Proving Hermite polynomials satisfy Hermite's equation

  • #1
92
5

Main Question or Discussion Point

My book (by Mary L Boas) introduces first the Hermite differential equation for Hermite functions:
$$y_n'' - x^2y_n=-(2n+1)y_n$$ and we find solutions like $$y_n=e^{x^2/2}D^n e^{-x^2}$$ where ##D^n=\frac{d^n}{dx^n}##

Now she says that multiplying ##y_n## by ##(-1)^ne^{x^2/2}## gives us what is known as Hermite polynomials: $$H_n(x)=(-1)^ne^{x^2}D^n e^{-x^2}$$which satisfies another equation: $$y''-2xy'+2ny=0 $$
So far so good until I try to prove if ##H_n(x)## does indeed satisfy the equation above. In her problem set she asks to check if ##e^{-x^2/2}H_n(x)## satisfies the Hermite polynomial equation and here I don't understand why is the extra factor ##e^{-x^2/2}## there?

Also, when I try to just check if ##H_n(x)## satisfies the equation I fail to get past this step which is:
$$H_n'(x)=(-1)^ne^{x^2}[D^{n+1}e^{-x^2}+2x \cdot D^ne^{-x^2}] $$ I don't know if this can be further simplified because when I try to calculate ##H_n''(x)## it gets even more complicated and I am unable to prove that ##H_n(x)## is a solution to the equation. My first thought was to find out what ##D^n e^{f(x)}## and I thought if ##f(x)=x^2## I would get a finite polynomial but that didn't happen.

All my questions can be summarised as follows:
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What is the purpose of ##e^{-x^2/2}## factor?
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How can I show ##H_n(x)## is a solution to the equation ##y''-2xy'+2ny=0 ##
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What is the relation between the differential equation of Hermite functions to that of Hermite polynomials? Are the two equations/solutions equivalent in some way?

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Answers and Replies

  • #2
18,199
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Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
 
  • #3
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Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
I have solved the 2nd question that is showing hermite polynomials do indeed satisfy the hermite equation. All I did was make a substitution: ##H_n(x) = e^{x^2} u, u=D^ne^{-x^2}##

I couldn't find answers to the rest unfortunately.
 
  • #4
Stephen Tashi
Science Advisor
7,242
1,329
All I did was make a substitution: ##H_n(x) = e^{x^2} u, u=D^ne^{-x^2}##
From that substitution, I'm curious how you got other terms with an "##n##" in them to cancel out the "##2ny##" term in the differential equation.

The substitution ##H_n(x) = (-1)^n e^{x^2/2} u ## with ##u = e^{x^2/2} D^n e^{-x^2} ## would allow us to express ##u''## as a function of ##u## since ##y = u## satisfies the first differential equation you gave: ## y''- x^2 y = -(2n+1)y ## , which implies ## u'' = x^2 u - (2n+1)u##.

I don't understand why is the extra factor ##e^{-x^2/2}## there?
A simple minded way to motivate a factor is that we want the Hermite polynomials to be polynomials in x. The result of ## D^n e^{-x^2} ## is going to have the factor ##e^{-x^2}## in it, which needs to be "canceled out" in order to produce a polynomial.
 

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