Differential Equations

2nd Order Equations

Chebyshev Differential Equation

Page 1
Theory
Page 2
Problems 1-2

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,

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

Substituting the expressions of derivatives into the differential equation gives:

\[\require{cancel}
{\left( {1 – {{\cos }^2}t} \right)\frac{1}{{{{\sin }^2}t}} \cdot}\kern0pt{ \left[ {\left( { – \frac{{\cos t}}{{\sin t}}} \right)\frac{{dy}}{{dt}} + \frac{{{d^2}y}}{{d{t^2}}}} \right] }
– {\cos t\left[ { – \frac{1}{{\sin t}}\frac{{dy}}{{dt}}} \right] }+{ {n^2}y = 0,\;\;}\Rightarrow
{{\frac{\cancel{{{\sin }^2}t}}{\cancel{{{\sin }^2}t}}\left[ { – \frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}} + \frac{{{d^2}y}}{{d{t^2}}}} \right] }+{ \frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}} + {n^2}y = 0,\;\;}}\Rightarrow
{{ – \cancel{\frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}}} + \frac{{{d^2}y}}{{d{t^2}}} }+{ \cancel{\frac{{\cos t}}{{\sin t}}\frac{{dy}}{{dt}}} + {n^2}y = 0.}}
\]

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

\[
{\cos \left( {\left( {n – 1} \right)t} \right) + \cos \left( {\left( {n + 1} \right)t} \right) }
= {2\cos \frac{{\left( {n – 1} \right)t + \left( {n + 1} \right)t}}{2} \cdot}\kern0pt{ \cos \frac{{\left( {n – 1} \right)t – \left( {n + 1} \right)t}}{2} }
= {2\cos \frac{{2nt}}{2}\cos \frac{{\left( { – 2t} \right)}}{2} }
= {2\cos \left( {nt} \right)\cos t,}
\]

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 Chebushev 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…

Chebyshev Polynomials of the Second Kind

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

\[
{{U_n}\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
1, \text{ if }n = 0 \\
2x, \text{ if }n = 1 \\
{2x{U_{n – 1}}\left( x \right) – {U_{n – 2}}\left( x \right),}\kern0pt{ \text{ if }n \le 2}
\end{cases}}
\]

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.

Chebyshev Polynomials of the First Kind

Figure 1.

Chebyshev Polynomials of the Second Kind

Figure 2.

Solved Problems

Click on problem description to see solution.

 Example 1

Find the general solution of the equation \(\left( {1 – {x^2}} \right)y^{\prime\prime} – xy’ + 2y \) \(= 0\) for \(\left| x \right| \lt 1.\)

 Example 2

Find the general solution of the differential equation \(\left( {1 – {x^2}} \right)y^{\prime\prime} – xy’ + 4y\) \( = 0\) for \(\left| x \right| \lt 1.\)

Page 1
Theory
Page 2
Problems 1-2