# Explaining results involving differential equations using the theorem for the existence and uniqueness of the solutions of a Cauchy Problem

Let $f(x,y) = 3(y-1)^{2/3y}$. I want to show that $y' = f(x,y)$ has two solutions for $y_0 = y(0) = 1$ and one solution for $y_0 = y(0) = 2$.

I am trying to understand the solution to this problem.

I understand all the computations that are done but I won't put all the details. just the idea of the solution

Let me start out by writing the theorem for the existence and uniqueness of the solution of the Cauchy Problem:

Let $U \in \mathbb{R}$ be an open subset. Let $f:U \rightarrow \mathbb{R^m}$ be a continuous function. $a,b \in \mathbb{R}_{>0}$ s.t. $A := \bar{B}_{\mathbb{R}}(x_0, a) \times \bar{B}_{\mathbb{R^n}}(y_0, b) \subset U.$ And $f\restriction_A$ is a lipschitz function. Then the Cauchy problem $y' = f(x,y), x(x_0) = y_0$ has one solution $\psi: [x_0-\alpha, x_0 + \alpha] \rightarrow \mathbb{R^m}$ where $\alpha = min(a, b/M), M= max_{x,y \in A} ||f(x,y)||_{\mathbb{R^m}}$.

• So to start the solution, they want to check that $f$ is a lipschitz function in the neighborhood of the the interval $(0,2)$

I suppose, A in the theorem is $(0,2)$ here, in which case, how was the interval chosen

• In order to . do so, let us consider the function $f$ on $[-1,1] \times [\dfrac{3}{2},\dfrac{5}{2}]$

Where do the intervals $[-1,1]$ and $[\dfrac{3}{2},\dfrac{5}{2}]$ come from?

• Now, in order to show that it's a lipschitz function, I apply the following theorem: Let $f: [a,b] \rightarrow \mathbb{R}$ continuous. $\exists \xi \in [a,b]$ s.t. $f(b)-f(a) = f'(\xi)(b-a)$ and $|f(b) - f(a)| \leq max_{[a,b]} f'(\xi)(b-a)$

• We apply this theorem and we get that $|f(x,y)-f(x,z)| \leq max_{[3/2,5/2]}\dfrac{1}{(w-1)^{3/2}}|y-z|$

The commputations I understand.

• Now let us show that $f$ is not a lipschitz function on $(0,1)$