Sturm comparison for scalar jacobi equations

The reduced Wronskian identity. [F1, given] Define $h(t):=u^{\mathrm{\prime }}(t)v(t)-u(t)v^{\mathrm{\prime }}(t)$ on $[0,L]$. By [F1] the product rule applies to both terms and the two differential equations give $h^{\mathrm{\prime }}=u^{\mathrm{\prime }\mathrm{\prime }}v+u^{\mathrm{\prime }}{v}^{\mathrm{\prime }}-u^{\mathrm{\prime }}{v}^{\mathrm{\prime }}-u{v}^{\mathrm{\prime }\mathrm{\prime }}=(-au)v-u(-bv)=(b-a)uv\mathrm{.}$ At the left endpoint, $h(0)=u^{\mathrm{\prime }}(0)v(0)-u(0)v^{\mathrm{\prime }}(0)=c\cdot 0-0\cdot c=0$.

Both solutions are positive immediately to the right of $0$, and the quotient has limit $1$. [F2, given] The functions $u^{\mathrm{\prime }}$ and $v^{\mathrm{\prime }}$ are continuous at $0$ with $u^{\mathrm{\prime }}(0)=v^{\mathrm{\prime }}(0)=c>0$, so [F2] yields $\delta >0$ with $u^{\mathrm{\prime }}\ge c\mathrm{/}2>0$ and $v^{\mathrm{\prime }}\ge c\mathrm{/}2>0$ on $[0,\delta ]$. For $0<t<\delta$ the mean value theorem in [F2] applied to $u$ and to $v$ on $[0,t]$ produces $\xi ,\eta \in (0,t)$ with $u(t)=u^{\mathrm{\prime }}(\xi )t>0$ and $v(t)=v^{\mathrm{\prime }}(\eta )t>0$. Hence the quotient $f:=u\mathrm{/}v$ is defined on $(0,\delta )$, and since $u(t)\mathrm{/}t\to u^{\mathrm{\prime }}(0)$ and $v(t)\mathrm{/}t\to v^{\mathrm{\prime }}(0)$ as $t\downarrow 0$ by the definition of the one-sided derivative at the endpoint, $f(t)=\frac{u(t)\mathrm{/}t}{v(t)\mathrm{/}t}⟶\frac{u^{\mathrm{\prime }}(0)}{v^{\mathrm{\prime }}(0)}=1(t\downarrow 0)\mathrm{.}$

The inequality propagates up to the supremum of the set of times where it holds. [F3, step 1.1, step 1.2] Put $A:=\{t\in (0,L]:u(s)\ge v(s)>0\text{ for every }s\in (0,t]\}\mathrm{.}$ This set is nonempty: on the initial interval of step 1.2 both solutions are positive, so step 1.1 gives $h^{\mathrm{\prime }}=(b-a)uv\ge 0$. The mean value theorem and $h(0)=0$ give $h\ge 0$ there. Therefore $(u\mathrm{/}v{)}^{\mathrm{\prime }}=h\mathrm{/}{v}^2\ge 0$, and the limit $u\mathrm{/}v\to 1$ gives $u\ge v$ throughout this initial interval. Any sufficiently small positive endpoint thus belongs to $A$. The set is bounded above by $L$, so $T:=\sup A$ exists in $\mathbb{R}$ by [F3] and $0<T\le L$. For every $s\in (0,T)$ choose $t\in A$ with $s<t$, which is possible because $T$ is the least upper bound of $A$; then $u(s)\ge v(s)>0$. Hence $uv\ge 0$ on $(0,T)$, and step 1.1 together with $a\le b$ gives $h^{\mathrm{\prime }}=(b-a)uv\ge 0$ on $(0,T)$. Since $h(0)=0$ by step 1.1, the mean value theorem in [F2] applied on $[0,s]$ yields $h(s)=h^{\mathrm{\prime }}(\xi )s\ge 0$ for every $s\in (0,T)$, so $h\ge 0$ on $(0,T)$. The denominator $v$ is positive on $(0,T)$, so the quotient rule in [F1] together with step 1.1 gives $f^{\mathrm{\prime }}=\frac{u^{\mathrm{\prime }}v-u{v}^{\mathrm{\prime }}}{v^2}=\frac{h}{v^2}\ge 0\text{on }(0,T),$ that is, $f$ is nondecreasing on $(0,T)$. For fixed $s\in (0,T)$ and $0<t<s$ this gives $f(s)\ge f(t)$, and letting $t\downarrow 0$ in the limit of step 1.2 yields $f(s)\ge 1$, that is, $u(s)\ge v(s)$.

