# Preface

Some differential equations may have polynomial solutions. This section discusses how to obtain polynomial solutions to some classes of differential equations.

# Some Famous Differential equations having Polynomial Solutions

Chebyshev polynomial solutions

The Chebyshev polynomials were discovered in 1854 by the Russian mathematician Pafnuty Chebyshev (1821--1894) in his paper Théorie des mécanismes connus sous le nom de parallélogrammes. Rather than viewing applications as the beneficiary of elegant mathematics, Chebyshev had a broader approach to describe the nature with mathematical tools. idea. In fact, he found many of his greatest theoretical mathematical discoveries by observing mechanical systems (click here to see the pictures). His discovery of Chebyshev polynomials was motivated by the mechanical problem how to convert a rotation into horizontal movement observed in a steam engine of train (nineteen century). Pafnuty also invented other polynomials now known as Hermite and Laguerre. For a more extensive account of the history of this discovery, see The theory of best approximation of functions.

By the way, Chebyshev used seven different spellings of his last name in Latin letters (including French transliteration Tchebycheff or Tchebychev of his name; that is why the letter ''T'' is used for his polynomials of the first kind) and two Russian spellings.

Legendre polynomial solutions

Hermite polynomial solutions

Laguerre and Sonin polynomial solutions

Let us start with the differential equation that was first considered by Edmond Nicolas Laguerre (1834--1886) in his 1879 article Sur l'intégrale int_x^(+infty)x^(-1)e^(-x)dx
$x\,u'' + \left( 1 +x \right) u' - n\,u = 0,$
Upon substituting the series
$u(x) = \sum_{k\ge 0} a_k x^k , \qquad y'(x) = \sum_{k\ge 0} a_{k+1} \left( k+1 \right) x^k$
into the given differential equation, we get a recuarrence
$k \left( k +1 \right) a_{k+1} + a_{k+1} \left( k +1 \right) + a_k k - n\,a_k = 0 .$
This yields
$a_{k+1} = \frac{n -k}{(k+1)^2}\, a_k , \qquad k=0,1,2,\ldots .$
So when n is a positive integer, the recurrence terminates at k = n and we get a polynomial solution.
$u(x) = \sum_{k=0}^n \binom{n}{k} \frac{x^k}{k!} , \qquad n=0,1,2,\ldots .$

Now we consider a differential equation
$x\,y'' + \left( 1 -x \right) y' + n\,y = 0,$
which was fist discovered by the Russian mathematician Pafnuty Chebyshev (1821--1894) in 1859. We solve this equation, called the Chebyshev--Laguerre equation, by power series method:
$y(x) = \sum_{k\ge 0} c_k x^k , \qquad y'(x) = \sum_{k\ge 0} a_{k+1} \left( k+1 \right) x^k .$
This yields the difference equation
$k \left( k +1 \right) c_{k+1} + c_{k+1} \left( k +1 \right) - c_k k+- n\,c_k = 0 .$
From it, it follows
$c_{k+1} = -\frac{n -k}{(k+1)^2}\, c_k , \qquad k=0,1,2,\ldots .$
So we get a polynomial solution, called the Laguerre polynomial:
$$\label{EqLaguerre.1} y = L_n (x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{x^k}{k!} , \qquad n=0,1,2,\ldots .$$
Although solutions of the Laguerre equation and the Chebyshev--Laguerre equation are expressed through similar formulas, their behaviors are completely different.

# Polynomial Solutions

This section is devoted to some second order differential equations that admit polynomial solutions. To motivate the reader, let us consider a generalized Hermite equation

$y'' -p\,x^M y' + p\,r\,x^{M-1} y =0 ,$
where p ≠ 0 and M and r are positive integers. The above equation becomes standard Hermite equation for M=1, p=2, and r a positive integer that admits the well-known Hermite polynomial solution.

