[proofplan] The proof reduces the equality case to a scalar comparison for the length $f(t)=|J(t)|_g$. Since there are no conjugate points, $J$ does not vanish on $(0,t_0]$, so we may write $J=fE$ with $E$ a unit field perpendicular to $\dot{\gamma}$. The Jacobi equation gives a differential inequality for $f$ against the model solution $s=\operatorname{sn}_k$, and the Wronskian of $f$ and $s$ is monotone. Equality at $t_0$ forces the monotone ratio $f/s$ to be identically equal to $1$, and then the nonnegative error terms in the scalar equation vanish, giving both parallelness of $E$ and equality of the radial sectional curvature. [/proofplan]

[step:Write the Jacobi field as its length times a unit perpendicular field] Let \begin{align*} s:[0,\infty)&\to\mathbb{R}\\ t&\mapsto \operatorname{sn}_k(t) \end{align*} denote the model solution in the statement. Since $D_tJ(0)=E_0$ and $|E_0|_g=1$, the [Jacobi field](/page/Jacobi%20Field) $J$ is not identically zero. Because $\gamma|_{[0,R]}$ has no [conjugate point](/page/Conjugate%20Point) to $\gamma(0)$, the nonzero Jacobi field $J$ cannot vanish at any $t\in(0,R]$. Hence the function \begin{align*} f:(0,t_0]&\to(0,\infty)\\ t&\mapsto |J(t)|_g \end{align*} is well-defined and positive. Define \begin{align*} E:(0,t_0]&\to TM\\ t&\mapsto \frac{J(t)}{f(t)}\in T_{\gamma(t)}M. \end{align*} Then $E$ is a smooth unit vector field along $\gamma|_{(0,t_0]}$, and since $J(t)\perp \dot{\gamma}(t)$, we have $E(t)\perp \dot{\gamma}(t)$ for every $t\in(0,t_0]$. Thus \begin{align*} J(t)=f(t)E(t) \end{align*} for every $t\in(0,t_0]$. [/step]

[step:Derive the scalar Jacobi inequality for the length]For $t\in(0,t_0]$, define the radial sectional curvature along the plane spanned by $\dot{\gamma}(t)$ and $E(t)$ by \begin{align*} K(t):=\operatorname{sec}_g(\dot{\gamma}(t),E(t)). \end{align*} Let $R$ denote the Riemann curvature tensor of $(M,g)$, viewed as the map \begin{align*} R:TM\times TM\times TM&\to TM\\ (X,Y,Z)&\mapsto R(X,Y)Z. \end{align*} Since $E(t)$ has unit length and is perpendicular to the unit vector $\dot{\gamma}(t)$, this is equivalently \begin{align*} K(t)=g(R(E(t),\dot{\gamma}(t))\dot{\gamma}(t),E(t)), \end{align*} with the curvature sign convention determined by the Jacobi equation \begin{align*} D_t^2J+R(J,\dot{\gamma})\dot{\gamma}=0. \end{align*} Because $g(E,E)=1$, differentiating gives $g(D_tE,E)=0$. Differentiating once more gives \begin{align*} g(D_t^2E,E)=-|D_tE|_g^2. \end{align*} Using $J=fE$, we compute \begin{align*} D_t^2J=f''E+2f'D_tE+fD_t^2E. \end{align*} Taking the $g$-[inner product](/page/Inner%20Product) with $E$ and using the Jacobi equation gives \begin{align*} 0 &=g(D_t^2J+R(J,\dot{\gamma})\dot{\gamma},E)\\ &=f''+2f'g(D_tE,E)+f\,g(D_t^2E,E)+f\,g(R(E,\dot{\gamma})\dot{\gamma},E)\\ &=f''-f|D_tE|_g^2+fK. \end{align*} Therefore \begin{align*} f''+k f=f\bigl(|D_tE|_g^2+k-K\bigr). \end{align*} The curvature hypothesis gives $K(t)\leq k$, so the function \begin{align*} q:(0,t_0]&\to[0,\infty)\\ t&\mapsto |D_tE(t)|_g^2+k-K(t) \end{align*} is nonnegative and satisfies \begin{align*} f''+k f=qf. \end{align*}[/step]

