The Method of Frobenius

**Aryth** · February 18th 2010, 04:34 PM

Hermite's Equation is an equation of the type:

$y''(x) - 2xy'(x) + 2ny(x) = 0, n \in \mathbb{N}_0$

Construct series solutions (about the origin) for n=1 and n=2.

I tried to solve this generally so I could just plug in n later for any other n's that I may need, but somewhere I may have messed up because my coefficients ended up all being zero for the n=1 series... So, all I need is someone to see if I messed up...

I started off by noting that the equation was in general form and that p(x) and q(x) are both analytical everywhere so x = 0 is a valid point for a convergent series solution.

Next (and this part is where I had my doubts because it is an ordinary point, not a singular one), I took the limits of xp(x) and $x^2q(x)$ and got zero for both. This lead to the roots:

$r_1 = 0$ and $r_2 = 1$

Now, I prepared a generalized solution:

$y(x) = x^r\sum_{i=0}^{\infty}a_ix^i=\sum_{i=0}^{\infty}a_ ix^{i+r}$

Plugging this into the DE yields:

$\sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r-2} - 2x\sum_{i=0}^{\infty}(i+r)a_ix^{i+r-1} + 2n\sum_{i=0}^{\infty}a_ix^{i+r}=0$

When the x in the second term is distributed, the second and third sum can be added together:

$\sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r-2} - 2(1 - n)\sum_{i=0}^{\infty}(i + r + 1)a_ix^{i+r} = 0$

Next I multiplied through by $x^2$ to eliminate negative indices:

$\sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r} - 2(1-n)\sum_{i=0}^{\infty}(i+r+1)a_ix^{i+r + 2} = 0$

Then I shifted the indices, the first sum gets the replacement k=i the second one gets k=i+2:

$\sum_{k=0}^{\infty}(k+r-1)(k+r)a_kx^{k+r} - 2(1-n)\sum_{k=2}^{\infty}(k+r-1)a_{k-2}x^{k+r} = 0$

Then I took out the first two terms of the first sum:

$a_0[r(r-1)]x^r + a_1[r(1+r)]x^{1 + r} + \sum_{k=2}^{\infty}(k+r-1)(k+r)a_kx^{k+r} -$ $2(1-n)\sum_{k=2}^{\infty}(k+r-1)a_{k-2}x^{k+r} = 0$

Then I combined the two sums:

$a_0[r(r-1)]x^r + a_1[r(1+r)]x^{1+r} + \sum_{k=2}^{\infty}(k+r-1)x^{k+r}[(k+r)a_k - 2(1-n)a_{k-2}] = 0$

Then I set the coefficients equal to 0 and used r = 1:

$a_0[1(0)] = 0$ -> $a_0$ is arbitrary.

$a_1[1(2)] = 0$ -> $a_1 = 0$

$(k+1)a_k - 2(1-n)a_{k-2} = 0$

$a_k = \frac{2(1-n)a_{k-2}}{(k+1)}, k \geq 2$

Now, when n=1, every coefficient is zero except $a_0$ so the solution for n=1 is:

$y(x) = a_0x$

When n=2, All of the odd numbered coefficients are zero so that the odd number powers remain, and I get:

$y(x) = a_0[x - \frac{2}{3}x^3 + \frac{4}{15}x^5 - \frac{8}{105}x^7 + ...]$

Thanks in advance for any help.

**HallsofIvy** · February 20th 2010, 08:47 AM

That equation is NOT singular at any point and you do not need to use
Frobenius' method. Just use a solution of the form $\sum_{n=0}^\infty a_nx^n$ .

Thread: The Method of Frobenius

LinkBack

Thread Tools

Display

The Method of Frobenius

Similar Math Help Forum Discussions

Second order DE (Frobenius Method)

Frobenius method

Frobenius Method

Method of Frobenius

power series method and frobenius method

Search Tags