Osgood's Criterion. Let $\omega(t,u)=\phi(t)\psi(u)$ where $\phi(t)\geq 0$ is continuous on the interval $(0,a)$ and $\psi(u)$ is continuous on $\mathbb R_{+}$, $\psi(0)=0$, $\psi(u)>0$ for $u>0$, and $$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$ for some $\epsilon>0$. Suppose that the function mapping $f:[0,a]\times B_R(x_0)\to \mathbb R^d$ satisfies the condition $$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$ for any $t\in(0,a]$ and any $x_1,x_2\in B_R(x_0)$. Then the initial value problem $$\dot{x}=f(t,x),\qquad x(0)=x_0$$ has at most one solution on the interval $[0,\delta]$ with some $\delta>0$.
Osgood's theorem allows for functions the mappings $f(t,x)$ which are discontinuous at $t=0$ t=0$. (in fact, Actually, the condition that$\phi(t)$is continuous on$(0,a)$can be replaced with a much weaker condition an assumption of integrability). mere integrability.) Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that$f$is continuous in$(t,x)$. Moreover, Wintner showed that Osgood's uniqueness condition implies the convergence of the successive Picard iterations to a local solution on a sufficiently small interval (A. Wintner, "On the Convergence of Successive Approximations", AmAmer. Journal of Math. Vol. 68 (1946), pp. 13-19). 2 added 330 characters in body; edited body Yes. This follows from the classical uniqueness theorem due to Osgood. Osgood's Criterion. Let$\omega(t,u)=\phi(t)\psi(u)$where$\phi(t)\geq 0$is continuous on the interval$(0,a)$and$\psi(u)$is continuous on$\mathbb R_{+}$,$\psi(0)=0$,$\psi(u)>0$for$u>0$, and $$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$ for some$\epsilon>0$. Suppose that the function$f:[0,a]\times B_R(x_0)\to \mathbb R^d$satisfies the condition $$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$ for any$t\in(0,a]$and$x_1,x_2\in B_R(x_0)$. Then the initial value problem $$\dot{x}=f(t,x),\qquad x(0)=x_0$$ has at most one solution on the interval$[0,\delta]$with some$\delta>0$. Osgood's theorem allows for functions$f(t,x)$which are discontinuous at$t=0$(in fact, the condition that$\phi(t)$is continuous can be replaced with a much weaker condition of integrability). Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that$f$is continuous in$(t,x)$. Wintner showed that Osgood's uniqueness condition implies the convergence of the successive Picard iterations to a local solution on a sufficiently small interval (A. Wintner, "On the Convergence of Successive Approximations", Am. Journal of Math. Vol. 68 (1946), pp. 13-19). 1 Yes. This follows from the classical uniqueness theorem due to Osgood. Osgood's Criterion. Let$\omega(t,u)=\phi(t)\psi(u)$where$\phi(t)\geq 0$is continuous on the interval$(0,a)$and$\psi(u)$is continuous on$\mathbb R_{+}$,$\psi(0)=0$,$\psi(u)>0$for$u>0$, and $$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$ for some$\epsilon>0$. Suppose that the function$f:[0,a]\times B_R(x_0)\to \mathbb R^d$satisfies the condition $$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$ for any$t\in(0,a]$and$x_1,x_2\in B_R(x_0)$. Then the initial value problem $$\dot{x}=f(t,x),\qquad x(0)=x_0$$ has at most one solution on the interval$[0,\delta]$with some$\delta>0$. Osgood's theorem allows for functions$f(t,x)$which are discontinuous at$t=0$(in fact, the condition that$\phi(t)$is continuous can be replaced with a much weaker condition of integrability). Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that$f$is continuous in$(t,x)\$.