 Solution

Use the period-one inner product $⟨Y,V⟩={\int }_0^{1}Y(\sigma )\cdot V(\sigma )d\sigma$. Periodic integration by parts gives ${\hat{L}}^{\ast }Y=-Y^{\prime }-DF(U{)}^{\mathsf{T}}Y$. Let $Y$ span this adjoint null space and normalize ${\int }_0^{1}Y\cdot U^{\prime }d\sigma =1$. For oscillator $i$, write $s=\sigma +{\theta }_i(\tau )$. Then ${\partial }_{\tau }{u}_{i,0}={\theta }_i^{\prime }{U}^{\prime }(s)$ and ${\partial }_{\sigma }{u}_{i,0}=U^{\prime }(s)$.

The Fredholm solvability condition makes the forcing in the order-$ϵ$ equation orthogonal to $Y(s)$. Its intrinsic and frequency terms integrate to ${\xi }_i-{\Omega }_1-{\theta }_i^{\prime }$, whereVoltage-only coupling contributesThus withthe complete phase equation isPeriodicity makes these formulae independent of the absolute initial phase. The adjoint function is a phase response function; it projects a small perturbation of each state component onto its accumulated phase change. The subtraction of $h(0)$ ensures zero coupling between identical phases, as required by the original voltage differences. The source's reference to its equation (1) is read as the uncoupled cycle equation, numbered (3) on the actual page.