It is well known that the Euler type DE $$ax^2 \frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$$ can be transformed to be of Cauchy-Euler type DE with constant coefficients using the substitution $x={\rm{e}}^t$, and so that it has the following form $$a \frac{d^2y}{dt^2}+(b-a)\frac{dy}{dt}+cy=0.$$
My problem(s):
1- Is it possible to transform the Legendre ODE $$(1-x^2) \frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0$$ into Cauchy-Euler type ODE with constant coefficients. I tried the same substitution but it doesn't works.
I wrote it as $$\left(\left(1-x^2\right)y^{\prime}\right)^{\prime}+cy=0,\qquad c=n(n+1).$$
But I don't what I have to do after that? could any one help?
The second question which is my important:
2- How to transform Legendre ODE to an equivalent LINEAR ODE with CONSTANT COEFFICIENTS regardless of its order.
Any advice?
Thanks all
Best Answer
In this case, I would say that the trick is to multiply the last equation you found with $(1-x^2)$. Then, you obtain \begin{equation} (1-x^2) \frac{\text{d}}{\text{d} x} \left[(1-x^2) \frac{\text{d}}{\text{d} x} y\right] + n(n+1)(1-x^2) y = 0. \end{equation} The idea is to find a new variable $z(x)$ such that \begin{equation} (1-x^2) \frac{\text{d}}{\text{d} x} = \frac{\text{d}}{\text{d} z} \end{equation} This means that $z(x)$ has to satisfy \begin{equation} \frac{\text{d} z}{\text{d} x} = \frac{1}{1-x^2}, \end{equation} so we obtain \begin{equation} z(x) = z_0 + \text{arctanh }x, \end{equation} which yields \begin{equation} \frac{\text{d}^2 y}{\text{d} z^2} + n(n+1)\, \text{sech}^2(z-z0)\,y = 0. \end{equation} So, the equation is simpler, but still not in the form you're looking for.
Suppose that there exists a coordinate transformation such that the Legendre equation can be written as a second order linear, constant coefficient ODE. Then, in terms of this new coordinate $\zeta(x)$, the solutions are given by \begin{equation} y = c_+e^{\zeta} + c_- e^{-\zeta}, \end{equation} because we can incorporate any constant in the exponential in the definition of $\zeta$. Can we find such a $\zeta$? Yes, that's actually not very difficult. You can, of course, substitute $y = e^{\pm\zeta(x)}$ in the original ODE and solve for $\zeta$. However, using the fact that the solutions to the Legendre equation are given by Legendre polynomials $P_n(x)$ and $Q_n(x)$, we can just write \begin{equation} \zeta(x) = c_1 \pm \log\left[c_2 P_n(x) + Q_n(x)\right], \end{equation} which turns out to be the solution to the ODE you would have obtained for $\zeta(x)$ if $e^{\pm \zeta}$ was substituted in the Legendre ODE.
The moral of the story is that coordinate transformations not always simplify the situation. In this case, the problem of finding the right coordinate transformation is as complicated as solving the original ODE.