An initial-value problem asks for a curve \(x:I\to\mathbb R^d\) satisfying \[ \dot x(t)=f(t,x(t)),\qquad x(t_0)=x_0. \] Existence, uniqueness, continuation, and dependence on data are separate questions. The hypotheses needed for each should not be conflated.
Classical solutions and the integral equation
Proposition. If \(f\) is continuous, a \(C^1\) curve \(x\) solves the initial-value problem on \(I\) if and only if \[ x(t)=x_0+\int_{t_0}^t f(s,x(s))\,\mathrm{d}s \quad(t\in I). \]
Proof. Integrating the differential equation gives the integral identity. Conversely, continuity of the integrand and the fundamental theorem of calculus show that the right side is \(C^1\) with derivative \(f(t,x(t))\).
The integral equation places the problem in the Banach space \(C([t_0-h,t_0+h];\mathbb R^d)\) and makes fixed-point and compactness methods available.
Peano existence
Theorem (Peano). Suppose \(f\) is continuous on the rectangle \[ R=[t_0-a,t_0+a]\times\overline{B(x_0,b)}. \] Let \(M=\sup_R\|f\|\) and choose \(h\le\min(a,b/M)\). Then the initial-value problem has at least one solution on \([t_0-h,t_0+h]\).
Proof. On \(X=C([t_0-h,t_0+h];\overline{B(x_0,b)})\), define \[ (Tx)(t)=x_0+\int_{t_0}^tf(s,x(s))\,\mathrm{d}s. \] The choice of \(h\) makes \(T(X)\subset X\). The image is uniformly bounded and equi-Lipschitz because \(\|(Tx)'(t)\|\le M\), so its closure is compact by Arzelà--Ascoli. Uniform continuity of \(f\) on \(R\) makes \(T\) continuous. The compact convex form of Schauder's fixed-point theorem gives \(Tx=x\), hence a solution.
Picard--Lindelöf uniqueness
Theorem. If \(f\) is continuous and locally Lipschitz in \(x\), uniformly for \(t\) on compact subrectangles, then there is a unique local solution through every \((t_0,x_0)\).
Existence and uniqueness follow by making the integral operator a contraction on a sufficiently short interval. The detailed iteration and estimates are on the next page.
Continuous dependence
Theorem. Suppose \(f,g\) are \(L\)-Lipschitz in state on a common region, and \(x,y\) solve \[ \dot x=f(t,x),\quad x(t_0)=x_0, \qquad \dot y=g(t,y),\quad y(t_0)=y_0. \] If \(\sup\|f-g\|\le\varepsilon\), then for \(t\ge t_0\), \[ \|x(t)-y(t)\| \le e^{L(t-t_0)}\|x_0-y_0\| +\frac{\varepsilon}{L}\{e^{L(t-t_0)}-1\}, \] with the obvious \(L=0\) limit.
Proof. The integral equations imply \[ \|x(t)-y(t)\| \le\|x_0-y_0\| +\int_{t_0}^t \{L\|x(s)-y(s)\|+\varepsilon\}\,\mathrm{d}s. \] Apply the inhomogeneous Gronwall inequality.
Existence without uniqueness
Example. The scalar problem \[ \dot x=2\sqrt{|x|},\qquad x(0)=0 \] has, for every \(c\ge0\), \[ x_c(t)= \begin{cases} 0,&t\le c,\\ (t-c)^2,&t\ge c. \end{cases} \] The vector field is continuous but not locally Lipschitz at zero. Peano existence holds, while uniqueness fails through arbitrary waiting time.
Parameters and differentiable dependence
If \(f(t,x,\theta)\) is \(C^1\) in \((x,\theta)\), then the solution is differentiable in \(\theta\) while it remains in a common compact region. The sensitivity \(Z=\partial_\theta x\) solves \[ \dot Z=f_x(t,x,\theta)Z+f_\theta(t,x,\theta), \qquad Z(t_0)=\partial_\theta x_0. \] This variational equation underlies identifiability, adjoint methods, and gradient-based parameter estimation.