2

I'm trying to learn about Perturbation, but feel like I'm confused before I've even started.

Right now I'm focused on using them to find solutions to polynomial equations.

The initial example I've been given has $x^3 - 4.001x + 0.002 = 0$, the numbers clearly lend towards $\epsilon$ = 0.001, and you can then have $x^3 - (4 + \epsilon)x + 2\epsilon = 0$

Where I'm confused, is how to apply this to a cubic, when there isn't an obvious value for $\epsilon$.

So, for example, IF we take away the .001 from the first equation & simplify, we have: $x^3 + 4x + 2 = 0$ as our starting equation, how would we then decide a reasonable value for $\epsilon$?

Do we just pick anything reasonably small? (though how small is reasonably?)

Cheers, Belle

MissGeek
  • 123
  • Are you reading some book in particular? – Will Jagy Mar 20 '15 at 04:24
  • I'm studying externally, working off the unit notes from my lecturer. I think I understand the next part of the process, but I'm stuck on the choice of ϵ. This isn't explained any further than what's above, it doesn't seem clear how we would go about this choice in other cases. – MissGeek Mar 20 '15 at 04:33

1 Answers1

0

the idea, as i understand it, is to identify a small or large parameter in the problem. you have identified in yours as $\epsilon.$ the non perturbed problem, that is with $\epsilon = 0$ has easy solution. in your case this is $x^3 - 4x = 0.$ the solutions are $$x = 0 , 2, -2$$ we can three roots of the perturbed problem $$x^3 -(4+\epsilon) x + 2\epsilon=0$$ near any one of the solution $x_0$ by looking for $$x = x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots$$ subbing in the equation and expanding we find $$(x_0+\epsilon x_1+\cdots)^3 -(4+\epsilon)(x_0+\epsilon x_1 + \cdots)+2\epsilon= 0$$ at order $\epsilon$ you get $$3x_0^2x_1-x_0 - 4x_1 + 2=0 \to x_1 = \frac{2-x_0}{4-3x_0^2} $$ the first order corrections to the roots are $$x = 0+\frac 12 \epsilon , 2+0\epsilon, -2-\frac 12\epsilon$$

abel
  • 29,170
  • So it's ok that introducing ϵ actually changes the value of the equation? (very slightly)

    As in, the example I was given above, setting ϵ to .001 didn't change the original equation at all as we adjusted the values to take it into account. In your example, we now have an extra ϵx, and multiplying 2 by ϵ will make it much smaller yeah?

     – MissGeek Mar 20 '15 at 04:36
  • @Belle, you are right. you definitely need to identify, usual the problem itself will suggest one, in the equation. too, you will have some idea of the solution to the unperturbed problem. it takes some for getting used to approximations. try some simple examples. find numerical solutions and see how the perturbation solution compares with the numerical ones to convince yourself that it is worth doing perturbation and that usually it it is much faster than numerical ones. – abel Mar 20 '15 at 04:42
  • Ok, thanks heaps for the help!! :)Externally studying this stuff is hard!! – MissGeek Mar 20 '15 at 06:28