The supremum reaches the first zero of $v$ or the domain endpoint. [step 2.1] Put ${\tau }_L:=\min \{\tau ,L\}$. We claim $T\ge {\tau }_L$. Suppose instead $T<{\tau }_L$, so that $T<L$ and $T<\tau$, and since $\tau$ is the first positive zero of $v$, one has $v>0$ on $(0,T]$; put $d:=v(T)>0$. From $u\ge v$ on $(0,T)$ and continuity of $u$ and $v$ at $T$ we get $u(T)\ge v(T)=d$. Choose $\epsilon >0$ with $T+\epsilon <L$ and with $u>d\mathrm{/}2>0$ and $v>d\mathrm{/}2>0$ on $[T,T+\epsilon ]$, possible by continuity. Then $uv\ge 0$ on $(0,T+\epsilon )$: on $(0,T)$ because $u\ge v>0$, and on $(T,T+\epsilon )$ because both factors are positive there. By step 1.1, $h^{\mathrm{\prime }}\ge 0$ on $(0,T+\epsilon )$, and from $h(0)=0$ the mean value theorem in [F2] gives $h\ge 0$ on $(0,T+\epsilon )$. Since $v>0$ on $(0,T+\epsilon )$, the quotient rule and step 1.1 give $f^{\mathrm{\prime }}=h\mathrm{/}{v}^2\ge 0$ on $(0,T+\epsilon )$; monotonicity of $f$ together with the limit $f(0+)=1$ of step 1.2 then gives $u(s)\ge v(s)$ for every $s\in (0,T+\epsilon )$. As $v>0$ on $(T,T+\epsilon )$, every $t$ with $T<t<T+\epsilon$ satisfies $t\in A$, contradicting the definition of $T$ as an upper bound of $A$. Therefore $T\ge {\tau }_L$. For every $t\in (0,{\tau }_L)$, step 2.1 gives $u(t)\ge v(t)>0$. If ${\tau }_L=L<\tau$, continuity extends $u\ge v>0$ to $t=L$; if ${\tau }_L=\tau \le L$, continuity gives $u(\tau )\ge v(\tau )=0$. This proves the stated conclusions on the domain $[0,L]$.

The first positive zero of $u$ is no earlier than $\tau$. [step 3.1] If $v$ has no zero in $(0,L]$, then $\tau =+\mathrm{\infty }$ and step 3.1 gives $u(t)\ge v(t)>0$ on all of $(0,L]$; in particular neither $u$ nor $v$ has a positive zero. If $v$ has a zero in $(0,L]$, let $\tau$ be its first one: by step 1.2 the zero set lies in $[\delta ,L]$ for some $\delta >0$; it is a nonempty closed subset of this compact interval, so it has a minimum $\tau$ with $v(\tau )=0$, and no zero lies in $(0,\tau )$. Since $u\ge v$ on $(0,\tau )$ by step 3.1 and $u$ is continuous, passing to the limit gives $u(\tau )\ge v(\tau )=0$. Hence $u>0$ on $(0,\tau )$, and its first positive zero, when it exists, is at least $\tau$.

Model case and boundary conventions. [F4, step 4.1] For $a\equiv b\equiv k>0$ and $u=v={\mathrm{sn}}_k$, [F4] shows that both equations hold with common initial slope $c=1$, and step 4.1 recovers ${\mathrm{sn}}_k>0$ on $(0,\pi \mathrm{/}\sqrt{k})$. For the comparison form, let $k>0$ and suppose the coefficient of a given solution $v$ satisfies $b(t)\le k$ for all $t$; take the pair of the theorem to be the given $v$ and $c{\mathrm{sn}}_k$: the first member solves $w^{\mathrm{\prime }\mathrm{\prime }}+bw=0$, the second solves $w^{\mathrm{\prime }\mathrm{\prime }}+kw=0$ by [F4] and has initial slope $c\cdot 1=c$, and $b\le k$ is the required pointwise ordering. Step 4.1, applied only on the common domain $[0,L]$, shows that $v$ has no zero on $(0,\min \{L,\pi \mathrm{/}\sqrt{k}\})$; if $L<\pi \mathrm{/}\sqrt{k}$, it is positive also at $L$, and if $L\ge \pi \mathrm{/}\sqrt{k}$, its first zero on $[0,L]$ is no earlier than $\pi \mathrm{/}\sqrt{k}$. The interval is one-sided at $0$, where the initial data and the one-sided derivatives live; the endpoint $L$ enters only through the convention for $\tau$. The degenerate coefficient $b\equiv 0$ admits $v(t)=ct$, which solves the equation and has no positive zero, so the statement is consistent in the flat case. The excluded case $c=0$ has the zero solution and no comparison content, and the empty interval is excluded by $L>0$. No manifold, no indexed family of solutions and no selection principle occur, so the argument is choice-free.