On the Efficiency of Equilibria in Mean-Field Oscillator Games

Abstract

A key question in the design of engineered competitive systems has been that of the efficiency loss of the associated equilibria. Yet, there is little known in this regard in the context of stochastic dynamic games, particularly in a large population regime. In this paper, we revisit a class of noncooperative games, arising from the synchronization of a large collection of heterogeneous oscillators. In Yin et al. (Proceedings of 2010 American control conference, pp. 1783–1790, 2010), we derived a PDE model for analyzing the associated equilibria in large population regimes through a mean field approximation. Here, we examine the efficiency of the associated mean-field equilibria with respect to a related welfare optimization problem. We construct constrained variational problems both for the noncooperative game and its centralized counterpart and derive the associated nonlinear eigenvalue problems. A relationship between the solutions of these eigenvalue problems is observed and allows for deriving an expression for efficiency loss. By applying bifurcation analysis, a local bound on efficiency loss is derived under an assumption that oscillators share the same frequency. Through numerical case studies, the analytical statements are illustrated in the homogeneous frequency regime; analogous numerical results are provided for the heterogeneous frequency regime.

Appendix

1.1 A.1 Proof of Lemma 1

Proof

Under Assumption 2, Eq. (11b) can be written as

$$\begin{aligned} (\omega-a)\partial_{\theta}p = -\partial_{\theta}[pu] + \frac {\sigma^2}{2}\partial^2_{\theta\theta}p. \end{aligned}$$

Integrating both sides of the equation with respect to θ,

$$\begin{aligned} u = \frac{\sigma^2}{2}\frac{\partial_{\theta}p}{p} + (a-\omega) + \frac{K}{p}, \end{aligned}$$

(48)

where K is a function of ω and is obtained as follows. Integrating both sides of the resulting equation (48) from 0 to 2π with respect to θ again, we obtain

$$\begin{aligned} \int_0^{2\pi} u \,\mathrm{d} \theta= \int _0^{2\pi}\frac{\sigma ^2}{2}\partial_{\theta} \ln p \,\mathrm{d} \theta+ K\int_0^{2\pi} \frac{1}{p}\,\mathrm{d} \theta+ (a-\omega)2\pi. \end{aligned}$$

From the assumption that h (thus u) and p are 2π-periodic in θ,

$$\begin{aligned} 0 = 0 + K\int_0^{2\pi}\frac{1}{p}\, \mathrm{d} \theta+ (a-\omega )2\pi\quad \Rightarrow\quad K = \frac{(\omega-a)2\pi}{\int_0^{2\pi}\frac{1}{p}\,\mathrm{d} \theta}. \end{aligned}$$

(49)

Finally, we get the result (7) by substituting K in (49) back to (48). □

1.2 A.2 Proof of Lemma 2

We consider the functional I[v]=I ₁[v]+I ₂[v]+I ₃[v] where

$$\begin{aligned} I_1[v] &= \int_0^{2\pi} \bar{c}v^2\,\mathrm{d} \theta, \\ I_2[v] &= \int_0^{2\pi} \biggl( \frac{R\sigma^4}{2}(\partial _{\theta}v)^2 - \lambda v^2 \biggr)\,\mathrm{d} \theta, \\ I_3[v] &= \int_0^{2\pi} \biggl( \frac{R}{2}(\omega-a)^2v^2 \biggl(1- \frac{2\pi}{v^2\int v^{-2}} \biggr)^2 \biggr)\,\mathrm{d} \theta, \end{aligned}$$

and derive its first variation. For I ₁[v],

$$\begin{aligned} DI_1[v]\cdot v' = \int_0^{2\pi}2 \bar{c}v\cdot v'\,\mathrm{d} \theta. \end{aligned}$$

(50)

For I ₂[v],

$$\begin{aligned} DI_2[v]\cdot v' = \int_0^{2\pi} { \bigl(-R\sigma^4\partial ^2_{\theta\theta}v \cdot v' - 2\lambda v\cdot v' \bigr)}\,\mathrm{d} \theta. \end{aligned}$$

(51)

A straightforward calculation gives

$$\begin{aligned} DI_3[v]\cdot v' &= \lim_{\epsilon\to0} \frac{I_3[v+\epsilon v'] - I_3[v]}{\epsilon} \\ &= \int_0^{2\pi} R(\omega-a)^2 \biggl(1- \biggl(\frac{2\pi}{v^2\int \frac{1}{v^2}} \biggr)^2 \biggr) v\cdot v' \,\mathrm{d} \theta. \end{aligned}$$

(52)

Using (50)–(52), we have obtain the nonlinear problem (15). Finally, (16) is the same constraint as (14).

