Integral equation for an initial value problem

Given the initial value problem

y^{'} (x) = f (x, y), y (x_{0}) = y_{0}

and any $C^{1}$ solution $y$ , the fundamental theorem of calculus gives $y (x) - y_{0} = \int_{x_{0}}^{x} y^{'} (t) d t$ , hence $y$ also solves

y (x) = y_{0} + \int_{x_{0}}^{x} f (t, y (t)) d t

The equivalence is exact in both directions: the integral form forces $y (x_{0}) = y_{0}$ at once, and differentiating the right-hand side by the fundamental theorem of calculus returns $y^{'} = f (x, y)$ .

The Picard iteration

The unknown is the function $y$ , appearing inside a nonlinear operator $K$ . Written as

y = y_{0} + K y, (K y) (x) = \int_{x_{0}}^{x} f (t, y (t)) d t

the problem becomes: find a fixed point of $K$ on the function space $C ([x_{0}, x_{0} + h])$ .

Define the successive approximations

y_{n} (x) = y_{0} + \int_{x_{0}}^{x} f (t, y_{n - 1} (t)) d t, y_{0} (t) \equiv y_{0}

so $y_{1} = y_{0} + \int_{x_{0}}^{x} f (t, y_{0}) d t$ , $y_{2} = y_{0} + \int_{x_{0}}^{x} f (t, y_{0} + \int_{x_{0}}^{t} f (s, y_{0}) d s) d t$ , and so on, with nested integrals. Under certain conditions, this should converge...

Numerical reading: quadrature is Picard, discretized

Runge-Kutta methods are Picard iterations in which the integral is replaced by a quadrature. Trapezoid on the integral form is the implicit trapezoid (Crank–Nicolson), midpoint is explicit RK2, Simpson is classical RK4.