The Interaction of Two Unsteady Point Vortex Sources in a Deformation Field in 2D Incompressible Flows

Abstract

Taking into account the coupling of the ocean with the atmosphere is essential to properly describe vortex dynamics in the ocean. The forcing of a circular eddy with the relative wind stress curl leads to an Ekman pumping with a nonzero area integral. This in turn creates a source or a sink in the eddy. We revisit the two point vortex-source interaction, now coupled with an unsteady wind, leading to a time-varying circulation and source strength. Firstly, we recover the various fixed points of the two vortex-source system, and we calculate their stability. Then we show the effect of a weak amplitude, subharmonic, or harmonic time variation of the wind, leading to a similar variation of the circulation and the source strength of the vortex sources. We use a multiple time scale expansion of the variables to calculate the long time variation of these vortex trajectories around neutral fixed points. We study the amplitude equation and obtain its solution. We compute numerically the unstable evolution of the vortex sources when the source and circulation have a finite periodic variation. We also assess the influence of this time variation on the dispersion of a passive tracer near these vortex sources.

APPENDIX A. EQUILIBRIUM POINTS AND STABILITY

This section is a reminder of the fixed points of the problem; it was addressed slightly differently in [4] and in [3]. Recall the various cases for equilibria here with our notations and in our specific cases. This is necessary to further study the vortex source evolution with unsteady circulation or source strength.

Recall that we have the condition $\Gamma_{0}\Omega<0$ and the formulas

$$\theta_{0}=-\frac{1}{2}\arctan\left[\frac{\Gamma_{0}+4\pi r_{0}^{2}\Omega}{S_{0}}\right],$$

(A.1)

or

$$\theta_{0}=-\frac{1}{2}\arctan\left[\frac{\Gamma_{0}+4\pi r_{0}^{2}\Omega}{S_{0}}\right]+\frac{\pi}{2},$$

(A.2)

and

$$r_{0}^{4}(\Omega^{2}-A^{2})+\frac{\Gamma_{0}\Omega}{2\pi}r_{0}^{2}+\frac{S_{0}^{2}+\Gamma_{0}^{2}}{16\pi^{2}}=0.$$

(A.3)

6.1. For $\Omega^{2}=A^{2}$:

Equilibrium. Starting from Eq. (A.3) with $\Omega^{2}=A^{2}$ and $\Gamma_{0}\Omega<0$, we have $r_{0}^{2}=-\frac{S_{0}^{2}+\Gamma_{0}^{2}}{8\pi\Gamma_{0}\Omega}>0$ and thanks to Eq. (A.1), we have

$$r_{0}=\sqrt{\frac{S_{0}^{2}+\Gamma_{0}^{2}}{8\pi(-\Gamma_{0}\Omega)}}\text{ and }\theta_{0}=\frac{1}{2}\arctan\left[\frac{S_{0}^{2}-\Gamma_{0}^{2}}{2S_{0}\Gamma_{0}}\right].$$

(A.4)

Stability. Is the equilibrium (A.4) stable? From the characteristic polynomial (3.7) of the differential matrix $\chi(X)=X^{2}-\frac{S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega}{4\pi^{2}r_{0}^{4}}$, we need to determine the sign of $\Delta_{0}$:

$$\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega=\frac{S_{0}^{2}+\Gamma_{0}^{2}}{2}>0.$$

(A.5)

So $\chi$ has two real roots: one positive and one negative. Then the equilibrium (A.4) is a saddle equilibrium point. We are not interested in this type of equilibrium.

6.2. For $\Omega^{2}\neq A^{2}$:

From the polynomial equation (A.3) in $r_{0}^{2}$:

$$\left(\Omega^{2}-A^{2}\right)X^{2}+\frac{\Gamma_{0}\Omega}{2\pi}X+\frac{S_{0}^{2}+\Gamma_{0}^{2}}{16\pi^{2}}=0,$$

(A.6)

we compute the discriminant

$$\Delta=\frac{1}{4\pi^{2}}\left[\Gamma_{0}^{2}\Omega^{2}-\left(S_{0}^{2}+\Gamma_{0}^{2}\right)\left(\Omega^{2}-A^{2}\right)\right],$$

(A.7)

and look at the sign of

$$\Delta^{\prime}=A^{2}\left(S_{0}^{2}+\Gamma_{0}^{2}\right)-S_{0}^{2}\Omega^{2},$$

(A.8)

$$\Delta^{\prime}=S_{0}^{2}\left(A^{2}-\Omega^{2}\right)+A^{2}\Gamma_{0}^{2}.$$

(A.9)

We want $\Delta^{\prime}$ to be positive because we want real (positive) solutions to Eq. (A.6). This brings three situations (we have already studied the situation $\Omega^{2}=A^{2}$):

• If $A^{2}>\Omega^{2}$, then $\Delta^{\prime}>0$ clearly from Eq. (A.9).

