Transport characteristics depend C^1 on the initial position

By [L2] and [L3], after shrinking to a neighborhood $U_0$ of ${\xi }_0$, every $\xi \in U_0$ has a unique characteristic $X(\cdot ;\xi )$ on the common compact interval $I$, and $(t,\xi )\mapsto X(t;\xi )$ is continuous there.

Fix a coordinate vector $e_j$ and a nonzero scalar $h$ with $\xi ,\xi +h{e}_j\in U_0$; by [L4], the difference quotient $Z_h(t):=\frac{X(t;\xi +h{e}_j)-X(t;\xi )}{h}$ satisfies $Z_h(t)=e_j+{\int }_{t_0}^t{A}_h(s,\xi )Z_h(s)ds$, where $A_h(s,\xi )={\int }_0^1{D}_{x}a(X(s;\xi )+\theta (X(s;\xi +h{e}_j)-X(s;\xi )),s)d\theta$, and because $a$ is $C^1$ on a compact neighborhood of the family of graphs, the matrices $A_h$ are uniformly bounded.

Apply [L6] to the integral equation in step 2.1. If $t\ge t_0$, then $\mathrm{\parallel }{Z}_h(t){\mathrm{\parallel }}_2\le 1+M{\int }_{t_0}^t\mathrm{\parallel }{Z}_h(s){\mathrm{\parallel }}_{2}ds$; if $t\le t_0$, rewrite step 2.1 as $Z_h(t)=e_j-{\int }_t^{t_0}{A}_h(s,\xi )Z_h(s)ds,$ so $\mathrm{\parallel }{Z}_h(t){\mathrm{\parallel }}_2\le 1+M{\int }_t^{t_0}\mathrm{\parallel }{Z}_h(s){\mathrm{\parallel }}_{2}ds$. Gronwall therefore gives $\mathrm{\parallel }{Z}_h(t){\mathrm{\parallel }}_2\le e^{M\mathrm{\mid }t-t_0\mathrm{\mid }}$ for every $t\in I$, uniformly in $h$. The continuity from step 1.1 together with the uniform continuity of $D_{x}a$ makes $A_h(\cdot ,\xi )\to D_{x}a(X(\cdot ;\xi ),\cdot )$ uniformly as $h\to 0$; comparing the equations for $Z_h$ and $Z_k$ on the forward or backward interval between $t_0$ and $t$ and applying [L5] again shows that $(Z_h)$ is Cauchy in $C(I;{\mathbb{R}}^n)$.

Let $Y_j(\cdot ,\xi )$ be the limit from step 3.1; passing to the limit in step 2.1 gives $Y_j(t,\xi )=e_j+{\int }_{t_0}^t{D}_{x}a(X(s;\xi ),s)Y_j(s,\xi )ds$, which [L4] rewrites as $Y_j^{\mathrm{\prime }}(t,\xi )=D_{x}a(X(t;\xi ),t)Y_j(t,\xi )$ with $Y_j(t_0,\xi )=e_j$, and doing this for every coordinate vector while invoking [L7] identifies the matrix $Y=(Y_1\cdots Y_n)$ with $D_{\xi }X$; its columns solve the displayed variational equation.

Fix $\xi$ and restrict nearby initial points $\eta$ to a closed ball contained in $U_0$. Put $A(t,\eta )=D_{x}a(X(t;\eta ),t)$ and $H={\max }_{t\in I}\mathrm{\mid }t-t_0\mathrm{\mid }$. Continuity from step 1.1 and the $C^1$ hypothesis make $A$ jointly continuous on this compact set, so it has a bound $M$ and $\delta (\eta ):={\sup }_{t\in I}\mathrm{\parallel }A(t,\eta )-A(t,\xi )\mathrm{\parallel }\to 0$ as $\eta \to \xi$. The column equations in step 4.1 and [L5]–[L6] give $\mathrm{\parallel }{Y}_j(t,\eta ){\mathrm{\parallel }}_2\le e^{MH}$. Subtract the column equations in step 4.1, splitting $A(t,\eta )Y_j(t,\eta )-A(t,\xi )Y_j(t,\xi )$ as $A(t,\eta )(Y_j(t,\eta )-Y_j(t,\xi ))+(A(t,\eta )-A(t,\xi ))Y_j(t,\xi )$. Reorient backward integrals over $[t,t_0]$ before applying [L6]; in either time direction [L5] yields ${\sup }_{t\in I}\mathrm{\parallel }{Y}_j(t,\eta )-Y_j(t,\xi ){\mathrm{\parallel }}_2\le H{e}^{2MH}\delta (\eta )⟶0.$ Thus every derivative column is continuous in $\xi$, uniformly on $I$; its time continuity from step 4.1 also gives joint continuity. Since $\xi$ was arbitrary, the first partial derivatives of $X(t;\cdot )$ are continuous on $U_0$, proving the asserted $C^1$ dependence and completing the variational-equation claim.