# Uniqueness of solutions to nonlinear ODEs

For a linear non-homogeneous ODE $$\dot{x}=f(t),\ \ x(0)=x_0, 0\leq t\leq 1,$$ if one has two different solutions then one has infinitely many solutions. The idea for showing such property is to use the fact that the solution space of $\dot{x}=0$ is a vector space.

Do we still have the same conclusion (locally) for the general nonlinear ODE, say $$\dot{x}=g(x),\ \ x(0)=x_0, 0\leq t\leq 1,$$ where $g$ is assumed to be a continuous function?

The classical example $\dot{x}=\sqrt{|x|}$, which demonstrates that uniqueness can fail, seems relevant. But I don't know how to generalize it find infinitely many solutions in some $[0,\delta]\subset [0,1]$. I just have a vague guess the point at which the two different solutions have different values may give two new initial conditions, which may give two more new IVPs. I don't see how to go on.

[Added:]It seems that if $\varphi(t)$ is a solution, then $$\varphi_c(t)=\begin{cases} x_0,\ &0\leq t\leq c;\\ \varphi(t-c),\ &c\leq t\leq 1. \end{cases}$$ gives a family of solutions where $0<c<1$. But how do I show that they must be different?

Case 1: $g(x_0)\ne 0$. Then the solution of IVP is unique in some interval of time. Indeed, let $x(t)$ be a solution with $x(0)=x_0$. Then $x'(t)=g(x(t))$ is continuous and nonzero in a neighborhood of $0$. Therefore, $x(t)$ is invertible in this neighborhood; let $t(x)$ be its inverse. By the inverse function theorem, $t(x)$ is differentiable with $t'(x)=g(x)$. This (together with $t(0)=0$) determines $t(x)$ uniquely: two $C^1$ functions with the same derivative are equal.
Case 2: $g(x_0)=0$. If uniqueness fails, there is a nonconstant solution $\varphi$ starting at $x_0$. As you correctly observed, shifting the solution in time produces an uncountable family of solutions of IVP. They are different because the time at which the solution leaves the stationary point $x_0$ is different.