• If $\Omega^{2}>A^{2}$, then $\Delta^{\prime}>0\iff\Omega^{2}<A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right)$.

• If $\Omega^{2}=A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right)$, then $\Delta^{\prime}=0$.

Because $A^{2}>\Omega^{2}$, we have $\Delta^{\prime}>0$ without any more condition, and we have two solutions to the polynomial equation (A.6):

$$X_{\pm}=\frac{\Gamma_{0}\Omega\pm\sqrt{\Delta^{\prime}}}{4\pi\left(A^{2}-\Omega^{2}\right)}.$$

(A.10)

Recall that we want only a nonnegative solution ($r_{0}^{2}>0$). Because we have supposed the condition $\Gamma_{0}\Omega<0$, this constraint removes $X_{-}$. The root $X_{+}$ is a nonnegative solution if and only if $\sqrt{\Delta^{\prime}}>-\Gamma_{0}\Omega>0$. This condition is valid because $\left(A^{2}-\Omega^{2}\right)\left(S_{0}^{2}+\Gamma_{0}^{2}\right)>0$ so $\Delta^{\prime}>\Gamma_{0}^{2}\Omega^{2}$.

Equilibrium for $X_{+}$. We have the following equilibrium point (with $\theta_{0}$ computed from Eq. (A.1)):

$$r_{0}=\sqrt{\frac{\Gamma_{0}\Omega+\sqrt{\Delta^{\prime}}}{4\pi\left(A^{2}-\Omega^{2}\right)}}\text{ and }\theta_{0}=-\frac{1}{2}\arctan\left[\frac{\Gamma_{0}A^{2}+\Omega\sqrt{\Delta^{\prime}}}{S_{0}\left(A^{2}-\Omega^{2}\right)}\right].$$

(A.11)

Stability for $X_{+}$. How is the equilibrium (A.11) stable? We need to know the sign of $\Delta_{0}$.

Proposition 1

Under all the conditions of this subsection, we have

$$\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega>0$$

and the equilibrium point (A.11) is a saddle equilibrium point.

Proof

Remember that we work under the assumption $A^{2}>\Omega^{2}$ and $\Gamma_{0}\Omega<0$. Then put $r_{0}^{2}$ in $\Delta_{0}$ and

$$\displaystyle\Delta_{0}>0\iff\left(S_{0}^{2}+\Gamma_{0}^{2}\right)\left(A^{2}-\Omega^{2}\right)+\Gamma_{0}^{2}\Omega^{2}>-\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}$$

$$\displaystyle\iff\left[S_{0}^{2}\left(A^{2}-\Omega^{2}\right)+\Gamma_{0}^{2}A^{2}\right]^{2}>\Gamma_{0}^{2}\Omega^{2}S_{0}^{2}\left(A^{2}-\Omega^{2}\right)+\Gamma_{0}^{4}\Omega^{2}A^{2}$$

$$\displaystyle\iff S_{0}^{4}\left(A^{2}-\Omega^{2}\right)+S_{0}^{2}\Gamma_{0}^{2}\left(2A^{2}-\Omega^{2}\right)+\Gamma_{0}^{4}A^{2}>0.$$

The right-hand side of the equivalence is true under the assumption $A^{2}>\Omega^{2}$. This concludes the proof of the proposition.     $\square$

We also have two roots of the polynomial (A.6):

$$X_{\pm}=\frac{-\Gamma_{0}\Omega\pm\sqrt{\Delta^{\prime}}}{4\pi\left(\Omega^{2}-A^{2}\right)}.$$

(A.12)

$X_{+}$ is clearly nonnegative. $X_{-}$ is also nonnegative because we have $-\Gamma_{0}\Omega>\sqrt{\Delta^{\prime}}>0$ (deduced from the hypothesis). So we have two situations to analyze:

Equilibrium and stability for $X_{+}$. We have the following equilibrium point (with $\theta_{0}$ computed from Eq. (A.1)):

$$r_{0}=\sqrt{\frac{-\Gamma_{0}\Omega+\sqrt{\Delta^{\prime}}}{4\pi\left(\Omega^{2}-A^{2}\right)}}\text{ and }\theta_{0}=\frac{1}{2}\arctan\left[\frac{\Gamma_{0}A^{2}-\Omega\sqrt{\Delta^{\prime}}}{S_{0}\left(\Omega^{2}-A^{2}\right)}\right].$$

(A.13)

How is this equilibrium (A.13) stable? We need to know the sign of $\Delta_{0}$.

