Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Answer to Problem 1P

The first solution of the differential equation is $\underset{\_}{y_1\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^n}{n!\left(n+1\right)!}}}$ and second solution is $\underset{\_}{y_2\left(x\right)=-y_1\left(x\right)\ln x+x^{-1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{H_n+H_{n-1}}{n!\left(n-1\right)!}{x}^n}\right]}$.

Explanation of Solution

Theorem used:

Let $r_1$ and $r_2$ be the roots of the indicial equation $F\left(r\right)=r\left(r-1\right)+p_{0}r+q_0$ with $r_1\ge r_2$.

If $r_1$ and $r_2$ are real, then in either the interval $-\rho <x<0$ or the interval $0<x<\rho$, there exists a solution of the form $y_1\left(x\right)={\left|x\right|}^{r_1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(r_1\right)x^n}\right]$ where $a_n\left(r_1\right)$ are the  given by recurrence relation with $a_0=1$ and $r=r_1$.

If $r_1-r_2$ is not zero or a positive integer, then in either the interval $-\rho <x<0$ or the interval $0<x<\rho$, there exists a second solution of the form $y_2\left(x\right)={\left|x\right|}^{r_2}\left[1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(r_2\right)x^n}\right]$ where $a_n\left(r_2\right)$ are also determined by the recurrence relation with $a_0=1$ and $r=r_2$.

Further the power series converges at least for $\left|x\right|<\rho$.

If $r_1=r_2$, then the second solution is $y_2\left(x\right)=y_1\left(x\right)\ln \left|x\right|+{\left|x\right|}^{r_1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{b}_n\left(r_1\right)x^n}\right]$.

If $r_1-r_2=N$, positive integer, then $y_2\left(x\right)=a{y}_1\left(x\right)\ln \left|x\right|+{\left|x\right|}^{r_2}\left[1+{\displaystyle \sum _{n=1}^{\infty }{c}_n\left(r_2\right)x^n}\right]$.

Calculation:

The given differential equation is given as $x^{2}y"+2xy\text{'}+xy=0$.

Compare the equation $x^{2}y"+2xy\text{'}+xy=0$ with $P\left(x\right)y"+Q\left(x\right)y\text{'}+R\left(x\right)y=0$ and note that $P\left(x\right)=x^2$, $Q\left(x\right)=2x$ and $R\left(x\right)=x$.

Note that the singular points occur when $P\left(x_0\right)=0$.

Thus, the singular point of the given equation is $x_0=0$.

If $x_0=0$ is a regular singular point, check whether the following limits are finite.

$p_0=\underset{x\to 0}{\lim }\frac{xQ\left(x\right)}{P\left(x\right)}$ and $q_0=\underset{x\to 0}{\lim }\frac{x^{2}R\left(x\right)}{P\left(x\right)}$

For $x_0=0$, obtain the value of $p_0$ as follows.

$\begin{array}{c}{p}_0=\underset{x\to 0}{\lim }x\frac{2x}{x^2}\\ =2\end{array}$

Similarly, for $x_0=0$, obtain the value of $q_0$ as follows.

$\begin{array}{c}{q}_0=\underset{x\to 0}{\lim }{x}^2\frac{x^2}{x^2}\\ =0\end{array}$.

Since both are finite, $x_0=0$ is a regular singular point of the given differential equation.

Since $p_0=2$ and $q_0=0$, the corresponding indicial equation is as follows:

$F\left(r\right)=r\left(r-1\right)+p_{0}r+q_0$

Substitute the value of $p_0=2$ and $q_0=0$ in $F\left(r\right)=r\left(r-1\right)+p_{0}r+q_0=0$.

$\begin{array}{c}F\left(r\right)=r\left(r-1\right)+2r+0\\ =r^2-r+2r\\ =r^2+r\end{array}$

Let $F\left(r\right)=0$ in the above equation it becomes as follows:

$\begin{array}{l}{r}^2+r=0\\ r\left(r+1\right)=0\end{array}$

Therefore, the roots of the equation $r^2-r+2r$ is $r=0,r=-1$.

Hence, $x_0$ is a regular singular point.

From the above Theorem, the first solution is given as $y_1\left(x\right)={\left|x\right|}^{r_1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(r_1\right)x^n}\right]$.

