Euclidean Ordinary Differential Equations with Smooth Dependence

Extend $C$ continuously to the open interval $(a-1,b+1)$ by setting $\tilde{C}(t):=C(a)$ for $t<a$, $\tilde{C}(t):=C(t)$ for $t\in I$, and $\tilde{C}(t):=C(b)$ for $t>b$. Regard $M_n(\mathbb{R})$ as ${\mathbb{R}}^{n^2}$. Then the right-hand side $\tilde{F}(t,Y):=\tilde{C}(t)Y$ is continuous on the open domain $(a-1,b+1)\times {\mathbb{R}}^{n^2}$ and is globally Lipschitz in $Y$ on every compact time interval, because $\mathrm{\parallel }\tilde{C}(t)(Y-Z){\mathrm{\parallel }}_2\le \mathrm{\parallel }\tilde{C}(t)\mathrm{\parallel }\mathrm{\parallel }Y-Z{\mathrm{\parallel }}_2$. Hence [F1] gives a unique local solution through $(t_0,Y_0)$.

Let $J=(\alpha ,\beta )$ be the maximal interval of existence of that solution in $(a-1,b+1)$. Since $C$ is continuous on compact $I$, [L2] supplies a bound $\mathrm{\parallel }C(t)\mathrm{\parallel }\le M$ on all of $I$. For $t\in J\cap [t_0,b]$, the Volterra equation from [F2] and the norm estimate [L3] give the forward inequality below.

For $t\in J\cap [a,t_0]$, the oriented Volterra equation gives $Y(t)=Y_0-{\int }_t^{t_0}C(s)Y(s)ds$, so [L3] gives the reflected inequality below.

Put $R:=\mathrm{\parallel }{Y}_0{\mathrm{\parallel }}_2{e}^{M(b-a)}$. If $\beta \le b$, then every sequence $t_j\uparrow \beta$ in $J$ has graph points in the compact set $[a,b]\times \bar{B}(0,R)\subseteq (a-1,b+1)\times M_n(\mathbb{R})$ by steps 2.1 and 2.2. So [L1] extends the solution past $\beta$, contradicting maximality. The same argument at the left endpoint rules out $\alpha \ge a$. Therefore $I\subseteq J$, and the restriction of the maximal solution to $I$ is defined on all of $I$.

Step 3.1 gives existence on all of $I$, and uniqueness is the local uniqueness already supplied by [F1].

Fix a compact subinterval $[u,v]\subseteq I$ containing $t_0$. By [F1] and [L1] the matrix ODE below has a unique solution.

Entrywise product differentiation from [L2] gives the identity below.

The same calculation applied to $A(t)B(t)$ gives $\frac{d}{dt}(A(t)B(t))=0$ and $A(t_0)B(t_0)=I_n$, hence $A(t)B(t)=I_n$ on $[u,v]$. Therefore $B(t)=A(t{)}^{-1}$ and $A(t)$ is invertible on $[u,v]$. Since $t\in I$ was arbitrary and lies in some compact subinterval containing $t_0$, every $A(t)$ is invertible on $I$.

By [L1], after shrinking $U_0$ if needed, every solution graph $(t,\mathrm{\Phi }(t,y))$ with $y\in U_0$ lies in one compact cylinder $K\subseteq D$. Fix $y\in U_0$, and let $A_y:I\to M_n(\mathbb{R})$ be the unique solution of the variational equation below.

Fix $y\in U_0$ and an increment $u\in {\mathbb{R}}^n$ with $y+u\in U_0$. By [F2], the difference $z_u(t):=\mathrm{\Phi }(t,y+u)-\mathrm{\Phi }(t,y)$ satisfies the Volterra equation below, and the state-variable mean-value formula gives the matrix field $B_u$.

Put ${\rho }_u(t):=z_u(t)-A_y(t)u$. Subtracting the Volterra equations for $z_u$ and $A_{y}u$ gives the identity below.

For $y,y^{\mathrm{\prime }}\in U_0$, write $E(t):=A_{y^{\mathrm{\prime }}}(t)-A_y(t)$ and $C_y(t):=D_{x}F(t,\mathrm{\Phi }(t,y))$. Then the continuity estimate below, together with the same Gronwall argument as in step 3.1, proves continuity of the derivative matrix.

Steps 3.1 and 4.1 prove that $\mathrm{\Phi }$ is $C^1$ in the initial-state variable and that its derivative matrix is the solution of the variational equation.

By [L1], the derivative $D_y\mathrm{\Phi }(t,y)$ exists and is the solution of the variational equation below.

For each $r\ge 1$, let $E_r$ be the finite-dimensional space of $r$-linear maps $({\mathbb{R}}^n{)}^r\to {\mathbb{R}}^n$, with $E_0:={\mathbb{R}}^n$. Repeated differentiation of the ODE or of its Volterra form produces the finite-dimensional jet system below.

