Integral equation form of an initial value problem

Statement

Let $J$ be an interval, let $U$ be an open subset of $\mathbb{R}^n$, let $f : J \times U \to \mathbb{R}^n$ be continuous, and let $t_0 \in J$ and $x_0 \in U$. Let $\gamma : J \to U$ be a continuous curve. Then $\gamma$ is differentiable on $J$ with $\gamma'(t) = f(t, \gamma(t))$ for each $t \in J$ and $\gamma(t_0) = x_0$ if and only if $$\gamma(t) = x_0 + \int_{t_0}^{t} f\left(s, \gamma(s)\right) ds \quad \text{for each } t \in J.$$ At an endpoint of $J$ that belongs to $J$, the derivative is one-sided.

Remarks

In other words, the solutions of the initial value problem on $J$ are the fixed points of the Picard operator. Proof: one direction is the fundamental theorem of calculus applied to $\gamma'$. For the other, the integrand $s \mapsto f(s, \gamma(s))$ is continuous, so the integral is differentiable in $t$. The integral form needs only continuity of $\gamma$. This makes it the right setting for fixed point arguments.

Explanations

L1 human

Integrate both sides of $\gamma' = f(t, \gamma)$ from $t_0$ to $t$. The initial condition and the differential equation become one equation with an integral.