For which of the following choices of a and b is the IVP below guaranteed a unique solution by the Picard- Lindelof theorem

$\frac{dy}{dt}=\sqrt{y^3-t^3}, y(a)=b$

  1. (-1,1)

  2. (1,-1)

  3. (4,5)

  4. (5,4)

how to solve this question i did't get any idea is that picard's theorem and Picard- Lindelof same even i cant solve the given differential equation please any help me how to solve this kind of problems thanks for your time

  • $\begingroup$ Can you show that $f(t,y) = \sqrt{y^{3} - t^{3}}$ is Lipschitz for some $t \in (c, d)$? $\endgroup$ Aug 24, 2017 at 3:44
  • $\begingroup$ @Mattos...$\frac{\partial f}{\partial t}=\frac{-3t^2}{2\sqrt{y^3-t^3}}$ $\endgroup$ Aug 24, 2017 at 3:54
  • $\begingroup$ You need to examine $f_{y}$ not $f_{t}$.. $\endgroup$ Aug 24, 2017 at 3:56
  • $\begingroup$ @Mattos...$\frac{\partial f}{\partial y}=\frac{3y^2}{2\sqrt{y^3-t^3}}$ $\endgroup$ Aug 24, 2017 at 4:00

1 Answer 1


If the function $f(y,t)=\sqrt{y^3-t^3}$ is Lipschitz, then there is a unique solution to IVP. This is the case whenever $b^3-a^3>0$. For a nonlinear system a unique solution is always guaranteed locally. For a global solution (i.e. it does not have finite escape time or so called blow-up) you need to perform further analysis such as finding a Lyapunov function.

Consider the Lyapunov function $V(t)=y^2(t)$ then $\dot{V}(t)=2y\dot{y}=2y\sqrt{y^3-t^3}$. To have a solution we already set $y^3-t^3>0$ which means that $y>0$ and thus $\dot{V}(t)>0$; this ensures that $V(t)$ goes to infinity as time passes and so does $y(t)$.


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