1.3 A.3 Proof of Lemma 3

Proof

The proof of the first half is same as Lemma 2. It remains to show that $\lambda^{*}(\omega,a) = \eta_{g}^{*}(\omega,a)$. Multiplying both sides of (15) with $\frac {R\sigma^{4} v^{*}}{2}$ and integrating from 0 to 2π with respect to θ, we obtain

$$\begin{aligned} &\int_0^{2\pi} \biggl[ \frac{R\sigma^4}{2}v^* \partial^2_{\theta \theta}v^* + \bigl(\lambda^* - \bar{c}^*\bigr) \bigl(v^*\bigr)^2 \\ & \quad{} - \frac{R}{2}(\omega- a)^2 \biggl( 1 - \biggl( \frac{2\pi}{(v^*)^2\int(v^*)^{-2}} \biggr)^2 \biggr) \bigl(v^*\bigr)^2 \biggr] \,\mathrm{d} \theta= 0. \end{aligned}$$

Because $\int_{0}^{2\pi} (v^{*})^{2} \,\mathrm{d} \theta= 1$,

$$\begin{aligned} \lambda^* &= \int_0^{2\pi} \biggl[ - \frac{R\sigma^4}{2}v^*\partial ^2_{\theta\theta} v^* + \bar{c}^* \bigl(v^*\bigr)^2 + \frac{R}{2}(\omega- a)^2 \biggl( 1 - \biggl(\frac{2\pi}{(v^*)^2\int(v^*)^{-2}} \biggr)^2 \biggr) \bigl(v^* \bigr)^2 \biggr] \,\mathrm{d} \theta \\ &= -\frac{R\sigma^4}{2}v^*\partial_{\theta}v^* |_0^{2\pi} + \int_0^{2\pi} \frac{R\sigma^4}{2}\bigl( \partial_{\theta}v^*\bigr)^2 \,\mathrm{d} \theta \\ &\quad\ {}+ \int_0^{2\pi} \biggl[\bar{c}^* \bigl(v^* \bigr)^2 + \frac{R}{2}(\omega- a)^2 \biggl( 1 - \biggl(\frac{2\pi}{(v^*)^2\int (v^*)^{-2}} \biggr)^2 \biggr) \bigl(v^* \bigr)^2 \biggr] \,\mathrm{d} \theta \\ &= \int_0^{2\pi} \biggl[ \frac{R\sigma^4}{2}\bigl( \partial_{\theta }v^*\bigr)^2 + \bar{c} ^* \bigl(v^* \bigr)^2 + \frac{R}{2}(\omega- a)^2 \biggl( 1 - \biggl(\frac{2\pi}{(v^*)^2\int (v^*)^{-2}} \biggr)^2 \biggr) \bigl(v^* \bigr)^2 \biggr] \,\mathrm{d} \theta \\ &= \int_0^{2\pi} \biggl[ \frac{R\sigma^4}{2}\bigl( \partial_{\theta }v^*\bigr)^2 + \bar{c} ^* \bigl(v^* \bigr)^2 + \frac{R}{2}(\omega- a)^2 \biggl( 1 - \frac{2\pi}{(v^*)^2\int (v^*)^{-2}} \biggr)^2 \bigl(v^*\bigr)^2 \biggr] \, \mathrm{d} \theta , \end{aligned}$$

(53)

where the second equality is obtained through integration by parts of the first term, and the third equality is obtained because v ^∗ is periodic function with period 2π. From definition (13), the right-hand side of (53) is $\eta_{g}^{*}(\omega,a)$. □

1.4 A.4 Proof of Proposition 1

Proof

(i) Let (h,p,η ^∗) be a solution to the PDE (11a)–(11c). Then the optimal control is given by $u^{*} = -\frac{1}{R}\partial_{\theta}h$. We have shown that u ^∗ also satisfies (7). Therefore, we obtain

$$\begin{aligned} \partial_{\theta}h = -\frac{R\sigma^2}{2}\partial_{\theta}\ln p - R(a-\omega) \biggl(1 - \frac{2\pi}{p\int p^{-1}} \biggr). \end{aligned}$$

(54)

