Existence, Uniqueness, etc.

Existence and Uniqueness

Theorem 1 (Picard–Lindelöf). Let $U \subseteq \R^{d+1}$ be open, $f \in C(U, \R^d)$ and $(t_0, x_0) \in U$. If $f$ is locally Lipschitz in space, then there exists a unique, $C^1$, local solution to the ODE $$\label{e:D}\tag{D} \dot x = f(t, x) \quad\text{with}\quad x(t_0) = x_0 \,. $$

Definition 2. Let $U \subseteq \R^{d+1}$ be open, $f \in C(U, \R^d)$. We say $f$ is locally Lipschitz in space on $U$ if for every $K \subseteq U$ compact, there exists a constant $L = L(K)$ such that for every $(t, x)$, $(t, y) \in K$ we have $$ \abs{f(t, x) - f(t, y)} \leq L \abs{x - y} \,. $$ We say $f$ is Lipschitz (or globally Lipschitz) in space if the constant $L$ above does not depend on $K$.

Definition 3. We say $x$ a local solution to the ODE \eqref{e:D} if there exists $T > 0$ such that $x$ satisfies \eqref{e:D} for all $t \in [t_0 - T, t_0 + T]$.

Proof sketch of Theorem 1. Let $T > 0$, find $V \subseteq \R^d$ open such that $x_0 \in V$ and $[t_0 - T, t_0 + T] \times V \Subset U$, let $X = C([t_0 - T, t_0 + T], \R^d)$, with the sup norm $\norm{x} = \sup_{t} \abs{x(t)}$. For all $x \in C( [t_0 - T, t_0 + T], \bar V )$, define $$ A x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds \,, $$ and check that $A$ is a contraction for $T$ small enough.

Remark 4. The solution to \eqref{e:D} may not be global (i.e. defined for all time).

Problem 5. Suppose $d = 1$, $f(t, x) = x^2$, and $x_0 \neq 0$. Show that the solution to \eqref{e:D} can not be defined for all $t \in \R$.

Proposition 6. If in Theorem 1 we had $f \in C^k$, then the solution $x$ is $C^{k+1}$.

Problem 7. Suppose in Theorem 1, instead of assuming $f \in C(U, \R^d)$ we only assumed that there exists $\delta > 0$, and an integrable function $L$ (defined in a neighborhood of $t_0$) such that for every $t$ and every $x, y \in B(x_0, \delta)$ we have $$ \abs{f(t, x) - f(t, y)} \leq L(t) \abs{x - y} \,. $$

  1. Show that the integral equation $$ x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds $$ has a unique local solution.
  2. Show that the solution may not be differentiable.

Continuous Dependence

Theorem 8. Suppose $U \subseteq \R^{d+1}$, $f, g \in C(U, \R^d)$ are two functions with (spatial) Lipschitz constant $L$. Let $x, y$ be solutions to the ODE $$ \dot x = f(t, x) \,,\quad \dot y = g(t, y) \,, $$ with initial conditions $x(t_0) = x_0$, $y(t_0) = y_0$. Then $$ \abs{x(t) - y(t)} \leq \abs{x_0 - y_0} e^{L |t - t_0|} + \frac{M}{L}\paren[\big]{ e^{L|t - t_0|} - 1} \quad\text{where} \quad M \defeq \sup_{(t, y) \in U} \abs{f(t, y) - g(t, y)} $$

Lemma 9 (Gronwall’s Lemma). Suppose $a > 0$, and $b, f \colon \R \to [0, \infty)$ are such that $$ f(t) \leq a + \int_0^t b(s) f(s) \, ds \,. $$ Then $$ f(t) \leq a \exp\paren[\bigg]{ \int_0^t b(s) \, ds } $$

Proof sketch. Suppose first $a > 0$, and check $\partial_t \ln( a + \int_0^t b f \, ds ) \leq b$, and integrate.

Proof sketch of Theorem 8. Set $d(t) = |x(t) - y(t)|$, and use Gronwall’s Lemma.

