I have been given the polynomial $$\epsilon x^3+x-2=0,$$ where epsilon is very small and I need to find the roots using perturbation theory. So far I have found the first root, 2, using the direct method and then have rescaled the x value with a new variable $x= \bar x \delta$ in order to find the other two roots. When this is put into the original equation, I get $$\epsilon\delta^3\bar x^3+\delta\bar x-2=0$$ I then looked at the coefficients of these terms and decided it was best to use $\delta=\epsilon^{-0.5}$ which gives the new equation $$\epsilon^{-0.5}\bar x^3+\epsilon^{-0.5}\bar x-2=0$$ but when I use the perturbation method with this, and equate the coefficients of different orders of epsilon, I get (for $O(\epsilon^0))$: $$\bar x_0^3+\bar x_0=0$$ which has roots 0, i and -i which surely can't be right? I don't have a clue where I'm going wrong. Any help would be greatly appreciated.
[Math] Stuck on perturbation theory for finding a root of polynomial, with rescaling
perturbation-theoryroots
Related Solutions
You get $y_0(t)=A_0e^t$ and $y_1(t)=A_1e^t-1$. If you combine the solutions, you get $$ y(t)=y_0(t)+ϵy_1(t)=(A_0+ϵA_1)e^t-ϵ $$ The combination of the constants is just a new constant, so that this solution is already the exact solution.
Mathieu's equations is a particularly tricky example since the value of the parameter $\delta$ is very important. I recommend starting off with some simpler equations requiring the method of multiple scales.
Is there an intuitive reason as to why we perform such power series expansion (are we assuming some perturbed $\tilde x(t)$ around the exact solution $x(t)$ using the perturbation parameter as $\epsilon$)?
We are looking for good approximations to $x(t)$ in the limit $\epsilon\to0$. It turns out that there are a range of related techniques to find asymptotic approximations to $x(t)$. Sometimes these give power series expansions, but often they do not (they often do not converge, for example, and include dependence on $\epsilon$ other than just positive powers). The simplest case is a simple power series, which works (surprisingly) often.
If I didn't know what the right-hand side of the above expansion, how could I have figured it out? Is it a standard expansion for such problems?
Yes it is fairly standard and can be figured out. There are two parts to this expansion, the sequence of gauge functions in the expansion (1, $\epsilon$, $\epsilon^2$, etc.) which is the simplest expansion we use, and also the multiple time-scales, $\xi=t$ and $\eta=\epsilon t$. This is known as the method of multiple scales. For motivation, look at any perturbation theory textbook, there is usually a worked example showing why an expansion like $x_0(t)+\epsilon x_1(t)+\ldots$ is insufficient in some cases and that something like $x_0(\xi,\eta)+\epsilon x_1(\xi,\eta)+\ldots$ is necessary.
I don't think I appreciate why we introduce a perturbation factor either. Is it simply because it makes an approximation in first-order convenient?
We don't usually introduce the perturbation factor, it's part of the problem. For example, an equation describes the motion of some object might have a relatively small amount of drag, and the scaling of the drag becomes the perturbation factor, i.e. $$ \frac{\mathrm d^2x}{\mathrm dt^2}=-1-\epsilon|x|^2$$ describes an object being thrown upwards which is affected by gravity and a small amount of air resistance (scaled by $\epsilon$). It's naturally a small effect and we expect a small change from the solution of $$ \frac{\mathrm d^2x}{\mathrm dt^2}=-1, $$ at least while $\epsilon$ is small.
Best Answer
The first root should be $2-8\epsilon$ as it should be up to order 1 not order 0, and what you have for $\bar {x_0}$ is correct except we reject the 0 value as that stops our approximation being of moderate size. You then need to also find $\bar {x_1}$
Yeah that's fine, note that your new direct perturbation should be in powers of $\epsilon^{\frac{1}{2}}$ for $\bar {x}$
You choose both $\bar{x_0}=i,-i$, in each case you get a different root, so by this method we get two more roots. i.e. we have a total of 3, which is good because the equation is cubic.