[guided]The goal is to isolate exactly where equality can fail in Rauch comparison. Since $J$ never vanishes on $(0,t_0]$, we can separate its size from its direction: \begin{align*} J(t)=f(t)E(t),\qquad f(t)=|J(t)|_g,\qquad |E(t)|_g=1. \end{align*} The field $E$ records the direction of $J$, while $f$ records its length. We now compute the scalar equation satisfied by $f$. First, $g(E,E)=1$, so differentiating along $\gamma$ gives \begin{align*} 0=\frac{d}{dt}g(E,E)=2g(D_tE,E), \end{align*} hence $g(D_tE,E)=0$. Differentiating this identity once more gives \begin{align*} 0=\frac{d}{dt}g(D_tE,E)=g(D_t^2E,E)+|D_tE|_g^2, \end{align*} so \begin{align*} g(D_t^2E,E)=-|D_tE|_g^2. \end{align*} Using $J=fE$, the covariant product rule gives \begin{align*} D_t^2J=f''E+2f'D_tE+fD_t^2E. \end{align*} The Jacobi equation is \begin{align*} D_t^2J+R(J,\dot{\gamma})\dot{\gamma}=0. \end{align*} Taking the inner product with $E$ gives \begin{align*} 0 &=g(D_t^2J+R(J,\dot{\gamma})\dot{\gamma},E)\\ &=f''+2f'g(D_tE,E)+f\,g(D_t^2E,E)+f\,g(R(E,\dot{\gamma})\dot{\gamma},E)\\ &=f''-f|D_tE|_g^2+fK(t), \end{align*} where \begin{align*} K(t):=\operatorname{sec}_g(\dot{\gamma}(t),E(t)). \end{align*} Thus \begin{align*} f''+k f=f\bigl(|D_tE|_g^2+k-K(t)\bigr). \end{align*} This identity displays the two possible sources of strict inequality: the direction $E$ may rotate, measured by $|D_tE|_g^2$, and the radial sectional curvature may be strictly smaller than $k$, measured by $k-K(t)$. Both terms are nonnegative because the curvature hypothesis says $K(t)\leq k$.[/guided]

[step:Use the Wronskian to prove monotonicity of the quotient $f/s$] Define the Wronskian-type function \begin{align*} W:(0,t_0]&\to\mathbb{R}\\ t&\mapsto f'(t)s(t)-f(t)s'(t). \end{align*} Since $s''+ks=0$ and $f''+kf=qf$, we have \begin{align*} W'(t) &=f''(t)s(t)+f'(t)s'(t)-f'(t)s'(t)-f(t)s''(t)\\ &=f''(t)s(t)-f(t)s''(t)\\ &=\bigl(q(t)f(t)-kf(t)\bigr)s(t)+kf(t)s(t)\\ &=q(t)f(t)s(t). \end{align*} For $t\in(0,t_0]$, the functions $q$, $f$, and $s$ are nonnegative, with $f(t)>0$ and $s(t)>0$. Hence \begin{align*} W'(t)\geq 0. \end{align*} Because $J(0)=0$ and $D_tJ(0)=E_0$, Taylor expansion along $\gamma$ gives functions $a,b:(0,t_0]\to\mathbb{R}$ satisfying \begin{align*} \lim_{t\downarrow0}a(t)=0,\qquad \lim_{t\downarrow0}b(t)=0, \end{align*} such that \begin{align*} f(t)=t\bigl(1+a(t)\bigr),\qquad f'(t)=1+b(t) \end{align*} as $t\downarrow0$. Since $s(0)=0$ and $s'(0)=1$, Taylor expansion for the model solution gives a function $c:(0,t_0]\to\mathbb{R}$ satisfying \begin{align*} \lim_{t\downarrow0}c(t)=0 \end{align*} such that \begin{align*} s(t)=t\bigl(1+c(t)\bigr),\qquad s'(t)\to1 \end{align*} as $t\downarrow0$. Therefore \begin{align*} \lim_{t\downarrow 0}\frac{f(t)}{s(t)}=1. \end{align*} Moreover, \begin{align*} \lim_{t\downarrow0}W(t) &=\lim_{t\downarrow0}\bigl(f'(t)s(t)-f(t)s'(t)\bigr)\\ &=0. \end{align*} For $t\in(0,t_0]$, \begin{align*} \frac{d}{dt}\left(\frac{f(t)}{s(t)}\right) =\frac{f'(t)s(t)-f(t)s'(t)}{s(t)^2} =\frac{W(t)}{s(t)^2}. \end{align*} Since $W$ is nondecreasing and has limit $0$ as $t\downarrow0$, we have $W(t)\geq0$. Thus $f/s$ is nondecreasing on $(0,t_0]$, and \begin{align*} \frac{f(t)}{s(t)}\geq1 \end{align*} for every $t\in(0,t_0]$. [/step]