Substitute the value of $r=0$ in $y_1\left(x\right)={\left|x\right|}^{r_1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(r_1\right)x^n}\right]$.

$y_1\left(x\right)=1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(0\right)x^n}$

Differentiate the equation $y_1\left(x\right)=1+{\displaystyle \sum _{n=1}^{\infty }{a}_n\left(0\right)x^n}$.

$y_{{}_1}^{\text{'}}\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\left(n+1\right)a_{n+1}\left(0\right)x^n}$

The coefficients $a_n\left(0\right)$, apart from $a_0\left(0\right)=1$ are determined by the recurrence relation $F\left(r+n\right)a_n+{\displaystyle \sum _{k=1}^{n-1}{a}_k\left[\left(r+k\right)p_{n-k}+q_{n-k}\right]=0}$ in case of $r=r_1$.

$n\left(n+1\right)a_n\left(0\right)+{\displaystyle \sum _{k=1}^{n-1}{a}_k\left(0\right)\left[k{p}_{n-k}+q_{n-k}\right]=0}$, where ${\displaystyle \sum _{n=0}^{\infty }{p}_n{x}^n=x\frac{Q\left(x\right)}{P\left(x\right)}=2}$ and ${\displaystyle \sum _{n=0}^{\infty }{q}_n{x}^n=x^2\frac{R\left(x\right)}{P\left(x\right)}=x}$.

Hence, the value of $p_0=2,p_n=0,\forall n\in N$ and $q_1=1,q_n=0,\forall n\in N_0$.

Factoring out $a_n\left(0\right)$ from the equation $n\left(n+1\right)a_n\left(0\right)+{\displaystyle \sum _{k=1}^{n-1}{a}_k\left(0\right)\left[k{p}_{n-k}+q_{n-k}\right]=0}$ is as follows:

$a_n\left(0\right)=\frac{-{\displaystyle \sum _{k=0}^{n-1}{a}_k\left(0\right)\left[k{p}_{n-k}+q_{n-k}\right]}}{n\left(n+1\right)}$ with first couple of elements is as follows:

$\begin{array}{l}{a}_1\left(0\right)=\frac{-a_0\left(0\right)\left[0+q_1\right]}{1\left(1+1\right)}=-\frac{1}{1\cdot 2}\\ a_2\left(0\right)=\frac{1}{1\cdot 2^2\cdot 3}\\ a_3\left(0\right)=-\frac{1}{1\cdot 2^2\cdot 3^2\cdot 4}\end{array}$

And the values continues as such.

Hence, the first solution is $\underset{\_}{y_1\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^n}{n!\left(n+1\right)!}}}$.

According to the same theorem, since $r_1-r_2=1$, the second solution is given by $y_2\left(x\right)=a{y}_1\left(x\right)\ln \left|x\right|+{\left|x\right|}^{r_2}\left[1+{\displaystyle \sum _{n=1}^{\infty }{c}_n\left(r_2\right)x^n}\right]$.

$\begin{array}{c}{y}_2\left(x\right)=a{y}_1\left(x\right)\ln \left|x\right|+x^{-1}\left[1+{\displaystyle \sum _{n=1}^{\infty }{c}_n\left(-1\right)x^n}\right]\\ =a{y}_1\left(x\right)\ln x+x^{-1}+{\displaystyle \sum _{n=1}^{\infty }{c}_{n+1}\left(-1\right)x^n}\end{array}$

Where $a=\underset{r\to r_2}{\lim }\left(r-r_2\right)a_{N\left(r\right)},N=r-r_2$ since $N=1$.

$a_{N\left(r\right)}=a_1\left(r\right)=\frac{-{\displaystyle \sum _{k=0}^{N-1}{a}_k\left(r\right)\left[\left(r+k\right)p_{N-k}+q_{N-k}\right]}}{\left(r+N\right)\left(r+N+1\right)}$

Substitute the limits in the above equation it becomes as follows:

$\begin{array}{c}{a}_1\left(r\right)=\frac{-a_0\left(r\right)\left[\left(r+0\right)p_1+q_1\right]}{\left(r+2\right)\left(r+1\right)}\\ =\frac{-\left[0+1\right]}{\left(r+2\right)\left(r+1\right)}\\ =\frac{-1}{\left(r+2\right)\left(r+1\right)}\end{array}$

Thus, substitute the value of $a=\underset{r\to -1}{\lim }\left(r+1\right)\frac{-1}{\left(r+2\right)\left(r+1\right)}=-1$.

