Let us look into the formal solution of Hermite polynomial equation.
Hermite differential equation is given by, $$ \frac{d^2y}{dx^2} - 2x \,\frac{dy}{dx} + 2ny = 0 $$
Using Frobenius method, we assume an infinite series solution of ,
$$ y(x) = \sum_{m=0}^{\infty} C_m x^{m+r} ........ where C_0 \neq 0 $$ Substituting this on our differential equation, our differential equation gets modified into $$ \sum_{m=0}^\infty \left[ (m+r)(m+r-1) C_m x^{m+r-2} + 2 [ n - (m+r)] C_m x^{m+r} \right] = 0 $$ This should be equal to zer0, which can be obtained only if and only if all the coefficients are zero.
Equating to zero, the coefficients of,
$$ (1) x^{r-2}, \ldots... r(r-1) = 0 \rightarrow r=0 \; or \; r=1 \; since \; C_0 \neq 0 $$ $$ (2) x^{r-1}, \ldots... (r+1)rC_1 = 0 .... \; if \;\; r=0, \; C_1\neq 0 ; and \; if \;\; r=1, \; then \; C_1=0 $$ $$ (3) x^{m+r}, \\~\\ for \; r=0 \dots......... \frac{C_{m+2}}{C_m} = \frac{-2(n-m)} {(m+1)(m+2)} \\~\\ for \; r=1 \dots...........\frac {C_{m+2}}{C_m} = \frac{-2[n-m-1]}{(m+2)(m+3)}$$
Considering r=0, the general expression for even coefficients expressed as,
$$ C_{2s} = \frac{(-1)^s 2^s n (n-2).....(n-2s+2)}{(2s!)} C_0 $$
and the odd coefficients are expressed as,
$$ C_{2s+1} = \frac{ (-1)^s 2^s (n-1) (n-3).....(n-2s+1)}{(2s+1)!} C_1$$
The general solution is therefore,
$$ y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right] + \\~\\ C_1x \left[ 1 + \sum_{s=1}^\infty \frac{(-1)^s 2^s (n-1)(n-3)....(n-2s+1)}{(2s+1)!} x^{2s+1} \right] $$
For r=1, if we proceed like the same, we will get a solution exactly similar to the second part of the series in the general solution with other coefficient. Since the coefficients are arbitrary, the series itself already inherent in the general solution, so we don't need to put separate attention on that.
That's it. We arrived to our general solution. Depending on the
nature of the problem, we can make the series to converge or stop by appropriately choosing the values of $ \; C_0 \;and \;C_1$
You can choose the constants in various ways, where all those solutions will obey our Hermite differential equation - It is true, because I have tried!
But all of the solutions satisfying our equation is not physically meaningful.
There is a conventional way of choosing the constants, such that the terms in the series have the properties known as orthogonality, completeness,etc. And those specific functions in our series is known as the Hermite polynomials.
In the general solution, we make the convergence test by taking the ratio of consecutive elements either in the odd series or in the even series, i.e. ratio of the terms $x^{2s}, x^{2s+2}$,
eg.In Even series $$\left|\frac{t_{s+1}}{t_s}\right| = \left| \frac{ C_{2s+2}}{C_{2s}} \right| = \left| \frac{-2 (n-2s)}{ (2s+2) (2s+1)} x^2 \right| $$
As "s" becomes very large,
$$ \lim_{s\rightarrow\infty} \left| \frac{C_{2s+2}}{C_{2s}}\right| = \left|\frac{4s}{4s^2} x^2\right| = \frac{x^2}{s} $$
[this is the same convergence of $ e^{x^2} $ series]. The series is an infinite series for all values of x and it will be meaningful only if it is terminated to finite terms. It can be achieved by choosing suitable values for "n".
Choosing the other constant as zero, we can always work with either odd or even series. Once we take an odd or even series, we choose the value of "n" such that,
for even series, the coefficients of $x^{2s+2}$ is zero. i.e. n = 2s
and $ C_1 = 0 $ the general series reduces to,
$$y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right]$$
Substituting, n = 2s, $$ y(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{(-1)^s 2^s (2s)(2s-2)...(2s-2s+2)} {(2s!)} x^{2s} \right] $$
$\rightarrow$ $$ y_{even}(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!} {2s!} x^{2s} \right]$$
for odd series, n = 2s+1 such that there is no terms involving power of x greater than 2s+1 and $C_0 = 0 $, then the series becomes,
$$ y_{odd}(x) = C_1x\left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!}{(2s+1)!} x^{2s+1} \right]$$
Now, the conventional choice states that the constants $C_0\,,C_1$ chosen in a way such that the power of $x^n$ has the coefficient $2^n$ , then,
$$ for \,\,C_{2n} \rightarrow C_0 \frac{(-1)^n 2^{2n} n!}{2n!}x^{2n} = 2^{2n} x^{2n} $$ which implies, $$ C_0 = \frac{2n! (-1)^n}{n!} $$
$$ for \,\,C_{2n+1} \rightarrow C_1 \frac{(-1)^n 2^{2n} n!}{(2n+1)!}x^{2n+1} = 2^{2n+1} x^{2n+1} $$$\rightarrow$ $$ C_1 = (-1)^n \frac{(2n+1)! \,(2)}{n!} $$
That is all we need to know!