Taking partial derivatives with respect to θ on both sides of (54), we obtain

$$\begin{aligned} \partial^2_{\theta\theta}h = -\frac{R\sigma^2}{2}\frac{p\partial ^2_{\theta\theta}p - (\partial_{\theta} p)^2}{p^2} - R(a-\omega)\frac{2\pi}{p\int p^{-1}}\frac{\partial _{\theta}p}{p}. \end{aligned}$$

(55)

Let $v = \sqrt{p}$, then

$$ \begin{aligned} \partial_{\theta}h &= - R\sigma^2 \frac{\partial_{\theta}v}{v} - R(a-\omega) \biggl(1 - \frac{2\pi}{v^2\int v^{-2}} \biggr), \\ \partial^2_{\theta\theta}h &= - R\sigma^2 \frac{\partial^2_{\theta \theta}v}{v} + R\sigma^2 \biggl(\frac{\partial_{\theta}v}{v} \biggr)^2 - 2R(a-\omega)\frac{2\pi}{v^2\int v^{-2}}\frac{\partial_{\theta}v}{v}. \end{aligned} $$

(56)

From the Assumption 2 and Eq. (11a),

$$\begin{aligned} (\omega-a)\partial_{\theta}h = \frac{1}{2R}(\partial_{\theta}h)^2 - \bar{c}+ \eta^* - \frac{\sigma^2}{2}\partial^2_{\theta\theta}h. \end{aligned}$$

(57)

Substituting (56) into (57), we obtain the left-hand side (LHS) of (57) as

$$\begin{aligned} -R\sigma^2(\omega-a)\frac{\partial_{\theta}v}{v} + R( \omega-a)^2 \biggl( 1- \frac{2\pi}{v^2\int v^{-2}} \biggr), \end{aligned}$$

and the right-hand side (RHS) of (57) as

$$\begin{aligned} \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} +\frac{R}{2}( \omega-a)^2 \biggl(1-\frac{2\pi}{v^2\int v^{-2}} \biggr)^2 - R \sigma^2(\omega-a)\frac{\partial_{\theta}v}{v} + \bigl(\eta^* - \bar{c}\bigr). \end{aligned}$$

So Eq. (57) becomes

$$\begin{aligned} 0 &= \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} +\frac{R}{2}( \omega-a)^2 \biggl(1-\frac{2\pi}{v^2\int v^{-2}} \biggr)^2 - R( \omega-a)^2 \biggl( 1- \frac{2\pi}{v^2\int v^{-2}} \biggr) + \bigl(\eta ^* - \bar{c}\bigr) \\ &= \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} -\frac{R}{2}( \omega-a)^2 \biggl(1-\frac{2\pi}{v^2\int v^{-2}} \biggr) \biggl(1+ \frac{2\pi}{v^2\int v^{-2}} \biggr) + \bigl(\eta^* - \bar{c}\bigr). \end{aligned}$$

(58)

Multiplying both sides of (58) with $\frac{2v}{R\sigma^{4}}$, one obtains the nonlinear equation (15). Finally, (16) is just the constraint for density function p=v ², and (17) is the same as (11c) under Assumption 2.

(ii) First multiplying both sides of (21) with $\frac{p}{R}$ and do a partial derivative with respect to θ, one obtains

$$\begin{aligned} \partial_{\theta} \biggl[\frac{p}{R}(\partial_{\theta}h) \biggr]&= -\frac{\sigma^2}{2}\partial^2_{\theta\theta}p+ \partial_{\theta } \biggl[(\omega-a) \biggl( 1- \frac {2\pi}{p\int_0^{2\pi}p^{-1}\,\mathrm{d} \theta}\biggr)p \biggr] \\ &= -\frac{\sigma^2}{2}\partial^2_{\theta\theta}p+ (\omega -a) \partial_{\theta} \biggl[p-\frac {2\pi}{\int_0^{2\pi}p^{-1}\,\mathrm{d} \theta} \biggr] \\ &= -\frac{\sigma^2}{2}\partial^2_{\theta\theta}p+ (\omega -a) \partial_{\theta}p, \end{aligned}$$

which gives

$$\begin{aligned} (\omega-a)\partial_{\theta}p= \frac{1}{R}\partial_{\theta} \bigl[p(\partial_{\theta}h) \bigr] + \frac{\sigma^2}{2}\partial^2_{\theta\theta}p. \end{aligned}$$

Since p(θ,t;ω)=v ²(θ−at;ω),

$$\partial_{t}p + \omega\partial_{\theta}p = (\omega-a)\partial _{\theta}p= \frac{1}{R}\partial_{\theta} \bigl[p( \partial_{\theta}h) \bigr] + \frac{\sigma^2}{2}\partial^2_{\theta\theta}p, $$

which gives (11b).