Proposition 2

Whatever the set of parameters we choose, if they satisfy the assumptions we made: $A^{2}<\Omega^{2}<A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right)$ and $\Gamma_{0}\Omega<0$ , then we have

$$\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega<0,$$

(A.14)

and the equilibrium point (A.13) is a neutral equilibrium point.

Proof

Consider $\Delta_{0}$ for the value $r_{0}$ we have in Eq. (A.13):

$$\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+\Gamma_{0}\Omega\left(\frac{-\Gamma_{0}\Omega+\sqrt{\Delta^{\prime}}}{\Omega^{2}-A^{2}}\right).$$

So

$$\displaystyle\Delta_{0}<0\iff\left(S_{0}^{2}+\Gamma_{0}^{2}\right)\left(\Omega^{2}-A^{2}\right)-\Gamma_{0}^{2}\Omega^{2}+\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}<0$$

$$\displaystyle\iff S_{0}^{2}\left(\Omega^{2}-A^{2}\right)+\Gamma_{0}^{2}A^{2}+\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}<0.$$

The right-hand side of the equivalence is true because $\Gamma_{0}\Omega<0$ and $\Omega^{2}-A^{2}<A^{2}\frac{\Gamma_{0}^{2}}{S_{0}^{2}}$ so $S_{0}^{2}\left(\Omega^{2}-A^{2}\right)+\Gamma_{0}^{2}A^{2}<0$. This concludes the proof of the proposition.     $\square$

Equilibrium and stability for $X_{-}$. We have the following equilibrium point (with $\theta_{0}$ computed from Eq. (A.1)):

$$r_{0}=\sqrt{\frac{-\Gamma_{0}\Omega-\sqrt{\Delta^{\prime}}}{4\pi\left(\Omega^{2}-A^{2}\right)}}\text{ and }\theta_{0}=\frac{1}{2}\arctan\left[\frac{\Gamma_{0}A^{2}+\Omega\sqrt{\Delta^{\prime}}}{S_{0}\left(\Omega^{2}-A^{2}\right)}\right].$$

(A.15)

Proposition 3

For the equilibrium (A.15) , $\Delta_{0}$ is nonnegative for every set of parameters such that $A^{2}<\Omega^{2}<A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right)$ and $\Gamma_{0}\Omega<0$ . So the equilibrium (A.15) is a saddle equilibrium point.

Proof

Look at the expression of $\Delta_{0}$:

$$\displaystyle\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega=S_{0}^{2}+\Gamma_{0}^{2}+\frac{-\Gamma_{0}^{2}\Omega^{2}-\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}}{\Omega^{2}-A^{2}},$$

(A.16)

in which the sign is the same as the sign of

$$\left(S_{0}^{2}+\Gamma_{0}^{2}\right)\left(\Omega^{2}-A^{2}\right)-\Gamma_{0}^{2}\Omega^{2}-\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}=\underbrace{S_{0}^{2}\left(\Omega^{2}-A^{2}\right)-\Gamma_{0}^{2}A^{2}}_{<0\text{ because }\Omega^{2}-A^{2}<A^{2}\frac{\Gamma_{0}^{2}}{S_{0}^{2}}}+\underbrace{\left(-\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}\right)}_{>0}.$$

(A.17)

So we have the following equivalences:

$$\displaystyle\Delta_{0}>0\iff-\Gamma_{0}\Omega\sqrt{\Delta^{\prime}}>\Gamma_{0}^{2}A^{2}-S_{0}^{2}\left(\Omega^{2}-A^{2}\right)$$

$$\displaystyle\iff\Gamma_{0}^{2}\Omega^{2}\Delta^{\prime}>\left(\Gamma_{0}^{2}A^{2}-S_{0}^{2}\left(\Omega^{2}-A^{2}\right)\right)^{2}$$

$$\displaystyle\iff\Gamma_{0}^{2}\Omega^{2}S_{0}^{2}\left(A^{2}-\Omega^{2}\right)+\Gamma_{0}^{4}A^{2}\left(\Omega^{2}-A^{2}\right)>-2\Gamma_{0}^{2}A^{2}S_{0}^{2}\left(\Omega^{2}-A^{2}\right)+S_{0}^{4}\left(\Omega^{2}-A^{2}\right)^{2}$$

$$\displaystyle\iff S_{0}^{4}\left(\Omega^{2}-A^{2}\right)+S_{0}^{2}\Gamma_{0}^{2}\left(\Omega^{2}-2A^{2}\right)-\Gamma_{0}^{4}A^{2}<0.$$

