# Legendre Differential Equation, $y_1,y_2$ linearly independent solutions

$$(1-x^2)y''-2xy'+p(p+1)y=0, p \in \mathbb{R} \text{ constant } \\ -1 < x<1$$

At the interval $(-1,1)$ the above differential equation can be written equivalently

$$y''+p(x)y'+q(x)y=0, -1<x<1 \text{ where } \\p(x)=\frac{-2x}{1-x^2} \\ q(x)= \frac{p(p+1)}{1-x^2}$$

$p,q$ can be written as power series $\sum_{n=0}^{\infty} p_n x^n, \sum_{n=0}^{\infty} q_n x^n$ respectively with centre $0$ and $\sum_{n=0}^{\infty} p_n x^n=p(x)$ and $\sum_{n=0}^{\infty} q_nx^n=q(x), \ \forall -1<x<1$

$$p(x)= \sum_{n=0}^{\infty} (-2) x^{2n+1}, -1<x<1$$

$$q(x)= \sum_{n=0}^{\infty} p(p+1) x^{2n}, \forall -1<x<1$$

Since $p,q$ can be written as power series with centre $0$ and radius of convergence $1$, it's logical to look for a solution of the differential equation of the form

$$y(x)=\sum_{n=0}^{\infty} a_n x^n \text{ with radius of convergence } R>0$$

$$-2xy'(x)= \sum_{n=1}^{\infty} -2n a_n x^n$$

$$y''(x)= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \\ -x^2y''(x)=\sum_{n=2}^{\infty} -n(n-1)a_nx^n$$

We have:

$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n\right]x^n=0, \forall x \in (-R,R)$$

It has to hold: $(n+2)(n+1)a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n=0, \forall n=0,1,2, \dots$

Thus: $$a_{n+2}=-\frac{(p-n)(p+n+1)}{(n+1)(n+2)}a_n, \forall n=0,1,2, \dots$$

So the solution is written as follows:

$$y(x)=a_0 \left[ 1- \frac{p(p+1)}{2!}x^2+ \frac{p(p-2)(p+1)(p+3)}{4!}x^4- \frac{p(p-2)(p-4)(p+1)(p+3)(p+5)}{6!}x^6+ \dots \right] +a_1 \left[ x- \frac{(p-1)(p+2)}{3!}x^3+ \frac{(p-1)(p-3)(p+2)(p+4)}{5!}x^5-\frac{(p-1)(p-3)(p-5)(p+2)(p+4)(p+6)}{7!}x^7+ \dots\right]$$

We will show that if $p \in \mathbb{R} \setminus{\mathbb{Z}}$ then the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$.

We have for $-1<x<1, x \neq 0$:

$$\left | \frac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}}\right |= \left |- \frac{(p-2n)(p+2n+1)}{(2n+1)(2n+2)}\right| |x|^2 \to |x|^2<1$$

So the series $\sum_{n=0}^{\infty} \overline{a_{2n}} x^{2n}$ converges for $-1<x<1$

In the same way, we show that the series $\sum_{n=0}^{\infty} \overline{a_{2n+1}}x^{2n+1}$ converges for $-1<x<1$.

According to the above, if $p \in \mathbb{R} \setminus{\mathbb{Z}}$ the power series at the right of $a_0, a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$ that are infinitely many times differentiable at $(-1,1)$.

Then we show that $y_1(x)$ converges and in the same way we could show that $y_2(x)$ converges.

But have we shown like that that the radius of convergence is $1$? And how do we deduce that the functions are infinitely many times differentiable?

Also what is meant with $\overline{a_{2n}}$?

Furthermore, what happens if $p \in \mathbb{Z}$?

EDIT: Couldn't we also write the solution $y$ in the following form?

$$y(x)= \sum_{k=0}^{\infty} a_{2k} x^{2k}+ \sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$$

where :

$$a_{2k}= \frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p)}{(2k)!}a_0$$

and

$$a_{2k+1}= \frac{\prod_{j=1}^{2k} (j+(-1)^j p)}{(2k+1)!}a_1$$

Or am I wrong?

If it is like that, applying the ratio test we would get:

For $n=2k$:

$$\left| \frac{\frac{\prod_{j=0}^{2k+1} (j+(-1)^{j+1} p) a_0 x^{2k+2}}{(2k+2)!}}{\frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p) a_0 x^{2k}}{(2k)!}}\right| = \left| \frac{(2k-p)(2k+1+p) x^2}{(2k+1)(2k+2)} \right| \to |x^2|<1$$

So the series $\sum_{k=0}^{\infty} a_{2k} x^{2k}$ converges for all $x$ such that $-1<x<1$.

Is it right so far?

If so,how do we deduce that if $p \in \mathbb{R} \setminus{\mathbb{Z}}$ then the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$ ?

EDIT 2: If $p$ is a positive integer, then one of the series terminates and becomes polynomial.

For example if $p=7$, then $a_7, a_9, a_{11}, \dots, a_{2n+1}=0$ and so $\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}= a_1 x+ a_3 x^3+ a_5 x^5$, right?

So does this mean that in such a case it doesn't hold that the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$?

(a) you need to do one of the tests, ratio or root test to determine the radius of convergence for each. it will follow from recurrence relation for $a_n$ that you have.

(b) i don't know what is meant by $\bar {a_n},$ there are no complex numbers here.

(c) if $p$ is a positive integer, then one of the series terminates and become legendre polynomial.

• You mean that if $p$ is a positive integer then one of $a_{2k}, a_{2k+1}$ will be equal to $0$ ?  If so then one of the series $\sum_{k=0}^{\infty} a_{2k} x^{2k}, \sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$ will terminate, right?  What do you mean that it will become Legendre polynomial? – evinda May 27 '15 at 16:23
• I think I see what you mean with: one of the series terminates and become legendre polynomial.  So in this case doesn't it hold that the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x),y_2(x)$, that are infinitely many times differentiable at (−1,1) ? – evinda May 27 '15 at 16:30
• Is the radius of convergence also defined for a polynomial or only for infinite series? – evinda Jun 3 '15 at 18:42