\cdots$$

Thus, in this case $A_0=0$, $A_1$ is arbitrary, and all succeeding coefficients are determined in terms of $A_1$. The solution becomes

$$y=A_1(x-\frac{x^2}{3}+\frac{x^2}{3\cdot 4}-\frac{x^3}{3\cdot 4\cdot 5}+\cdots )$$

If this solution is put in the form

$$\begin{array}{rl}y=& \frac{2A_1}{x}(\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\cdots )\\ =& \frac{2A_1}{x}[x-1+(1-\frac{x^1}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\cdots )]\end{array}$$

The series in parentheses in the final form is recognized as the expansion of $e^{-x}$ , and, writing $2A_1=c$, the solution obtained may be put in the closed form

$$y=c\frac{e^{-x}-1+x}{x}$$

$x\ne 0$ for the analytic solution, but in the power series form, it can be seen that $x=0⟹y=0$, it means $x=0$ is meaningful, thus

$$y=\{\begin{array}{ll}c\frac{e^{-x}-1+x}{x}& x\ne 0\\ 0& x=0\end{array}$$

Method of Frobenius

The Method of Frobenius

• The series solutions is valid at $x=0$ if it is the ordinary point, or it's the singular point, where the power series is diverges
• If $x=0$ is the singular point for the ODE, the normal power series may fail.

For regular singular point: $xP(x)$ and $x^{2}Q(x)$ is analytic at $x=0$ where $y^{″}+P(x)y^{\prime }+Q(x)y=0$

Let us restrict attention to solutions valid about the point $x=0$ . Solutions valid near a more general point $x=x_0$ may be obtained in an analogous way

Suppose the equation has been put in the form

$$Ly\equiv R(x)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+\frac{1}{x}P(x)\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{x^2}Q(x)y=0.$$

where $R(x)$ does not vanish in some interval including $x=0$.

We also suppose that $P(x),Q(x)$, and $R(x)$ are regular at $x=0$, therefore, based on the equation above, we can write

$$x{a}_1(x)=\frac{P(x)}{R(x)},x^2{a}_2(x)=\frac{Q(x)}{R(x)}$$

It is convenient to suppose also that the original equation has been divided through by a suitable constant so that $R(0)=1$. Then we may write

$$\begin{array}{rl}& P(x)=P_0+P_{1}x+P_2{x}^2+\cdots \\ & Q(x)=Q_0+Q_{1}x+Q_2{x}^2+\cdots \\ & R(x)=1+R_{1}x+R_2{x}^2+\cdots \end{array}$$

The series converge in some interval including $x=0$.

We attempt to find nontrivial solutions which are in the form of a power series in $x$ multiplied by a power of $x$,

$$y=x^s\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k=A_0{x}^s+A_1{x}^{s+1}+A_2{x}^{s+2}+\cdots ,$$

where $s$ is to be determined.

Substituting the above expressions into the original ODE gives

$$\begin{array}{rl}Ly\equiv & (1+R_{1}x+R_2{x}^2+\cdots )\times [s(s-1)A_0{x}^{s-2}\\ & +(s+1)s{A}_1{x}^{s-1}+(s+2)(s+1)A_2{x}^s+\cdots ]\\ & +(P_0+P_{1}x+P_2{x}^2+\cdots )\times [s{A}_0{x}^{s-2}+(s+1)A_1{x}^{s-1}\\ & +(s+2)A_2{x}^s+\cdots ]\\ & +(Q_0+Q_{1}x+Q_2{x}^2+\cdots )\times [A_0{x}^{s-2}+A_1{x}^{s-1}+A_2{x}^s+\cdots ]\end{array}$$

multiplying term by term and collecting the coefficients of successive powers of $x$

$$\begin{array}{rl}Ly\equiv & [s(s-1)+P_{0}s+Q_0]A_0{x}^{s-2}\\ & +\{[(s+1)s+P_0(s+1)+Q_0]A_1\\ & +[R_{1}s(s-1)+P_{1}s+Q_1]A_0\}{x}^{s-1}\\ & +\{[(s+2)(s+1)+P_0(s+2)+Q_0]A_2\\ & +[R_1(s+1)s+P_1(s+1)+Q_1]A_1\\ & +[R_{2}s(s-1)+P_{2}s+Q_2]A_0\}{x}^s+\cdots \end{array}$$