[step:Force every nonnegative error term to vanish from equality at $t_0$] The hypothesis $|J(t_0)|_g=s(t_0)$ says \begin{align*} \frac{f(t_0)}{s(t_0)}=1. \end{align*} Since $f/s$ is nondecreasing on $(0,t_0]$ and has limit $1$ as $t\downarrow0$, it follows that \begin{align*} \frac{f(t)}{s(t)}=1 \end{align*} for every $t\in(0,t_0]$. Hence \begin{align*} f(t)=s(t) \end{align*} for every $t\in(0,t_0]$. The derivative of $f/s$ is identically zero, so $W(t)=0$ for every $t\in(0,t_0]$. Since \begin{align*} W'(t)=q(t)f(t)s(t) \end{align*} and $f(t)>0$, $s(t)>0$ on $(0,t_0]$, we obtain \begin{align*} q(t)=0 \end{align*} for every $t\in(0,t_0]$. By the definition of $q$, \begin{align*} |D_tE(t)|_g^2+k-K(t)=0. \end{align*} Both summands are nonnegative, so \begin{align*} |D_tE(t)|_g^2=0,\qquad K(t)=k \end{align*} for every $t\in(0,t_0]$. Therefore $D_tE(t)=0$ on $(0,t_0]$, and \begin{align*} \operatorname{sec}_g(\dot{\gamma}(t),E(t))=k \end{align*} for every $t\in(0,t_0]$. [/step]

[step:Extend the parallel direction to the initial point and identify $J$] Since $D_tE=0$ on $(0,t_0]$, the field $E$ is parallel along $\gamma|_{(0,t_0]}$. Parallel transport preserves length and orthogonality to $\dot{\gamma}$, because $D_t\dot{\gamma}=0$ and $D_tE=0$. We have already shown $J(t)=s(t)E(t)$ for $t\in(0,t_0]$. Since $s'(0)=1$ and $D_tJ(0)=E_0$, taking the limit as $t\downarrow0$ gives \begin{align*} \lim_{t\downarrow0}E(t) =\lim_{t\downarrow0}\frac{J(t)}{s(t)} =E_0. \end{align*} Define $E(0):=E_0$. Then $E$ is the parallel extension of $E_0$ along $\gamma|_{[0,t_0]}$, has unit length, remains perpendicular to $\dot{\gamma}$, and satisfies \begin{align*} J(t)=s(t)E(t)=\operatorname{sn}_k(t)E(t) \end{align*} for every $t\in[0,t_0]$. The curvature equality proved above gives \begin{align*} \operatorname{sec}_g(\dot{\gamma}(t),E(t))=k \end{align*} for every $t\in(0,t_0]$. This is precisely the asserted equality case. [/step]