Existence and Uniqueness Theorem Question 1 DE

Question

I am currently struggling with this question:

Explain what the Uniqueness and Existence Theorems say about solutions to this differential equation: $\frac{dy}{dt} = t\sqrt y$

I know that $f$ is continuous if $y>0$, any $t$ which satisfies the existence theorem. $\frac{\partial f}{\partial y}$ is $\frac{t}{2\sqrt y}$ and is continuous if $y>0$ or $y$ is not equal to zero. I think that solutions are unique, given that they satisfy the conditions for continuity as stated above. I'm just not sure if I am on the right track in terms of how I am thinking about this. I've only started learning about this so I'm just not sure about this whole thing. Thank you!

Answer 1

The given function does not satisfy the Lipschitz condition in the neighborhood of zero. So it cannot have a unique solution.

Answer 2

You are right about existence. Uniqueness is a different matter, since $t\,\sqrt y$ is not Lipschitz near $y=0$. In fact, there is no uniqueness for the initial condition $y(0)=0$.

Answer 3

If you restrict yourself to $\{(t, y) \in \mathbb{R}^2 : t \in \mathbb{R}, \, y > 0\}$, then yes, the Picard-Lindelof states that as long as a solution stays there, it is unique. And the standard theorem on separable equations gives that its solution is $y(t) = (\frac{1}{4} t^2 + C)^2$.

But both $y_1 \equiv 0$ and $y_2(t) = \frac{1}{16}t^4$, $t \ge 0$, satisfy the initial value problem $y' = t y^{1/2}$, $y(0) = 0$.

Please, use MathJax!