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Would the Picard–Lindelöf theorem still be true if the requirement that f be Lipschitz continuous in y was replaced with the requirement that f be almost Lipschitz in y?

If not, are there any moduli of uniform continuity weaker than Lipschitz continuity that it is known suffice, or results indicating that there can't be any?

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The book "Uniqueness and nonuniqueness criteria for ordinary differential equations" by Ratan Prakash Agarwal, Ravi P. Agarwal, V. Lakshmikantham is a great reference. – AppliedSide Feb 14 '11 at 4:23
up vote 16 down vote accepted

Yes. This follows from the classical uniqueness theorem due to Osgood (the original paper appeared in 1898).

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 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 the mappings $f(t,x)$ which are discontinuous at $t=0$. (Actually, the condition that $\phi(t)$ is continuous on $(0,a)$ can be replaced with an assumption of 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 successive Picard iterations to a local solution on a sufficiently small interval (A. Wintner, "On the Convergence of Successive Approximations", Amer. Journal of Math. Vol. 68 (1946), pp. 13-19).

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