We have to study the sign of a second-degree polynomial in $S_{0}^{2}$ for which the discriminant is

$$\delta=\Gamma_{0}^{4}\left(\Omega^{2}-2A^{2}\right)^{2}+4\Gamma_{0}^{4}A^{2}\left(\Omega^{2}-A^{2}\right)=\Gamma_{0}^{4}\Omega^{4}>0.$$

(A.18)

The two roots are

$$\frac{-\Gamma_{0}^{2}\left(\Omega^{2}-2A^{2}\right)+\Gamma_{0}^{2}\Omega^{2}}{2\left(\Omega^{2}-A^{2}\right)}=\frac{A^{2}\Gamma_{0}^{2}}{\Omega^{2}-A^{2}}>0,$$

(A.19)

and

$$\frac{-\Gamma_{0}^{2}\left(\Omega^{2}-2A^{2}\right)-\Gamma_{0}^{2}\Omega^{2}}{2\left(\Omega^{2}-A^{2}\right)}=-\Gamma_{0}^{2}<0.$$

(A.20)

Because $S_{0}^{2}>0$, to have $\Delta_{0}>0$, we need $S_{0}^{2}$ to be smaller than the largest root, but this is not an additional constraint because

$$\displaystyle S_{0}^{2}<\frac{A^{2}\Gamma_{0}^{2}}{\Omega^{2}-A^{2}}\iff\left(\Omega^{2}-A^{2}\right)S_{0}^{2}<A^{2}\Gamma_{0}^{2}$$

$$\displaystyle\iff\Omega^{2}<A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right).$$

So the polynomial $\left(\Omega^{2}-A^{2}\right)X^{2}+\Gamma_{0}^{2}\left(\Omega^{2}-2A^{2}\right)X-\Gamma_{0}^{4}A^{2}$ is nonpositive for every value between $0$ and $\frac{A^{2}\Gamma_{0}^{2}}{\Omega^{2}-A^{2}}$. Because $S_{0}^{2}$ is in this interval, we have $\Delta_{0}>0$ for every set of parameters such that $A^{2}<\Omega^{2}<A^{2}\left(1+\frac{\Gamma_{0}^{2}}{S_{0}^{2}}\right)$ and $\Gamma_{0}\Omega<0$.     $\square$

In this section, we have $\Delta^{\prime}=0$. Then there is only one solution to Eq. (A.6):

$$X=\frac{-\Gamma_{0}\Omega}{4\pi\left(\Omega^{2}-A^{2}\right)}=\frac{\Gamma_{0}^{2}+S_{0}^{2}}{4\pi\left(-\Gamma_{0}\Omega\right)}>0.$$

(A.21)

This gives the following equilibrium point:

$$r_{0}=\sqrt{\frac{\Gamma_{0}^{2}+S_{0}^{2}}{4\pi\left(-\Gamma_{0}\Omega\right)}}\text{ and }\theta_{0}=\frac{1}{2}\arctan\left(\frac{S_{0}}{\Gamma_{0}}\right).$$

(A.22)

To know the type of stability, we compute $\Delta_{0}$:

$$\Delta_{0}=S_{0}^{2}+\Gamma_{0}^{2}+4\pi r_{0}^{2}\Gamma_{0}\Omega=S_{0}^{2}+\Gamma_{0}^{2}+\Gamma_{0}\Omega\frac{\Gamma_{0}^{2}+S_{0}^{2}}{\left(-\Gamma_{0}\Omega\right)}=0.$$

(A.23)

So we cannot conclude about the stability of the equilibrium (A.22).

APPENDIX B. MULTIPLE TIME SCALE DEVELOPMENT

The multiple time scale method is here expanded for the subharmonic case. The harmonic case is similar.

6.1. Order $\varepsilon^{1}$

We have the following system at order $\varepsilon^{1}$, computed from Eqs. (4.5) and (4.8):

$$\left\{\begin{matrix}\partial_{t_{0}}r_{1}=-ar_{1}-b(r_{0}\theta_{1})\\ \partial_{t_{0}}\left(r_{0}\theta_{1}\right)=-cr_{1}+a(r_{0}\theta_{1}).\end{matrix}\right.$$

(B.1)

So

$$\begin{cases}r_{1}=C_{1,1}\left(t_{2},t_{3}\right)\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}\\ r_{0}\theta_{1}=D_{1,1}(t_{2},t_{3})\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}\textbf{}\end{cases}$$

(B.2)

with

$$D_{1,1}(t_{2},t_{3})=\mu_{1}C_{1,1}(t_{2},t_{3}),$$

(B.3)

and $\mu_{1}=-\frac{a+i\omega_{0}}{b}$.