Now, we will derive the first few Hermite polynomials as an example,
n=0 $$ H_0(x) = C_0 \,\,where\,\, C_0 = \frac{2(0)!}{0!}(-1)^0 = 1 $$ Therefore , $$ H_0(x) = 1 $$
n=1 $$ H_1(x) = C_1x \,\,where\,\, C_1 = \frac{(-1)^0 1! 2}{0!} =2 $$ $\rightarrow$$$ H_1(x) = 2x $$
n=2 $$ H_2(x) = C_0 + C_0 \frac{(-1) 2^2 1!}{2!} x^2 \,\,where\,\, C_0 = \frac{2(1)! (-1)^1}{1!} = -2 $$$\rightarrow$ $$ H_2(x) = -2 + (-2) (-2x^2) = 4x^2 -2 $$
n=3, $$ H_3(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 $$ and $$ C_1 = \frac{(-1)^1 3! 2}{1!} = -12 $$ $\rightarrow$ $$ H_3(x) = -12 x + (-12) \frac{(-4)}{(6)}x^3 = 8x^3 - 12x $$
n=4, $$ H_4(x) = C_0 + \frac{(-1) 2^2 1!}{2!} x^2 C_0 + \frac{(-1)^2 2^4 2! }{4!} x^4 C_0 $$ and $$ C_0 = \frac{2(2)! (-1)^2}{2!} = 12 $$ $ \rightarrow$ $$ H_4(x) = 12 + 12 (-2) x^2 + 12 \frac{4}{3} x^4 = 16x^4 - 24x^2 +12 $$
n=5 $$ H_5(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 + C_1 \frac{(-1)^2 2^4 2!}{5!}x^5 $$ and $$ C_1 = \frac{(-1)^2 5! 2}{2!} = 120 $$ $\rightarrow $ $$ H_5(x) = 120 x + (-1) 80 x^3 + 32x^5 = 32x^5 - 80x^3 +120x $$
Thus we can find all the Hermite polynomials.
You may ask "Why this choice of Hermite polynomials?"
It is because of the special properties followed by these polynomials. For eg. These polynomials can be simply written in one line using a formula known as Rodrigues formula. And there is a specific generating function for these and orthogonality property for specific conditions and etc. All these make Hermite polynomials more special in real life.
Hermite differential equation is given by, $$ \frac{d^2y}{dx^2} - 2x \,\frac{dy}{dx} + 2ny = 0 $$
Using Frobenius method, we assume an infinite series solution of ,
$$ y(x) = \sum_{m=0}^{\infty} C_m x^{m+r} ........ where C_0 \neq 0 $$ Substituting this on our differential equation, our differential equation gets modified into $$ \sum_{m=0}^\infty \left[ (m+r)(m+r-1) C_m x^{m+r-2} + 2 [ n - (m+r)] C_m x^{m+r} \right] = 0 $$ This should be equal to zer0, which can be obtained only if and only if all the coefficients are zero.
Equating to zero, the coefficients of,
$$ (1) x^{r-2}, \ldots... r(r-1) = 0 \rightarrow r=0 \; or \; r=1 \; since \; C_0 \neq 0 $$ $$ (2) x^{r-1}, \ldots... (r+1)rC_1 = 0 .... \; if \;\; r=0, \; C_1\neq 0 ; and \; if \;\; r=1, \; then \; C_1=0 $$ $$ (3) x^{m+r}, \\~\\ for \; r=0 \dots......... \frac{C_{m+2}}{C_m} = \frac{-2(n-m)} {(m+1)(m+2)} \\~\\ for \; r=1 \dots...........\frac {C_{m+2}}{C_m} = \frac{-2[n-m-1]}{(m+2)(m+3)}$$
Considering r=0, the general expression for even coefficients expressed as,
$$ C_{2s} = \frac{(-1)^s 2^s n (n-2).....(n-2s+2)}{(2s!)} C_0 $$
and the odd coefficients are expressed as,
$$ C_{2s+1} = \frac{ (-1)^s 2^s (n-1) (n-3).....(n-2s+1)}{(2s+1)!} C_1$$
The general solution is therefore,
$$ y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right] + \\~\\ C_1x \left[ 1 + \sum_{s=1}^\infty \frac{(-1)^s 2^s (n-1)(n-3)....(n-2s+1)}{(2s+1)!} x^{2s+1} \right] $$
For r=1, if we proceed like the same, we will get a solution exactly similar to the second part of the series in the general solution with other coefficient. Since the coefficients are arbitrary, the series itself already inherent in the general solution, so we don't need to put separate attention on that.
That's it. We arrived to our general solution. Depending on the
nature of the problem, we can make the series to converge or stop by appropriately choosing the values of $ \; C_0 \;and \;C_1$
You can choose the constants in various ways, where all those solutions will obey our Hermite differential equation - It is true, because I have tried!
