Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to learn frobenius method by solving some problems (ODEs).

For example:

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

Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$. Then, I took derivatives and put into the equation: $$\sum\limits_{n=0}^\infty a_n(n+r)^2x^{n+r-1}+2\sum\limits_{n=0}^\infty a_n(n+r+1)x^{n+r}+\sum\limits_{n=0}^\infty a_nx^{n+r+1}=0$$ After I shifted to make their orders same: $$\sum\limits_{k=-2}^\infty a_{k+2}(k+r+2)^2x^{k+r+1}+2\sum\limits_{k=-1}^\infty a_{k+1}(k+r+2)x^{k+r+1}+\sum\limits_{k=0}^\infty a_kx^{k+r+1}$$ And if I leave first $k=-2, k=-1$ parts, I can find relationship among 3 coefficient: $$\sum\limits_{k=0}^\infty [x^{k+r+1}(a_{k+2}(k+r+2)^2+a_{k+1}(k+r+2)+a_k)]=0$$ Now, here I could find relationship with $a_{k+2},a_{k+1},a_k$. But, in this method we should find proportionality between 2 coefficients, not 3. For example, this: Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$

Can you, please, suggest a solution?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

You state:

After I shifted to make their orders same: $$ \sum_{k = -2}^\infty a_{k + 2}(k + r + 2)^2 x^{k + r + 1} + 2 \sum_{k = -1}^\infty a_{k + 1}(k + r + 2)x^{k + r + 1}+\sum_{k = 0}^\infty a_k x^{k + r + 1} $$

What you do is the following. Given that $$ \sum_{k = 0}^\infty a_k (k + r)^2 x^{k + r - 1} + \sum_{k = 0}^\infty a_k(2 k + 2 r + 1)x^{k + r} + \sum_{k = 0}^\infty a_k x^{k + r + 1} $$ (note that your second sum was incorrectly calculated)

you need to separate the necessary terms of the sums in order to group the powers of $x$ correctly, i.e:

\begin{multline} \sum_{n = 0}^\infty a_n (n + r)^2 x^{n + r - 1} + 2 \sum_{n = 0}^\infty a_n (n + r + 1) x^{n + r} + \sum_{n = 0}^\infty a_n x^{n + r + 1} = \\ a_0 r^2 x^{r-1} + a_1 (r + 1)^2 x^r + \sum_{n=2}^\infty a_n (n + r)^2 x^{n + r - 1} + (2 r + 1) a_0 x^r + \\ \sum\limits_{n = 1}^\infty a_n (2 n + 2 r+ 1)x^{n + r} +\sum_{n = 0}^\infty a_n x^{n + r + 1} = 0 \end{multline}

Regrouping orders, you have

\begin{multline} a_0 r^2 x^{r-1} + [a_1 (r + 1) + a_0 (2 r + 1)] x^r + \\ \sum_{k = 0}^\infty \left\{ a_{k + 2}(k + r + 2)^2 + 2 a_{k + 1} (k + r + 2) + a_k \right\} x^{k + r + 1} = 0 \end{multline}

Each power of $x$ needs to vanish, hence $r^2 = 0$. This is the indicial polynomial (details here). This means that $r = 0$ and

\begin{align} a_1 + a_0 &= 0 \\ a_{k + 2}(k + 2)^2 + 2 a_{k + 1} (2 k + 3) + a_k &= 0 \end{align}

which closes the recurrence relation. The first tree terms are \begin{align} a_1 &= -a_0\\ a_2 &= \frac{1}{2!}a_0\\ a_3 &= -\frac{1}{3!}a_0 \end{align} and it's clear that a relationship is forming. By induction, the whole solution can be computed.

Note that, assuming that $y$ is somehow well behaved, for $x \sim 0$,

$$ x y'' + (2x + 1) y' + (x + 1) y = 0 \quad \sim \quad y' + y = 0. $$

Proposing the anzats $y(x) = e^{-x} z(x)$ and substituting in the original ode,

$$ x y'' + (2x + 1) y' + (x + 1) y = e^{-x}\left(x z'' + z'\right) = 0, $$

and it's easily verified that $z = c_1 \log x + c_2$. Hence

$$ y(x) = e^{-x}\left(c_1 \log x + c_2\right) $$

Cool trick ha?

share|improve this answer
    
Nice trick! When you said: > "which closes the relation." You mean that it's impossible to use Frobenius method here? –  Bek Abdik May 16 '13 at 6:05
    
Then, I think we should make $a_{-1}=0$ in the beginning. I'm so happy now that I got it. Thanks. –  Bek Abdik May 16 '13 at 6:18
    
There was a mistake in the recurrence relation. See the edit for details. I'm not sure what you mean by making $a_{-1} = 0$ in the beginning. The point is that there is no $a_{-1}$, that's why you have to let go the first terms of the sums before shifting the indices, which in turn closes the recurrence relation. –  Pragabhava May 16 '13 at 6:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.