# Frobenius method for Hermite Equation

• sunquick
In summary, the student is trying to solve a homework equation using the Frobenius method and is having difficulty understanding why the Frobenius method would be useful around ordinary points.
sunquick

## Homework Statement

I'm asked to solve the Hermite Differential Equation
$$y''(x) - 2 x y'(x) + \lambda y(x) = 0$$

using the Frobenius method

2. Homework Equations

I am to assume the solution is in the form
$$y(x) = \sum a_n x^{n+r}$$
where r are the roots of the indicial equation that in this case reads
$$r(r-1)=0$$

## The Attempt at a Solution

Solving the indicial equation, I get r=0 and r=1. If r=0, then the Frobenius method is just the ordinary power series method (expanding around an ordinary point) and I every worked example of the same problem I'm attempting to solve do use the ordinary power series method without mentioning the indicial equation or the name of Frobenius.

Indeed going with r = 0 I get
$$y(x) = a_0 \left( 1 - \frac{\lambda}{2!} x^2 - \frac{(4-\lambda) \lambda}{4!} x^4 - \frac{(8-\lambda)(4- \lambda)\lambda}{6!} x^6 - ...\right) + a_1 \left( x + \frac{2-\lambda}{3!} x^3 + \frac{(6-\lambda)(2-\lambda)}{5!} x^5 + ...\right)$$

but going with r =1 I get
$$\sum (n+1) n a_n x^{n-1} - 2 \sum (n+1) a_n x^{n+1} + \sum \lambda x^{n+1} = 0$$
$$a_{n+2} = \frac{2(n+1)- \lambda}{(n+2)(n+1)}$$
$$y(x) = a_0 \left(x + \frac{2 - \lambda}{2!} x^3 + \frac{(6-\lambda)(2-\lambda)}{4!} x^5 + ...\right) + a_1 \left( x^2 + \frac{4-\lambda}{3!} x^4 + \frac{(6-\lambda)(4-\lambda)}{5!} x^6 + ... \right)$$

Sou now I'm confused
1. Because I get four different linearly independent solutions to a second order equation, even though they're very similar which makes me think they might be related
2. I'm not sure how the method of Frobenius is supposed to work around ordinary points. I thought it was a more general power series method good for both ordinary and regular singular points, but now I'm not so sure if it's supposed to be used around ordinary points.
3. Arfken&Weber says I should have an even solution for r=0, with all powers of x even in the power expansion, and for r=1 I should take only odd powers of x for the power series expansion. That might make sense, but I can't understand where that comes from, it seems just pulled out of hat.Why can't we have both even and odd powers for r=0 and r=1?

sunquick said:
the indicial equation that in this case reads
$$r(r-1)=0$$
Are you sure?

Well now I'm not ..

## 1. What is the Frobenius method for Hermite Equation?

The Frobenius method is a technique used to solve certain types of second-order differential equations, including the Hermite equation. It involves assuming a power series solution and solving for the coefficients using a recursion relation.

## 2. When is the Frobenius method applicable?

The Frobenius method is applicable when the differential equation has a regular singular point, meaning that all the coefficients of the equation are analytic functions near the point of interest. In the case of the Hermite equation, the regular singular point is at x=0.

## 3. What is the general form of the Hermite equation that can be solved using the Frobenius method?

The general form of the Hermite equation that can be solved using the Frobenius method is: y'' - 2xy' + λy = 0, where λ is a constant. This equation arises in many areas of physics, including quantum mechanics and heat transfer.

## 4. Can the Frobenius method be used to find all solutions to the Hermite equation?

No, the Frobenius method can only be used to find one solution to the Hermite equation. This solution is known as the "Frobenius solution" and is valid only near the singular point. Other techniques, such as the power series method, may be used to find the general solution.

## 5. Are there any special cases where the Frobenius method for Hermite equation can be simplified?

Yes, when the constant λ is an integer, the Frobenius method can be simplified. In this case, the recursion relation for the coefficients of the power series solution becomes a finite recursion, and the solution can be expressed in terms of Hermite polynomials. This simplification is known as the "polynomial case" of the Frobenius method.

• Calculus and Beyond Homework Help
Replies
7
Views
714
• Calculus and Beyond Homework Help
Replies
3
Views
626
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
512
• Calculus and Beyond Homework Help
Replies
5
Views
660
• Calculus and Beyond Homework Help
Replies
2
Views
903
• Calculus and Beyond Homework Help
Replies
10
Views
774
• Calculus and Beyond Homework Help
Replies
6
Views
462
• Calculus and Beyond Homework Help
Replies
2
Views
440
• Calculus and Beyond Homework Help
Replies
1
Views
834