2.1.274 Problem 277

Internal problem ID [9444]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 277
Date solved : Friday, April 25, 2025 at 06:08:28 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{u}^{\prime \prime }+\frac{2u^{\prime }}{x}+a^{2}u& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.163 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& u^{\prime \prime }+\frac{2u^{\prime }}{x}+a^{2}u& =0\text{(2)}& A{u}^{\prime \prime }+B{u}^{\prime }+Cu& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =\frac{2}{x}C& =a^2\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =u{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-a^2}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-a^2\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-a^2)z(x)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $u$ is found using the inverse transformation

$$\begin{array}{rl}u& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.274: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=-a^2$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)={\mathrm{e}}^{\sqrt{-a^2}x}$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $u$ is found from

$$\begin{array}{rl}{u}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{\frac{2}{x}}{1}dx}\\ & =z_1{e}^{-\ln \left(x\right)}\\ & =z_1\left(\frac{1}{x}\right)\end{array}$$

Which simplifies to

$$u_1=\frac{{\mathrm{e}}^{\sqrt{-a^2}x}}{x}$$

The second solution $u_2$ to the original ode is found using reduction of order

$$u_2=u_1\int \frac{e^{\int -\frac{B}{A}dx}}{u_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{u}_2& =u_1\int \frac{e^{\int -\frac{\frac{2}{x}}{1}dx}}{{\left(u_1\right)}^2}dx\\ & =u_1\int \frac{e^{-2\ln \left(x\right)}}{{\left(u_1\right)}^2}dx\\ & =u_1\left(\frac{\sqrt{-a^2}{\mathrm{e}}^{-2\sqrt{-a^2}x}}{2a^2}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}u& =c_1{u}_1+c_2{u}_2\\ & =c_1\left(\frac{{\mathrm{e}}^{\sqrt{-a^2}x}}{x}\right)+c_2\left(\frac{{\mathrm{e}}^{\sqrt{-a^2}x}}{x}\left(\frac{\sqrt{-a^2}{\mathrm{e}}^{-2\sqrt{-a^2}x}}{2a^2}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 21

ode:=diff(diff(u(x),x),x)+2/x*diff(u(x),x)+a^2*u(x) = 0; 
dsolve(ode,u(x), singsol=all);
 

$$u=\frac{c_1\sin \left(ax\right)+c_2\cos \left(ax\right)}{x}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)+\frac{2\left(\frac{d}{dx}u\left(x\right)\right)}{x}+a^{2}u\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{2}{x},P_3\left(x\right)=a^2]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=2\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=0\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & a^{2}u\left(x\right)x+\left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)x+2\frac{d}{dx}u\left(x\right)=0\\ \bullet & & \text{Assume series solution for}u\left(x\right)\\ & & u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}x\cdot u\left(x\right)\text{to series expansion}\\ & & x\cdot u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+1}\\ & \circ & \text{Shift index using}k->k-1\\ & & x\cdot u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=1}}{a}_{k-1}{x}^{k+r}\\ & \circ & \text{Convert}\frac{d}{dx}u\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{dx}u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)x^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(1+r)x^{-1+r}+a_1(1+r)(2+r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+r+1)(k+2+r)+a^2{a}_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-1,0\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(2+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+r+1)(k+2+r)+a^2{a}_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & a_{k+2}(k+2+r)(k+3+r)+a^2{a}_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{a^2{a}_k}{(k+2+r)(k+3+r)}\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+2}=-\frac{a^2{a}_k}{(k+1)(k+2)}\\ \bullet & & \text{Solution for}r=-1\\ & & [u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1},a_{k+2}=-\frac{a^2{a}_k}{(k+1)(k+2)},0=0]\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=-\frac{a^2{a}_k}{(k+2)(k+3)}\\ \bullet & & \text{Solution for}r=0\\ & & [u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=-\frac{a^2{a}_k}{(k+2)(k+3)},2a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [u\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-1})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{c}_k{x}^k),b_{k+2}=-\frac{a^2{b}_k}{(k+1)(k+2)},0=0,c_{k+2}=-\frac{a^2{c}_k}{(k+2)(k+3)},2c_1=0]\end{array}$$

✓ Mathematica. Time used: 0.032 (sec). Leaf size: 42

ode=D[u[x],{x,2}]+2/x*D[u[x],x]+a^2*u[x]==0; 
ic={}; 
DSolve[{ode,ic},u[x],x,IncludeSingularSolutions->True]
 

$$u(x)\to \frac{e^{-iax}(2c_1-\frac{i{c}_2{e}^{2iax}}{a})}{2x}$$

✓ Sympy. Time used: 0.237 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
a = symbols("a") 
u = Function("u") 
ode = Eq(a**2*u(x) + Derivative(u(x), (x, 2)) + 2*Derivative(u(x), x)/x,0) 
ics = {} 
dsolve(ode,func=u(x),ics=ics)
 

$$u\left(x\right)=\frac{C_1{J}_{\frac{1}{2}}\left(ax\right)+C_2{Y}_{\frac{1}{2}}\left(ax\right)}{\sqrt{x}}$$