Local flows for time-measurable fields with integrable local bounds

Fix an initial pair $(s,x_0)$ and a closed ball $K=\bar{B}(x_0,r)$. On a sufficiently short one-sided interval $J$ adjoining $s$, absolute continuity of the integrals of $m_K,L_K$ gives ${\int }_J{m}_K\le r$ and ${\int }_J{L}_K<1$. On continuous curves taking values in $K$, define $\mathrm{\Phi }(X)(t)=x_0+{\int }_s^{t}b(q,X(q))dq,$ with the oriented integral when J lies to the left of s. The integrand is measurable because $q\mapsto (q,X(q))$ is continuous and b is Borel; it is integrable because $\mathrm{\mid }b(q,X(q))\mathrm{\mid }\le m_K(q)$ almost everywhere. The map preserves the closed sup-norm ball of radius r about the constant curve $x_0$, and $\mathrm{\parallel }\mathrm{\Phi }(X)-\mathrm{\Phi }(Y){\mathrm{\parallel }}_{\mathrm{\infty }}\le ({\int }_J{L}_K)\mathrm{\parallel }X-Y{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F1] it has a unique fixed point. The integral FTC [F2] makes this curve absolutely continuous and a solution almost everywhere.

We first prove the scalar zero-derivative implication needed for uniqueness. Let $H:[a,c]\to \mathbb{R}$ be absolutely continuous with $H^{\mathrm{\prime }}=0$ outside a null set $N$ containing the endpoints. For $\epsilon >0$ and $n\ge 1$, let $E_n$ consist of all $x\in [a,c]$ such that $\mathrm{\mid }H(y)-H(x)\mathrm{\mid }\le \epsilon \mathrm{\mid }y-x\mathrm{\mid }$ whenever $y\in [a,c]$ and $\mathrm{\mid }y-x\mathrm{\mid }<1\mathrm{/}n$. Continuity of $H$ makes $E_n$ closed: pass to the limit in this inequality for each fixed y with $\mathrm{\mid }y-x\mathrm{\mid }<1\mathrm{/}n$. These compact sets increase with n and cover $[a,c]\setminus N$. Partition $[a,c]$ into finitely many closed intervals of length less than $1\mathrm{/}n$. On each interval I, the compact image $H(E_n\cap I)$ has diameter at most $\epsilon \mathrm{\mid }I\mathrm{\mid }$, so covering it by its smallest containing interval gives $\lambda (H(E_n))\le \epsilon (c-a)$. By [F8], the increasing union of the compact images has measure at most $\epsilon (c-a)$, hence ${\lambda }^{\ast }(H([a,c]\setminus N))\le \epsilon (c-a)$. Letting $\epsilon \downarrow 0$ makes this image null; [F7] makes $H(N)$ null as well. Thus the whole range of H is null, and the intermediate value theorem [F8] makes H constant. The singleton interval is immediate.

Any absolutely continuous solution Y on a compact interval with image in K satisfies its integral equation. Indeed, $f(q)=b(q,Y(q))$ is measurable and bounded almost everywhere by $m_K$, hence integrable. By [F2], $I(t)={\int }_a^{t}f(q)dq$ is absolutely continuous and $I^{\mathrm{\prime }}=f$ almost everywhere. Each component of $Y-I$ is absolutely continuous by the finite-family definition (split a given tolerance between the two summands), and has derivative zero almost everywhere. Step 1.2 makes it constant. Therefore $Y(t)-Y(a)={\int }_a^{t}b(q,Y(q))dq$ for every t, and subtraction gives the oriented version with any base time.

If two solutions have the same value at one time and remain in a compact K on a common compact interval, partition that interval into finitely many subintervals on which $\int L_K<1$. By step 2.1 both satisfy their integral equations. On each subinterval beginning at a time where the solutions agree, these equations bound their sup-norm difference by $(\int L_K)$ times itself, so the difference is zero. Induction over the partition proves uniqueness throughout the common interval. Their compatible union therefore defines a unique maximal locally absolutely continuous solution: every compact subinterval of its domain is covered by finitely many local solution intervals, on which the integral equation holds. Suppose a maximal endpoint $\tau$ lies in $(0,T)$ and there are times $t_n\uparrow \tau$ with $X(t_n)$ in a compact set K. Choose R with K contained in the ball $B_R(0)$, put $K_1={\bar{B}}_{R+1}(0)$, and take n large enough that ${\int }_{t_n}^{\tau }{m}_{K_1}<1$. If the path first reaches the boundary of $K_1$ at some $r\in (t_n,\tau )$, it stays in $K_1$ on $[t_n,r]$, so $\mathrm{\mid }X(r)-X(t_n)\mathrm{\mid }\le {\int }_{t_n}^r{m}_{K_1}<1$, while $\mathrm{\mid }X(r)-X(t_n)\mathrm{\mid }\ge \mathrm{\mid }X(r)\mathrm{\mid }-\mathrm{\mid }X(t_n)\mathrm{\mid }\ge 1$, a contradiction. Thus it stays in $K_1$ up to $\tau$; the same integrable bound makes it Cauchy there, and its finite limit lets step 1.1 extend it past $\tau$, again a contradiction. Hence the path eventually leaves every compact set at an interior right maximal endpoint. Time reversal gives the same conclusion at an interior left maximal endpoint.