Next, substituting p(θ,t;ω)=v ²(θ−at;ω) into (21), one obtains

$$\begin{aligned} \partial_{\theta}h &= - R\sigma^2\frac{\partial_{\theta}v}{v} + R( \omega-a) \biggl(1 - \frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr). \end{aligned}$$

(59)

So

$$\begin{aligned} \partial_{t}h +\omega\partial_{\theta}h &=(\omega-a)\partial _{\theta}h = - R\sigma^2(\omega-a) \frac{\partial_{\theta}v}{v} + R( \omega-a)^2 \biggl(1 - \frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr), \end{aligned}$$

(60)

$$\begin{aligned} (\partial_{\theta}h )^2 &= R^2\sigma^4 \biggl(\frac {\partial_{\theta} v}{v} \biggr)^2 + R^2( \omega-a)^2 \biggl(1-\frac{2\pi}{v^2\int_0^{2\pi}\,\mathrm{d} \theta } \biggr)^2 \\ &\quad\ {}- 2R^2\sigma^2(\omega- a)\frac{\partial_{\theta }v}{v} \biggl( 1 - \frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr), \end{aligned}$$

(61)

$$\begin{aligned} \partial^2_{\theta\theta}h &= -R\sigma^2 \frac{\partial^2_{\theta \theta}v}{v}+ R\sigma^2 \biggl(\frac {\partial_{\theta}v}{v} \biggr)^2 + 2 R(\omega-a) \frac{2\pi}{v^2\int_0^{2\pi} v^{-2}\,\mathrm{d} \theta} \frac{\partial_{\theta}v}{v}. \end{aligned}$$

(62)

Multiplying both sides of (61) with $\frac{1}{2R}$, those of (62) with $-\frac{\sigma^{2}}{2}$ and adding them together, one obtains

$$\begin{aligned} \frac{1}{2R} (\partial_{\theta}h )^2-\frac{\sigma ^2}{2} \partial^2_{\theta\theta}h &= \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} + \frac{R}{2}(\omega-a)^2 \biggl(1 - \frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta } \biggr)^2 - R\sigma^2(\omega-a)\frac{\partial_{\theta}v}{v}. \end{aligned}$$

(63)

Multiplying both sides of (15) with $\frac{R\sigma^{4}}{2v}$, one obtains

$$\begin{aligned} \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} + \eta^* - \bar{c} -\frac{R}{2}( \omega-a)^2 \biggl(1 - \biggl(\frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr)^2 \biggr) = 0, \end{aligned}$$

which gives

$$\begin{aligned} \frac{R\sigma^4}{2}\frac{\partial^2_{\theta\theta}v}{v} &= - \bigl(\eta^* - \bar{c}\bigr) + \frac{R}{2}(\omega-a)^2 \biggl(1 - \biggl(\frac{2\pi}{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr)^2 \biggr). \end{aligned}$$

(64)

Substituting (64) into (63), one obtains

$$\begin{aligned} \frac{1}{2R} (\partial_{\theta}h )^2-\frac{\sigma ^2}{2} \partial^2_{\theta\theta}h &= -\bigl(\eta^* - \bar{c}\bigr) + R( \omega-a)^2 \biggl(1 - \frac{2\pi }{v^2\int_0^{2\pi}v^{-2}\,\mathrm{d} \theta} \biggr) - R \sigma^2(\omega-a) \frac{\partial_{\theta}v}{v}\\ &= -\bigl(\eta^* - \bar{c}\bigr) + \partial_{t}h + \omega \partial_{\theta}h, \end{aligned}$$

where the last equality comes from (60). Rearranging the last equation, one obtains

$$\begin{aligned} \partial_{t}h + \omega\partial_{\theta}h &= \frac{1}{2R} (\partial_{\theta}h )^2 - \bar{c}+ \eta^* -\frac{\sigma^2}{2} \partial^2_{\theta\theta}h, \end{aligned}$$

(65)

which gives (11a). Finally, (11c) is obtained from (17) under Assumption 2. □

1.5 A.5 Proof of Lemma 4

Proof

The Euler–Lagrange equation (24) is obtained from considering the first variation of (22)–(23), which can be derived in a fashion similar to that in Lemma 2. Comparing equation (22) with (13), the only difference in the integrand is the first term: the latter is $\bar{c}v^{2}$ and the former is $\mathcal{C} [v] v^{2}$. So we derive its first variation here as

$$\begin{aligned} DI_1[v]\cdot v' &= \int_{\varOmega}\int _0^{2\pi}2\mathcal{C}[v] v\cdot v'\, \mathrm{d} \theta g(\omega)\,\mathrm{d} \omega \end{aligned}$$