6.2. Order $\varepsilon^{2}$

With Eqs. (4.6) and (4.9) and because $\partial_{t_{1}}r_{1}=\partial_{t_{1}}\left(r_{0}\theta_{1}\right)=0$, we have the following system in $(r_{2},r_{0}\theta_{2})$:

$$\begin{cases}\partial_{t_{0}}r_{2}=-ar_{2}-b(r_{0}\theta_{2})+f_{2}(t_{0},t_{1},t_{2},t_{3})\\ \partial_{t_{0}}\left(r_{0}\theta_{2}\right)=-cr_{2}+a(r_{0}\theta_{2})+g_{2}(t_{0},t_{1},t_{2},t_{3}),\end{cases}$$

(B.4)

where

$$\begin{cases}f_{2}(t_{0},t_{1},t_{2},t_{3})=\frac{a\delta r_{0}}{2}\cos(2\omega_{0}t_{0})+\frac{a}{2r_{0}}r_{1}^{2}+\frac{a}{r_{0}}\left(r_{0}\theta_{1}\right)^{2}-\frac{b}{r_{0}}r_{1}\left(r_{0}\theta_{1}\right)\\ g_{2}(t_{0},t_{1},t_{2},t_{3})=\frac{c\delta r_{0}}{2}\cos(2\omega_{0}t_{0})+\frac{3c}{2r_{0}}r_{1}^{2}+\frac{b}{r_{0}}\left(r_{0}\theta_{1}\right)^{2}.\end{cases}$$

(B.5)

The system (B.4) gives:

• For $r_{2}$:

  $$\displaystyle\partial_{t_{0}}^{2}r_{2}=-a\left(-ar_{2}-b\left(r_{0}\theta_{2}\right)+f_{2}\right)-b\left(-cr_{2}+a\left(r_{0}\theta_{2}\right)+g_{2}\right)+\partial_{t_{0}}f_{2}$$
  $$\displaystyle=\left(a^{2}+bc\right)r_{2}+h_{2}(t_{0},t_{1},t_{2},t_{3}).$$

  (B.6)

• For $r_{0}\theta_{2}$:

  $$\displaystyle\partial_{t_{0}}^{2}\left(r_{0}\theta_{2}\right)=-c\left(-ar_{2}-b\left(r_{0}\theta_{2}\right)+f_{2}\right)+a\left(-cr_{2}+a\left(r_{0}\theta_{2}\right)+g_{2}\right)+\partial_{t_{0}}g_{2}$$
  $$\displaystyle=\left(bc+a^{2}\right)\left(r_{0}\theta_{2}\right)+k_{2}(t_{0},t_{1},t_{2},t_{3}),$$

  (B.7)

where

$$\begin{cases}h_{2}(t_{0},t_{1},t_{2},t_{3})=\left(-af_{2}-bg_{2}+\partial_{t_{0}}f_{2}\right)(t_{0},t_{1},t_{2},t_{3})\\ k_{2}(t_{0},t_{1},t_{2},t_{3})=\left(-cf_{2}+ag_{2}+\partial_{t_{0}}g_{2}\right)(t_{0},t_{1},t_{2},t_{3}).\end{cases}$$

(B.8)

• Development of $f_{2}$:

  $$\displaystyle f_{2}=\frac{a}{r_{0}}\left[3-\frac{2c}{b}\right]|C_{1,1}|^{2}$$
  $$\displaystyle+\left[\frac{ar_{0}}{4}+\frac{C_{1,1}^{2}}{r_{0}}\left(\frac{3a}{2}+\frac{a\left(a+i\omega_{0}\right)^{2}}{b^{2}}+i\omega_{0}\right)\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}$$
  $$\displaystyle f_{2}=F_{2,0}|C_{1,1}|^{2}+\left[F_{2,2,1}+F_{2,2,2}C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}.$$

  (B.9)

• Development of $g_{2}$:

  $$\displaystyle g_{2}=\frac{c}{r_{0}}|C_{1,1}|^{2}+\left[\frac{cr_{0}}{4}+\frac{C_{1,1}^{2}}{r_{0}}\left(\frac{3c}{2}+\frac{\left(a+i\omega_{0}\right)^{2}}{b}\right)\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}$$
  $$\displaystyle g_{2}=G_{2,0}|C_{1,1}|^{2}+\left[G_{2,2,1}+G_{2,2,2}C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}.$$

