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.

Introduce the augmented variable $(x,\lambda )\in {\mathbb{R}}^{n+m}$ and define the autonomous-in-parameter system below.

The augmented right-hand side is smooth in the initial data $(y,\lambda )$, so [L1] applies on a common compact local time interval and makes the augmented solution map smooth in $(y,\lambda )$. Projecting to the $x$-component preserves that smoothness, which gives the claimed smooth dependence of solutions on initial state and parameter.

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 $\mathrm{\Omega }$.

The curve $u\mapsto \mathrm{\Phi }(u+s,x_0)$ solves the ODE and at time $u=0$ has value [F1, L1] $\mathrm{\Phi }(s,x_0)$. The curve $u\mapsto \mathrm{\Phi }(u,\mathrm{\Phi }(s,x_0))$ also solves the same ODE and has the same value at time $0$. Because [F1] makes both curves defined on an open interval about every common time where they exist, [L1] applies and forces them to agree on their common domain.

Evaluating the equality from step 1.1 at the time $u=t$ gives [step 1.1] $\mathrm{\Phi }(t,\mathrm{\Phi }(s,x_0))=\mathrm{\Phi }(t+s,x_0)$ whenever both sides are defined.

The vector field $V(x)=x^2$ is smooth on $\mathbb{R}$. The explicit curve [F1, L1] $x(t)=1\mathrm{/}(1-t)$ satisfies $x(0)=1$ and $x^{\mathrm{\prime }}(t)=1\mathrm{/}(1-t{)}^2=x(t{)}^2$ for $t<1$, so [L1] identifies it with the unique local solution through $1$.

This solution cannot be extended past $t=1$ as a real-valued solution, [step 1.1] because $x(t)\to +\mathrm{\infty }$ as $t\uparrow 1$. Hence the maximal solution is not defined for all time.

Therefore a smooth Euclidean vector field need not be complete.

For each initial value $x(0)=a>0$, the unique solution is [L2] $x_a(t)=a\mathrm{/}(1-at)$, defined only for $t<1\mathrm{/}a$. Thus every initial point has a local solution by [L2].

If one positive time $h$ worked for all initial values, then taking [L1, step 1.1, assume-hyp] $a>1\mathrm{/}h$ would give a solution through $a$ defined on $[-h,h]$. But step 1.1 shows the maximal positive existence time is $1\mathrm{/}a<h$, a contradiction. This does not conflict with [L1], because [L1] is a neighbourhood theorem, not a global one over all of $\mathbb{R}$.

Therefore pointwise local existence does not imply one uniform time [step 2.1] interval for all initial data.

The explicit solution $x(t)=1\mathrm{/}(1-t)$ solves $x^{\mathrm{\prime }}=x^2$ for all $t<1$ and [given] cannot be extended through $t=1$, so its maximal interval is $(-\mathrm{\infty },1)$.

This domain is open and not closed. That matches [L1] and [F1], and it [F1, L1, step 1.1] directly refutes the false claim.

For each parameter $\lambda \in \mathbb{R}$, the unique solution is [L1] $x(t;\lambda )=\mathrm{\mid }\lambda \mathrm{\mid }t$. This depends continuously on $\lambda$ for every fixed $t$.