(66)

$$\begin{aligned} &\quad\ {}+ \int_{\varOmega}\int_0^{2\pi} \biggl(v^2(\theta;\omega)\int_{\varOmega}\int _0^{2\pi}c^\bullet(\theta-\vartheta) 2 v \bigl(\vartheta; \omega'\bigr) \\ &\quad\ {}\cdot v'\bigl(\vartheta; \omega'\bigr) \,\mathrm{d} \vartheta g\bigl(\omega' \bigr)\,\mathrm{d} \omega' \biggr) \,\mathrm{d} \theta g(\omega )\, \mathrm{d} \omega \end{aligned}$$

(67)

$$\begin{aligned} &=:DI_{11}[v] \cdot v' + DI_{12}[v] \cdot v'. \end{aligned}$$

(68)

Note the integrand of (66) is same as that of (50). Since c ^•(⋅) is even, (67) can be written as

$$\begin{aligned} DI_{12}[v]\cdot v' &= \int_{\varOmega}\int _0^{2\pi} \biggl(v^2(\theta ;\omega)\int _{\varOmega}\int_0^{2\pi}c^\bullet( \vartheta-\theta) 2 v\bigl(\vartheta; \omega'\bigr) \\ &\quad\ {}\cdot v'\bigl(\vartheta;\omega'\bigr)\,\mathrm{d} \vartheta g\bigl(\omega'\bigr)\,\mathrm{d} \omega' \biggr)\, \mathrm{d} \theta g(\omega)\,\mathrm{d} \omega \end{aligned}$$

(69)

$$\begin{aligned} &= \int_{\varOmega}\int_0^{2\pi} \biggl(v^2\bigl(\vartheta;\omega'\bigr)\int _{\varOmega}\int_0^{2\pi}c^\bullet( \theta-\vartheta) 2 v(\theta; \omega) \\ &\quad\ {}\cdot v'(\theta;\omega)\, \mathrm{d} \theta g(\omega)\,\mathrm{d} \omega \biggr)\,\mathrm{d} \vartheta g \bigl(\omega'\bigr)\,\mathrm{d} \omega ' \end{aligned}$$

(70)

$$\begin{aligned} &= \int_{\varOmega}\int_0^{2\pi} \biggl(\int_{\varOmega}\int_0^{2\pi }c^\bullet( \theta-\vartheta) v^2\bigl(\vartheta;\omega'\bigr) \, \mathrm{d} \vartheta g\bigl(\omega'\bigr)\,\mathrm{d} \omega' \biggr) 2 v(\theta; \omega) \\ &\quad\ {}\cdot v'(\theta; \omega)\,\mathrm{d} \theta g(\omega)\,\mathrm{d} \omega \end{aligned}$$

(71)

$$\begin{aligned} &=\int_{\varOmega}\int_0^{2\pi} \mathcal{C}[v](\theta) 2 v(\theta ;\omega )\cdot v'(\theta;\omega)\, \mathrm{d} \theta g(\omega)\,\mathrm{d} \omega = DI_{11}[v] \cdot v', \end{aligned}$$

(72)

where (70) is obtained by switching variable θ with ϑ and ω with ω′ in (69), (71) is obtained by rearrangement of (70), and (72) is obtained from definition of $\mathcal{C}[v]$ in (17). So we obtain

$$\begin{aligned} DI_1[v] \cdot v'= \int_{\varOmega}\int _0^{2\pi}4\mathcal{C}[v] v\cdot v'\, \mathrm{d} \theta g(\omega)\,\mathrm{d} \omega, \end{aligned}$$

where the integrand is as twice as that in (50), which leads to the difference between (24) and (15).

