Appendix A: Proof of Lemma 1

Let $\Psi_k(t)=\phi_k(t)-\theta_{1k}(t)$. Rearranging Eq. (7), we obtain

$$A_k(t)=S_k(t)\cos\Psi_k(t)-S_k(t)\cot\theta_k(t)\sin\Psi_k(t).$$ (36)

Differentiating both terms in t in the above equation, we obtain

$$y'(t)+\frac{P'(t)}{P(t)}y(t)=\frac{Q'(t)}{P(t)}-\frac{C_{k,R}(t)}{P(t)},$$ (37)

where $y(t)=\cot\theta_k(t)$, $P(t)=S_k(t)\sin\Psi_k(t)$, and $Q(t)=S_k(t)\cos\Psi_k(t)$. Since the above equation is a linear differential equation, a general solution y(t) is determined by

$$y(t)=\frac{1}{P(t)}\left(Q(t)-\int C_{k,R}(t)dt+C\right),$$ (38)

where C is an undermined coefficient. Hence, from

$$\cot\theta_{k}(t)=\frac{S_k(t)\cos\Psi_k(t)-\int C_{k,R}(t)dt+C}{S_k(t)\sin\Psi_k(t)},$$ (39)

we obtain

$$\theta_{k}(t)=\arctan\left(\frac{S_k(t)\sin(\phi_k(t)-\theta_{1k}(t))}{S_k(t)\cos(\phi_k(t)-\theta_{1k}(t)+C_k(t)}\right),$$ (40)

where $C_k(t)=-\int C_{k,R}(t)dt+C$. On the other hand, applying Eq. (36) into Eq. (40), we obtain

$$S_k(t)\left(\cos\Psi_k(t)-\cos\Psi_k(t)-\frac{C_k(t)}{S_k(t)}\right)$$ A_k(t) =

= -C_k(t). (41)

Considering that C=-C_k,0, that is the same result as integrating Eq. (9).