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I am trying to find the solution to the following equation, $\epsilon x^3 -x^2 +x-\epsilon^{\frac{1}{2}}=0$, for the first two non-zero solutions as $\epsilon \to 0^+$.

I have used the principal of dominant balance (by a Kruskal Newton Graph) to find that my ansatz should be of the form $x(\epsilon)=\dfrac{z(\epsilon)}{\epsilon^\alpha}$ where $\alpha=0,\dfrac{1}{2},1$.

Now if we substitute these into our equation we get, for each $\alpha$:

$\alpha=0 : \ \epsilon x^3 -x^2 +x-\epsilon^{\frac{1}{2}}=0 $ - noting $x=z$;

$\alpha=\dfrac{1}{2}: z\epsilon^{\frac{1}{2}}-z^2+z\epsilon^\frac{1}{2}-\epsilon^\frac{3}{2}$;

$\alpha=1: \ z^3-z^2+z\epsilon-\epsilon^\frac{1}{2}$.

Now it is clear that if we perturb our equations in the form $z(\epsilon)=z_0+\epsilon z_1 +...$ then we have, for $\alpha=0 \ \& \ 1$, an $\epsilon^\frac{1}{2}$ term which can only be equated to zero. Does this mean that we do not have a valid solution for these choices of $\alpha$?

Whereas, if we consider $\alpha=\dfrac{1}{2}$ then we have an equation of the form: $(z_0+\epsilon z_1)^3\epsilon^\frac{1}{2} -(z_0+\epsilon z_1)^2 +(z_0+\epsilon z_1)\epsilon^\frac{1}{2}-\epsilon^\frac{3}{2}=0$ whose coefficients can all be equated, thus valid solutions.

As a result is the only case where we can find solutions at $\alpha=\dfrac{1}{2}$?

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    $\begingroup$ There are 6 possible balances, three involve $\alpha<0$. $\endgroup$ – David Oct 31 '18 at 21:55
  • $\begingroup$ Yes, you're right - I missed $\alpha =-0.25,-0.5,\dfrac{1}{6}$. I will update this post sometime in the next two days. I assume that I should then get a balance for each order of \epsilon somewhere? $\endgroup$ – KieranSQ Oct 31 '18 at 22:01
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Your original equation is $$\epsilon x^3 -x^2 +x-\epsilon^{\frac{1}{2}}=0. \tag1$$ To simplify a bit we define $\, \delta := \sqrt{\epsilon}.\,$ The equation now becomes $$ \delta^2 x^3 - x^2 + x - \delta = 0. \tag2$$

For any $\, \delta \ne 0, \,$ there are three rots of the cubic. If we try a small value for $\, \delta, \,$ such as $\, \delta = 0.01, \,$ the three roots are easily determined to be approximately $$ x_1 \approx \delta + \delta^2, \quad x_2 \approx 1 - \delta, \quad x_3 \approx \delta^{-2} - 1. \tag3$$

It is now easy to find further terms of the Puiseux series expansions. Note that for any $\, \epsilon>0, \,$ there are two values for its square root $\, \delta,\,$ and this gives six roots of equation $\,(1)\,$ since its roots in terms of $\, \delta \,$ are different if we take the negative square root.

You may be interested in ways to increase precision of approximations to the roots. For example, for the root $\,x_1,\,$ the function $\, f_1(t) := \delta + t^2 - \delta^2t^3 \,$ has $\,x_1\,$ as a fixed point. Using iteration we get the sequence of approximations to $\,x_1\,$ as follows: $$ \delta + O(\delta^2) \, \mapsto \, \delta + \delta^2 + O(\delta^3) \, \mapsto \, \delta + \delta^2 + 2\delta^3 + O(\delta^4) \, \dots. $$

Similarly, for the root $\,x_2,\,$ the function $\, f_2(t) := (1+\sqrt{1 - 4\delta + 4\delta^2 t^3})/2 \,$ has $\,x_2\,$ as a fixed point. Using iteration we get the sequence of approximations to $\,x_2\,$ as follows: $$ 1 + O(\delta) \, \mapsto \, 1 - \delta + O(\delta^3) \, \mapsto \, 1 - \delta - 3 \delta^3 - 3\delta^4 + O(\delta^5) \, \dots.$$

The remaining root $\, x_3 = \delta^{-2} - x_1 - x_2 = \delta^{-2} - 1 - \delta^2 + \delta^3 - 2\delta^4 + O(\delta^5). \,$

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Following your approach $x=ϵ^{-α}z$ gives the equation for $z$ as $$ ϵ^{1-α}z^3-z^2+ϵ^{α}z-ϵ^{2α+\frac12} $$ The corners of the concave curve $\min\{1-α,0,α,2α+\frac12\}$ are at $α=-\frac12, 0, 1$. plot of the minimum curve

  • For $α=-\frac12$ we get in separation of medium and small terms $$ z=1-ϵ^2z^3+ϵ^{\frac12}z^2 $$ which successively gives $z_0=1$, $z_1=1+ϵ^{\frac12}$, $z_2=1+ϵ^{\frac12}+2ϵ$, ... so that $$x=ϵ^{\frac12}+ϵ+2ϵ^{\frac32}+\dots$$

  • For $α=0$, $$z=1-ϵ^{\frac12}z^{-1}+ϵz^2$$ so that again adding a term per iteration $z_0=1$, $z_1=1-ϵ^{\frac12}$, $$x=z_2+O(ϵ^2)=1-ϵ^{\frac12}-3ϵ^{\frac32}+...$$

  • For $α=1$, $$z=1-ϵz^{-1}+ϵ^{\frac52}z^{-2}$$ so that $z_1=1-ϵ$, $z_2=1-ϵ-ϵ^2$, $$x=ϵ^{-1}-1-ϵ+...$$

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