# Perturbation method for a phase coexistence system

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The differential equation with $k$ dimensions has the form $\frac{d\mathbf{F}}{d\mathbf{x}} = \mathbf{F}(\mathbf{x},\epsilon)$, where $\mathbf{x}$ is the variable vector and $\epsilon$ is a parameter.  I focus on the steady state, which yields the solution of $\mathbf{F}(\mathbf{x},\epsilon)=0$. Assume that the equation has a perturbation solution $\mathbf{x} = \mathbf{a}\epsilon +\mathcal{O}(\epsilon^2)$ for the parameter $\epsilon\rightarrow 0$ and another perturbation solution $\mathbf{x} = \mathbf{b}_0-\mathbf{b}_1\epsilon^{-1} +\mathcal{O}(\epsilon^{-2})$ for the parameter $\epsilon\rightarrow \infty$.

Now I already know (from numerical simulations) that the above two kinds of solutions i.e., $x_i\rightarrow 0$ for $i\in I$ and $x_j\rightarrow b_0$ for $j\in J$, coexists for the parameter $\epsilon$ of a specific value interval. ($K=\{1,2,..., k\}$ and the sets $I,J\subseteq K$, $I\cap J=\emptyset$).
My question is how to identify the solution sets $I$ and $J$ by perturbation method? or I should consider its Hamiltonian?

asked Jul 19, 2019
edited Jul 19, 2019

## 1 Answer

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Make an Laurent series ansatz $x=\sum_{k\in Z} z_k \varepsilon^{k}$, insert it into the differential equation, expand into a Laurent series and compare coeffiicients to get differential equations for the $z_k$. Then you can analyze a truncated version of this system to find out which coefficients can be set to zero without impairing solvability.

answered Jul 19, 2019 by (14,537 points)
edited Jul 21, 2019 by Dilaton

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