How Can the Frobenius Method Be Used to Solve Complex ODEs?

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Discussion Overview

The discussion revolves around the application of the Frobenius method to solve a specific ordinary differential equation (ODE) of the form xy'' + (1-2x)y' + (x-1)y = 0, which has a regular singularity at x = 0. Participants explore the series expansion around this point and the implications of arbitrary coefficients in the solution.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using the Frobenius method and derives a recurrence relation for the coefficients a_m in the power series solution.
  • Another participant questions whether any non-zero values for a_0 and a_1 would yield valid solutions, suggesting that this leads to an infinite number of basis solutions.
  • Some participants discuss the implications of setting specific values for a_0 and a_1, noting that it affects the ease of finding a second independent solution.
  • A later reply introduces a method for finding a second solution independent of the first, referencing a known particular solution and its relationship to the general solution.
  • There is a contention regarding the necessity of setting a_0 equal to a_1 for the solution to be valid, with some arguing that it leads to contradictions when considering the general solution's behavior at x=0.
  • Another participant raises a different ODE and inquires about the selection of coefficients and the process for finding additional solutions.

Areas of Agreement / Disagreement

Participants express differing views on the implications of choosing arbitrary coefficients a_0 and a_1, with some asserting that any choice (as long as a_0 is non-zero) is valid, while others argue that specific relationships between these coefficients are necessary for the solution's validity. The discussion remains unresolved regarding the best approach to selecting these coefficients and the implications for finding additional solutions.

Contextual Notes

Participants note that the Frobenius method applied at x=0 may not provide information about certain solutions that are not analytic at that point, which complicates the analysis of the general solution.

Who May Find This Useful

This discussion may be useful for students and researchers interested in advanced methods for solving ordinary differential equations, particularly those involving series solutions and the Frobenius method.

Usagi
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Ok here's a funny ODE to solve:

xy'' + (1-2x)y' + (x-1)y = 0

clearly a straight forward power series substitution won't work here since we have a regular singularity at x = 0

so try the frobenius method by expanding around x = 0.

Assume y = \sum_{m=0}^{\infty} a_mx^{m+r} is a solution where r is some constant.

So we have y' = \sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1} and y'' = \sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}

Put this back in:

x \left(\sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}\right) + (1-2x) \left(\sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}\right) + (x-1) \left(\sum_{m=0}^{\infty} a_mx^{m+r}\right) = 0

after some algebra and stuff:

\sum_{m=0}^{\infty} (m+r)^2 a_m x^{m+r-1} - \sum_{m=0}^{\infty} [2(m+r)+1] a_m x^{m+r} + \sum_{m=0}^{\infty} a_m x^{m+r+1} = 0

clearly lowest term is x^{r-1} with it's coefficient asr^2a_0 hence r^2a_0 = 0

Now a_0 \neq 0, so r^2 = 0 \implies r = 0

Now we find the coefficients of the term x^s where s is some constant, this gives:

(s+1)^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0

rearranging gives:

a_{s+1} = \frac{(2s+1)a_s - a_{s-1}}{(s+1)^2} for s = 1, 2, etc

Thus we found a recurrence relationship with a_0 and a_1 as arbitrary initial values.

A bit of playing around quickly shows that:

a_2 = \frac{3a_1 - a_0}{4}

a_3 = \frac{11a_1 - 5a_0}{36}

a_4 = \frac{25a_1 - 13a_0}{288}

Thus we have one of the solutions to be y = a_0 + a_1x + (\frac{3a_1 - a_0}{4})x^2 + (\frac{11a_1 - 5a_0}{36})x^3 + (\frac{25a_1 - 13a_0}{288})x^4 + ...

However because a_0 and a_1 are arbitrary, let us pick... a_0 = a_1 = 1, now magically we have:

y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+ ... = e^x

So y = e^x is one of the basis for the general solution of this ode.

Now I was wondering, since a_0 and a_1 are arbitrary, then would ANY a_0 (\neq 0) and a_1 work? Say a_0 = 4 and a_1 = 3 which then implies that there is an "infinite" number of different basis for the general solution of this ode?

Thanks
 
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Usagi said:
Ok here's a funny ODE to solve:

xy'' + (1-2x)y' + (x-1)y = 0

clearly a straight forward power series substitution won't work here since we have a regular singularity at x = 0

so try the frobenius method by expanding around x = 0.

