# Chebyshev Differential Equation

## Definition and General Solution

The differential equation of type

$\left( {1 - {x^2}} \right)y^{\prime\prime} - xy' + {n^2}y = 0,$

where $$\left| x \right| \lt 1$$ and $$n$$ is a real number, is called the Chebyshev equation after the famous Russian mathematician Pafnuty Chebyshev.

This equation can be converted to a simpler form using the substitution $$x = \cos t.$$ Indeed, in this case we have

$x = \cos t,\;\; \Rightarrow dx = - \sin tdt,\;\; \Rightarrow \frac{{dt}}{{dx}} = - \frac{1}{{\sin t}}.$

Hence,

$y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}\frac{{dt}}{{dx}} = - \frac{1}{{\sin t}}\frac{{dy}}{{dt}},$
$y^{\prime\prime} = \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\frac{{dt}}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = - \frac{1}{{\sin t}}\frac{d}{{dt}}\left( { - \frac{1}{{\sin t}}\frac{{dy}}{{dt}}} \right) = \frac{1}{{\sin t}}\left[ {\frac{d}{{dt}}\left( {\frac{1}{{\sin t}}} \right)\frac{{dy}}{{dt}} + \frac{1}{{\sin t}}\frac{{{d^2}y}}{{d{t^2}}}} \right] = \frac{1}{{\sin t}}\left[ {\left( { - \frac{{\cos t}}{{{{\sin }^2}t}}} \right)\frac{{dy}}{{dt}} + \frac{1}{{\sin t}}\frac{{{d^2}y}}{{d{t^2}}}} \right] = \frac{1}{{{{\sin }^2}t}}\left[ {\left( { - \frac{{\cos t}}{{\sin t}}} \right)\frac{{dy}}{{dt}} + \frac{{{d^2}y}}{{d{t^2}}}} \right].$

Substituting the expressions of derivatives into the differential equation gives:

$\left( {1 - {{\cos }^2}t} \right)\frac{1}{{{{\sin }^2}t}} \left[ {\left( { - \frac{{\cos t}}{{\sin t}}} \right)\frac{{dy}}{{dt}} + \frac{{{d^2}y}}{{d{t^2}}}} \right] - \cos t\left[ { - \frac{1}{{\sin t}}\frac{{dy}}{{dt}}} \right] + {n^2}y = 0,\;\; \Rightarrow \frac{\cancel{{{\sin }^2}t}}{\cancel{{{\sin }^2}t}}\left[ { - \frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}} + \frac{{{d^2}y}}{{d{t^2}}}} \right] + \frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}} + {n^2}y = 0,\;\; \Rightarrow - \cancel{\frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}}} + \frac{{{d^2}y}}{{d{t^2}}} + \cancel{\frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}}} + {n^2}y = 0.$

As a result we can write the equation in the compact form:

$\frac{{{d^2}y}}{{d{t^2}}} + {n^2}y = 0.$

The general solution of the last equation is given by the formula

$y\left( t \right) = {C_1}\cos \left( {nt} \right) + {C_2}\sin\left( {nt} \right),$

which can be also written as

$y\left( t \right) = C\cos \left( {nt + \alpha } \right).$

Here $${C_1},$$ $${C_2},$$ and $$C,$$ $$\alpha$$ are arbitrary real numbers. For simplicity, we can set $$\alpha = 0.$$ Then the general solution of the original Chebyshev equation will be given by the formula:

$y\left( x \right) = C\cos \left( {n\arccos x} \right).$

In this expression, $$n$$ may be any real number. But if $$n$$ is an integer, the given function is the Chebyshev polynomial of the first kind.

## Chebyshev Polynomials of the First Kind

The Chebyshev polynomial of the first kind is called the function

${T_n}\left( x \right) = \cos \left( {n\arccos x} \right),$

where $$\left| x \right| \le 1$$ and $$n = 0,1,2,3, \ldots$$. Next, we show that this function is really an algebraic polynomial. For $$n = 0$$ and $$n = 1$$ we have

${T_0}\left( x \right) = \cos 0 = 1,$
${T_1}\left( x \right) = \cos \left( {\arccos x} \right) = x.$

By setting $$x = \cos t,$$ we can write:

${T_1}\left( t \right) = \cos \left( {\arccos \left( {\cos t} \right)} \right) = \cos t,$
${T_n}\left( t \right) = \cos \left( {n\arccos \left( {\cos t} \right)} \right) = \cos \left( {nt} \right),$
${T_{n - 1}}\left( t \right) = \cos \left( {\left( {n - 1} \right)\arccos \left( {\cos t} \right)} \right) = \cos \left( {\left( {n - 1} \right)t} \right),$
${T_{n + 1}}\left( t \right) = \cos \left( {\left( {n + 1} \right)\arccos \left( {\cos t} \right)} \right) = \cos \left( {\left( {n + 1} \right)t} \right).$

Since

$\cos \left( {\left( {n - 1} \right)t} \right) + \cos \left( {\left( {n + 1} \right)t} \right) = 2\cos \frac{{\left( {n - 1} \right)t + \left( {n + 1} \right)t}}{2} \cos \frac{{\left( {n - 1} \right)t - \left( {n + 1} \right)t}}{2} = 2\cos \frac{{2nt}}{2}\cos \frac{{\left( { - 2t} \right)}}{2} = 2\cos \left( {nt} \right)\cos t,$

we obtain the following recursive relationship for the Chebyshev polynomials of the first kind:

${T_{n - 1}} + {T_{n + 1}} = 2{T_n}{T_1},\;\; \Rightarrow {T_{n + 1}} = 2{T_n}x - {T_{n - 1}}.$

Now we can easily calculate the Chebyshev polynomials of higher orders:

${T_2} = 2{T_1}\left( x \right)x - {T_0} = 2{x^2} - 1,$
${T_3} = 2{T_2}\left( x \right)x - {T_1} = 2\left( {2{x^2} - 1} \right)x - x = 4{x^3} - 3x,$
${T_4} = 2{T_3}\left( x \right)x - {T_2} = 2\left( {4{x^3} - 3x} \right)x - \left( {2{x^2} - 1} \right) = 8{x^4} - 8{x^2} + 1,$
${T_5} = 2{T_4}\left( x \right)x - {T_3} = 2\left( {8{x^4} - 8{x^2} + 1} \right)x - \left( {4{x^3} - 3x} \right) = 16{x^5} - 20{x^3} + 5x,$

and so on$$\ldots$$

## Chebyshev Polynomials of the Second Kind

The Chebyshev polynomials of the second kind can be also defined by the recursive relationship:

${U_n}\left( x \right) = \begin{cases} 1, \text{ if }n = 0 \\[0.4em] 2x, \text{ if }n = 1 \\[0.4em] {2x{U_{n - 1}}\left( x \right) - {U_{n - 2}}\left( x \right), \text{ if }n \le 2} \end{cases}.$

The polynomials of the second kind are solutions to the Chebyshev differential equation of the type

$\left( {1 - {x^2}} \right)y^{\prime\prime} - 3xy' + n\left( {n + 2} \right)y = 0.$

The graphs of the Chebyshev polynomials of the 1st and $$2$$nd kind are shown in Figures $$1$$ and $$2,$$ respectively.

See solved problems on Page 2.