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.
Figure 1. Figure 2.