Assume y = \sum_{m=0}^{\infty} a_mx^{m+r} is a solution where r is some constant.

So we have y' = \sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1} and y'' = \sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}

Put this back in:

x \left(\sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}\right) + (1-2x) \left(\sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}\right) + (x-1) \left(\sum_{m=0}^{\infty} a_mx^{m+r}\right) = 0

after some algebra and stuff:

\sum_{m=0}^{\infty} (m+r)^2 a_m x^{m+r-1} - \sum_{m=0}^{\infty} [2(m+r)+1] a_m x^{m+r} + \sum_{m=0}^{\infty} a_m x^{m+r+1} = 0

clearly lowest term is x^{r-1} with it's coefficient asr^2a_0 hence r^2a_0 = 0

Now a_0 \neq 0, so r^2 = 0 \implies r = 0

Now we find the coefficients of the term x^s where s is some constant, this gives:

(s+1)^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0

rearranging gives:

a_{s+1} = \frac{(2s+1)a_s - a_{s-1}}{(s+1)^2} for s = 1, 2, etc

Thus we found a recurrence relationship with a_0 and a_1 as arbitrary initial values.

A bit of playing around quickly shows that:

a_2 = \frac{3a_1 - a_0}{4}

a_3 = \frac{11a_1 - 5a_0}{36}

a_4 = \frac{25a_1 - 13a_0}{288}

Thus we have one of the solutions to be y = a_0 + a_1x + (\frac{3a_1 - a_0}{4})x^2 + (\frac{11a_1 - 5a_0}{36})x^3 + (\frac{25a_1 - 13a_0}{288})x^4 + ...

However because a_0 and a_1 are arbitrary, let us pick... a_0 = a_1 = 1, now magically we have:

y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+ ... = e^x

So y = e^x is one of the basis for the general solution of this ode.

Now I was wondering, since a_0 and a_1 are arbitrary, then would ANY a_0 (\neq 0) and a_1 work? Say a_0 = 4 and a_1 = 3 which then implies that there is an "infinite" number of different basis for the general solution of this ode?

Thanks

In...

http://www.mathhelpboards.com/f17/another-second-order-non-homogeneous-ode-2237/#post10243

... it has been explained, given a second order ODE like...

$\displaystyle y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y =0$ (1)

... a particular solution of which that we call v(x) is known, how to find another solution u(x) independent from v(x)...

$\displaystyle u= v\ \int \frac{e^{-\int p(x)\ dx}}{v^{2}}\ dx$ (2)

In Your case is $\displaystyle p(x)= \frac{1-2\ x}{x}$ and You found $v(x) = e^{x}$, so that is...

$\displaystyle e^{- \int p(x)\ dx} = \frac{e^{2\ x}}{x} \implies u(x) = e^{x}\ \int \frac{dx}{x} = e^{x}\ \ln x$ (2)

Now it is clear that u(x) is not analytic in x=0 and the Frobenoius method, applied in x=0, can't give any information about it... Kind regards $\chi$ $\sigma$
 
ok thanks for that, but what if i picked a_0 = 1 and a_1 = 3?

Then the first solution would be an infinite sum and not e^x, how do i then find the second solution?
 
chisigma said:
In...

http://www.mathhelpboards.com/f17/another-second-order-non-homogeneous-ode-2237/#post10243

... it has been explained, given a second order ODE like...

$\displaystyle y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y =0$ (1)

... a particular solution of which that we call v(x) is known, how to find another solution u(x) independent from v(x)...

$\displaystyle u= v\ \int \frac{e^{-\int p(x)\ dx}}{v^{2}}\ dx$ (2)

In Your case is $\displaystyle p(x)= \frac{1-2\ x}{x}$ and You found $v(x) = e^{x}$, so that is...

$\displaystyle e^{- \int p(x)\ dx} = \frac{e^{2\ x}}{x} \implies u(x) = e^{x}\ \int \frac{dx}{x} = e^{x}\ \ln x$ (2)

Now it is clear that u(x) is not analytic in x=0 and the Frobenoius method, applied in x=0, can't give any information about it... Kind regards $\chi$ $\sigma$

... and for supply an as complete as possible answer, the general solution of the proposed ODE is...

