Question

(a) When \(\alpha = n\) is a nonnegative integer, Hermite’s differential equation always possesses a polynomial solution of degree \(n\). Use \({y_1}(x)\), given in Problem 53, to find polynomial solutions for \(n = 0,n = 2,\)and\(n = 4\). Then use \({y_2}(x)\) to find polynomial solutions for \(n = 1,n = 3,\)and\(n = 5\).

(b) A Hermite polynomial \({H_n}(x)\) is defined to be the \({n^{th}}\) degree polynomial solution of Hermite’s equation multiplied by an appropriate constant so that the coefficient of \({x^n}\) in \({H_n}(x)\) is \({2^n}\). Use the polynomial solutions in part (a) to show that the first six Hermite polynomials are

\(\begin{aligned}{}{H_0}(x) = 1\\{H_1}(x) = 2x\\{H_2}(x) = 4{x^2} - 2\\{H_3}(x) = 8{x^3} - 12x\\{H_4}(x) = 16{x^4} - 18{x^2} + 12\\{H_5}(x) = 32{x^5} - 160{x^3} + 120x\end{aligned}\)

Short answer

(a)The polynomial solutions are,

For\({y_1}(x)\),

\(\begin{aligned}{l}n = 0,{y_1}(x) = 1\\n = 2,{y_1}(x) = 1 - 2{x^2}\\n = 4,{y_1}(x) = 1 - 4{x^2} + \frac{4}{3}{x^4}\end{aligned}\)

For \({y_2}(x)\),

\(\begin{aligned}{}n = 1,{y_2}(x) = x\\n = 3,{y_2}(x) = x - \frac{2}{3}{x^3}\\n = 5,{y_2}(x) = x - \frac{4}{3}{x^3} + \frac{4}{{15}}{x^5}\end{aligned}\)

(b) Hence proved that the first six Hermite polynomials.

Step-by-step solution

Define Hermite polynomial of degree \(n\).

The Hermite Polynomial of degree \(n\) is defined to be the \({n^{th}}\) degree polynomial solution of Hermite’s equation multiplied by a constant so that the coefficient of \({x^n}\) is \({2^n}\).

Determine the two linear independent solution.

(a)

Let the form of the solution of the givendifferential equation be\(y = \sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)… (1), where\({x_0} = 0\).

The first and second derivation of the equation (1) is given by,

\(\begin{aligned}{}{y_1}(x) &= 1 + \sum\limits_{k = 1}^\infty {{{( - 1)}^k}} \frac{{{2^k}n(n - 2)L(n - 2k + 2)}}{{(2k)!}}{x^{2k}}\\ &= 1 - \frac{{2n}}{{2!}}{x^2} + \frac{{{2^2}n(n - 2)}}{{4!}}{x^4} - \frac{{{2^3}n(n - 2)(n - 4)}}{{6!}}{x^6} + \ldots \\{y_2}(x) &= \quad x + \sum\limits_{k = 1}^\infty {{{( - 1)}^k}} \frac{{{2^k}(n - 1)(n - 3)L(n - 2k + 1)}}{{(2k + 1)!}}{x^{2k + 1}}\\ &= x - \frac{{2(n - 1)}}{{3!}}{x^3} + \frac{{{2^2}(n - 1)(n - 3)}}{{5!}}{x^5} - \frac{{{2^3}(n - 1)(n - 3)(n - 5)}}{{7!}}{x^7} + \ldots \end{aligned}\)

Determine the first three polynomial solution.

For even number of\(n\),\({y_1}(x)\)is a polynomial solution and\({y_2}(x)\)is a series solution and for the odd number of\(n\), vice versa.

For\({y_1}(x)\),

\(\begin{aligned}{}n = 0,{y_1}(x) = 1 + 0 + 0 + \cdots + 0 = 1 - \frac{{2 \cdot 2}}{{2!}}{x^2} + 0 + 0 + \cdots + 0\\n = 2,{y_1}(x) = 1 - 2{x^2} = 1 - \frac{{2 \cdot 4}}{{2!}}{x^2} + \frac{{{2^2} \cdot 4 \cdot (4 - 2)}}{{4!}}{x^4} + 0 + 0 + \cdots + 0\\n = 4,{y_1}(x) = 1 - 4{x^2} + \frac{4}{3}{x^4}\end{aligned}\)

For \({y_2}(x)\),

\(\begin{aligned}{}n = 1,{y_2}(x) = x + 0 + 0 + \cdots + 0 = x\\n = 3,{y_2}(x) = x - \frac{{2 \cdot 2}}{{3!}}{x^3} + 0 + ... + 0 = x - \frac{2}{3}{x^3}\\n = 5,{y_2}(x) = x - \frac{{2 \cdot 4}}{{3!}}{x^3} + \frac{{{2^2} \cdot 4 \cdot 2}}{{5!}}{x^5} + 0 + ... + 0 = x - \frac{4}{3}{x^3} + \frac{4}{{15}}{x^5}\end{aligned}\)

Derive the six Hermite polynomials.

(b)

For\({H_0}(x)\), then\(n = 0\).

The zeroth-degree polynomial solution of Hermite’s equation is,

Using the same analysis, the first, second, third, fourth, fifth and sixth polynomials are,

\(\begin{aligned}{}{H_2}(x) &= {a_2}(1 - 2{x^2})\\ &= - 2(1 - 2{x^2})\\ &= 4{x^2} - 2\end{aligned}\)

\(\begin{aligned}{}{H_3}(x) &= \quad {a_3}\left( {x - \frac{2}{3}{x^3}} \right)\\ &= - 3 \times {2^2}\left( {x - \frac{2}{3}{x^3}} \right)\\ &= 8{x^3} - 12x\\{H_4}(x) = {a_4}\left( {1 - 4{x^2} + \frac{4}{3}{x^4}} \right)\\ &= {2^2} \times 3\left( {1 - 4{x^2} + \frac{4}{3}{x^4}} \right)\\ &= 16{x^4} - 48{x^2} + 12\\{H_5}(x) &= {a_5}\left( {x - \frac{4}{3}{x^3} + \frac{4}{{15}}{x^5}} \right)\\ &= 15 \times {2^3}\left( {x - \frac{4}{3}{x^3} + \frac{4}{{15}}{x^5}} \right)\\ &= 32{x^5} - 160{x^3} + 120x\end{aligned}\)

Hence proved.