 Solution

At a locked state choose the rotating frame with ${\theta }_i^{\prime }=0$. For small phase differences, Taylor expansion gives $h({\theta }_j-{\theta }_i)-h(0)=h^{\prime }(0)({\theta }_j-{\theta }_i)+O(|{\theta }_j-{\theta }_i{|}^2)$. Let the graph Laplacian have $L_{ii}={\sum }_j{d}_{ij}$ and $L_{ij}=-d_{ij}$ for $i\ne j$. The leading locked equations areIf coupling is balanced, ${\sum }_j{d}_{ij}={\sum }_j{d}_{ji}$ for every $i$, so $1^{\mathsf{T}}L=0$. Summing over cells cancels all linear coupling terms and givesIn particular symmetric reciprocal gap-junction conductances satisfy this condition. To leading order the collective physical frequency is $1+ϵN^{-1}{\sum }_i{\xi }_i$.

The arithmetic-mean conclusion requires this extra balance assumption. The printed condition $d_{ii}=0$ alone is insufficient. For a connected directed coupling network let ${\pi }^{\mathsf{T}}L=0$, normalized by ${\pi }^{\mathsf{T}}1=1$. The valid weighted collective oscillator frequency result isFor two cells with $d_{12}=p>0$, $d_{21}=q>0$, one has $\pi =(q,p)/(p+q)$ and ${\Omega }_1=(q{\xi }_1+p{\xi }_2)/(p+q)$ at linear order. For example $p=1,q=2,{\xi }_1=0,{\xi }_2=\delta$ gives ${\Omega }_1=\delta /3$, rather than $\delta /2$, while the locked phase difference is $O(\delta )$ when $h^{\prime }(0)\ne 0$. This supplies a counterexample with arbitrarily small phase differences. It also identifies exactly why reciprocal coupling recovers the requested mean: the left null space is then spanned by the uniform vector.