For the sake of convenience, let us take

$$f(s)=s(s-1)+P_{0}s+Q_0=s^2+(P_0-1)s+Q_0$$

and

$$\begin{array}{rl}{g}_n(s)& =R_n(s-n)(s-n-1)+P_n(s-n)+Q_n\\ & =R_n(s-n{)}^2+(P_n-R_n)(s-n)+Q_n\end{array}$$

Hence, we have

$$\begin{array}{rl}Ly\equiv & f(s)A_0{x}^{s-2}+[f(s+1)A_1+g_1(s+1)A_0]x^{s-1}\\ & +[f(s+2)A_2+g_1(s+2)A_1+g_2(s+2)A_0]x^s+\cdots \\ & +[f(s+k)A_k+\sum _{n=1}^k{g}_n(s+k)A_{k-n}]x^{s+k-2}+\cdots \end{array}$$

The vanishing of the coefficient of the lowest power $x^{s-2}$ gives the requirement

$$f(s)=0⟹s^2+(P_0-1)s+Q_0=0$$

This equation determines two values of $s$ (which may however be equal) and is called the indicial equation.

For each such value of $s$, the vanishing of the coefficient of $x^{s-1}$ gives the requirement

$$f(s+1)A_1=-g_1(s+1)A_0$$

and hence determines $A_1$ in terms of $A_0$ if $f(s+1)\ne 0$. Next, the vanishing of the coefficient of $x^s$ determines $A_2$ in terms of $A_1$ and $A_0$,

$$f(s+2)A_2=-g_1(s+2)A_1-g_2(s+2)A_0,$$

and hence in terms of $A_0$, if $f(s+2)\ne 0$. In general, the vanishing of the coefficient of $x^{s+k-2}$ gives the recurrence formula

$$f(s+k)A_k=-\sum _{n=1}^k{g}_n(s+k)A_{k-n}(k\geqq 1),$$

which determines each $A_k$ in terms of the preceding $A$ 's, and hence in terms of $A_0$, if for each $k$ the quantity $f(s+k)$ is not zero.

Exceptional cases for the method of Frobenius

$$\begin{array}{rl}& f(s)=0⟹s^2+(P_0-1)s+Q_0=0\\ ⟹& \{\begin{array}{l}{s}_1=\frac{1-P_0}{2}+\frac{1}{2}\sqrt{(1-P_0{)}^2-4Q_0}\\ s_2=\frac{1-P_0}{2}-\frac{1}{2}\sqrt{(1-P_0{)}^2-4Q_0}\end{array}\end{array}$$

as the root of

$$y=x^s\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k=A_0{x}^s+A_1{x}^{s+1}+A_2{x}^{s+2}+\cdots ,$$

• Two roots to the equation $f(s)=0$ are equal

$$\begin{array}{rl}{y}_1(x)& =\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^{k+s_1}\equiv A_0{u}_1(x)\\ y_2(x)& =C{u}_1(x)\log x+\sum _{k=0}^{\mathrm{\infty }}{B}_k{x}^{k+s_2}\end{array}$$

• Two distinctive roots, one is imaginary and the other is real (?) For two roots $s_1$ and $s_2$, series solution $y=x^s\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k,(s=s_1,s_2)$ can also be constructed

$$y=c_1{y}_1(x)+c_2{y}_2(x).$$

Rethinking Example II

$$Ly\equiv x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+(x^2+x)\frac{\mathrm{d}y}{\mathrm{d}x}-y=0.$$

Assuming a solution of the form

$$y=x^s\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^k=\sum _{k=0}^{\mathrm{\infty }}{A}_k{x}^{k+s}.$$

By direct substitution, there follows

$$Ly=(s^2-1)A_0{x}^s+\sum _{k=1}^{\mathrm{\infty }}\{[(s+k{)}^2-1]A_k+(s+k-1)A_{k-1}\}{x}^{k+s}$$