But all of the solutions satisfying our equation is not physically meaningful.
There is a conventional way of choosing the constants, such that the terms in the series have the properties known as orthogonality, completeness,etc. And those specific functions in our series is known as the Hermite polynomials.
In the general solution, we make the convergence test by taking the ratio of consecutive elements either in the odd series or in the even series, i.e. ratio of the terms $x^{2s}, x^{2s+2}$,
eg.In Even series $$\left|\frac{t_{s+1}}{t_s}\right| = \left| \frac{ C_{2s+2}}{C_{2s}} \right| = \left| \frac{-2 (n-2s)}{ (2s+2) (2s+1)} x^2 \right| $$
As "s" becomes very large,
$$ \lim_{s\rightarrow\infty} \left| \frac{C_{2s+2}}{C_{2s}}\right| = \left|\frac{4s}{4s^2} x^2\right| = \frac{x^2}{s} $$
[this is the same convergence of $ e^{x^2} $ series]. The series is an infinite series for all values of x and it will be meaningful only if it is terminated to finite terms. It can be achieved by choosing suitable values for "n".
Choosing the other constant as zero, we can always work with either odd or even series. Once we take an odd or even series, we choose the value of "n" such that,
for even series, the coefficients of $x^{2s+2}$ is zero. i.e. n = 2s
and $ C_1 = 0 $ the general series reduces to,
$$y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right]$$
Substituting, n = 2s, $$ y(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{(-1)^s 2^s (2s)(2s-2)...(2s-2s+2)} {(2s!)} x^{2s} \right] $$
$\rightarrow$ $$ y_{even}(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!} {2s!} x^{2s} \right]$$
for odd series, n = 2s+1 such that there is no terms involving power of x greater than 2s+1 and $C_0 = 0 $, then the series becomes,
$$ y_{odd}(x) = C_1x\left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!}{(2s+1)!} x^{2s+1} \right]$$
Now, the conventional choice states that the constants $C_0\,,C_1$ chosen in a way such that the power of $x^n$ has the coefficient $2^n$ , then,
$$ for \,\,C_{2n} \rightarrow C_0 \frac{(-1)^n 2^{2n} n!}{2n!}x^{2n} = 2^{2n} x^{2n} $$ which implies, $$ C_0 = \frac{2n! (-1)^n}{n!} $$
$$ for \,\,C_{2n+1} \rightarrow C_1 \frac{(-1)^n 2^{2n} n!}{(2n+1)!}x^{2n+1} = 2^{2n+1} x^{2n+1} $$$\rightarrow$ $$ C_1 = (-1)^n \frac{(2n+1)! \,(2)}{n!} $$
That is all we need to know!
Now, we will derive the first few Hermite polynomials as an example,
n=0 $$ H_0(x) = C_0 \,\,where\,\, C_0 = \frac{2(0)!}{0!}(-1)^0 = 1 $$ Therefore , $$ H_0(x) = 1 $$
n=1 $$ H_1(x) = C_1x \,\,where\,\, C_1 = \frac{(-1)^0 1! 2}{0!} =2 $$ $\rightarrow$$$ H_1(x) = 2x $$
n=2 $$ H_2(x) = C_0 + C_0 \frac{(-1) 2^2 1!}{2!} x^2 \,\,where\,\, C_0 = \frac{2(1)! (-1)^1}{1!} = -2 $$$\rightarrow$ $$ H_2(x) = -2 + (-2) (-2x^2) = 4x^2 -2 $$
n=3, $$ H_3(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 $$ and $$ C_1 = \frac{(-1)^1 3! 2}{1!} = -12 $$ $\rightarrow$ $$ H_3(x) = -12 x + (-12) \frac{(-4)}{(6)}x^3 = 8x^3 - 12x $$
n=4, $$ H_4(x) = C_0 + \frac{(-1) 2^2 1!}{2!} x^2 C_0 + \frac{(-1)^2 2^4 2! }{4!} x^4 C_0 $$ and $$ C_0 = \frac{2(2)! (-1)^2}{2!} = 12 $$ $ \rightarrow$ $$ H_4(x) = 12 + 12 (-2) x^2 + 12 \frac{4}{3} x^4 = 16x^4 - 24x^2 +12 $$
n=5 $$ H_5(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 + C_1 \frac{(-1)^2 2^4 2!}{5!}x^5 $$ and $$ C_1 = \frac{(-1)^2 5! 2}{2!} = 120 $$ $\rightarrow $ $$ H_5(x) = 120 x + (-1) 80 x^3 + 32x^5 = 32x^5 - 80x^3 +120x $$
Thus we can find all the Hermite polynomials.
You may ask "Why this choice of Hermite polynomials?"
It is because of the special properties followed by these polynomials. For eg. These polynomials can be simply written in one line using a formula known as Rodrigues formula. And there is a specific generating function for these and orthogonality property for specific conditions and etc. All these make Hermite polynomials more special in real life.
No comments:
Post a Comment
Let everyone know what you think about this