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

[Usagi]

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 as$$r^2a_0$$ 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

 

[chisigma]

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 as$$r^2a_0$$ 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 $\chi$ $\sigma$

 

[Usagi]

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]

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

$\chi$ $\sigma$

 

[Usagi]

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?

 

[chisigma]

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

$\chi$ $\sigma$

 

[Usagi]

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?

 

[chisigma]

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

$\chi$ $\sigma$