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Chebyshev Differential Equation

(1-x^2){d^2y\over dx^2} - x{dy\over dx} + m^2y = 0
\end{displaymath} (1)

for $\vert x\vert < 1$. The Chebyshev differential equation has regular Singularities at $-1$, 1, and $\infty$. It can be solved by series solution using the expansions
$\displaystyle y$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty a_n x^n$ (2)
$\displaystyle y'$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=1}^\infty n a_n x^{n-1}$  
  $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+1) a_{n+1} x^n$ (3)
$\displaystyle y''$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+1)n a_{n+1} x^{n-1} = \sum_{n=1}^\infty (n+1)n a_{n+1} x^{n-1}$  
  $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n.$ (4)

Now, plug (2-4) into the original equation (1) to obtain
$(1-x^2)\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$
$ -x\sum_{n=0}^\infty (n+1)n_{n+1}x^n+m^2\sum_{n=0}^\infty a_nx^n=0\quad$ (5)
$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n+2}$
$-\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1} + m^2\sum_{n=0}^\infty a_nx^n=0\qquad$ (6)
$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty n(n-1)a_nx^{n+2}$
$ -\sum_{n=1}^\infty na_nx^n+ m^2\sum_{n=0}^\infty a_nx^n=0\quad$ (7)

$2\cdot 1 a_2+3\cdot 2 a_3x-1\cdot ax+m^2a_0+m^2a_1x$
$+\sum_{n=2}^\infty [(n+2)(n+1)a_{n+2} -n(n-1)a_n-na_n+m^2a_n]x^n=0\quad$ (8)
$(2a_2+m^2 a_0)+[(m^2-1)a_1+6a_3]x$
$ +\sum_{n=2}^\infty [(n+2)(n+1)a_{n+2}+(m^2-n^2)a_n]x^n=0,\quad$ (9)

2a_2+m^2 a_0=0
\end{displaymath} (10)

\end{displaymath} (11)

a_{n+2} ={n^2-m^2\over(n+1)(n+2)} a_n \qquad \hbox{for }n=2,3,\ldots.
\end{displaymath} (12)

The first two are special cases of the third, so the general recurrence relation is
a_{n+2} ={n^2-m^2\over(n+1)(n+2)} a_n \qquad \hbox{for }n=0,1,\ldots.
\end{displaymath} (13)

From this, we obtain for the Even Coefficients
$\displaystyle a_2$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 2}}m^2a_0$ (14)
$\displaystyle a_4$ $\textstyle =$ $\displaystyle {2^2-m^2\over 3\cdot 4} a_2 = {(2^2-m^2)(-m^2)\over 1\cdot 2\cdot 3\cdot 4} a_0$ (15)
$\displaystyle a_{2n}$ $\textstyle =$ $\displaystyle {[(2n)^2-m^2][(2n-2)^2-m^2]\cdots[-m^2]\over(2n)!} a_0,$  

and for the Odd Coefficients
$\displaystyle a_3$ $\textstyle =$ $\displaystyle {1-m^2\over 6}a_0$ (17)
$\displaystyle a_5$ $\textstyle =$ $\displaystyle {3^2-m^2\over 4\cdot 5} a_3 = {(3^2-m^2)(1^2-m^2)\over 5!} a_1$ (18)
$\displaystyle a_{2n-1}$ $\textstyle =$ $\displaystyle {[(2n-1)^2-m^2][(2n-3)^2-m^2]\cdots[1^2-m^2]\over(2n+1)!} a_1.$  

So the general solution is

$y=a_0\left[{1+ \sum_{k=2, 4, \ldots}^\infty {[k^2-m^2][(k-2)^2-m^2]\cdots[-m^2]\over k!} x^k}\right]$
$ +a_1\left[{x+\sum_{k=3, 5, \ldots}^\infty{[(k-2)^2-m^2][(k-2)^2-m^2]\cdots[1^2-m^2]\over k!} x^k}\right]\quad$ (20)

If $n$ is Even, then $y_1$ terminates and is a Polynomial solution, whereas if $n$ is Odd, then $y_2$ terminates and is a Polynomial solution. The Polynomial solutions defined here are known as Chebyshev Polynomials of the First Kind. The definition of the Chebyshev Polynomial of the Second Kind gives a similar, but distinct, recurrence relation

a_{n+2}' ={(n+1)^2-m^2\over(n+2)(n+3)} a_n' \qquad \hbox{for }n=0,1,\ldots.
\end{displaymath} (21)

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© 1996-9 Eric W. Weisstein