Let $a<b$, let $I\subset\mathbb{R}$ be an open interval containing $[a,b]$, let $U\subset\mathbb{R}^n$ be open, let $V\subset\mathbb{R}^n$ be open, and let $D\subset [a,b]\times U$ be a relatively [open set](/page/Open%20Set). Let $L:I\times U\times V\to\mathbb{R}$ be $C^2$. Suppose there are maps
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\begin{align*}
q:D\to V
\end{align*}
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and
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\begin{align*}
S:D\to\mathbb{R}
\end{align*}
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which extend to $C^1$ and $C^2$ maps, respectively, on an open neighbourhood of $D$ in $I\times U$. Assume that the Hilbert field identities
Let $A,B\in U$, and let $y_0\in C^1([a,b];U)$ satisfy $y_0(a)=A$, $y_0(b)=B$, have graph contained in $D$, have $y_0'(x)\in V$ for every $x\in[a,b]$, and solve the characteristic equation
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\begin{align*}
y_0'(x)=q(x,y_0(x))
\end{align*}
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for every $x\in[a,b]$. Then $y_0$ minimizes the action
among all competitors $y\in C^1([a,b];U)$ such that $y(a)=A$, $y(b)=B$, the graph of $y$ is contained in $D$, and $y'(x)\in V$ for every $x\in[a,b]$, where $\mathcal{L}^1$ denotes one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure).
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If, in addition, $E(x,y,v)>0$ whenever $(x,y)\in D$, $v\in V$, and $v\ne q(x,y)$, and if the initial value problem $z'(x)=q(x,z(x))$, $z(a)=A$, has at most one solution $z\in C^1([a,b];U)$ whose graph lies in $D$, then $y_0$ is the unique minimizer in this competitor class.