$\displaystyle y(x)= c_{1}\ e^{x} + c_{2}\ e^{x}\ \ln x$ (1)

... if You intend to set the coefficients $a_{0}$ and $a_{1}$, necessarly must be $a_{0}=a_{1}=a$, and the reason of that is evident observing (1)...

Kind regards

$\chi$ $\sigma$
 
Thanks, however I was told that as long as we don't set a_0 to be 0, then a_0 and a_1 can be anything and it will still be a perfectly valid solution for the ode? It's just that it makes finding the second solution harder than if we set a_0 = a_1?
 
Usagi said:
Thanks, however I was told that as long as we don't set a_0 to be 0, then a_0 and a_1 can be anything and it will still be a perfectly valid solution for the ode? It's just that it makes finding the second solution harder than if we set a_0 = a_1?

If y(x) is the solution You are searching and is $\displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}$, then it must be necessarly $a_{0}= y(0)$ and $a_{1}= y^{\ '}(0)$... right?... but if the general solution is...

$y(x)= c_{1}\ e^{x} + c_{2}\ e^{x}\ \ln x$ (1)

... y(x) and its derivative don't exist in x=0 unless is $c_{2}=0$, but if so, then il must be necessarly $y(0)=y^{\ '}(0)$...

The conclusion is : imposing $a_{0} \ne a{1}$ is a contradiction...

Kind regards

$\chi$ $\sigma$
 
thanks that makes sense!

What about this one:

y'' + (x-1)y = 0

\sum_{m=0}^{\infty} (m+r)(m+r-1)a_m x^{m+r-2} + (x-1) \sum_{m=0}^{\infty} a_m x^{m+r} = 0

\sum_{m=0}^{\infty} (m+r)(m+r-1)a_m x^{m+r-2} + \sum_{m=0}^{\infty} a_m x^{m+r+1} - \sum_{m=0}^{\infty} a_m x^{m+r} = 0

so r(r-1) = 0 -> r= 0, 1

Consider r = 1:

\sum_{m=0}^{\infty} (m+1)m a_m x^{m-1} + \sum_{m=0}^{\infty} a_m x^{m+2} - \sum_{m=0}^{\infty} a_m x^{m+1} = 0

Let us find the coefficient of x^s

(s+2)(s+1)a_{s+1} + a_{s-2} - a_{s-1} = 0

so a_{s+1} = \frac{a_{s-1} - a_{s-2}}{(s+2)(s+1)} for s = 2, 3, 4, ...

so then how do i know what values to set a_0, a_1 and a_2?

and also how would i find the other solution for the basis?
 
Usagi said:
thanks that makes sense!

What about this one:

y'' + (x-1)y = 0

\sum_{m=0}^{\infty} (m+r)(m+r-1)a_m x^{m+r-2} + (x-1) \sum_{m=0}^{\infty} a_m x^{m+r} = 0

\sum_{m=0}^{\infty} (m+r)(m+r-1)a_m x^{m+r-2} + \sum_{m=0}^{\infty} a_m x^{m+r+1} - \sum_{m=0}^{\infty} a_m x^{m+r} = 0

so r(r-1) = 0 -> r= 0, 1

Consider r = 1:

\sum_{m=0}^{\infty} (m+1)m a_m x^{m-1} + \sum_{m=0}^{\infty} a_m x^{m+2} - \sum_{m=0}^{\infty} a_m x^{m+1} = 0

Let us find the coefficient of x^s

(s+2)(s+1)a_{s+1} + a_{s-2} - a_{s-1} = 0

so a_{s+1} = \frac{a_{s-1} - a_{s-2}}{(s+2)(s+1)} for s = 2, 3, 4, ...

so then how do i know what values to set a_0, a_1 and a_2?

and also how would i find the other solution for the basis?

Althogh it may seem 'simple' the second order linear ODE...

$\displaystyle y^{\ ''} + (x-1)\ y =0$ (1)

... is not very comfortable and its general solution is...

$\displaystyle y= c_{1}\ \text{Ai} (1-x) + c_{2}\ \text{Bi} (1-x)$ (2)... where Ai(*) and Bi (*) are the so called 'Airy Functions'...Airy Functions -- from Wolfram MathWorld

... and may be that the Frobenius approach is not the best way to attack (1)...

Kind regards

$\chi$ $\sigma$
 

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