Ordinary Differential Equations/The Picard–Lindelöf theorem

In this section, our aim is to prove several closely related results, all of which are occasionally called "Picard-Lindelöf theorem". This type of result is often used when it comes to arguing for the existence and uniqueness of a certain ordinary differential equation, given that some boundary conditions are satisfied.

Local results
Picard–Lindelöf Theorem (Banach fixed-point theorem version):

Let $$I := [a, b]$$ be an interval, let $$f: I \times \mathbb R^n \to \mathbb R^n$$ be a continuous function, and let
 * $$x'(t) = f(t, x(t))$$

be the associated ordinary differential equation. If $$f$$ is Lipschitz continuous in the second argument, then this ODE possesses a unique solution on $$[a, a+\epsilon]$$ for each possible initial value $$x(0) = x_0 \in \mathbb R^n$$, where $$\epsilon < 1/L$$, $$L$$ being the Lipschitz constant of the second argument of $$f$$.

Proof:

We first rewrite the problem as a fixed-point problem. Indeed, using the fundamental theorem of calculus, one can show that the simultaneous equations
 * $$\begin{cases}

x'(t) = f(t, x(t)) & t \in [a, a+\epsilon] \\ x(0) = x_0 & \end{cases}$$ are equivalent to the single equation
 * $$\forall t \in [a, a + \epsilon]: x(t) = x_0 + \int_a^t f(s, x(s)) ds$$,

where $$\epsilon$$ is to be determined at a later stage. This means that the function $$x(t)$$ is a fixed point of the function
 * $$T: \mathcal C([a, a + \epsilon]) \to \mathcal C([a, a + \epsilon]), T(x)(t) := x_0 + \int_a^t f(s, x(s)) ds$$.

Now $$T$$ satisfies a Lipschitz condition as follows:
 * $$\begin{align}

\left\| T(x)(t) - T(y)(t) \right\| & = \left\| \int_a^t f(s, x(s)) ds - \int_a^t f(s, y(s)) ds \right\| \\ & \le \int_a^t \|f(s, x(s)) - f(s, y(s))\| ds \\ & \le \int_a^t L \|x(s) - y(s)\| ds \\ & \le (t - a) L \|x - y\|_\infty \le \epsilon L \|x - y\|_\infty, \end{align}$$ where we took the norm on $$\mathcal C([a, a+ \epsilon])$$ to be the supremum norm. If now $$\epsilon < \frac{1}{L}$$, then $$T$$ is a contraction, and hence the Banach fixed-point theorem is applicable, giving us both existence and uniqueness.

Replacing the fixed-point principle by summation techniques, we get a slightly better result in the sense that the domain of definition of the function $$f$$ does not have to be all of $$[a, b] \times \mathbb R^n$$.

Picard–Lindelöf theorem (telescopic series version):

Let $$f: [a, b] \times \Omega \to \mathbb R^n$$ be a function which is continuous and Lipschitz continuous in the second argument, where $$\Omega \subseteq \mathbb R^n$$, and let $$(t_0, x_0) \in \mathbb R \times \Omega$$ with the property that $$[t_0 - s, t_0 + s] \times \overline{B_r(x_0)} \subseteq [a, b] \times \Omega$$ for some $$s,r > 0$$. If in this case $$\gamma \le \min\{s, r/M\}$$, where $$M := \underset{(t,x) \in [t_0-s,t_0+s] \times B_r(x_0)}{\sup \| f(t, x) \|}$$, then the initial value problem
 * $$\begin{cases}

x'(t) = f(t, x(t)) & t \in [t_0 - \gamma, t_0 + \gamma] \\ x(t_0) = x_0 & \end{cases}$$

possesses a unique solution.

Proof:

We first prove uniqueness. To do so, we use Gronwall's inequalities. Suppose $$x, y$$ are both solutions to the problem. Then
 * $$\left\| x(t) - y(t) \right\| = \left\| \int_{t_0}^t f(t, x(t)) - f(t, y(t)) dt \right\| \le \int_{t_0}^t \left\|f(t, x(t)) - f(t, y(t))\right\| dt \le 0 + \int_{t_0}^t L \left\| x(t) - y(t) \right\| dt$$,

and hence by Gronwall's inequalities
 * $$\left\| x(t) - y(t) \right\| \le 0 \cdot e^{\int_{t_0}^t L dt} = 0$$

for both $$t \in [t_0, t_0 + s]$$ (right Gronwall's inequality) and $$t \in [t_0 - s, t_0]$$ (left Gronwall's inequality).

