📂Odinary Differential Equations

Picard Method

Theorem¹

If $E$ is open in $\mathbb{R}^{n}$ and the following initial value problem is given for $f \in C^{1} (E)$,

$$\begin{cases} \dot{ \phi } = f ( \phi ) \\ \phi (0) = \phi_{0} \end{cases}$$

let’s define the sequence of functions $\left\{ u_{k} (t) \right\} _{ k =0}^{ \infty }$ as follows.

$$\begin{cases} u_{0} (t) = \phi_{0} \\ u_{k+1} (t) = \phi_{0} + \int_{0}^{t} f \left( u_{k} (s) \right) ds \end{cases}$$

Then the continuous function $u (t) := \lim_{k \to \infty} u_{k} (t)$ is the solution to the given initial value problem.

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• $C^{1}$ is the set of functions with continuous derivatives.

Explanation

It is naturally assumed that $u$ exists, and it would be meaningless if it didn’t. Also, $u$ must be continuous, but $u_{k}$ does not need to be. Thus, the rigorous parts of mathematics that seem sloppy are usually compensated for in the theorem where this method is used.

Proof

$$\begin{align*} & u (t) \\ =& \lim_{k \to \infty} u_{k+1} (t) \\ =& \lim_{k \to \infty} \left( \phi_{0} + \int_{0}^{t} f \left( u_{k} (s) \right) ds \right) \\ =& \phi_{0} + \int_{0}^{t} \lim_{k \to \infty} f \left( u_{k} (s) \right) ds \\ =& \phi_{0} + \int_{0}^{t} f \left( \lim_{k \to \infty} u_{k} (s) \right) ds & \because \text{continuity of } f \\ =& \phi_{0} + \int_{0}^{t} f \left( u(s) \right) ds \end{align*}$$

If $t = 0$, then

$$u (0) = \phi_{0} + \int_{0}^{0} f \left( u(s) \right) ds = \phi_{0} + 0 = \phi_{0}$$

According to the Fundamental Theorem of Calculus, if the function $f$ is continuous on the closed interval $[a,b]$, then function $F(x) = \int_{a}^{x} f(t) dt$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $${{dF(x)} \over {dx}} = f(x)$$

Since $u$ is continuous, $( f \circ u )$ is also continuous, and by the Fundamental Theorem of Calculus,

$$\dot{u } (t) = \left( \phi_{0} + \int_{0}^{t} f \left( u(s) \right) ds \right)' = f \left( u (t) \right)$$

Therefore, it is known that $u$ is the solution to the given initial value problem.

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