Remark 10. If in Gronwall’s Lemma $a = a(t)$ is a nonnegative increasing function of $t$, then the conclusion still holds with $a$ replaced by $a(t)$.

Analogous to the principal matrix for linear equations, we now look at the flow of \eqref{e:D}.

Definition 11. We say $\phi$ is the flow map of \eqref{e:D} if $$ \partial_t \phi_{s, t}(x) = f(t, \phi_{s, t}(x)) \quad\text{and}\quad \phi_{s, s}(x) = x \,. $$

When $f$ is locally Lipschitz, the same uniqueness argument from before gives $$ \phi_{s, t} \circ \phi_{r, s} = \phi_{r, t} \,. $$ If $f$ is more regular, then $\phi$ is also more regular; time differentiability is guaranteed through Proposition 6. Regularity in space requires proof.

Theorem 12 (Differentiability). If $f \in C^1(U, \R^d)$, then the flow map $\phi$ is also $C^1$.

Proof sketch. Let $D$ denote the derivative in space alone. If $\phi_{t_0, t}$ is $C^1$, then $V(t, x) = D \phi_{t_0, t}(x)$ would satisfy $$\label{e:V}\tag{V} \partial_t V = Df(t, \phi_{t_0, t}) V \quad\text{with}\quad V(t_0, x) = I \,. $$ Since $f \in C^1$, \eqref{e:V} has a unique local solution. We will show that this solution is the derivative of $\phi$.

For this, let $$ R(t, h) = \phi_{t_0, t}(x+h) - \phi_{t_0, t}(x) - V(t, x)h \,, $$ and explicitly show $\abs{R(t, h)} / \abs{h} \to 0$ as $h \to 0$.

Proposition 13. If $f \in C^k(U, \R^d)$, then the flow map $\phi$ is also $C^k$.

Proposition 14 (Liouville Formula). If $f \in C^1(U, \R^d)$ show that $\det \phi$ satisfies the Liouville equation $$ \partial_t \det(D \phi_{t_0, t}) = \dv f(t, \phi_{t_0, t}) \cdot \det(D \phi_{t_0, t}) \,. $$

Extensibility

Remark 15. Consider \eqref{e:D} with $f$ satisfying the assumptions of Theorem 1. Suppose $\phi_1$ and $\phi_2$ are two solutions to \eqref{e:D} defined on $(t_0 - T, t_0 + T)$ with $\phi_1(t_0) = \phi_2(t_0)$. By Theorem 1, there is some small $\delta > 0$ such that $\phi_1 = \phi_2$ on $(t_0 - \delta, t_0 + \delta)$. By iterating Theorem 1 it’s easy to see $\delta = T$.

Definition 16. Fix $t_0 \in \R$, $x_0 \in U$. The maximal existence time of the solution to \eqref{e:D} with initial data $x(t_0) = x_0$, denoted by $T_{+}$ is the supremum of all times $T$ for which there exists a solution to \eqref{e:D} with initial data $x(t_0) = x_0$ defined on the entire interval $[t_0, T]$.

Proposition 17. Let $f \in C(U, \R^d)$ be locally Lipschitz in space, $(t_0, x_0) \in U$, and $T_{+}$ be the maximal existence time of \eqref{e:D}. Then either $T_{+} = \infty$, or for every compact set $K \subseteq U$, there exists $t < T_{+}$ such that $(t, x(t)) \not\in K$.

Theorem 18. If $f \in C(\R^{d+1}, \R^d)$ is locally Lipschitz in space, and there exists nonnegative functions $a, b$, such that $a$ is increasing, $b$ is nonnegative, and for all $x \in \R^d$, $t \geq t_0$ we have $$ \abs{f(t, x)} \leq a(t) + b(t) \abs{x} \,. $$ Then the maximal time of existence of \eqref{e:D} with initial data $x(t_0) = x_0$ is infinite.

