# Question on legendre equation - part 2

I would like to know if is possible to have regular solutions of Legendre equation when the constant $l$ in the Legendre equation $(1-x^2)u''-2xu''+l(l+1)u=0$ is a non integer number?

I am interested in polynomial solutions for non integer.

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You'll not get polynomials, you'll get the Legendre Functions of the first and second kind. –  Pragabhava Nov 3 '12 at 20:11
Legendre functions of first kind are polynomials. I think if you have, for example, $l=\frac{\sqrt{13}-1}{2}$ then $l(l+1)=3$ and you will have polynomial solution to Legendre equation for non integer $l$. Is it right? –  TheStudent Nov 3 '12 at 20:45

Where does the restriction to integers comes from?

Taking $$(1 - x^2) u'' - 2 x u' + \nu u = 0$$ and $u = \sum_{k=0}^\infty a_k x^{k + s}$, the indicial equation is $s (s - 1) = 0$, which means only one solution can be generated by this ansat; $s = 0$ leads to the recurrence relation $$a_{k + 2} = \frac{k (k + 1) - \nu}{(k + 1) (k + 2)} a_k$$ where $k \in \mathbb{Z}$, and here lies the (im)possibility of polynomial solutions.

The only way for $u$ to be a polynomial is that $$k (k + 1) - \nu = 0$$ has a solution in $\mathbb{Z}$.

The second solution is never a polynomial.

From MathWorld:

If $l$ is an integer, the function of the first kind reduces to a polynomial known as the Legendre polynomial.

If $l = \frac{\sqrt{13}-1}{2}$ the solution is not a polynomial.

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