Multiplying both sides of (24) by $\frac{\sigma^{4} Rv}{2}$ and integrating from 0 to 2π, we obtain the following:

$$\begin{aligned} \lambda^*(\omega,a) &= \int_0^{2\pi} \frac{R\sigma ^4}{2}\bigl(\partial_{\theta}v^*\bigr)^2 + 2 \mathcal{C}\bigl[v^*\bigr] \bigl(v^*\bigr)^2 \\ &\quad\ {}+ \frac{R}{2}( \omega-a)^2 \biggl( 1 - \biggl(\frac{2\pi }{(v^*)^2\int (v^*)^{-2}} \biggr)^2 \biggr) \bigl(v^*\bigr)^2 \,\mathrm{d} \theta \\ &= \int_0^{2\pi} \frac{R\sigma^4}{2}\bigl( \partial_{\theta }v^*\bigr)^2 + 2 \mathcal{C}\bigl[v^*\bigr] \bigl(v^*\bigr)^2 + \frac{R}{2}(\omega-a)^2 \biggl( 1 - \frac{2\pi}{(v^*)^2\int (v^*)^{-2}} \biggr)^2 \bigl(v^*\bigr)^2 \, \mathrm{d} \theta. \end{aligned}$$

Taking expectations on both sides, we obtain the result (26). □

1.6 A.6 Proof of Lemma 5

Equation (32) is rewritten as

$$ \sigma^4 R \partial^2_{\theta\theta}v + 2 \biggl( \lambda- \alpha \int_0^{2\pi} c^\bullet( \theta,\vartheta) v^2(\vartheta) \,\mathrm{d} \vartheta \biggr)v = 0. $$

(73)

We substitute the expansion (36) into (73) and the normalization condition ∫v ² dθ=1, and collect the terms according to different orders of x.

At O(1), we have the steady state solution

$$\begin{aligned} v_0 = \frac{1}{\sqrt{2\pi}},\quad\quad\lambda_0 = \alpha C_0^\bullet= \frac{\alpha}{2\pi}\int_0^{2\pi} c^\bullet(\theta-\vartheta) \,\mathrm{d} \vartheta. \end{aligned}$$

At O(x),

$$\begin{aligned} 0&=\sigma^4r_0 \partial^2_{\theta\theta}v_1 + 2v_0 \biggl( \lambda _1 - \alpha\int _0^{2\pi}c^\bullet(\theta- \vartheta)2v_0v_1(\vartheta) \,\mathrm{d} \vartheta \biggr), \end{aligned}$$

(74)

$$\begin{aligned} 0&=\int_0^{2\pi} v_1(\theta)\, \mathrm{d} \theta. \end{aligned}$$

(75)

Suppose we have the Fourier expansion for the function v ₁(θ)

$$\begin{aligned} v_1(\theta) = \sum_k v_{1k} e^{ik\theta}. \end{aligned}$$

(76)

Substitute (76) into (75),

$$\begin{aligned} \int_0^{2\pi} v_{10}\,\mathrm{d} \theta= 0 \quad\Rightarrow\quad v_{10} = 0. \end{aligned}$$

Substitute (76) into (74) to obtain

$$\begin{aligned} \sum_{k} \bigl( - k^2 \sigma^4 r_0 - 8\alpha\pi v_0^2 C_k^\bullet \bigr) v_{1k} e^{ik\theta} + 2v_0\lambda_1 = 0. \end{aligned}$$

We collect the terms with respect to e ^ikθ. When k=0,

$$\begin{aligned} - 8\alpha\pi v_0^2 C_0^\bullet v_{10} + 2v_0 \lambda_1 = 0 \quad \Rightarrow \quad \lambda_1 = 0 \text{ since } v_{10} = 0. \end{aligned}$$

When k=1, $(-\sigma^{4}r_{0} - 8\alpha\pi v_{0}^{2}C_{1}^{\bullet}) v_{11} = 0$. If v ₁₁≠0,

$$r_0 = -\frac{8\alpha\pi v_0^2 C_1^\bullet}{\sigma^4} = \frac{\alpha}{2\sigma^4} = R_c^\alpha. $$

When k≥2, $C_{k}^{\bullet}= 0$,

$$-k^2\sigma^4r_0 v_{1k} = 0 \quad \Rightarrow\quad v_{1k} = 0. $$

When k<0, it is similar as k>0. The existence of bifurcation implies v ₁≠0, so $v_{11} = \bar{v}_{1,-1} \neq0$. So we obtain

$$\begin{aligned} v_1 &= v_{11}e^{i\theta} + \mathrm{c.c} = 2|v_{11}|\cos(\theta+ \angle v_{11}), \end{aligned}$$

(77)

where |v ₁₁| and ∠v ₁₁ are the amplitude and phase angle, respectively, of v ₁₁.

