Differential Equation from a generating function related to legendre polynomial

Question

Consider the generating function $$\frac1{1-2tx+t^2}=\sum_{n=0}^{\infty}y_n(x)t^n$$. I wish to find a second order differential equation of the form $$p(x)y_n''(x)+q(x)y_n'(x)+\lambda_ny_n(x)=0$$ and a recurrence relation satisfied by $y_n(x)$ of the form $$a_ny_{n+1}(x)+b_ny_n(x)+c_ny_{n-1}(x)=xy_n(x)$$. How should I proceed? Note that differentiating with respect to $t$ or $x$ eventually lands me with an expression involving coefficients of $x$ or $t$ respectively. The generating function is just square of the one for Legendre's polynomial. Does that have anything got to simplify the problem? Any hints. Thanks beforehand.

Answer 1

After a little search, the generating function is just the one for Chebyshev polynomials of the second kind, $U_n(x)$. Thus from Wikipedia, the desired differential equation and recurrence relation are:

Recurrence relation $$y_{n+1}(x)-xy_n(x)+y_{n-1}(x)=0$$ and

Differential equation $$(1-x^2)y_n''(x)+3xy_n'(x)-n(n+2)y_n(x)=0$$.

The derivation for even a more general kind of functions, called Gugenbauer polynomials or ultraspherical polynomials is given in Stein and Weiss book on Introduction to Fourier Analysis

Answer 2

The three-term relation is easy in this case: simply multiply both sides by $1-2xt+t^2$, and then we have $$ 1 = \sum_{n=0}^{\infty} (1-2xt+t^2)y_n(x)t^n, $$ and equating powers of $t^{n+1}$ gives $$ y_{n-1}(x) -2xy_n(x)+y_{n+1}(x) = 0 $$ for $n \geq 1$ (this falls apart at the bottom $n$, of course, as one would expect).

The differential equation is harder. We can find the differentiated terms by differentiating the generating function with respect to $x$, since $$ \frac{\partial}{\partial x} \sum_{n=0}^{\infty} y_n(x)t^n = \sum_{n=0}^{\infty} y_n'(x)t^n. $$ Thus $p(x)y_n''(x)+q(x)y_n'(x)$ is the coefficient of $t^n$ in the series expansion of $$ \left( p(x) \frac{\partial^2}{\partial x^2} + q(x) \frac{\partial}{\partial x} \right) \frac{1}{1-2xt+t^2} = \frac{2qt + (8p-4qx)t^2 + 2qt^3}{(1-2xt+t^2)}. $$ We need to match this with a $\lambda_n y_n$ term. The right way to do this is to differentiate with respect to $t$ and then multiply by $t$ to keep the powers the same, which gives $$ t\frac{\partial}{\partial t} \sum_{n=0} y_n(x)t^n = \sum_{n=0}^{\infty} ny_n(x) t^n.$$ We then match up the coefficients of $t$ in the generating function expression with those we derived by differentiating with respect to $x$ to make them vanish and, therefore find a polynomial in $n$ that works. A general rule is that two differentiations should be sufficient (giving a quadratic in $n$), so here we have $$ \left( A \left( t\frac{\partial}{\partial t} \right)^2 + B t\frac{\partial}{\partial t} + C \right) \frac{1}{1-2xt+t^2} = \\ \frac{1}{(1-2xt+t^2)^3}(C + (2 A x + 2 B x - 4 C x)t + (-4 A - 2 B + 2 C + 4 A x^2 - 4 B x^2 + 4 C x^2) t^2 + (-6 A x + 6 B x - 4 C x) t^3 + (4 A - 2 B + C) t^4 ) $$

Now equating coefficients of $t$ in $$p\partial_{x}^2 G+q \partial_x G + (A(t\partial_t)^2+B(t\partial_t)+C)G = 0$$ gives \begin{align} 0 + C &= 0 \\ 2q + 2 A x + 2 B x - 4 C x &= 0 \\ 8p-4qx -4 A - 2 B + 2 C + 4 A x^2 - 4 B x^2 + 4 C x^2 &= 0 \\ 2q -6 A x + 6 B x - 4 C x &= 0 \\ 0 + 4 A - 2 B + C &= 0 \end{align} Looks bad, but isn't since the outer equations are simple. We immediately find $C=0$ and $B=2A$, simplifying to \begin{align} q + 3 A x &= 0 \\ 2p-qx -2 A - A x^2 &= 0 \\ q +3 A x &= 0 \\ \end{align} Hence $q=-3Ax$, and then $p=A(1-x^2)$, which gives $$ (1-x^2)y_n''-3xy_n' + n(n+2) y_n = 0. $$