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I have set myself the following singular perturbation problem:

For small values of $\varepsilon > 0$ find the two roots closest to $x=0$ for the equation.

$${x^4} - \,\,{x^2} + \,\,\varepsilon (x + 1)\,\, = \,\,0$$

The two roots in question have to be found using a singular perturbation approach. What are the different ways this can be approached? How would you obtain the first two non-zero terms of the expansions which give the roots in terms of $\varepsilon$?

I have tried a rescaling with "dominent terms" approach, but to no avail.

Thank you!

Mike

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    $\begingroup$ Use $\displaystyle x^{2} = x^{4} + \epsilon\left(x + 1\right)\quad \stackrel{{\rm as}\ x\ \to\ 0}{\sim} \quad \epsilon\quad\implies x = \pm\sqrt{\epsilon}$ $\endgroup$
    – Felix Marin
    Commented May 14 at 17:11
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    $\begingroup$ Another useful observation is that $x=−1$ is a root of both the original polynomial and its perturbation. Hence $x+1$ should factor out of both, and indeed $$x^4−x^2+\epsilon(x+1)=(x+1)[x^3−x^2+\epsilon]=0$$ So it suffices to consider the simpler (but still singular) perturbation problem $x^3−x^2+\epsilon=0$. $\endgroup$
    – Semiclassical
    Commented Jun 10 at 7:49

2 Answers 2

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As shown by Felix Marin in his comment, the two roots of the equation $$ x^4-x^2+\varepsilon(x+1)=0 \tag{1} $$ closest to $x=0$ are, to lowest order in $\varepsilon$, $x_1=-\sqrt{\varepsilon}$ and $x_2=\sqrt{\varepsilon}$. Let me show how to obtain the next order term for $x_1$. Substituting $x=-\sqrt{\varepsilon}+\delta$ in $(1)$ and expanding in powers of $\delta$, we get $$ (\varepsilon^2-\varepsilon^{3/2})+(-4\varepsilon^{3/2}+2\varepsilon^{1/2}+\varepsilon)\delta+(6\varepsilon-1)\delta^2-4\varepsilon^{1/2}\delta^3+\delta^4=0. \tag{2} $$ Assuming $|\delta| \ll \sqrt{\varepsilon} \ll 1$, we may rewrite $(2)$ as $$ [-\varepsilon^{3/2}+o(\varepsilon^{3/2})]+[2\varepsilon^{1/2}+o(\varepsilon^{1/2})]\delta=0 \implies \delta=\frac{\varepsilon}{2}+o(\varepsilon). \tag{3} $$ Therefore, $$ x_1=-\sqrt{\varepsilon}+\frac{\varepsilon}{2}+o(\varepsilon)\qquad(\varepsilon\ll 1). \tag{4} $$ A similar calculation shows that $$ x_2=\sqrt{\varepsilon}+\frac{\varepsilon}{2}+o(\varepsilon)\qquad(\varepsilon\ll 1). \tag{5} $$

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    $\begingroup$ After some Mathematica manipulations I noticed the following. If you substitute $\epsilon=16t^2$, then the two solutions can be written as $$x_{1,2}=4\left(\mp t+2t^2\mp 10 t^3+64t^4\mp\cdots\right)$$ with $x_1$ taking the minus signs. The resulting coefficient sequence, sans minus signs, is $1, 2, 10, 64, 462, 3584, 29172, 245760...$ which already seems to be extent on OEIS as A078531. I imagine the characterization given there is equivalent to the one here. $\endgroup$
    – Semiclassical
    Commented Jun 10 at 7:33
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As per the comment of @Semiclassical, you can cancel the factor $(x+1)$. Then the remaining equation can be transformed into fixed-point form $$ x=\pm\sqrt{\frac{ε}{1-x}}=\pm\sqrt{ε}\left(1+\frac{x}2+\frac{3}{8}x^2+...\right). $$ The iterates are $x_0=\pm \sqrt{ε}$, $$ x_1=\pm\sqrt{ε}\left(1\pm\frac{\sqrt{ε}}2+\frac{3}{8}ε^2+...\right) $$ etc., the next ones are better done using a CAS.

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