Prufer angle for the interval equation and strict monotonicity in the energy

[The angle identity] Put $R(x):=(u(x{)}^2+u^{\mathrm{\prime }}(x{)}^2{)}^{1\mathrm{/}2}$. Since $(u,u^{\mathrm{\prime }})(0)=(0,1)$, $R(0)=1$; and $R$ never vanishes: if $u(x_0)=u^{\mathrm{\prime }}(x_0)=0$ for some $x_0$, uniqueness in [F1] forces $u\equiv 0$, contradicting $u^{\mathrm{\prime }}(0)=1$. Define $A(x):=u(x)\cos \theta (x)-u^{\mathrm{\prime }}(x)\sin \theta (x);$ we show $A\equiv 0$. Differentiating and using ${\theta }^{\mathrm{\prime }}=(u^{\mathrm{\prime }2}-u{u}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}(u^2+u^{\mathrm{\prime }2})$, $D:=u^2+u^{\mathrm{\prime }2}$, $A^{\mathrm{\prime }}=u^{\mathrm{\prime }}\cos \theta -u\sin \theta {\theta }^{\mathrm{\prime }}-u^{\mathrm{\prime }\mathrm{\prime }}\sin \theta -u^{\mathrm{\prime }}\cos \theta {\theta }^{\mathrm{\prime }}=u^{\mathrm{\prime }}\cos \theta (1-{\theta }^{\mathrm{\prime }})-(u^{\mathrm{\prime }\mathrm{\prime }}+u{\theta }^{\mathrm{\prime }})\sin \theta \mathrm{.}$ Now $1-{\theta }^{\mathrm{\prime }}=(D-u^{\mathrm{\prime }2}+u{u}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}D=u(u+u^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}D$ and $u^{\mathrm{\prime }\mathrm{\prime }}+u{\theta }^{\mathrm{\prime }}=u^{\mathrm{\prime }\mathrm{\prime }}+\frac{u(u^{\mathrm{\prime }2}-u{u}^{\mathrm{\prime }\mathrm{\prime }})}{D}=\frac{u^{\mathrm{\prime }\mathrm{\prime }}D+u{u}^{\mathrm{\prime }2}-u^2{u}^{\mathrm{\prime }\mathrm{\prime }}}{D}=\frac{u^{\mathrm{\prime }2}(u^{\mathrm{\prime }\mathrm{\prime }}+u)}{D}$. Hence $A^{\mathrm{\prime }}=\frac{u^{\mathrm{\prime }}(u+u^{\mathrm{\prime }\mathrm{\prime }})}{D}(u\cos \theta -u^{\mathrm{\prime }}\sin \theta )=\frac{u^{\mathrm{\prime }}(u+u^{\mathrm{\prime }\mathrm{\prime }})}{D}A=:c(x)A$ with $c$ continuous, and $A(0)=0\cdot 1-1\cdot 0=0$; by [F2], $A(x)={\int }_0^{x}cA$, so $\mathrm{\mid }A(x)\mathrm{\mid }\le {\int }_0^x\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }A\mathrm{\mid }$ and [F4] with $a=0$ gives $A\equiv 0$. Therefore $u=R\sin \theta$ and $u^{\mathrm{\prime }}=R\cos \theta$: the rotation matrix identity $\left(\begin{array}{c}u\\ u^{\mathrm{\prime }}\end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}A\\ B\end{array}\right)$ with $B:=u\sin \theta +u^{\mathrm{\prime }}\cos \theta$ gives $u=B\sin \theta$, $u^{\mathrm{\prime }}=B\cos \theta$ and $B^2=R^2-A^2=R^2$; since $B$ is continuous, $B^2=R^2>0$ and $B(0)=u(0)\sin 0+u^{\mathrm{\prime }}(0)\cos 0=1=R(0)$, the continuous function $B\mathrm{/}R$ has square $1$ and value $1$ at $0$, so $B\mathrm{/}R\equiv 1$ and $B=R$. This proves the first part of (a).

[Differentiability in $E$] Let $R>0$ and consider $[0,L]\times [-R,R]$. By [F1] the maps $(x,E)\mapsto u,u^{\mathrm{\prime }}$ are continuous, so $(x,E)\mapsto q(x,E):=(u^{\mathrm{\prime }2}-u{u}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}(u^2+u^{\mathrm{\prime }2})$ is continuous on the compact set and, since $u^2+u^{\mathrm{\prime }2}$ is continuous and strictly positive there, its denominator has a positive lower bound. By [F1] $u,u^{\mathrm{\prime }}$ are differentiable in $E$ with derivatives $v,v^{\mathrm{\prime }}$ continuous on $[0,L]\times [-R,R]$; hence ${\mathrm{\partial }}_{E}q$ is continuous on the compact set, and the difference quotients of $q$ in $E$ converge uniformly to ${\mathrm{\partial }}_{E}q$. Since $\theta (x,E)={\int }_0^{x}q(t,E)dt$, the standard differentiation under the integral sign (difference quotients, [F2] for the inner integration, [F3] for the limit) gives ${\mathrm{\partial }}_E\theta (x,E)={\int }_0^x{\mathrm{\partial }}_{E}q(t,E)dt\mathrm{.}$

[The Prufer equation] From the integral definition, $\theta$ is $C^1$ in $x$ with $\theta (0)=0$ and ${\theta }^{\mathrm{\prime }}=(u^{\mathrm{\prime }2}-u{u}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}(u^2+u^{\mathrm{\prime }2})$. Substituting $u{u}^{\mathrm{\prime }\mathrm{\prime }}=u(V-E)u=(V-E)u^2$ and the identities of step 1.1, ${\theta }^{\mathrm{\prime }}=\frac{u^{\mathrm{\prime }2}+(E-V)u^2}{u^2+u^{\mathrm{\prime }2}}={\cos }^2\theta +(E-V){\sin }^2\theta ,$ which completes (a).