Fix $r\ge 1$. If $\mathrm{\Phi }$ is $C^r$ in $y$, then its jet $J_r(t,y)=(\mathrm{\Phi },D_y\mathrm{\Phi },\dots ,D_y^r\mathrm{\Phi })(t,y)$ is a continuous solution of the system from step 2.1 on the same compact interval $I$, with initial data $X_0(t_0)=y$, $X_1(t_0)=I_n$, and $X_m(t_0)=0$ for $m\ge 2$. The highest-jet component is linear in $X_r$ once the lower jets are fixed, so [L2] supplies its unique evolution on $I$. Applying [L1] to this enlarged smooth system shows that $J_r$ depends $C^1$ on the initial value $y$. In particular its first component $\mathrm{\Phi }$ is $C^{r+1}$ in $y$.

Step 1.1 is the base case $r=1$. Step 3.1 upgrades $C^r$ regularity of $\mathrm{\Phi }$ to $C^{r+1}$ for every $r\ge 1$, so by induction $\mathrm{\Phi }$ is $C^m$ in $y$ for every $m$. Therefore $\mathrm{\Phi }$ is smooth in the initial-state variable.

Hence the solution map depends smoothly on initial data.

The field $G$ is jointly smooth on the open time-state domain $D$. Its last $m$ equations are ${\lambda }^{\mathrm{\prime }}=0$, so the parameter is constant along each augmented solution. Thus projection to the first $n$ coordinates identifies augmented solutions through $(y,\lambda )$ with solutions of the displayed original IVP, in both directions.

Choose $a,r>0$ so $[t_0-a,t_0+a]\times \bar{B}(z_0,2r)\subseteq D$. By [F1], $G$ and $D_{z}G$ have bounds $M\ge 1$ and $L\ge 1$ on this cylinder, and the mean-value inequality on each open state ball gives the same state-Lipschitz bound $L$. Choose $H>0$ with $H<a$, $HM<r$ and $HL<1$. For every initial state $z\in B(z_0,r)$, the radius-$r$ cylinder centred at $z$ lies in the larger cylinder. Hence [F2] gives a unique augmented solution $Z(t,z)$ on $J=[t_0-H,t_0+H]$, all of whose graphs lie in that one compact cylinder. This establishes the common interval needed below.

Now [L1] applies to the jointly smooth field $G$ on the common interval $J$ of step 2.1, and gives smooth dependence of $Z(t,z)$ on $z$ near $z_0$. We record why this dependence has the joint regularity needed for time derivatives. At first order, [F4] gives $X_1=D_{z}Z$ and $X_1^{\mathrm{\prime }}=D_{z}G(t,Z)X_1$, $X_1(t_0)=I_d$. The coefficient is jointly continuous; the uniform-in-time estimates in [F4], combined with continuity of the solution of this linear equation in time, make $X_1$ jointly continuous. The solution $Z$ itself is jointly continuous by the same initial-data estimates and its continuity in time.

The higher-order argument does not require a new existence assumption. For $r\ge 1$, use the finite-dimensional jet variables $X_0,\dots ,X_r$, where $X_j$ is a $j$-linear map and $X_0\in {\mathbb{R}}^d$. Repeated chain rules in [F3] give $X_0^{\mathrm{\prime }}=G(t,X_0)$ and $X_j^{\mathrm{\prime }}=D_{z}G(t,X_0)X_j+P_j(t,X_0,\dots ,X_{j-1})$ for $j\ge 1$, with $P_1=0$. Here $P_j$ is the finite sum over partitions of the $j$ differentiation directions into at least two nonempty blocks, applying the corresponding higher derivative of $G$ to the lower jets indexed by those blocks. Initial values are $X_0(t_0)=z$, $X_1(t_0)=I_d$ and $X_j(t_0)=0$ for $j\ge 2$. This jet field is jointly smooth. For nearby full jet initial values, $X_0$ exists on $J$ by step 2.1; successively each higher component is affine linear with bounded continuous coefficients and forcing on $J$, so [F5] gives its evolution there. The same linear integral estimates bound each component uniformly for nearby initial values, giving a common compact jet cylinder for each finite order. Thus [F4] legitimately applies to this enlarged system on $J$. Starting with its first-order variational equation, differentiation of the Volterra identities and uniqueness in [F5] identify the prescribed jet with $(Z,D_{z}Z,\dots ,D_z^{r}Z)$. Its highest component is $C^1$ in the initial state, so these identities upgrade $C^r$ to $C^{r+1}$; this uses the highest component, not merely the first one. Induction, with the uniform estimates of [F4], gives joint continuity of all state derivatives.

The equation ${\mathrm{\partial }}_{t}Z=G(t,Z)$ now supplies all mixed derivatives: differentiating it in state variables gives the chain-rule expressions in the jointly continuous state jets, and their displayed jet equations permit subsequent time differentiation. Induction on the time order, using joint smoothness of $G$, proves joint smoothness on the open time interval $(t_0-H,t_0+H)$ and the initial-state neighbourhood. Put $I=[t_0-H\mathrm{/}2,t_0+H\mathrm{/}2]$; the solution map on $I$ therefore extends smoothly to an open time neighbourhood. Projection as in step 1.1 gives the claimed common local solutions and smooth dependence on $(y,\lambda )$, including the endpoints of $I$. All solutions and all jet evolutions are uniquely determined by their IVPs; the Picard sequences use prescribed iteration and the argument makes no family selection or additional Choice assumption.