From Example II

$$Ly\equiv \sum _{k=0}^{\mathrm{\infty }}((k+s{)}^2-1)A_k{x}^{k+s}+\sum _{k=1}^{\mathrm{\infty }}(k+s-1)A_{k-1}{x}^{k+s}=0$$

Hence the indicial equation is

$$s^2-1=0$$

and the recurrence formula is

$$[(s+k{)}^2-1]A_k+(s+k-1)A_{k-1}=0$$

or

$$(s+k-1)[(s+k+1)A_k+A_{k-1}]=0(k\geqq 1)$$

The exponents $s_1$ and $s_2$ are $+1$ and $-1$ , respectively. Since they differ by an integer, a solution of the required type is assured only when $s$ has the larger value +1 .

With $s=+1$, the recurrence formula becomes

$$k[(k+2)A_k+A_{k-1}]=0$$

or, since $k\ne 0$,

$$A_k=-\frac{A_{k-1}}{k+2}(k\geqq 1)$$

Thus, one has

$$A_1=-\frac{A_0}{3},A_2=\frac{A_0}{3\cdot 4},A_3=-\frac{A_0}{3\cdot 4\cdot 5},\dots$$

and hence the solution corresponding to $s=1$ is

$$\begin{array}{rl}y& =x[A_0-\frac{A_0}{3}x+\frac{A_0}{3\cdot 4}{x}^2-\frac{A_0}{3\cdot 4\cdot 5}{x}^3+\cdots ]\\ & =A_0(x-\frac{x^2}{3}+\frac{x^3}{3\cdot 4}-\frac{x^4}{3\cdot 4\cdot 5}+\cdots )\\ & =2A_0\left(\frac{e^{-x}-1+x}{x}\right)\end{array}$$

in accordance with the result obtained previously. With $s=-1$, the recurrence formula becomes

$$(k-2)(k{A}_k+A_{k-1})-0(k\geqq 1).$$

It is important to notice that the factor $k-2$ cannot be canceled except on the understanding that $k\ne 2$. That is, when $k=2$, the correct form of the recurrence formula is $0=0$. If $k=1$, there follows

$$A_1=-A_0$$

If $k=2$, the recurrence formula is identically satisfied, so that $A_2$ is arbitrary.

If $k\geqq 3$, the recurrence formula can be written in the form

$$A_k=-\frac{A_{k-1}}{k}(k\geqq 3).$$

If we take $A_2=0$, then there follows $A_3=A_4=\cdots =0$, and the solution corresponding to $s=-1$ becomes simply

$$y=x^{-1}(A_0-A_{0}x)=A_0\frac{1-x}{x}.$$

The general solution of the given equation is then a linear combination of the two solutions so obtained, and hence can be taken conveniently in the form

$$y=c_1\frac{e^{-x}-1+x}{x}+c_2\frac{1-x}{x},$$

or, alternatively,

$$y=C_1\frac{e^{-x}}{x}+C_2\frac{1-x}{x},$$

where $C_1=c_1$, and $C_2=c_2-c_1$. The only solution regular at $x=0$ is the one formerly obtained,

$$y=c\frac{e^{-x}-1+x}{x}.$$

It can be verified that if $A_2$ is left arbitrary, the solution corresponding to the exponent $-1$ is the sum of $A_0$ times the two-term solution obtained and $A_2$ times the infinite series solution corresponding to the exponent $+1$ . That means, When $A_2$ is left arbiary, the solution is contained in the solution when $s_1=+1$.

Special functions derived from the series solutions of ODEs

• The Bessel functions of order $p$ by solving

$$x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+x\frac{\mathrm{d}y}{\mathrm{d}x}+(x^2-p^2)y=0$$

• The Ber and Bei functions by solving

$$x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+x\frac{\mathrm{d}y}{\mathrm{d}x}-(p^2+i{x}^2)y=0$$

• The Legendre functions of order $p$ by solving

$$(1-x^2)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}-2x\frac{\mathrm{d}y}{\mathrm{d}x}+p(p+1)y=0$$

• The hypergeometric function by solving

$$x(1-x)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+[\gamma -(\alpha +\beta +1)x]\frac{\mathrm{d}y}{\mathrm{d}x}-\alpha \beta y=0$$