Now on to existence. Once again, we inductively define
 * $$x_0(t) := x_0$$ (the constant function),
 * $$x_{n+1}(t) := x_0 + \int_{t_0}^t f(\tau, x_n(\tau)) d\tau$$.

Since $$f$$ is not necessarily defined on any larger set than $$[t_0 - s, t_0 + s] \times B_r(x_0)$$, we have to prove that this definition always makes sense, i.e. that $$f(t, x_n(t))$$ is defined for all $$n$$ and $$t \in [t_0 - \gamma, t_0 + \gamma]$$, that is, $$x_n(t) \in B_r(x_0)$$ for $$t \in [t_0 - \gamma, t_0 + \gamma]$$. We prove this by induction.

For $$n = 0$$, this is trivial.

Assume now that $$x_n(t) \in B_r(x_0)$$ for $$t \in [t_0, t_0 + \gamma]$$. Then
 * $$\begin{align}

\|x_{n+1}(t) - x_0\| &= \left\| \int_{t_0}^t f(\tau, x_n(\tau)) d\tau \right\| \\ &\le \int_{t_0}^t \|f(\tau, x_n(\tau))\| d\tau \\ &\le M \cdot \gamma \\ & \le M \cdot r/M = r. \end{align}$$ For $$t \in [t_0-\gamma,t_0]$$ we obtain an analogous bound.

By the telescopic sum, we have
 * $$x_n(t) - x_0 = \sum_{j=1}^n (x_j(t) - x_{j-1}(t))$$.

Furthermore, for $$t \in [t_0, t_0 + \gamma]$$ and $$j \ge 1$$,
 * $$\begin{align}

\|x_{j+1}(t) - x_j(t)\| & = \left\| x_0 + \int_{t_0}^t f(\tau, x_j(\tau)) d\tau - \left( x_0 + \int_{t_0}^t f(\tau, x_{j-1}(\tau)) d\tau \right) \right\| \\ & \le \int_{t_0}^t \|f(\tau, x_j(\tau)) - f(\tau, x_{j-1}(\tau)) \| d\tau \\ & \le \int_{t_0}^t L \|x_j(\tau) - x_{j-1}(\tau)\| d\tau. \end{align}$$ Hence, by induction,
 * $$\|x_{j+1}(t) - x_j(t)\| \le \int_{t_0}^t Lr \frac{|\tau - t_0|^{j-1} L^{j-1}}{(j-1)!}d\tau \le r \frac{|t - t_0|^j L^j}{j!} \le r\frac{\gamma^jL^j}{j!}$$.

Again, by the very same argument, an analogous bound holds for $$t \in [t_0 - \gamma, t_0]$$.

Thus, by the Weierstraß M-test, the telescopic sum
 * $$x_n(t) - x_0 = \sum_{j=1}^n (x_j(t) - x_{j-1}(t))$$

converges uniformly; in particular, $$x_n$$ converges.

It is now possible to interchange differentiation and summation in the latter sum; for, on the one hand, we are uniformly convergent, and on the other hand,
 * $$\sum_{j=1}^n (x_j'(t) - x_{j-1}'(t)) = \sum_{j=2}^n (f(t, x_{j-1}(t)) - f(t, x_{j-2}(t))) + f(t, x_0) = f(t, x_{n-1}(t))$$,

which converges to $$f(t, x(t))$$ for $$n \to \infty$$ due to theorem 2.5 and the convergence of $$x_n$$; note that the image of each $$x_n$$ is contained within the compact set $$\overline{B_r(x_0)}$$, the closure of $$B_r(x_0)$$. Hence indeed
 * $$x'(t) = \left( \sum_{j=1}^\infty (x_j(t) - x_{j-1}(t)) \right)' = \sum_{j=1}^\infty (x_j'(t) - x_{j-1}'(t)) = f(t, x(t))$$

on $$[t_0 - \gamma, t_0 + \gamma]$$.