Proposition 19. If in Theorem 1 we instead assumed $$ \abs{f(t, x)} \leq a(t) + b(t) \abs{x}^{\alpha} \,, $$ for some $\alpha > 1$, then the maximal existence time could be finite.

Problem 20. Let $U \subseteq \R^d$, and suppose $f \in C(\R \times U, \R^d)$ is locally Lipschitz in space. Let $g \colon U \to \R$ be $C^1$, and suppose $K = \set{g \leq 0} \subseteq U$ is compact. Suppose further, $\grad g \neq 0$ on $\set{g = 0}$. If for all $x \in \set{g = 0}$, $t \in \R$, $$ f(t, x) \cdot \grad g(x) \leq 0 \,, $$ then show that for any $x_0 \in K$, $t_0 \in \R$ the maximal (forward) existence time of \eqref{e:D} is infinite.

Beyond Locally Lipschitz coefficients

Problem 21. For $\alpha \in (0, 1)$, find infinitely many solutions to the ODE $\dot x = x^\alpha$ with initial data $x(0) = 0$.

Theorem 22 (Peano). Let $\delta > 0$, $x_0 \in \R^d$, $t_0 \in \R$, $T > 0$, $U = (t_0, t_0 + T) \times B(x_0, \delta)$ and suppose $f \in C(\bar U, \R^d)$. There exists a solution to \eqref{e:D} with initial data $x(t_0) = x_0$.

Proof sketch. Fix $h > 0$ and define $$\label{e:E}\tag{E} t_{n+1} = t_n + h\,, \quad x_h(t_{n+1}) = x_h(t_n) + f(t_n, x_h(t_n)) h \,, $$ and for times between $t_n$ and $t_{n+1}$ define $x_h$ by linear interpolation. Use Arzelà–Ascoli to get a uniformly convergent subsequence $(x_{h_k})_{k \in \N}$. Now use uniform continuity of $f$ to ensure $f(t, x_{h_k}(t))$ converges uniformly and hence verify \eqref{e:D}.

Remark 23. Usually when computing solutions numerically, one uses \eqref{e:E}. (This is a forward Euler scheme.)

Definition 24. Let $\omega\colon [0, \infty) \to [0, \infty)$ be a continuous, increasing function with $\omega(0) = 0$. We $\omega$ is a modulus of continuity for $f$ if for every $x, y$, $$ \abs{f(x) - f(y)} \leq \omega(\abs{x - y}) \,. $$

Theorem 25 (Osgood). Suppose $U \subseteq \R^{d+1}$ is open, $f \in C(U, \R^d)$ has spatial modulus of continuity $\omega$. That is, suppose for every $(t, x), (t, y) \in U$, $$ \abs{f(t, x) - f(t, y)} \leq \omega(\abs{x - y}) \,. $$ If $$ \int_0^1 \frac{dr}{\omega(r)} = +\infty\,, $$ then solutions to the initial value problem for \eqref{e:D} are locally unique.

Proof sketch. Let $x, y$ be two solutions with $x(t_0) = y(t_0) = x_0$, and set $r(t) = \abs{x(t) - y(t)}$. For $t > t_0$ where $r(t) > 0$, compute $\partial_t (r^2(t))$: $$ \partial_t (r^2(t)) = 2 (x(t) - y(t)) \cdot (\dot x(t) - \dot y(t)) \leq 2 r(t) \omega(r(t)) \,. $$ If $r(t_1) > 0$ for some $t_1 > t_0$, set $t_* = \sup \{ t \in [t_0, t_1] : r(t) = 0 \}$. Separating variables on $(t_*, t_1]$ yields $$ \int_{t_*}^{t_1} \frac{\partial_t (r(t)^2) }{2 r(t) \omega(r(t))} \, dt \leq t_1 - t_* \quad\implies\quad \int_0^{r(t_1)} \frac{dr}{\omega(r)} \leq t_1 - t_* < \infty \,, $$ which is a contradiction.