External Lorentz ODE local well-posedness

Choose a chart $(U,\phi )$ with $x_0\in U\subseteq O$ and write its coordinates as $y^a={\phi }^a(x)$. The state domain $\phi (U)\times {\mathbb{R}}^4$ is open and the equation is ${\dot{y}}^a=U^a$, ${\dot{U}}^a=-{\mathrm{\Gamma }}^a{}_{bc}(y)U^b{U}^c+(q\mathrm{/}m)F^a{}_b(y)U^b$, where $F^a{}_b:=g^{ac}{F}_{cb}$ by the definition of $F^{\mathrm{♯}}$. To extend the scalar rules [F5] to this state domain, let $h,k$ be $C^1$ scalar functions on a Euclidean open set and write $\mathrm{\Delta }h=h(w+\eta )-h(w)=Dh(w)\eta +r_h(\eta )$ with $r_h(\eta )=o(\mathrm{\parallel }\eta {\mathrm{\parallel }}_2)$, and similarly for $k$. The identity $\mathrm{\Delta }(hk)=k(w)\mathrm{\Delta }h+h(w)\mathrm{\Delta }k+\mathrm{\Delta }h\mathrm{\Delta }k$ gives $D(hk)(w)=k(w)Dh(w)+h(w)Dk(w)$, since $\mathrm{\Delta }h,\mathrm{\Delta }k=O(\mathrm{\parallel }\eta {\mathrm{\parallel }}_2)$ and their product is $o(\mathrm{\parallel }\eta {\mathrm{\parallel }}_2)$; adding expansions gives $D(h+k)=Dh+Dk$. These derivatives are continuous: their entries are finite sums and products of continuous scalar functions, and continuity of a product follows directly from $\mathrm{\mid }ab-a_0{b}_0\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b-b_0\mathrm{\mid }+\mathrm{\mid }{b}_0\mathrm{\mid }\mathrm{\mid }a-a_0\mathrm{\mid }$ with $a$ locally bounded by continuity. The coordinate projections are linear with constant derivatives; composing the $C^1$ coefficients $g^{ac},{\mathrm{\Gamma }}^a{}_{bc},F_{cb}$ with the projection $(y,U)\mapsto y$ uses [F6]. Thus these local sum and product rules show the eight-component right-hand side $H$ is $C^1$, with continuous derivative matrix formed from its component derivatives.

Choose a nondegenerate closed convex coordinate box $B$ about $(\phi (x_0),U_0^a)$ contained in the state domain. It is compact by [F9], so [F8] bounds the absolute values of all entries of $DH$ by one finite $M\ge 0$ on $B$. For $w,w^{\mathrm{\prime }}\in B$, apply [F6] and the one-variable mean value theorem [F4] to $t\mapsto H^i(w+t(w^{\mathrm{\prime }}-w))$ on $[0,1]$, for each of the eight components. It gives $\mathrm{\mid }{H}^i(w^{\mathrm{\prime }})-H^i(w)\mathrm{\mid }\le M{\sum }_{j=1}^8\mathrm{\mid }{w}_j^{\mathrm{\prime }}-w_j\mathrm{\mid }\le 8M\mathrm{\parallel }{w}^{\mathrm{\prime }}-w{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }H(w^{\mathrm{\prime }})-H(w){\mathrm{\parallel }}_2\le 8\sqrt{8}M\mathrm{\parallel }{w}^{\mathrm{\prime }}-w{\mathrm{\parallel }}_2$. Since the box has the initial state in its interior, the field is locally Lipschitz there; the same construction applies at every state.

Apply [F3] to this locally Lipschitz coordinate vector field to obtain a unique local solution. Because the vector field is $C^1$, composing it with the $C^1$ solution and applying [F6] shows its derivative is $C^1$, so the curve is $C^2$. The coordinate equation is equivalent to the covariant equation by the Christoffel coefficients of [F2], so coordinate changes carry solutions to solutions. For any two geometric solutions with the same initial data on a connected common interval, their equality set is nonempty and closed by continuity; it is relatively open because at each equality point a common chart and [F3] give local uniqueness. Connectedness makes the equality set the whole interval.

Metric compatibility and antisymmetry give $\frac{d}{d\tau }g(U,U)=2g(U,{\mathrm{\nabla }}_{U}U)=2(q\mathrm{/}m)g(U,F^{\mathrm{♯}}(U))=2(q\mathrm{/}m)F(U,U)=0$. Thus the initial norm $-c^2$ is preserved, so $U$ stays timelike. Let $N$ be the smooth timelike field defining the time orientation. By [F7], $g(U,N)$ is nonzero throughout the solution and the initial future direction means it is initially negative; continuity therefore keeps it negative on the connected local interval. Again by [F7], $U$ remains future-directed.