# Clarification of Frobenius method roots

The frobenius method states that for repeated roots or roots that differ by an integer, an alternative method must be used to find the second solution once one is found. When they say "roots that differ by an integer", does this mean any integer? Or only by the integer one?

Because upon solving the differential equation $y'' - \frac{6y}{z^2} = 0$ I received roots $\sigma = 3, -2$. these both solved the equation as $y = c_1 z^3 + c_2 z^{-2}$ and this alternative method for a second solution was not required.

$$y(z) = \sum_{n = 0}^{ \infty} a_n z^{\sigma + n}$$

$$y'' = \sum_{n = 0}^{ \infty} (\sigma + n)(\sigma + n - 1) a_n z^{\sigma + n - 2}$$

Plugging these in and setting $n = 0 \forall n >0$ I find the relation for $\sigma$

$$(\sigma(\sigma - 1) - 6)a_0 = 0$$

Because we demand that $a_0 \neq 0$ We find $\sigma = 3, -2$

• Yes, they can differ by any integer or $|\sigma_1-\sigma_2|\in\Bbb N$. – Mikasa Jan 4 '13 at 20:46
• @Cactus BAMF, When there is a better answer in your question after you accept an answer, you can unaccept the answer and reaccept another answer. – doraemonpaul Jan 10 '13 at 13:30

In fact the statement "The frobenius method states that for repeated roots or roots that differ by an integer, an alternative method must be used to find the second solution once one is found." is always true for multiple roots cases but not always true for the cases that some of the roots are differed by integers. Since for the cases that some of the roots from the indical equations are differed by integers, in some situations some of the roots from the indical equations can find more than one group of the linearly independent solutions:

$1.$ For the cases that having more than one indical equations, when some of the roots from the indical equations are common, as taking the common ones can eliminate more than one indical equations at the same time, so we can find more than one group of the linearly independent solutions at the same time. For example the question no. $5$ in http://tw.knowledge.yahoo.com/question/question?qid=1511121805995 and Can I use Frobenius method here? excluding the case of $n=3$ .

$2.$ When some of the roots from the indical equations can obtain the self-fitful recurrence relations. For example for the recurrence relations of the form $f(n)a_n=g(n)a_{n-k}$ where $k\in\mathbb{N}$ , when there exist some non-negative integers $n$ such that $f(n)=0$ , they are self-fitful. So we can find more than one group of the linearly independent solutions from each of the recurrence relations. For example the question no. $1$ in http://tw.knowledge.yahoo.com/question/question?qid=1512050807030 and http://faculty.pccu.edu.tw/~meng/Math3.pdf#page=6.

Now for this problem,

Let $y=\sum\limits_{n=0}^\infty a_nz^{n+\sigma}$ ,

Then $y'=\sum\limits_{n=0}^\infty(n+\sigma)a_nz^{n+\sigma-1}$

$y'=\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}$

$\therefore\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}-\dfrac{6}{z^2}\sum\limits_{n=0}^\infty a_nz^{n+\sigma}=0$

$\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}-\sum\limits_{n=0}^\infty6a_nz^{n+\sigma-2}=0$

$\sum\limits_{n=0}^\infty((n+\sigma)(n+\sigma-1)-6)a_nz^{n+\sigma-2}=0$

Since there are not any indical equations present, that means $\sigma$ can be chosen as any complex number.

However, in fact, take $\sigma=3$ or $\sigma=-2$ will bring the above equation most simplified.

Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $\sigma=-2$ :

$\sum\limits_{n=0}^\infty((n-2)(n-3)-6)a_nz^{n-4}=0$

$\sum\limits_{n=0}^\infty n(n-5)a_nz^{n-4}=0$

$\therefore n(n-5)a_n=0$

$\therefore\begin{cases}a_0=a_0\\a_5=a_5\\a_n=0~\forall n\in\mathbb{N}\setminus\{5\}\end{cases}$

Hence $y=c_1z^{-2}+c_2z^3$

Therefore in this case the alternative method for a second solution of course is not required.

• @Your answer made my short hint very clear. +1 – Mikasa Jan 7 '13 at 5:49

If our indical equation has two solutions $\sigma_1,\sigma_2$ and we find $|\sigma_1-\sigma_2|$ in $\Bbb N$, then it may be the solutions $y_1(x),y_2(x)$ which are due to $\sigma_1,\sigma_2$, are not independent. In your OE, we see that they are independent on $\Bbb R^*$ and indeed the OE is a Cauchy-Euler equation. If $y_1(x),y_2(x)$ are not independent we can see that $y_1(x)=x^{\sigma_1}\sum_{n=0}^{\infty}a_n(\sigma_1)x^n$ and the other $y_2(x)$ would be found by taking $y_2(x)=v(x)y_1(x)$ and using variation of parameters.

• Of course. Thank you so much that makes perfect sense now – Cactus BAMF Jan 4 '13 at 21:24
• (-1): Not the real explanations of this problem. – doraemonpaul Jan 10 '13 at 13:13