  (B.10)

• Development of $h_{2}$:

  $$\displaystyle h_{2}=\left[-aF_{2,0}-bG_{2,0}\right]|C_{1,1}|^{2}+\left[\left(-bG_{2,2,1}+\left(-a+2i\omega_{0}\right)F_{2,2,1}\right)\right.$$
  $$\displaystyle\left.+\left(-bG_{2,2,2}+\left(-a+2i\omega_{0}\right)F_{2,2,2}\right)C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}$$
  $$\displaystyle h_{2}=H_{2,0}|C_{1,1}|^{2}+\left[H_{2,2,1}+H_{2,2,2}C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}.$$

  (B.11)

• Development of $k_{2}$:

  $$\displaystyle k_{2}=\left[-cF_{2,0}+aG_{2,0}\right]|C_{1,1}|^{2}+\left[\left(-cF_{2,2,1}+\left(a+2i\omega_{0}\right)G_{2,2,1}\right)\right.$$
  $$\displaystyle\left.+\left(-cF_{2,2,2}+\left(a+2i\omega_{0}\right)G_{2,2,2}\right)C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}$$
  $$\displaystyle k_{2}=K_{2,0}|C_{1,1}|^{2}+\left[K_{2,2,1}+K_{2,2,2}C_{1,1}^{2}\right]\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}.$$

  (B.12)

Then from Eqs. (B.6) and (B.7) we have

• The homogeneous solutions:

  $$\begin{cases}r_{2}=C_{2,1}\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}\\ \left(r_{0}\theta_{2}\right)=D_{2,1}\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}\end{cases}$$

  (B.13)

• The particular solutions for the constant terms:

  $$\begin{cases}r_{2}=\frac{H_{2,0}}{\omega_{0}^{2}}|C_{1,1}|^{2}\\ \left(r_{0}\theta_{2}\right)=\frac{K_{2,0}}{\omega_{0}^{2}}|C_{1,1}|^{2}\end{cases}$$

  (B.14)

• The particular solutions for $\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}$:

  $$\begin{cases}r_{2}=-\frac{H_{2,2,1}+H_{2,2,2}C_{1,1}^{2}}{3\omega_{0}^{2}}\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}\\ \left(r_{0}\theta_{2}\right)=-\frac{K_{2,2,1}+K_{2,2,2}C_{1,1}^{2}}{3\omega_{0}^{2}}\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}.\end{cases}$$

  (B.15)

So the total solution of Eqs. (B.6) and (B.7) is

$$\left\{\begin{matrix}r_{2}&=C_{2,0}|C_{1,1}|^{2}+C_{2,1}\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}+\left(C_{2,2,1}+C_{2,2,2}C_{1,1}^{2}\right)\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}\\ \left(r_{0}\theta_{2}\right)&=D_{2,0}|C_{1,1}|^{2}+D_{2,1}\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}+\left(D_{2,2,1}+D_{2,2,2}C_{1,1}^{2}\right)\mathrm{e}^{2i\omega_{0}t_{0}}+\mathrm{c.c}\end{matrix}\right.$$

(B.16)

with (for $i=1,2$)

$$C_{2,0}=\frac{H_{2,0}}{\omega_{0}^{2}},C_{2,2,i}=-\frac{H_{2,2,i}}{3\omega_{0}^{2}},D_{2,0}=\frac{K_{2,0}}{\omega_{0}^{2}},D_{2,2,i}=-\frac{K_{2,2,i}}{3\omega_{0}^{2}}.$$

(B.17)

6.3. Order $\varepsilon^{3}$

With Eqs. (4.7) and (4.10), we have the following system at the order $\varepsilon^{3}$:

$$\begin{cases}\partial_{t_{0}}r_{3}=-ar_{3}-b(r_{0}\theta_{3})+f_{3}(t_{0},t_{1},t_{2},t_{3})\\ \partial_{t_{0}}(r_{0}\theta_{3})=-cr_{3}+a(r_{0}\theta_{3})+g_{3}(t_{0},t_{1},t_{2},t_{3}),\end{cases}$$

