Solving a Perturbed Cubic Equation

Question

Consider a cubic equation

$(1 + \epsilon)x^3 - 2ax^2 + (a - 3\epsilon)x + 2\epsilon = 0$

where $\epsilon > 0$ and $a \gg 1$.

In the limit of $\epsilon \rightarrow 0$,

$x(x^2 - 2ax + a) = 0$

so that

$ x_0 = a \pm \sqrt{a^2 - a} \approx 2a$ or $1/2 + 1/(8a)$.

I want to find approximate solutions to the cubic equation, which is accurate even for $\epsilon \sim a$. For $x_0 \approx 2a \gg 1$, the balancing between the third and second order terms just gives

$x \approx 2a/(1+\epsilon)$

which is found to be quite accurate when $\epsilon \sim a$.

For the other solution, I tried expanding the solution from $x_0$.

$x = x_0 + x_1$

and kept the terms first order in $x_1$ to obtain

$3\epsilon x_0^2x_1 - 3\epsilon x_1 -4a x_0 x_1 + a x_1 + 3 x_0^2 x_1 + 2\epsilon = 0$

or since $a \gg 1$, $x_0 \approx 1/2$ so $x_1 = 2\epsilon/a$.

Comparing $x = x_0 + x_1$ with the exact solution, however, I found that the approximation becomes inaccurate when $\epsilon \sim a$.

Can anyone help me how to attack this problem?

Answer 1

Hint

Why not considering $$F(x,\epsilon)=(1 + \epsilon)x^3 - 2ax^2 + (a - 3\epsilon)x + 2\epsilon = 0$$ and use implicit differentiation to obtain the value of $\frac{dx}{d\epsilon}$ and compute its value for $\epsilon=0$ and then use $$x \simeq x_0+\frac{dx}{d\epsilon} \Delta \epsilon$$

For sure, you must do the above for each of the roots $$x_1=0$$ $$x_2=a-\sqrt{a^2-a}$$ $$x_3=a+\sqrt{a^2-a}$$ I strongly recommend that you leave the roots as they are above, compute the corresponding $\frac{dx_i}{d\epsilon}$ for each of the roots (setting $\epsilon=0$) and expand from this point.

For sure, you could do what alex.jordan suggested; his solution is more rigorous than mine but it will be quite tedious even to have the first order expansion.

Answer 2

You could find all three roots exactly using the cubic formula/cubic-solving process, expand the roots as series in $\epsilon$ and then cut them off at whatever order you like.

The formula for one of the real roots is here. To get the other two roots multiply the giant cube roots by $\omega$ and $\omega^2$ respectively, and then in the other order, where $\omega$ is a complex root of unity.

Answer 3

Since we are told that $a\gg1$ and $\epsilon$ can be about anything one could write $a:={1\over q}$ with $0<q\ll1$, where $q$ is considered a small parameter. Writing $t$ instead of $\epsilon$ the equation then becomes $$x-2x^2+\bigl((1+t)x^3-3tx+2t\bigr)\>q=0\ .\tag{1}$$ Letting $q=0$ here we obtain the two solutions $x=0$ and $x={1\over2}$. We therefore introduce the expansions $$x(q)=\sum_{k\geq1} c_k q^k,\qquad y(q)={1\over2}+\sum_{k\geq1} d_k q^k\ .$$ with undetermined coefficients $c_k$, resp. $d_k\>$. Entering these into $(1)$ and comparing coefficients (I used Mathematica for this) one obtains $$\eqalign{x(q)&=-2 tq + 2 t^2q^2 - 10 t^3 q^3 + (8 t^3 + 66 t^4) q^4 + (-64 t^4 - 434 t^5) q^5\ +?q^6\ ,\cr y(q) &= {1\over2} + {1 + 5 t\over8}q + {1 - 2 t - 35 t^2\over16} q^2 + {5 - 9 t + 111 t^2 + 1405 t^3\over128}q^3 \ + ?q^4\ .\cr}$$ The third root, whose order of magnitude is $a$, cannot be obtained in this way. Maybe writing $x:=a\>\xi$ with a new unknown $\xi$ will do the job. I leave this to you.