Consider the Chebyshev differential equation. (1-x^2)y''-xy'+\lambda^2y=0. where \lambda is a constant

Chesley 2021-08-31 Answered
Consider the Chebyshev differential equation
\(\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}\)
where \(\displaystyle\lambda\) is a constant. Find the power series solutions of this differential equation about \(\displaystyle{x}_{{0}}={0}\). For what values of the constant \(\displaystyle\lambda\) these solutions reduce to polynomials, known as Chebyshev polynomials. Compute first four of the Chebyshev polynomials.

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Expert Answer

bahaistag
Answered 2021-09-01 Author has 22497 answers
The objective is to find power series solution to Chebyshev differential equation
\(\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}\)
about the point \(\displaystyle{x}_{{0}}={0}\) and \(\displaystyle\lambda\) is a constant.
Also, we have to find the values of \(\displaystyle\lambda\) such that solution reduces to Chebyshev polynomials and we have to estimate first four Chebyshev polynomials.
Compare the equation \(\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}\) with \(\displaystyle{P}{\left({x}\right)}{y}{''}+{Q}{\left({x}\right)}{y}'+{y}={0}\).
Here, \(\displaystyle{P}{\left({x}\right)}={1}-{x}^{{2}}\) and \(\displaystyle{Q}{\left({x}\right)}=-{x}\)
Both the functions \(\displaystyle{P}{\left({x}\right)}={1}-{x}^{{2}}\) and \(\displaystyle{Q}{\left({x}\right)}=-{x}\) are analytic at \(\displaystyle{x}_{{0}}={0}\).
Now, we look for solution of the form \(\displaystyle{y}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{x}^{{n}}\)
Substitute the solution in the original equation \(\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}\)
\(\displaystyle{y}'{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{n}{a}_{{n}}{x}^{{{n}-{1}}}\)
\(\displaystyle{y}{''}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{{n}-{2}}}\)
Therefore we have,
\(\displaystyle{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{{n}-{2}}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{\sum_{{{n}={1}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+{\sum_{{{n}={0}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}\)
Now replace n by n+2 in first term and simplify
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{n}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{\sum_{{{n}={1}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+{\sum_{{{n}={0}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}\)
\(\displaystyle{2}{a}_{{2}}+{6}{a}_{{3}}{x}+{\sum_{{{n}={2}}}^{\infty}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{n}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{a}_{{1}}{x}-{\sum_{{{n}={2}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+\lambda^{{2}}{a}_{{0}}+{x}\lambda^{{2}}{a}_{{0}}+{\sum_{{{n}={2}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}\)
\(\displaystyle{2}{a}_{{2}}+\lambda^{{2}}{a}_{{0}}+{\left({6}{a}_{{3}}-{a}_{{1}}+\lambda^{{1}}{a}_{{0}}\right)}{x}+{\sum_{{{n}={2}}}^{\infty}}{\left[{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}-{\left[{n}{\left({n}-{1}\right)}+{n}-\lambda^{{2}}\right]}{a}_{{n}}\right]}{x}^{{n}}={0}\)
Now, equate the powers of x from both the sides,
\(\displaystyle{2}{a}_{{2}}+\lambda^{{2}}{a}_{{0}}={0}\)
\(\displaystyle{6}{a}_{{3}}-{a}_{{1}}+\lambda^{{2}}{a}_{{0}}={0}\)
\(\displaystyle{\left[{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}-{\left[{n}{\left({n}-{1}\right)}+{n}-\lambda^{{2}}\right]}{a}_{{n}}\right]}={0}\quad{n}={2},{3},\ldots\)
Now, we solve the values obtained in above equations,
\(\displaystyle{a}_{{2}}=-{\frac{{\lambda^{{2}}}}{{{2}}}}{a}_{{0}}\)
\(\displaystyle{a}_{{3}}={\frac{{{1}-\lambda^{{2}}}}{{{3}!}}}{a}_{{1}}\)
\(\displaystyle{a}_{{{n}+{2}}}={\frac{{{n}^{{2}}-\lambda^{{2}}}}{{{\left({n}+{2}\right)}{\left({n}+{1}\right)}}}}{a}_{{n}}\quad{n}={2},{3},\ldots\) With the help of general formula, the few more terms are,
\(\displaystyle{a}_{{4}}={\frac{{{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{4}!}}}{a}_{{0}}\)
\(\displaystyle{a}_{{5}}={\frac{{{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{5}!}}}{a}_{{1}}\)
\(\displaystyle{a}_{{6}}={\frac{{{\left({4}^{{2}}-\lambda^{{2}}\right)}{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{6}!}}}{a}_{{0}}\) Therefore, the pattern can be set up as,
\(\displaystyle{a}_{{{2}{n}}}={\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{a}_{{0}},\quad{n}={1},{2},\ldots\)
\(\displaystyle{a}_{{{2}{n}+{1}}}={\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{a}_{{1}},\quad{n}={1},{2},\ldots\)
Since, our solution was of the form \(\displaystyle{y}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{x}\). Thus, the power series solution is
\(\displaystyle{y}={1}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{x}^{{{2}{n}}}+{x}+\)
\(\displaystyle+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{x}^{{{2}{n}+{1}}}\)
\(\displaystyle={u}{\left({x}\right)}+{v}{\left({x}\right)}\)
where,
\(\displaystyle{u}{\left({x}\right)}={1}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{x}^{{{2}{n}}}\)
\(\displaystyle{v}{\left({x}\right)}={x}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{x}^{{{2}{n}+{1}}}\)
Also, we have to find the values of \(\displaystyle\lambda\) such that solution reduces to Chebyshev polynomials and we have to estimate first four Chebyshev polynomials.
When \(\displaystyle\lambda={0},{1},{2},{3}\) we get four Chebyshev polynomials.
When \(\displaystyle\lambda={0}\) and \(\displaystyle{a}_{{{2}{n}}}={0}\ \text{ for }\ \geq{1}\) so,
\(\displaystyle{u}{\left({x}\right)}={1}\)
When \(\displaystyle\lambda={1}\) and \(\displaystyle{a}_{{{2}{n}+{1}}}={0}\ \text{ for }\ {n}\geq{1}\) so,
\(\displaystyle{v}{\left({x}\right)}={x}\)
When \(\displaystyle\lambda={2}\) and \(\displaystyle{a}_{{{2}{n}}}={0}\ \text{ for }\ {n}\geq{2}\) so,
\(\displaystyle{u}{\left({x}\right)}={1}-{2}{x}^{{2}}\)
When \(\displaystyle\lambda={3}\) and \(\displaystyle{a}_{{{2}{n}+{1}}}={0}\ \text{ for }\ {n}\geq{2}\) so,
\(\displaystyle{v}{\left({x}\right)}={x}-{\frac{{{4}}}{{{5}}}}{x}^{{3}}\)
So, first four Chebyshev polynomials are \(\displaystyle{1},{x},{1}-{2}{x}^{{2}}\) and \(\displaystyle{x}-{\frac{{{4}}}{{{5}}}}{x}^{{3}}\)
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