# Picking the correct Ansatz for valid solutions in Asymptotic Methods

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 $$\alpha=\dfrac{1}{2}$$?

• There are 6 possible balances, three involve $\alpha<0$. – David Oct 31 '18 at 21:55
• 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? – KieranSQ Oct 31 '18 at 22:01

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

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 $$α=-\frac12, 0, 1$$.

• 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 $$z_1=1-ϵ$$, $$z_2=1-ϵ-ϵ^2$$, $$x=ϵ^{-1}-1-ϵ+...$$