To calculate the value of $c_n\left(0\right)$ substitute the value of $y_2\left(x\right)=a{y}_1\left(x\right)\ln x+x^{-1}+{\displaystyle \sum _{n=1}^{\infty }{c}_{n+1}\left(-1\right)x^n}$ into the initial differential equation.

First differentiate the equation $y_2\left(x\right)=a{y}_1\left(x\right)\ln x+x^{-1}+{\displaystyle \sum _{n=1}^{\infty }{c}_{n+1}\left(-1\right)x^n}$.

$y_{{}_2}^{\text{'}}\left(x\right)=a{y}_{{}_1}^{\text{'}}\left(x\right)\ln x+a\frac{y_1\left(x\right)}{x}-x^{-2}+{\displaystyle \sum _{n=1}^{\infty }n{c}_{n+1}\left(-1\right)x^{n-1}}$

Again, differentiate the equation $y_{{}_2}^{\text{'}}\left(x\right)=a{y}_{{}_1}^{\text{'}}\left(x\right)\ln x+a\frac{y_1\left(x\right)}{x}-x^{-2}+{\displaystyle \sum _{n=1}^{\infty }n{c}_{n+1}\left(-1\right)x^{n-1}}$.

$y_{{}_2}^{\text{'}\text{'}}\left(x\right)=a{y}_{{}_1}^{\text{'}\text{'}}\left(x\right)\ln x+2a\frac{y_1^{\text{'}}\left(x\right)}{x}-a\frac{y_1\left(x\right)}{x}+2x^{-3}+{\displaystyle \sum _{n=1}^{\infty }n\left(n-1\right)c_{n+1}\left(-1\right)x^{n-2}}$

Now, the given differential equation becomes as follows:

$\begin{array}{l}{x}^2\left[a{y}_{{}_1}^{\text{'}\text{'}}\left(x\right)\ln x+2a\frac{y_1^{\text{'}}\left(x\right)}{x}-a\frac{y_1\left(x\right)}{x}+2x^{-3}+{\displaystyle \sum _{n=1}^{\infty }n\left(n-1\right)c_{n+1}\left(-1\right)x^{n-2}}\right]\\ +2x\left[a{y}_{{}_1}^{\text{'}}\left(x\right)\ln x+a\frac{y_1\left(x\right)}{x}-x^{-2}+{\displaystyle \sum _{n=1}^{\infty }n{c}_{n+1}\left(-1\right)x^{n-1}}\right]\\ +x\left[a{y}_1\left(x\right)\ln x+x^{-1}+{\displaystyle \sum _{n=1}^{\infty }{c}_{n+1}\left(-1\right)x^n}\right]=0\\ \left[2c_2\left(-1\right)+c_1\left(-1\right)-3a_1\left(0\right)\right]x\\ +{\displaystyle \sum _{n=1}^{\infty }\left[-\left(2n+1\right)a_n\left(0\right)+n\left(n+1\right)\right]c_{n+1}\left(-1\right)c_n\left(-1\right)x^n}=0\end{array}$

Equating the coefficient on both sides in the above equation.

$\{\begin{array}{l}2c_2\left(-1\right)+c_1\left(-1\right)-3a_1\left(0\right)=0\\ -\left(2n+1\right)a_n\left(0\right)+n\left(n+1\right)c_{n+1}\left(-1\right)c_n\left(-1\right)=0\end{array}$

Let $c_2\left(-1\right)=1$ in the above equation this implies as follows:

$\begin{array}{l}{c}_2\left(-1\right)=\frac{3a_1\left(0\right)-c_1\left(-1\right)}{2}=-\frac{H_2+H_1}{2!}\\ c_3\left(-1\right)=\frac{H_3+H_2}{2\cdot 3}\end{array}$

$c_n\left(-1\right)={\left(-1\right)}^{n+1}\frac{H_n+H_{n-1}}{n!\left(n-1\right)!}$

Hence, the second solution is $\underset{\_}{y_2\left(x\right)=-y_1\left(x\right)\ln x+x^{-1}+\left[1+{\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{H_n+H_{n-1}}{n!\left(n-1\right)!}{x}^n}\right]}$.