At O(x ²),

$$\begin{aligned} &\sigma^4r_0 \partial^2_{\theta\theta}v_2 + \sigma^4r_1 \partial ^2_{\theta\theta}v_1 \\ &\quad{}+ 2 v_0 \biggl(\lambda_2 - \alpha \int _0^{2\pi} c^\bullet(\theta-\vartheta) \bigl(v_1^2(\vartheta) + 2v_0 v_2( \vartheta)\bigr)\,\mathrm{d} \vartheta \biggr) \\ &\quad{}- 4\alpha v_0v_1(\theta)\int_0^{2\pi} c^\bullet(\theta-\vartheta )v_1(\vartheta) \,\mathrm{d} \vartheta= 0, \end{aligned}$$

(78)

$$\begin{aligned} &\int_0^{2\pi}v_1^2(\theta) + 2v_0v_2(\theta) \,\mathrm{d} \theta= 0. \end{aligned}$$

(79)

Suppose v ₂(θ) also has the Fourier expansion

$$\begin{aligned} v_2(\theta) = \sum_k v_{2k} e^{ik\theta}. \end{aligned}$$

(80)

Substitute (76) and (80) into (79),

$$\begin{aligned} v_{11}v_{1,-1} + v_0 v_{2,0} = 0, \quad\text{or}\quad v_{2,0} = - \frac{v_{11}v_{1,-1}}{v_0}. \end{aligned}$$

(81)

Substitute (76) and (80) into (78),

We collect the terms of e ^ikθ for different values of k. When k=0,

$$\begin{aligned} 0 & = -8\alpha\pi v_0^2 C_0^\bullet v_{20} - 4\alpha\pi v_0 C_0^\bullet(2v_{11}v_{1,-1}) \\ &\quad\ {}- 8\alpha\pi v_0 \bigl(v_{11}v_{1,-1}C_{-1}^\bullet + v_{1,-1}v_{11}C_1^\bullet \bigr) + 2v_0\lambda_2, \\ &\quad\Rightarrow\quad -8\alpha\pi v_0^2 C_0^\bullet v_{2,0} + 2v_0\lambda_2 =0, \\ &\quad\Rightarrow\quad \lambda_2 = \frac{8\alpha\pi v_0^2 C_0^\bullet v_{2,0}}{2v_0} = 4\alpha\pi v_0 C_0^\bullet \biggl(-\frac{v_{11}v_{1,-1}}{v_0} \biggr) \\ & = -4\alpha\pi C_0^\bullet v_{11}v_{1,-1} = -\alpha\pi v_{11} v_{1,-1}. \end{aligned}$$

When k=1, −σ ⁴ r ₁ v ₁₁=0,⇒r ₁=0. When k=2,

$$\begin{aligned} -4\sigma^4 r_0 v_{2,2} - 8\alpha\pi v_0 v_{11}^2 C_1^\bullet= 0,\quad \Rightarrow\quad v_{2,2} &= -\frac{2\alpha\pi v_0}{\sigma^4 r_0} v_{11}^2C_1^\bullet \\ &= \frac{1}{2}\pi v_0 v_{11}^2. \end{aligned}$$

When k>2, v _2k =0. For k<0, it is similar. So we obtain

$$\begin{aligned} v_2 &= \frac{1}{2} v_{20} + v_{21}e^{i\theta} + v_{22}e^{i2\theta} + \mathrm{c.c} \\ &= v_{20} + v_0\pi|v_{11}|^2\cos2( \theta+ \angle v_{11}) + 2|v_{21}|\cos(\theta+ \angle v_{21}) \end{aligned}$$

(82)

At O(x ³),

$$\begin{aligned} &\sigma^4r_0 \partial^2_{\theta\theta}v_3 + \sigma^4 r_2 \partial ^2_{\theta\theta}v_1 \\ &\quad{} + 2v_0 \biggl( \lambda_3 - \alpha\int c^\bullet( \theta- \vartheta ) \bigl(2v_0 v_3(\vartheta) + 2v_1(\vartheta) v_2(\vartheta)\bigr)\,\mathrm{d} \vartheta \biggr) \\ &\quad{}+ 2v_1(\theta) \biggl( \lambda_2 - \alpha\int c^\bullet(\theta- \vartheta) \bigl( v_1^2(\theta) + 2v_0 v_2(\vartheta)\bigr)\,\mathrm{d} \vartheta \biggr) \\ &\quad{}- 4\alpha v_0 v_2(\theta)\int c^\bullet(\theta- \vartheta)v_1(\vartheta)\,\mathrm{d} \vartheta= 0, \end{aligned}$$