Assume the spatial derivative hypothesis. On a short interval and compact ball as in step 1.1, solutions from nearby initial points remain in a common compact ball. The contraction estimate gives $\mathrm{\parallel }X(\cdot ;s,x_0+h)-X(\cdot ;s,x_0){\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\mid }h\mathrm{\mid }$. Write ${\mathrm{\Delta }}_h(t)=X(t;s,x_0+h)-X(t;s,x_0)$. The state-variable fundamental theorem of calculus [F3] gives ${\mathrm{\Delta }}_h(t)=h+{\int }_s^t{A}_h(q){\mathrm{\Delta }}_h(q)dq,A_h(q)={\int }_0^1{D}_{x}b(q,X(q;s,x_0)+\theta {\mathrm{\Delta }}_h(q))d\theta \mathrm{.}$ Put $A(q)=D_{x}b(q,X(q;s,x_0))$ at the almost-everywhere times where the spatial differential exists, with value zero on its exceptional time set. The matrix equation $V(t)=I_d+{\int }_s^{t}A(q)V(q)dq$ has a unique continuous solution by the contraction argument of step 1.1, since ${\int }_J{L}_K<1$. Subtracting its product with h from the equation for ${\mathrm{\Delta }}_h$ gives ${\mathrm{\Delta }}_h(t)-V(t)h={\int }_s^t{A}_h(q)({\mathrm{\Delta }}_h(q)-V(q)h)dq+{\int }_s^t(A_h(q)-A(q))V(q)hdq\mathrm{.}$ Taking sup norms and absorbing the first term bounds the left side by $C\mathrm{\mid }h\mathrm{\mid }{\int }_J\mathrm{\parallel }{A}_h-A\mathrm{\parallel }dq=o(\mathrm{\mid }h\mathrm{\mid })$. Thus $D_{x_0}X=V$; dominated convergence in the same integral equation proves continuity of V in $x_0$. The integral FTC [F2] yields $\dot{V}=A(t)V$ almost everywhere. Repeating on a finite cover by overlapping short intervals gives $C^1$ dependence throughout each common flow domain and continuous dependence in the uniform path norm on trajectories existing through the full interval.

Uniqueness makes the time map from s to t injective. Continuous dependence makes its domain open; solving backward from time t shows its image is open and its inverse is the backward flow, which is $C^1$ by step 3.1. The chain rule [F5] says the derivative of the inverse times V is the identity. Thus the time map is a $C^1$ diffeomorphism between its open domain and image.

Fix a compact flow tube and a compact set K of states it visits. For a fixed target time t, choose a compact time interval J containing t and the varying start times, with all trajectories in the flow identity remaining in K. For trajectories with the same starting time, the integral equation gives $\mathrm{\mid }X(r;s,y)-X(r;s,z)\mathrm{\mid }\le \mathrm{\mid }y-z\mathrm{\mid }+{\int }_{\min \{s,r\}}^{\max \{s,r\}}{L}_K(q)\mathrm{\mid }X(q;s,y)-X(q;s,z)\mathrm{\mid }dq$; iterating this inequality yields the spatial Lipschitz bound $\mathrm{\mid }X(t;s,y)-X(t;s,z)\mathrm{\mid }\le e^{{\int }_J{L}_K}\mathrm{\mid }y-z\mathrm{\mid }$. For $s,s+h\in J$, the flow identity and the integral equation along the short backward segment give $X(t;s+h,x)=X(t;s,X(s;s+h,x))$ and $\mathrm{\mid }X(s;s+h,x)-x\mathrm{\mid }\le {\int }_{[s,s+h]}{m}_K(q)dq$, where the integral is over the unoriented interval. Hence $\mathrm{\mid }X(t;s+h,x)-X(t;s,x)\mathrm{\mid }\le e^{{\int }_J{L}_K}{\int }_{[s,s+h]}{m}_K(q)dq$. Summing over any disjoint adjacent intervals shows that $s\mapsto X(t;s,x)$ is locally absolutely continuous, uniformly for x in the tube. To identify its derivative, choose a countable dense subset Q of K. Apply [F2] to the indefinite integrals of $m_K,L_K$ and each component of $s\mapsto b(s,q)$ for $q\in Q$; their differentiability gives the corresponding one-sided interval averages at almost every s. By Countable Choice select exceptional null sets for this countable family, and take their union with the exceptional set for the spatial Lipschitz bound. Outside this union all these averages converge to the values at s; [F6] also gives the centered-average conclusion there after enlarging the null set. Approximation by q and the common Lipschitz bound show that the averages of $b(r,x)$ over intervals shrinking to s converge to $b(s,x)$ uniformly for $x\in K$. For the solution started at x at time $s+h$, the magnitude bound gives $\mathrm{\mid }X(r;s+h,x)-x\mathrm{\mid }\le {\int }_{[s,s+h]}{m}_K=O(\mathrm{\mid }h\mathrm{\mid })$ uniformly between s and s+h at these Lebesgue points. Replacing $b(r,x)$ by $b(r,X(r;s+h,x))$ in the average contributes at most $\mathrm{\mid }h{\mathrm{\mid }}^{-1}O(\mathrm{\mid }h\mathrm{\mid }){\int }_{[s,s+h]}{L}_K(r)dr=O(\mathrm{\mid }h\mathrm{\mid })$. The integral equation therefore gives $[X(s;s+h,x)-x]\mathrm{/}h\to -b(s,x)$ uniformly on K. The flow identity $X(t;s+h,x)=X(t;s,X(s;s+h,x))$ and the spatial derivative from step 4.1 yield ${\mathrm{\partial }}_{s}X(t;s,x)=-D_{x}X(t;s,x)b(s,x)$ almost everywhere on the tube. The right side has the integrable local bound $e^{{\int }_J{L}_K}{m}_K(s)$, consistent with and also giving the local absolute continuity.