Theorem: The equation $$y'' -p\,x^M y' + p\,r\,x^{M-1} y =0$$ has a polynomial solution of degree r if and only if $$r = k\left( M+1 \right)$$ or $$r = k\left( M+1 \right) +1$$ for some k = 0,1,2,....    ⦻⧫

Formal substitution of $$y(x) = \sum_{n\ge 0} a_n x^n$$ into the given equation gives the recursive formula:
$a_{n+M+1} = \frac{p \left( n-r \right) a_n}{\left( n+M+1 \right)\left( n+M \right)} , \qquad n=0,1,2,\ldots .$
Now suppose that a polynomial solution of degree r exists. If r=1, then $$r= 0 \left( M+1 \right) +1 .$$ If r≥2, then $$r \ge M+1 .$$
Example: Consider the differential equation $$y'' - x^4 y' + 5\,x^3 y =0 ,$$ which corresponds to p=1, M=4, and r=5.

According to Theorem, this differential equation admits a polynomial solution. Substituting the series representations

\begin{align*} y(x) &= \sum_{n\ge 0} a_n x^n \quad \Longrightarrow \quad x^3 y = \sum_{n\ge 3} a_{n-3} x^{n} , \\ y' (x) &= \sum_{n\ge 1} n\,a_n x^{n-1} \quad \Longrightarrow \quad x^4 y' = x^4 \sum_{n\ge 1} n\,a_n x^{n-1} = \sum_{n\ge 4} (n-3)\, a_{n-3} x^{n} , \\ y'' (x) &= \sum_{n\ge 2} n(n-1)\,a_n x^{n-2} \quad \Longrightarrow \quad y'' = \sum_{n\ge 0} (n+2)(n+1)\,a_{n+2} x^{n} , \end{align*}
into the given equation, we obtain the recurrence for its coefficients:
$(n+2)(n+1) a_{n+2} - (n+3)\, a_{n+3} + 5a_{n-3} =0 \quad \Longrightarrow \quad a_{n+6} = \frac{1}{n+6} \left[ (n+4)(n+5)\,a_{n+5} + 5\,a_{n} \right] , \quad n=3,4,\ldots .$
Initially, we have
$a_2 = a_3 = a_4 =0, \quad a_5 = - \frac{1}{4}\, a_0 , \quad a_6 = - \frac{1}{6}\, a_1 ,$
and
$a_k = \frac{1}{k} \left[ (k-1)(k-2)\, a_{k-1} + 5\,a_{k-6} \right] =0, \quad k = 7,8,\ldots .$
So we get the polynomial solution
$y = a_0 \left( 1 - \frac{1}{5}\, x^5 \right) + a_1 x \left( 1 - \frac{1}{6}\, x^5 \right) .$
Example: Consider the differential equation $$y'' - 3\,x^5 y' + 39\,x^4 y =0 ,$$ which corresponds to p=3, M=5, and r=7.

According to Theorem, this differential equation admits a polynomial solution. Substituting the series $$y = \sum_{n\ge 0} a_n x^n$$ into the given equation, we obtain

$y = a_0 \left( 1 - \frac{13}{10}\, x^6 \right) + a_1 x \left( 1 - \frac{9}{7}\, x^6 \right) .$
Example: Consider the differential equation $$y'' - x^2 y' + 5\,x\, y =0 ,$$ which corresponds to p=3, M=5, and r=7.

According to Theorem, this differential equation does not have a polynomial solution because there is no such integer k that satisfies either 5 = k(2+1) = 3k or 5 = k(2+1) +1 = 3k+1. ■

# Nonhomogeneous Hermite equations

Now we turn our attention to nonhomogeneous equations.

Theorem: The initial value problem for the nonhomogeneous Hermite equation

$y'' -2x\,y' + 2 \left( 2m \right) y = x^{2n+1} , \qquad y(0) =a, \quad y' (0) =b ,$
has a polynomial solution if and only if
$b = -(2n+1) \prod_{i=0}^n \frac{1}{4i+2-4m} .$
⧫
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