Bessel functions in resolution of second order ODE We have the following equation:
$$
v''(r)+\frac{1}{r} v'(r) - w v(r)=0, \quad 0 < r < ε; \\ v'(0)=0,
$$
where $w$ is an positive constant.
To resolve this problem, we introduce a new independent variable $\xi= \sqrt{w} r$ and a new function $z(\xi)= v(r)$. Then 
$$
v'(r)=\sqrt{w} z'(\xi), \quad v''(r)= w z''(\xi),\\
z''(\xi)+\dfrac{1}{\xi} z'(\xi)- z(\xi)=0, \quad 0 < \xi < ε \sqrt{w},\\
z'(0)=0.
$$
Let $\zeta= i \xi (i=\sqrt{-1})$. Then 
$$
y''(\zeta)+ \dfrac{1}{\zeta} y'(\zeta)+ y(\zeta)=0, \quad 0 < |\zeta| < ε \sqrt{w}.
$$
We have 
$$
y(\zeta)= C J_0(\zeta),
$$
where $J_0$ is the Bessel function of the zero order. We will use the asymptotic expansion 
$$
J_0(\zeta)=1- \dfrac{\zeta^2}{2^2}+ \frac{\zeta^4}{2^2 \cdot 4^2}- \cdots
$$
Solution of the problem writes 
$$
z(\zeta)= y(\zeta)= C J_0(i \xi)= C \left(1+\frac{\xi^2}{2^2}+ \dfrac{\xi^4}{2^2 \cdot 4^2}+\cdots \right).
$$
My questions are: i don't understand the method used to resolve this problem and why and how we introduce the Bessel functions?
 A: Bessel's differential equation can be solved through Frobenius' power-series method and a solution for $n=0$ is given by the entire function 
$$ J_0(z)=\sum_{m\geq 0}\frac{(-1)^m\,x^{2m}}{4^m m!^2} $$
related with the Fourier series of $\arcsin(x)$. That is not an elementary function strictly speaking, but it is a really important function in dealing with some differential equations (especially some differential equations involving the Laplacian operator) and it deserves for sure to be studied, like the $\Gamma$ or Beta function.
That said, the shown solution just perform a change of variable in order to turn the given differential equation into a Bessel differential equation. The solutions depend on $J_0$ and there is nothing we may change about that.
A: Your equation may be written as
$$
           r^2 v''(r)+rv'(r)-wr^2 v(r) = 0.
$$
You are interested in solutions for $w > 0$. It turns out that, if you find a solution $f(r)$ for $w=1$, then $v(r)=f(\sqrt{w}r)$ is a solution for the general $w > 0$. You can see this by substituting $v(r)=f(\sqrt{w}r)$ into your equation to obtain
$$
         r^2 wf''(\sqrt{w}r)+r\sqrt{w}f'(\sqrt{w}r)-wr^2 f(\sqrt{w}r) = 0 \\
         (\sqrt{w}r)^2 f''(\sqrt{w}r)+(\sqrt{w}r)f'(\sqrt{w}r)-(\sqrt{w}r)^2 f(\sqrt{w}r) = 0  \\
           s^2 f''(s)+sf'(s)-s^2f(s) = 0,\;\; s = \sqrt{w}r.
$$
This is a standard trick used to study the Bessel equation. The conclusion is that, if you know a solution $f(r)$ of your equation where $w=1$, then $y(r)=f(\sqrt{w}r)$ is a solution of your equation for a general $w > 0$.
The reduced equation is very close to the reduced Bessel equation of order $0$, which is
$$
           \rho^2 R''(\rho)+\rho R'(\rho)+\rho^2 R(\rho) = 0.
$$
The only obvious change between your equation and the reduced Bessel equation is the sign change for the $R$ term. The Bessel equation solutions extend into the complex plane, which is something you have to know in order to make sense of a substitution with $i=\sqrt{-1}$ in it. What you are really doing is solving the above equation for a function that is holomorphic in the slitted plane $\{ re^{i\theta} : -\pi < \theta < \pi, r > 0 \}$. Knowing that such a solution exists, then you can replace $\rho$ by $iz$, and the equation will still hold by the identity principle for holomorphic functions. However, I would say that the idea you can make a substitution $\rho = it$ for $t > 0$ is non-sensical without knowing a lot about the solutions of the equation, and about Complex Function Theory. Frankly, I think treating this as a "substitution" is sloppy Math, but that's my opinion. However, knowing that there are solutions that are holomorphic in the slitted plane, you can substitute $\rho = it$ where $t$ is real, and the resulting equation must continue to hold, i.e,
$$
                (it)^2 R''(it)+(it)R'(it) + (it)^2 R(it) = 0 \\
          t^2 \frac{d^2}{dt^2}R(it)+t\frac{d}{dt}R(it)-t^2R(it)=0
$$
That is, $S(t)=R(it)$ does satisfy
$$
                t^2 S''(t)+tS'(t)-t^2 S(t) = 0,
$$
which is your reduced equation.
