📂Odinary Differential Equations

Picard Method

Theorem¹

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

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

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 a solution to the given initial value problem.

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

Explanation

Naturally, $u$ is assumed to exist; if it does not, the theorem is meaningless. Also, $u$ must be continuous, but $u_{k}$ need not be continuous. Such parts, too sloppy to be called rigorous mathematics, are usually made up for in the theorems 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

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

Fundamental theorem of calculus: If a function $f$ is continuous on the closed interval $[a,b]$, then the function $F(x) = \int_{a}^{x} f(t) dt$ is continuous on $[a,b]$, 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, we can see that $u$ is a solution to the given initial value problem.

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