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{\bf Question}

Show that the small $\varepsilon$ expansion of the roots of

$$x^2-x-2+\ds\frac{1}{2}\varepsilon(x^3+2x+3)=0$$

are

$$x=\left\{\begin{array}{c}
-1\\2-\ds\frac{5}{2}\varepsilon+O(\varepsilon^2)\\
-\ds\frac{2}{\varepsilon}-1+O(\varepsilon)\end{array}\right.$$

Sketch the behaviour of the roots as $\varepsilon\to 0^+$.




\vspace{.5in}

{\bf Answer}

$x^2-x-2+\ds\frac{\varepsilon}{2}(x^32+2x+3)=0$

Substitute $x=-1$ to check it's a root:
$1+1-2+\ds\frac{\varepsilon}{2}(-1-2+3)=0 \surd$

When $\varepsilon=0,\ x^2-x-2=0 \Rightarrow x=-1$ or +2, non
degenerate.

Therefore can try $x=x_0+\varepsilon x+O(\varepsilon^32)$ as
ansatz.

Note $\varepsilon$ multiplies highest power $\Rightarrow$ singular
perturbation.

Therefore try $x-\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon)$ as
ansatz.

This completes our 3 roots:

\un{Regular}:

$[x_0+\varepsilon x_1+O(\varepsilon^2)]^2-[x_0+\varepsilon x_1
+O(\varepsilon^2)]-2+\ds\frac{\varepsilon}{2}[x_0+\varepsilon x_1
+O(\varepsilon^2)]^3+\varepsilon(x_0+\varepsilon
x_1+O(\varepsilon^2)) +\ds\frac{3}{2}\varepsilon=0$

$x_0^2+2\varepsilon x_0x_1-x_0-\varepsilon
x_1-2+\ds\frac{\varepsilon}{2}x_0^3+\varepsilon
x_0+\ds\frac{3}{2}\varepsilon+O(\varepsilon^2)=0$

Therefore
$(x_0^2-x_0-2)+\varepsilon\left(2x_0x_1-x_1+\ds\frac{x_0^3}{2}+
x_0+\ds\frac{3}{2}\right)+O(\varepsilon^2)=0$

Balance at

$O(\varepsilon^0): x_0^2-x_0-2 \rightarrow x_0=-1$ or $x_0=+2$ as
above

$O(\varepsilon^1):
2x_0x_1-x_1+\ds\frac{x_0^3}{2}+x_0+\ds\frac{3}{2}=0 \Rightarrow
x_0=-1 \Rightarrow x_1=0$ (as expected since $x_0=-1$ is EXACT, or
$x_0=2 \Rightarrow x_0=\ds\frac{5}{2}$

$\Rightarrow
\left\{\begin{array}{l}x=-1\\x=2-\ds\frac{5}{2}\varepsilon+O(\varepsilon^2)
\end{array} \right.$

Singular root:

$\left[\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon)\right]^2-
\left[\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon)\right]-2$

$+\ds\frac{1}{2}\varepsilon
\left[\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon)\right]^3+\ds\frac{\varepsilon}{2}\cdot
2\left[\ds\frac{x_0}{\varepsilon}+x_1+O(\varepsilon)\right]+3\ds\frac{\varepsilon}{2}=0$

$\ds\frac{x_0^2}{\varepsilon^2}+2\ds\frac{x_0x_1}{\varepsilon}-\ds\frac{x_0}{\varepsilon}
+\ds\frac{x_0^3}{2^2}+\ds\frac{3}{2}\ds\frac{x_0^2x_1}{\varepsilon}+O(\varepsilon^0)=0$

Balance at:

$O\left(\ds\frac{1}{\varepsilon^2}\right):
x_0^2+\ds\frac{x_0^3}{2}=0 \Rightarrow$ either
$\left.\begin{array}{l}x_0=0\\x_0=0 \end{array}\right\}$ (non
singular roots) discard, or $x_0=-2$

$O\left(\ds\frac{1}{\varepsilon}\right):
2\un{x_0x_1}-x_0+\ds\frac{3}{2}x_0^2x_1=0 \rightarrow x_0=-2
\Rightarrow x_0=-2 \Rightarrow x_0=-1$

Therefore root is $x=-\ds\frac{2}{\varepsilon}-1+O(\varepsilon),\
\varepsilon\to 0^+$

\un{From $-\infty$}

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