Solving Differential Equation with Fröbenius Method

Thanks for the help!In summary, the conversation discusses solving a differential equation using the Fröbenius method and finding a solution in terms of Bessel functions. The speaker also discusses finding another linear independent solution and ultimately rewrites their recursion formula in terms of Bessel functions.
  • #1
DevoBoy
8
0
I'm trying to solve the differential equation:

[tex]x^2y''+2xy'+(x^2-2)y=0[/tex]

using the Fröbenius method.

So I want a solution on the form

[tex]y=\sum_{n=0}^\infty a_{n}x^{n+s}[/tex]

After finding derivatives of y, inserting into my ODE, and after some rearranging:

[tex]\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}[/tex]

Looking at n=0, assuming [tex]a_{0}[/tex] different from zero, I get two possible values for s:

[tex]s_{1}=1[/tex]
[tex]s_{2}=-2[/tex]

Both giving [tex]a_{1}=0[/tex]

Choosing s=1, I get

[tex]a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}[/tex]

I know [tex]a_{n}=0[/tex] for odd n, so I'm interrested in finding [tex]a_{n}[/tex] for even n, expressed by [tex]a_{0}[/tex]. Best thing I can come up with is

[tex]a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}[/tex]

which seems overly complicated given the simple recursive formula. Any ideas?


Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for [tex]a_{n}[/tex].
 
Last edited:
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  • #2
Your equation is almost certainly some kind of Bessel equation. You should seek solutions in terms of Bessel functions.

Edit:
In fact, it is a Bessel function. Specifically, a spherical Bessel function apparently. There are also general solutions on the page. Remember that in order to obtain the correct expansion, you must use the boundary values of the problem.
 
Last edited:
  • #3
DevoBoy said:
I'm trying to solve the differential equation:

[tex]x^2y''+2xy'+(x^2-2)y=0[/tex]

using the Fröbenius method.

So I want a solution on the form

[tex]y=\sum_{n=0}^\infty a_{n}x^{n+s}[/tex]

After finding derivatives of y, inserting into my ODE, and after some rearranging:

[tex]\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}[/tex]

Looking at n=0, assuming [tex]a_{0}[/tex] different from zero, I get two possible values for s:

[tex]s_{1}=1[/tex]
[tex]s_{2}=-2[/tex]

Both giving [tex]a_{1}=0[/tex]

Choosing s=1, I get

[tex]a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}[/tex]

I know [tex]a_{n}=0[/tex] for odd n, so I'm interrested in finding [tex]a_{n}[/tex] for even n, expressed by [tex]a_{0}[/tex]. Best thing I can come up with is

[tex]a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}[/tex]

which seems overly complicated given the simple recursive formula. Any ideas?


Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for [tex]a_{n}[/tex].
I don't see how you could get a sum like that. Starting with
[tex]a_n= \frac{-a_{n-2}}{n(n+3)}[/tex]
You get, for the first few terms, [itex]a_2= -a_0/(2(5))[/itex], [itex]a_4= a_0/(2(5)(4)(7))[/itex], [itex]a_6= -a_0/(2(4)(5)(7)(6)(9)[/itex]

Okay, it looks like
[tex]a_{2n}= \frac{3(-1)^n a_0}{(2n)!(2(n+1)+ 1)}[/tex]
but there is no sum!
 
  • #4
HallsofIvy said:
I don't see how you could get a sum like that. Starting with
[tex]a_n= \frac{-a_{n-2}}{n(n+3)}[/tex]
You get, for the first few terms, [itex]a_2= -a_0/(2(5))[/itex], [itex]a_4= a_0/(2(5)(4)(7))[/itex], [itex]a_6= -a_0/(2(4)(5)(7)(6)(9)[/itex]

Okay, it looks like
[tex]a_{2n}= \frac{3(-1)^n a_0}{(2n)!(2(n+1)+ 1)}[/tex]
but there is no sum!

You're absolutely right. My mistake.

After reading up on Bessel functions (thanks ObsessiveMathsFreak!), I've rewritten my recursion formula as

[tex]a_{2n}=\frac{(-1)^{n}a_{0}\Gamma(1+3/2)}{2^{2n}n!\Gamma(n+5/2)}[/tex]

which seems to expand beautifully.
 

Related to Solving Differential Equation with Fröbenius Method

1. What is the Fröbenius method?

The Fröbenius method is a technique used to solve differential equations that cannot be solved by traditional methods. It involves assuming a solution in the form of a power series and using recursion to determine the coefficients of the series.

2. When is the Fröbenius method used?

The Fröbenius method is used when the differential equation has a regular singular point, meaning that the equation is not well-behaved at that point. This method is particularly useful for solving equations in physics and engineering that involve boundary conditions at a singular point.

3. How does the Fröbenius method work?

The Fröbenius method involves substituting the assumed power series solution into the differential equation and solving for the coefficients using recursion. The recursion formula is derived by equating the coefficients of each term in the power series and solving for them in terms of the initial coefficients.

4. What are the limitations of the Fröbenius method?

The Fröbenius method is limited to solving differential equations with regular singular points. It also requires the differential equation to have a certain form, known as the indicial equation, in order to determine the series solution. Additionally, the convergence of the series solution may be an issue if the coefficients are not well-behaved.

5. Are there any applications of the Fröbenius method?

Yes, the Fröbenius method has many applications in physics and engineering, particularly in solving differential equations that arise in quantum mechanics, electromagnetism, and fluid mechanics. It is also used in signal processing and image reconstruction, as well as in the study of singularities in complex analysis.

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