# 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,

${y’ = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}\frac{{dt}}{{dx}} }={ – \frac{1}{{\sin t}}\frac{{dy}}{{dt}},}$

${y^{\prime\prime} = \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) } = {\frac{d}{{dt}}\frac{{dt}}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) } = { – \frac{1}{{\sin t}}\frac{d}{{dt}}\left( { – \frac{1}{{\sin t}}\frac{{dy}}{{dt}}} \right) } = {{\frac{1}{{\sin t}}\left[ {\frac{d}{{dt}}\left( {\frac{1}{{\sin t}}} \right)\frac{{dy}}{{dt}} }\right.}+{\left.{ \frac{1}{{\sin t}}\frac{{{d^2}y}}{{d{t^2}}}} \right] }} = {{\frac{1}{{\sin t}}\left[ {\left( { – \frac{{\cos t}}{{{{\sin }^2}t}}} \right)\frac{{dy}}{{dt}} }\right.}+{\left.{ \frac{1}{{\sin t}}\frac{{{d^2}y}}{{d{t^2}}}} \right] }} = {{\frac{1}{{{{\sin }^2}t}}\left[ {\left( { – \frac{{\cos t}}{{\sin t}}} \right)\frac{{dy}}{{dt}} }\right.}+{\left.{ \frac{{{d^2}y}}{{d{t^2}}}} \right].}}$

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 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…

### 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.

## Solved Problems

Click or tap a problem to see the 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 $$\left| x \right| \lt 1.$$
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Concept
Page 2
Problems 1-2