Hermite Differential Equation

Hermite Differential Equation:

Differential equation occurs in various branch of physics, among them Hermite differential equation occur while solving quantum harmonic ossilator in quantum physics. It helps to find energy value and various kind of wave function.

The form of Hermite differential equation

$$y''-2xy'+2ny=0\dotsm(1)$$

the above equation has \(p(x)= -2x\) and \(Q(x)= 2n\) which is clearly analytic about origin. i.e \(x\to 0\) . So solution of the equation (1) can be written in series solution.

The series solution is in the form as,

$$ y=\sum_{r=0}^\infty a_r x^{k+r}, a_\circ\ne0\dotsm(2)$$

Where the value of k should be determined.

Since equation (2) is a solution of equation (1) so it should satisfy equation(1). We evaluate y' and y''

$$y'=\sum_{r=0}^\infty a_r (k+r) x^{x+r-1}$$

and

$$y''=\sum_{r=0}^\infty a_r(k+r) (k+r-1) x^{k+r-2}$$

Now putting the value of y',y'' and y in equation (1) we get,

$$or,\;\;\; \sum_{r=0}^\infty a_r (k+r) (k+r-1) x^{x+r-2} - 2x\sum_{r=0}^\infty a_r (k+r) x^{k+r-1} + 2n\sum_{r=0}^\infty a_r x^{k+r}=0$$

$$or,\;\;\; \sum_{r=0}^\infty\biggl[ a_r(k+r) (k+r-1) x^{k+r-2}+ 2(n-k-r) a_r x^{k+r}\biggr]=0$$

Equation the coefficient of \((x^{k-2}\)) i.e. put r=0 with '0' we get,

$$ a_\circ k(k-1)=0$$

Since, \(a_\circ\ne0\) so, either k=0 or 1

Now equating the coefficient of \(x^{k-1}(r=1)\) to zero.

we get,

$$ a_1(k+1)k=0$$

\(a_1\) is arbitrary may not equal to zero.

Equating the coefficient of \(x^{k+r}\) with zero we get,

$$a_{r+2} (k+r+2) (k+r+1) + 2 (n-k - r) a_r=0$$

this is the recurrence relation of coefficient.

$$or,\;\;\; a_{r+2}=\frac{-2(n-k-r)}{(k+r+1)(k+r+2)} a_r\dotsm(3)$$

Case I:

$$ k=0$$

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

put, r=0

$$ a_2= \frac{-2n}{1\times 2}a_\circ = \frac {-2n}{2!} a_\circ$$

put, r=2

$$ a_4= \frac{-2(n-2)}{3\times 4}a_2$$

$$= \frac{-2(n-2)}{3\times 4}\times \frac{-2n}{2!}$$

$$= \frac{(-2)^2 n(n-2)}{4!}$$

put, r=4

$$a_6= \frac{-2(n-4)}{6\cdot 6}a_4$$

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

$$=\frac{(-2)^3 n(n-2)(n-4)}{6!}a_\circ$$

put r=2r( Generalized this term is always even)

$$a_2r= \frac{(-2)^r n(n-2) (n-4)\dotsm(n-2r+2) a_\circ}{2r!}$$

For odd items:

put r=1

$$a_3=\frac{-2(n-1)a_1}{2\cdot 3}$$

$$= \frac{-2(n-1)a_1}{3!}$$

put r=3

$$a_5= \frac{-2(n-3) a_3}{4\cdot5}$$

$$=\frac{(-2)^2(n-1) (n-3) a_1}{5!}$$

put r=5

$$ a_7= \frac{(-2)(n-7) a_5}{6\cdot 7}$$

$$= \frac{(-2)^3(n-1)(n-3)(n-5)a_1}{7!}$$

Similarly for,

$$ a_{2r+1}=\frac{(-2)^r(n-1)(n-3)(n-5)\dotsm(n-2r+1)a_1}{(2r+1!)}$$

Now complete solution for both term is

$$y=\sum_{r=0}^\infty a_r x^{k+4}=\sum_{r=0}^\infty a_ rx^r= a_\circ x^0+ a_1 x^1+ a_2x^2+ a_3x^3+ a_4x^4+a_5x^5+\dotsm$$

$$y=(a_\circ+ a_2 x^2+ a_4 x^4+ a_6 x^6+\dotsm)(a_1 x^1+ a_3 x^3+ a_5 x^5 + a_7 x^7+ \dotsm)$$

$$=\biggl( a_\circ + \frac{(-2n)a_\circ x^2}{2!}+\frac{(-2)^2 n(n-2)a_\circ x4}{4!}+\dotsm+\frac{(-2)^r n(n-2)-(n-2r+2)a_\circ}{(2r)!}\biggr)+ \biggl( a_1 xx+ \frac{ (-2) (n-1) a_1 x^3}{3!}+\dotsm+ \frac{(-2)^r(n-1)(n-5)\dotsm(n-2r+1)a_1}{(2r+1)!}$$

This is the required solution.

Case II:

For k=1 all odd term vanishes \(a_1=a_3=a_5=a_7\dotsm=0\)

We have,

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

For k=1,

$$ a_{r+2}=\frac{-2(n-1-r) a_r}{(r+2)(r+3)}$$

Put r=0

$$ a_2=\frac{-2(n-1)a_\circ}{2\cdot 3}$$

$$=\frac{-2(n-1)}{3!}a_\circ$$

Put r=2,

$$ a_4=\frac{-2(n-3)}{4\cdot 5}\times \frac{-2(n-1)}{3\times 2}a_\circ$$

$$=\frac{(2)^2(n-1) (n-3) a_\circ}{5!}$$

Put r=4,

$$a_6=\frac{-2(n-5) a_4}{6\cdot 7}$$

$$= \frac{-2(n-5)}{6\times 7}\times \frac{(-2)^2 (n-1) (n-3) a_\circ}{5!}$$

$$\frac{(-2)^3 (n-1) (n-3) (n-5) a_\circ}{7!}$$

Similarly,

$$ a_{2r}=\frac{(-2)^r (n-1) (n-3) (n-5) \dotsm (n-2r+1) a_\circ}{(2r+1)!}$$

The solution of Hermite differential equation for k=1 is

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

$$=\sum_r=0^\infty a_r x^{1+r}$$

$$= a_\circ x^1 + a_1 x^2+ a_2 x^3+ a_3 x^4 + a_4 x^5+ a_5 x^6+ \dotsm$$

$$=\sum_{r=0}^\infty a_{2r}x^{2r+1}$$

$$=\sum_{r=0}^\infty \frac{(-2)^r (n-1) (n-3) (n-5) \dot (n-r+1) a_\circ x^{2r+1}}{(2r+1)!}$$

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
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5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.