= Solution
Use the period-one inner product $\langle Y,V\rangle=\int_0^1Y(\sigma)\cdot V(\sigma)\,d\sigma$. Periodic <integration by parts> gives $\widehat L^*Y=-Y'-DF(U)^{\mathsf T}Y$. Let $Y$ span this adjoint <null space> and normalize $\int_0^1Y\cdot U'\,d\sigma=1$. For oscillator $i$, write $s=\sigma+\theta_i(\tau)$. Then $\partial_\tau u_{i,0}=\theta_i'U'(s)$ and $\partial_\sigma u_{i,0}=U'(s)$.
The <Fredholm solvability condition> makes the forcing in the order-$\epsilon$ equation orthogonal to $Y(s)$. Its intrinsic and <frequency> terms integrate to $\xi_i-\Omega_1-\theta_i'$, where
$$
\boxed{\xi_i=\int_0^1\sum_{a=1}^mY^a(s)G_i^a(U(s))\,ds.}
$$
Voltage-only coupling contributes
$$
\sum_jd_{ij}\int_0^1Y^1(s)[U^1(s+\theta_j-\theta_i)-U^1(s)]\,ds.
$$
Thus with
$$
\boxed{h(\phi)=\int_0^1Y^1(s)U^1(s+\phi)\,ds}
$$
the complete phase equation is
$$
\boxed{\theta_i'=\xi_i-\Omega_1+\sum_jd_{ij}[h(\theta_j-\theta_i)-h(0)].}
$$
Periodicity 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.
Back to article page
