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Is there someone that maybe can explain where come from the substitution $y(z) = (1-z^2)^{m/2}\cdot u(z)$ in the associated legendre equation? It come from series power? I can't manage the complete solution of this equation. Actually i need only to understand why we use this substitution. Sorry my rusty english.

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It will be better if I shall provide some references, rather to answer the question directly. It will provide you more information about this context.

$1.~~$ "Special Functions for Scientists and Engineers" by W. Bell

(Chapter $~3~$, article $~3.8~$, Page No. $~62~$)

$2.~~$ "Special Functions & Their Applications" by N. N. Lebedev & Richard A. Silverman

(Chapter $~7~$, article $~7.12~$, Page No. $~192~$)

Good Luck.

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I'm not sure what you would like to know, but I will try to give you some explanantions regarding the relation between Legendre equation and associated Legendre equation.

As you know, Legendre differential equation is,

\begin{equation} \begin{aligned} (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0 \end{aligned} \end{equation}

If you differentiate the above equation once, you obtain, \begin{equation} \begin{aligned} (1-x^2)\frac{d^3y}{dx^3}-4x\frac{d^2y}{dx^2}+[n(n+1)-2]\frac{dy}{dx}=0 \end{aligned} \end{equation}

Differentiating twice, you obtain, \begin{equation} \begin{aligned} (1-x^2)\frac{d^4y}{dx^4}-6x\frac{d^3y}{dx^3}+[n(n+1)-6]\frac{d^2y}{dx^2}=0 \end{aligned} \end{equation} and so on.

Differentiating $m$ times $(n>m)$, you will find that the general equation can be obtained as,

\begin{equation} \begin{aligned} (1-x^2)\frac{d^{m+2}y}{dx^{m+2}}-2 (m+1)x\frac{d^{m+1}y}{dx^{m+1}}+[n(n+1)-m(m+1)]\frac{d^my}{dx^m}=0 \end{aligned} \end{equation}

You can prove the above relation by the induction method.

Now, if you transform, $y=(1-x^{2})^{m/2}z$ and put it into the differential equation, known as the associated Legendre differential equation below,

\begin{equation} \begin{aligned} (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\left[n(n+1)-\frac{m^2}{1-x^2}\right]y=0 \end{aligned} \end{equation}

you will see that the equation becomes,

\begin{equation} \begin{aligned} (1-x^2)^{m/2} \left [(1-x^2)\frac{d^{2}z}{dx^{2}}-2 (m+1)x\frac{dz}{dx}+[n(n+1)-m(m+1)]z\right]=0 \end{aligned} \end{equation}

So this leads us to the relation,

\begin{equation} \begin{aligned} z=\frac{d^my}{dx^m} \end{aligned} \end{equation}

which indicates that, if y is the solution of Legendre differential equation $P_n(x)$,

\begin{equation} \begin{aligned} P_n^m(x)=(1-x^{2})^{m/2}\frac{d^my}{dx^m}=(1-x^{2})^{m/2}\frac{d^mP_n(x)}{dx^m} \end{aligned} \end{equation}

will be that of the associated Legendre differential equation.

In this note, please be reminded that I have omitted the discussion of the function of second kind, $Q_n(x)$ and $Q_n^m(x)$, for simplicity. But the story I believe is basically the same.

Hopefully, my explanation serves your question.

*P.S. I have added some replies in response to your comment.

I don't inderstand what is meant by "artificial". The function that satisfies the differential equation written above, i.e., \begin{equation} \begin{aligned} (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\left[n(n+1)-\frac{m^2}{1-x^2}\right]y=0 \end{aligned} \end{equation}

which is a more general form of Legendre equation, is the associated Legendre function, $P_n^m(x)$ expressed here in the note.

Of course, since the Legendre function is polynomial and can be expressed as, for example,

\begin{equation} \begin{aligned} P_n(x)=\sum^{[n/2]}_r(-1)^r\frac{(2n-2r)!}{2^n r!(n-r)!(n-2r)!}x^{n-2r} =\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1) \end{aligned} \end{equation}

(ther are several other ways to express the summation form) you can easily see that the associated Legendre function expressed as $(1-x^2)^{m/2}P_n(x)$ is also a polynomial function. For more detail about the form of the polynomial, see for example, "Handbook of Mathematical Functions", by M. Abramowitz and I. A. Stegun. If you prefer to use the polynomial form, you can do that, but it's just doing the same thing as I have explained.

In the equation, \begin{equation} \begin{aligned} (1-x^2)\frac{d^{m+2}y}{dx^{m+2}}-2 (m+1)x\frac{d^{m+1}y}{dx^{m+1}}+[n(n+1)-m(m+1)]\frac{d^my}{dx^m}=0 \end{aligned} \end{equation}

\begin{equation} \begin{aligned} \frac{d^{m+2}y}{dx^{m+2}}=\frac{d^2}{dx^2}\left(\frac{d^{m}y}{dx^{m}}\right) =\frac{d^2}{dx^2}z \end{aligned} \end{equation} and \begin{equation} \begin{aligned} \frac{d^{m+1}y}{dx^{m+1}}=\frac{d}{dx}\left(\frac{d^{m}y}{dx^{m}}\right) =\frac{d}{dx}z \end{aligned} \end{equation} and so on, which thereby indicates that the function $(1-x^2)^{m/2}z$ satisfies the associated Legendre differential equation and as a whole the above equation becomes, \begin{equation} \begin{aligned} (1-x^2)^{m/2} \left [(1-x^2)\frac{d^{2}z}{dx^{2}}-2 (m+1)x\frac{dz}{dx}+[n(n+1)-m(m+1)]z\right]=0 \end{aligned} \end{equation}

as I have explained in the note.

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  • $\begingroup$ Why did you write z=d(^m)y/dx(^m)? $\endgroup$ – Junior Cossi Aug 15 at 1:41
  • $\begingroup$ By the way, i think your explanation very artificial. There must be a more natural one, perhaps using power series. $\endgroup$ – Junior Cossi Aug 15 at 1:47
  • $\begingroup$ See the replies I have uploaded in addition to my note. $\endgroup$ – 萬雄彦 Aug 15 at 5:30

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