(1) show that $f(x, y) = y^{1/2}$ does not satisfy a Lipschitz condition on the rectangle $x ∈ [−1, 1], y ∈ [0, 1]$;
(2) show that $f(x, y) = y^{1/2}$ does satisfy the Lipschitz condition on the rectangle $x ∈ [−1, 1], y ∈ [c, d]$, where $0 < c < d$.
(3) What can you say about the solutions of the differential equation $y' = \sqrt{y}$, equipped with an initial condition $y(x_0) = y_0$ when $(x_0, y_0)$ belong to the closed rectangles in (1) and (2)?
We have $f(x,y) = y^{1/2}$ and the rectangle $x ∈ [−1, 1], y ∈ [0, 1]$. By the definition of a rectangle, $f(x,y)$ is defined on it, since $y \geq 0$. Working over the Lipschitz facts :
$$|f(x,y_2) - f(x,y_1)| = |y_2^{1/2} - y_1^{1/2}|$$
Checking to see if there exists such a Lipschitz bound, let's work over MVT. Considering the function $g(y) = y^{1/2}$, which is differentiable in $[0,1]$, means you can apply the MVT, which means that there exists $ξ \in (y_1,y_2) :$
$$g'(ξ) = \frac{y_2^{1/2} - y_1^{1/2}}{y_2-y_1} = \frac{1}{2}ξ^{-1/2}$$
Taking absolute values over the expression, we can yield :
$$|y_2^{1/2}-y_1^{1/2}|=\frac{1}{2}|ξ^{-1/2}||y_2-y_1|$$
But that's an unbounded expression which can range everywhere, depending on the values of $y_2,y_1$ and of course $ξ$. Thus, the given $f(x,y)$ does not satisfy a Lipschitz condition within the given rectangle.
You can work in a similar way to prove that in the second case, you can get a Lipschitz bound.
Note - General Tip: Please notice the difference between $\sqrt{x}$ and $x^{1/2}$. They are only equal in the case of $x\geq 0$. For negative/other numbers, the equality does not hold and in such cases you'd need to clear out which branch of the root are you working over.
For the third and last part, we have the IVP : $y' = \sqrt{y}, \space y(x_0) = y_0$ with $(x_0,y_0)$ belonging in the rectangles above. Note that you have an IVP of the form :
$$y' = f(x,y), \space y(x_0)=y_0$$
This is a classic case of Piccard's - Peano's existence theorems (and uniqueness) and applications via Lipschitz conditions. More specifically :
The function $f(x,y) = \sqrt{y}$ is continuous in for $y\in [0,1]$. Then, this means that it will also be continuous in a domain of the form :
$$D= \{(x,y) \in \mathbb R^2 : |y-y_0|<a, |x-x_0| < b\}, \space a,b>0$$
which means that by the theorems stated above, that there exists a solution to the IVPs. Moreover though, if $f(x,y)$ is Lipschitz continuous on the rectangle (domain such as the above one) that you are testing, then this means that the solution will be unique in that domain. On the other hand, if Lipschitz conditions cannot be fulfilled, it just means that you cannot make a safe conclusion about the uniqueness or non-uniqueness (the solution may be unique, but you cannot be sure - the solution may not be unique - a solution may not even exist IF the function won't be continuous in such a $D$, which is not the case here)
I think you can make conclusions about the exercise now and proceed on a complete and thorough solution.
