Series solution and Hermite polynomial of Hermite differential equation

Series solution :

For series solution (k=0)

we have recurrence relation as,

$$ a_{r+2}=\frac{-2 (n-r) a_r}{(r+1) (r+2)}$$

$$ a_r= \frac{-(r+1) (r+2)}{2(n-r) a_{r+2}}{2(n-r)}$$

Put r= n - 2

$$ a_{n-2}=\frac{-n(n-1) a_n}{2\cdot 2}$$

Put r=n-4,

$$ a_{n-4}=\frac{-(n-3) (n-2) a_{n-2}}{2\cdot 4}$$

$$=\frac{-(n-3)(n-2)}{2\cdot 4}\cdot \frac{(-n) (n-1) a_n}{2\cdot 2}$$

$$=(-1)^2 \frac{n(n-1)(n-2)(n-3)a_n}{2^2 \cdot 2\cdot 4}$$

Put r=n-6

$$ a_{n-6}=\frac{-(n-4)(n-5)a_{n-4}}{2\cdot 6}$$

$$= \frac{-(n-4)(n-5)}{2\cdot 6}\cdot \frac{(-1)^2 n(n-1) (n-2) (n-3) a_n}{2^2 \cdot 2\cdot 4}$$

$$=\frac{(-1)^3 n(n-1)(n-2) (n-3) (n-4) (n-5) \dotsm a_n}{(2)^3\cdot 2\cdot 4\cdot 6}$$

$$ a_{n-2J}=\frac{(-1)^Jn(n-1)(n-2)(n-3)(n-4)(n-5)\dotsm (n-2J+1) a_n}{2^J\cdot 2\cdot 4\cdot 6\dotsm 2J}$$

$$=\frac{(-1)^Jn(n-1)(n-2)(n-3)\dotsm (n-2J+1) a_n}{2^J\cdot 2\cdot 4\cdot 6\dotsm 2J}\times \frac{(2-2j-2)\dotsm 3.2.1}{(n-2J)(n_2J-1)(n-2J-2),\dotsm 3,2,1}$$

$$=\frac{(-1)^J\cdot n! a_n}{2^J \cdot 2 \cdot 4\cdot 6\dotsm 2J(n-2J)!}$$

$$=\frac{(-1)^J\cdot n! a_n}{2^J\cdot 2^J J!(n-2J)!}$$

$$=\frac{(-1)^J\cdot n! a_n}{2^{2J}\cdot J!(n-2J)!}$$

Now, solution of Hermite for even term is

$$ y= \sum_{r=0}^\infty a_r x^{k+r}$$

$$=\sum_{r=0}^\infty a_r x^r$$

$$=\sum_{j=0}^{\frac n2} a_{n-2J} x^{n-2J}$$

$$=\sum_{j=0}^{\frac n2} \frac{(-1)^J n! a_n}{2^{2J} J! (n-2J)!} x^{n-2J}$$

Now, for odd term, same solution is obtained but limit vary from 0 to \(\frac{n-1}{2}\) i.e.

$$y=\sum_{J=0}^{\frac{n-1}{2}} \frac{(-1)^J n! a_n x^{n-2J}}{2^{2J} J! (n-2J)!}$$

So, the complete solution is

$$y=\sum_{J=0}^N\frac{(-1)^J n! a_n x^{n-2J}}{2^{2J} J! (n-2J)!}$$

Where N= n/2 for n\(\to\) even

=n-1/2 for n\(\to\) odd

This is the series solution of Hermite differential equation.

For Hermite polynomial \(a_n= 2^n\) then above equation becomes

$$H_n(x)=y=\sum_{J=0}^N\frac{(-1)^J n! (2 x)^{n-2J}}{ J! (n-2J)!}$$

Where N= n/2 for n\(\to\) even

=n-1/2 for n\(\to\) odd

First four Hermite Polynomial:

We have Hermite polynomial as,

$$H_n(x)=y=\sum_{J=0}^N\frac{(-1)^J n! (2 x)^{n-2J}}{ J! (n-2J)!}$$

Where N= n/2 for n\(\to\) even

=n-1/2 for n\(\to\) odd

put, n=0 (even)

$$N=\frac02=0$$

$$H_0(x)=\sum_{J=0}^0\frac{(-1)^J 0! (2x)^{0-2J}}{(0-2J)!J!}$$

$$=\frac{(-1)^0 0! (2x)^0 }{0! 0!}=1$$

Put n=1 (odd)

$$ N=\frac{n-1}{2}=\frac{1-1}{2}=0$$

$$H_1(x)=\sum_{J=0}^0 \frac{(-1)^J 1! (2x)^{1-2J}}{(1-2J)! J!}$$

$$=\frac{(-1)^0 1! (2x)^{1-0}}{1! 0! }= 2x$$

Put n=2 (even)

$$N= \frac n2=1$$

$$H_2(x)=\sum_{J=0}^1\frac{(-1)^J 2! (2x)^{2-2J}}{(1- 2J)! J!}$$

$$=\frac{(-1)^0 2! (2x)^{2-0}}{(2-2\times 0) ! 0!} + \frac{(-1)^1 2! (2x)^{2-2\times 1}}{(2-2\times 1) ! 1!}$$

$$= 4x^2-2$$

Put n=3 (odd)

$$N=\frac{n-1}{2}$$

$$=\frac{3-1}{2}$$

$$=1$$

$$H_3(x)=\sum_{j=0}^1\frac{(-1)^J 3! (2x)^{3-2j)}}{(3-2J)! J!}$$

$$= \frac{(-1)^0 3! (2x)^3}{ 3! 0!}+\frac{(-1)^1 3! (2x)^1}{(3-2)! 1!}$$

$$= 8x^3- 12x$$

These are the first four polynomials of Hermite differential equation .

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.