[Wronskian identity] Fix $E$ and let $v:={\mathrm{\partial }}_{E}u(\cdot ,E)$, so $-v^{\mathrm{\prime }\mathrm{\prime }}+Vv-Ev=u$ and $v(0)=v^{\mathrm{\prime }}(0)=0$ by [F1]. Then $(u^{\mathrm{\prime }}v-u{v}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=u^{\mathrm{\prime }\mathrm{\prime }}v-u{v}^{\mathrm{\prime }\mathrm{\prime }}=(V-E)uv-u((V-E)v-u)=u^2,$ using $u^{\mathrm{\prime }\mathrm{\prime }}=(V-E)u$ and $v^{\mathrm{\prime }\mathrm{\prime }}=(V-E)v-u$. Since $u(0)=v(0)=0$, $u^{\mathrm{\prime }}(x)v(x)-u(x)v^{\mathrm{\prime }}(x)={\int }_0^{x}u(t{)}^{2}dt(x\in [0,L])\mathrm{.}$

[Energy derivative] Differentiate the relations $u=r\sin \theta$ and $u^{\mathrm{\prime }}=r\cos \theta$ of step 1.1 in $E$: writing $\dot{r}={\mathrm{\partial }}_{E}r$ and $\dot{\theta }={\mathrm{\partial }}_E\theta$, $v=\dot{r}\sin \theta +r\cos \theta \dot{\theta },v^{\mathrm{\prime }}=\dot{r}\cos \theta -r\sin \theta \dot{\theta }\mathrm{.}$ Multiplying the first identity by $u^{\mathrm{\prime }}=r\cos \theta$ and the second by $-u=-r\sin \theta$ and adding eliminates $\dot{r}$ and gives $u^{\mathrm{\prime }}v-u{v}^{\mathrm{\prime }}=r^2\dot{\theta }$, so by step 2.2 ${\mathrm{\partial }}_E\theta (x,E)=\frac{u^{\mathrm{\prime }}(x)v(x)-u(x)v^{\mathrm{\prime }}(x)}{u(x{)}^2+u^{\mathrm{\prime }}(x{)}^2}=\frac{{\int }_0^{x}u(t,E{)}^{2}dt}{u(x,E{)}^2+u^{\mathrm{\prime }}(x,E{)}^2}\mathrm{.}$ The denominator is the strictly positive endpoint value $r(x,E{)}^2$, independent of the integration variable. The numerator has nonnegative integrand and is positive for every $x>0$: $u(t,E)=t+O(t^2)$ near $t=0$ because $u(0)=0$, $u^{\mathrm{\prime }}(0)=1$ and $u^{\mathrm{\prime }}$ is continuous, so $u(t,E{)}^2>0$ for small $t>0$. Hence ${\mathrm{\partial }}_E\theta (x,E)>0$ for every $x\in (0,L]$: $E\mapsto \theta (x,E)$ is strictly increasing, and it is continuous on $[0,L]\times \mathbb{R}$ by [F1] and step 1.2. This proves (b).

[Crossing structure] By (a), at a point with $\theta (x_0,E)=k\pi$ one has $\sin \theta =0$, $\cos \theta =\pm 1$, hence ${\mathrm{\partial }}_x\theta (x_0,E)={\cos }^2\theta =1>0$. Also $\theta (0,E)=0$ and ${\mathrm{\partial }}_x\theta (0,E)=1>0$, so $\theta (x,E)>0$ for all small $x>0$. We claim $\theta (x,E)>0$ for every $x\in (0,L]$: otherwise let $x_1$ be the first point in $(0,L]$ with $\theta (x_1,E)=0$; then $\theta >0$ on $(0,x_1)$ and continuity gives $\theta (x_1)=0$, so the left difference quotient of $\theta$ at $x_1$ is at most $0$, whence ${\mathrm{\partial }}_x\theta (x_1,E)\le 0$, contradicting ${\mathrm{\partial }}_x\theta (x_1,E)=1$. The same argument at level $k\pi$ shows that once $\theta$ has crossed the level $k\pi$ from below it stays strictly above: suppose $\theta (x_0,E)=k\pi$ and there is $x_2>x_0$ with $\theta (x_2,E)\le k\pi$; let $x_1:=\inf \{x>x_0:\theta (x,E)\le k\pi \}$. Then $\theta (x_1,E)=k\pi$ and $\theta >k\pi$ on $(x_0,x_1)$, so the left difference quotient at $x_1$ is $\le 0$ and ${\mathrm{\partial }}_x\theta (x_1,E)\le 0$, contradicting ${\mathrm{\partial }}_x\theta (x_1,E)=1$. Hence $\theta (x,E)>k\pi$ for every $x\in (x_0,L]$. If $0<k\pi <\theta (L,E)$, the intermediate value theorem applied to the continuous function $\theta (\cdot ,E)$ on $[0,L]$ gives some $x$ with $\theta (x,E)=k\pi$, and the just-proved strict monotonicity above the level makes this $x$ unique. This proves (c).