(B.18)

where $f_{3}$ and $g_{3}$ are the following given functions:

$$\left\{\begin{matrix}f_{3}&=-\partial_{t_{2}}r_{1}+\frac{ar_{1}r_{2}}{r_{0}}-\frac{ar_{1}^{3}}{2r_{0}^{2}}-\frac{ar_{1}\cos(2\omega_{0}t_{0})}{2}-\frac{b\left(r_{2}(r_{0}\theta_{1})+r_{1}\left(r_{0}\theta_{2}\right)\right)}{r_{0}}+\frac{ar_{1}(r_{0}\theta_{1})^{2}}{r_{0}^{2}}\\ &+\frac{2a(r_{0}\theta_{1})(r_{0}\theta_{2})}{r_{0}}+\frac{2b(r_{0}\theta_{1})^{3}}{3r_{0}^{2}}-ar_{0}\omega_{0}t_{1}\sin(2\omega_{0}t_{0})\\ g_{3}&=-\partial_{t_{2}}(r_{0}\theta_{1})+\frac{3cr_{1}r_{2}}{r_{0}}-\frac{2cr_{1}^{3}}{r_{0}^{2}}-cr_{1}\cos(2\omega_{0}t_{0})-\frac{2a(r_{0}\theta_{1})^{3}}{3r_{0}^{2}}+\frac{2b(r_{0}\theta_{1})(r_{0}\theta_{2})}{r_{0}}\\ &-c\omega_{0}t_{1}\sin(2\omega_{0}t_{0}).\end{matrix}\right.$$

(B.19)

The system (B.18) gives:

• For $r_{3}$:

  $$\displaystyle\partial_{t_{0}}^{2}r_{3}=-a\left(-ar_{3}-b\left(r_{0}\theta_{3}\right)+f_{3}\right)-b\left(-cr_{3}+a\left(r_{0}\theta_{3}\right)+g_{3}\right)+\partial_{t_{0}}f_{3}$$
  $$\displaystyle=\left(a^{2}+bc\right)r_{3}+h_{3}.$$

  (B.20)

• For $r_{0}\theta_{3}$:

  $$\displaystyle\partial_{t_{0}}^{2}\left(r_{0}\theta_{3}\right)=-c\left(-ar_{3}-b\left(r_{0}\theta_{3}\right)+f_{3}\right)+a\left(-cr_{3}+a\left(r_{0}\theta_{3}\right)+g_{3}\right)+\partial_{t_{0}}g_{3}$$
  $$\displaystyle=\left(bc+a^{2}\right)\left(r_{0}\theta_{3}\right)+k_{3}.$$

  (B.21)

We do not develop $f_{3},\ g_{3},\ h_{3}$ and $k_{3}$ as we did for the order $\varepsilon^{2}$. We only introduce the following notations:

$$\begin{cases}f_{3}=F_{3,0}+F_{3,1}\mathrm{e}^{i\omega_{0}t_{0}}+F_{3,2}\mathrm{e}^{2i\omega_{0}t_{0}}+F_{3,3}\mathrm{e}^{3i\omega_{0}t_{0}}+\mathrm{c.c}\\ g_{3}=G_{3,0}+G_{3,1}\mathrm{e}^{i\omega_{0}t_{0}}+G_{3,2}\mathrm{e}^{2i\omega_{0}t_{0}}+G_{3,3}\mathrm{e}^{3i\omega_{0}t_{0}}+\mathrm{c.c}\\ h_{3}=H_{3,0}+H_{3,1}\mathrm{e}^{i\omega_{0}t_{0}}+H_{3,2}\mathrm{e}^{2i\omega_{0}t_{0}}+H_{3,3}\mathrm{e}^{3i\omega_{0}t_{0}}+\mathrm{c.c}\\ k_{3}=K_{3,0}+K_{3,1}\mathrm{e}^{i\omega_{0}t_{0}}+K_{3,2}\mathrm{e}^{2i\omega_{0}t_{0}}+K_{3,3}\mathrm{e}^{3i\omega_{0}t_{0}}+\mathrm{c.c}.\end{cases}$$

(B.22)

Then, if we denote by $L$ the self-adjoint linear operator $\partial_{t_{0}}^{2}+\omega_{0}^{2}$, we have $r_{1}^{\star}Lr_{3}=r_{1}^{\star}h_{3}=r_{1}^{\star}L^{\star}r_{3}=0=\langle r_{1},h_{3}\rangle$. But $\langle e^{in\omega_{0}t_{0}},e^{ip\omega_{0}t_{0}}\rangle=\delta_{n,p}$ (Kronecker symbol) for $n,p\in\mathbf{Z}$ and because $r_{1}=C_{1,1}\mathrm{e}^{i\omega_{0}t_{0}}+\mathrm{c.c}$, we have

$$\langle r_{1},h_{3}\rangle=C_{1,1}H_{3,1}+\mathrm{c.c}=0.$$

(B.23)

Because $H_{3,1}=(-a+i\omega_{0})F_{3,1}-bG_{3,1}$, we deduce the amplitude equation

$$(-a+i\omega_{0})F_{3,1}-bG_{3,1}=0.$$

(B.24)

