Lemma 23.2.401D02D96B9AF2

Proof. Suppose $k_2>k_1$, and put $R=C_2-C_1$. At a zero of $R$, the equation gives $R'=(k_2-k_1)C_1^{1-m}>0$. Such zeros are isolated, so there are finitely many on the circle. A periodic differentiable function cannot cross zero only upwards, since returning around the circle would require a downward crossing. Hence $R$ has no zero and a constant strict sign. The identity $\int _0^2C_ i\, \mathrm dr=-\int _0^2F\, \mathrm dr/k_ i$ shows that this sign is positive, and the formula for $L(C)$ in (23.2.39) then gives $L(C_2)>L(C_1)$.

For equal $k$, uniqueness for the smooth scalar equation $C'=kC^{1-m}+FC^{-m}$ on $C>0$ shows that two solutions which meet coincide. Two solutions which never meet are strictly ordered, which contradicts the equality of their integrals. This proves uniqueness for fixed $k$. $\square$