Fix $t_{\ast }\in (\alpha ,\beta )$. If $t\ge t_{\ast }$, then [F2] and [L1] give the first inequality below, and if $t\le t_{\ast }$ they give the reflected inequality.

Suppose $\beta <\mathrm{\infty }$. Then step 1.1 shows that for $t$ close to $\beta$ the graph point $(t,x(t))$ stays inside the compact box $[\beta -1,\beta ]\times \bar{B}(x(t_{\ast }),M(\beta -t_{\ast })+1)$ of the ODE domain $\mathbb{R}\times {\mathbb{R}}^n$, contradicting [L2]. Thus $\beta =\mathrm{\infty }$. The same argument at the left endpoint gives $\alpha =-\mathrm{\infty }$.

Therefore every maximal solution is global, so the vector field is complete.

The norm function $x\mapsto \mathrm{\parallel }V(x){\mathrm{\parallel }}_2$ is continuous on the compact support [L2] $K$, so [L2] gives a bound $M$ there. Outside $K$ the vector field vanishes by definition of support. Hence $\mathrm{\parallel }V(x){\mathrm{\parallel }}_2\le M$ for every $x\in {\mathbb{R}}^n$.

Smooth maps are locally Lipschitz on Euclidean open sets, so step 1.1 makes [L1, step 1.1] $V$ a bounded locally Lipschitz vector field. Therefore [L1] applies and yields completeness.

Since a smooth vector field is continuous and locally Lipschitz, [L1] applies [L1, L2, choose] at $(0,x_0)$. The uniform local existence result [L2] therefore gives $h>0$ and an open neighbourhood $W$ of $x_0$ such that every $y\in W$ has a unique solution on $[-h,h]$, and all these solution graphs stay inside one compact cylinder in $\mathbb{R}\times U$.

Define $\mathrm{\Phi }(t,y)$ to be that unique solution value at time $t$. By [L4] the [L4, step 1.1] map is smooth in the initial state variable $y$ on $(-h,h)\times W$. For each fixed $y$, the curve $t\mapsto \mathrm{\Phi }(t,y)$ is a solution, so it is differentiable in $t$ and satisfies ${\mathrm{\partial }}_t\mathrm{\Phi }(t,y)=V(\mathrm{\Phi }(t,y))$ with $\mathrm{\Phi }(0,y)=y$.

The uniqueness in step 1.1 is exactly the local uniqueness from [L1], and [F1, L1, L3, step 2.1] the maximal-solution theorem [L3] records that these local flows are the local pieces of unique maximal trajectories rather than unrelated solution branches.

Introduce an extra variable $s$ and consider the autonomous system below on $\mathbb{R}\times {\mathbb{R}}^n$.

The first equation forces $s(t)=s_0+t$ when $s(0)=s_0$, so the second equation becomes exactly the original nonautonomous system with initial time $s_0$. Reading the initial value $(s_0,x_0)$ as a parameter, [L2] makes the resulting solution map smooth in $(s_0,x_0)$. This is precisely the claimed local theorem for nonautonomous smooth ODEs.

Let $(t_{\ast },x_{\ast })\in \mathrm{\Omega }$ and put $p:=\mathrm{\Phi }(t_{\ast },x_{\ast })$. By [L1], applied at the [L1, choose] state point $p$, there exist $\delta >0$ and an open neighbourhood $W\subseteq U$ of $p$ such that every $y\in W$ has a unique solution on $[-\delta ,\delta ]$, and these solutions vary smoothly with $y$.

By [L3], for initial states $x$ sufficiently close to $x_{\ast }$, the solution [L3, step 1.1] $\mathrm{\Phi }(\cdot ,x)$ is defined at least on a compact interval around $t_{\ast }$ and its value at time $t_{\ast }$ lies in $W$. Therefore for every such $x$ and every $\mathrm{\mid }s\mathrm{\mid }<\delta$, the solution continued from time $t_{\ast }$ by the local flow of step 1.1 is defined at time $t_{\ast }+s$. Hence all pairs $(t_{\ast }+s,x)$ with $\mathrm{\mid }s\mathrm{\mid }<\delta$ and $x$ near $x_{\ast }$ lie in $\mathrm{\Omega }$.

Step 2.1 gives an open neighbourhood of $(t_{\ast },x_{\ast })$ contained in $\mathrm{\Omega }$, [L1, L2, L3, step 2.1] so $\mathrm{\Omega }$ is open. On that neighbourhood, the evaluation map is the composite of the continuous time-$t_{\ast }$ map with the local smooth flow from step 1.1, hence is jointly continuous and smooth in the state variable. Since $(t_{\ast },x_{\ast })$ was arbitrary, the same holds on all of $\mathrm{\Omega }$.