(83)

$$\begin{aligned} &\int v_0 v_3(\theta) + v_1( \theta)v_2(\theta)\,\mathrm{d} \theta= 0. \end{aligned}$$

(84)

Suppose v ₃(θ) has the Fourier expansion

$$\begin{aligned} v_3(\theta) = \sum_k v_{3k}e^{ik\theta}. \end{aligned}$$

(85)

Substitute (76), (80), and (85) into (84),

$$\begin{aligned} v_0 v_{3,0} + v_{11}v_{2,-1} + v_{1,-1}v_{2,1} = 0. \end{aligned}$$

Substitute (76), (80), and (85) into (84),

$$\begin{aligned} &\sum_{k} \biggl\{ \bigl(-k^2\bigr) \sigma^4r_0 v_{3k} - \sigma^4r_2 k^2 v_{1k} + 2\lambda_2 v_{1k} \\ &\quad{}- 8\alpha\pi v_0 C_k^\bullet \biggl(\sum _{m+l=k} v_{1m}v_{2l} + v_0v_{3k} \biggr) \\ &\quad{}- 4\alpha\pi\sum_{m+l = k}v_{1m} C_l^\bullet \biggl(\sum_{a+b=l} v_{1a}v_{1b} + 2v_0v_{2l} \biggr) \\ &\quad{}- 8\alpha\pi v_0\sum_{m+l=k}v_{2m}C_l^\bullet v_{1l} \biggr\} e^{ik\theta} + 2v_0 \lambda_3 = 0. \end{aligned}$$

(86)

We collect the terms of e ^ikθ for different values of k. When k=0,

$$\begin{aligned} 0 &= - 8\alpha\pi v_0 C_0^\bullet(v_{11}v_{2,-1} + v_{1,-1}v_{21} + v_0v_{30}) \\ &\quad\ {}- 4\alpha\pi \bigl(v_{11} C_{-1}^\bullet(2v_0 v_{2,-1}) + v_{1,-1}C_1^\bullet 2v_0v_{21} \bigr) \\ &\quad\ {}- 8\alpha\pi v_0 (v_{21}C_{-1}^\bullet v_{1,-1} + v_{2,-1}C_1^\bullet v_{11} + 2v_0\lambda_3, \\ &\quad\Rightarrow\quad \lambda_3 = \alpha\pi( v_{11}v_{2,-1} + v_{1,-1}v_{2,1}). \end{aligned}$$

When k=1,

$$\begin{aligned} 0 &= -\sigma^4r_0 v_{31} - \sigma^4 r_2 v_{11} + 2\lambda_2 v_{11} \\ &\quad\ {}- 8\alpha\pi v_0\bigl(2C_1^\bullet v_{11}v_{20} + 2C_1^\bullet v_{1,-1}v_{22} + C_1^\bullet v_0 v_{31}\bigr), \\ &\quad\Rightarrow\quad r_2 = -\frac{7\alpha}{2\sigma^4}\pi v_{11}v_{1,-1}. \end{aligned}$$

In all, we obtain the formula

$$\begin{aligned} R &= r_0 + x^2 r_2 + o\bigl(x^2 \bigr) = r_0 - \frac{7\alpha}{2\sigma^4}\pi |v_{11}|^2x^2 + o\bigl(x^2\bigr), \end{aligned}$$

(87)

$$\begin{aligned} \lambda&= \lambda_0 + x^2 \lambda_2 + o \bigl(x^2\bigr) = \lambda_0 - \alpha\pi|v_{11}|^2x^2 + o\bigl(x^2\bigr), \end{aligned}$$

(88)

$$\begin{aligned} v &= v_0 + x v_1 + x^2 v_2 + o \bigl(x^2\bigr) \\ &= v_0 + x|v_{11}|2\cos(\theta+ \angle v_{11}) + x^2\bigl(-\sqrt{2\pi}|v_{11}|^2 \\ &\quad\ {}+ \pi v_0|v_{11}|^2\cos2(\theta+\angle v_{11}) + 2|v_{21}|\cos(\theta+\angle v_{11}) \bigr) + o\bigl(x^2\bigr) \\ &= v_0 + 2\cos(\theta+ \angle v_{11})|v_{11}|x \\ &\quad\ {}+ \bigl(-\sqrt{2\pi} + v_0\pi\cos2(\theta+ \angle v_{11}) \bigr)|v_{11}|^2x^2 + o \bigl(x^2\bigr). \end{aligned}$$

(89)