So we only have to compute $F_{3,1}$ and $G_{3,1}$ from Eq. (B.19): writing

$$\begin{cases}F_{3,1}=-\partial_{t_{2}}C_{1,1}+\mathrm{I}\overline{C_{1,1}}+\underline{\overline{\mathrm{II}}}|C_{1,1}|^{2}C_{1,1}\\ G_{3,1}=\frac{a+i\omega_{0}}{b}\partial_{t_{2}}C_{1,1}+\underline{\overline{\mathrm{III}}}\overline{C_{1,1}}+\underline{\overline{\mathrm{IV}}}|C_{1,1}|^{2}C_{1,1},\end{cases}$$

(B.25)

we have:

$$\displaystyle\mathrm{I}=-\frac{a}{4}+\frac{C_{2,2,1}}{r_{0}}\left(2a-i\omega_{0}\right)+\frac{D_{2,2,1}}{r_{0}}\left(-b-\frac{2a^{2}}{b}+\frac{2ai\omega_{0}}{b}\right),$$

(B.26)

$$\displaystyle\underline{\overline{\mathrm{II}}}=\frac{1}{r_{0}^{2}}\left[a\left(-\frac{3}{2}+2\frac{a^{2}}{b^{2}}+\frac{c}{b}\right)+2i\omega_{0}\left(\frac{a^{2}}{b^{2}}+\frac{c}{b}\right)\right]$$

$$\displaystyle+\frac{C_{2,0}}{r_{0}}\left(2a+i\omega_{0}\right)+\frac{D_{2,0}}{r_{0}}\left(-b-\frac{2a^{2}}{b}-\frac{2ai\omega_{0}}{b}\right)$$

$$\displaystyle+\frac{C_{2,2,2}}{r_{0}}\left(2a-i\omega_{0}\right)+\frac{D_{2,2,2}}{r_{0}}\left(-b-\frac{2a^{2}}{b}+\frac{2ai\omega_{0}}{b}\right),$$

(B.27)

$$\displaystyle\underline{\overline{\mathrm{III}}}=-\frac{c}{2}+\frac{3c}{r_{0}}C_{2,2,1}+\frac{2D_{2,2,1}}{r_{0}}\left(-a+i\omega_{0}\right),$$

(B.28)

$$\displaystyle\underline{\overline{\mathrm{IV}}}=-\frac{6c}{r_{0}^{2}}-\frac{2ac}{r_{0}^{2}b^{2}}\left(a+i\omega_{0}\right)+\frac{3c}{r_{0}}\left(C_{2,0}+C_{2,2,2}\right)$$

$$\displaystyle-\frac{2}{r_{0}}\left(\left(a+i\omega_{0}\right)D_{2,0}+\left(a-i\omega_{0}\right)D_{2,2,2}\right).$$

(B.29)

From Eq. (B.24) we obtain the amplitude equation

$$\partial_{t_{2}}C_{1,1}=\left(\underline{\overline{\mathrm{V}}}+i\underline{\overline{\mathrm{VI}}}\right)\overline{C_{1,1}}+\left(\underline{\overline{\mathrm{VII}}}+i\underline{\overline{\mathrm{VIII}}}\right)|C_{1,1}|^{2}C_{1,1},$$

(B.30)

with

$$\left\{\begin{aligned} &\displaystyle\underline{\overline{\mathrm{V}}}=\mathop{\rm Re}\nolimits\left(-\dfrac{(a-i\omega_{0})\mathrm{I}+b\underline{\overline{\mathrm{III}}}}{2i\omega_{0}}\right)\\ &\displaystyle\underline{\overline{\mathrm{VI}}}=\mathop{\rm Im}\nolimits\left(-\dfrac{(a-i\omega_{0})\mathrm{I}+b\underline{\overline{\mathrm{III}}}}{2i\omega_{0}}\right)\\ &\displaystyle\underline{\overline{\mathrm{VII}}}=\mathop{\rm Re}\nolimits\left(-\dfrac{(a-i\omega_{0})\underline{\overline{\mathrm{II}}}+b\underline{\overline{\mathrm{IV}}}}{2i\omega_{0}}\right)\\ &\displaystyle\underline{\overline{\mathrm{VIII}}}=\mathop{\rm Im}\nolimits\left(-\dfrac{(a-i\omega_{0})\underline{\overline{\mathrm{II}}}+b\underline{\overline{\mathrm{IV}}}}{2i\omega_{0}}\right).\end{aligned}\right.$$

(B.31)