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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 $(0,1)$

Once again where does $(0,1)$ come from and why do we have to show that f is not a lipschitz function on this interval$

  • Then the last step is simply to show that $y' = f(x,y)$ has two solutions for $y_0 = y(0) = 1$ and one solution for $y_0 = y(0) = 2$.

Could someone try to explain and maybe claify the different steps of the solution

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Your main confusion is to think that $(0,1)$ and $(0,2)$, given in your explanation, are intervals. In fact, they are the points of the initial condition of the Cauchy Problem, i.e. $(x_0,y_0) = (0,1)$ and $(x_0,y_0) = (0,2)$ for both questions in your statement.

For the second case, the one that you want to prove the uniqueness of the Cauchy Problem $$ \begin{cases} f(x,y) = 3(y-1)^{2/3y} \\ y_0 = y(0) = 2 \end{cases}, $$ you must find the set $A$ of the theorem to use it. This set is $A = [-1,1] \times \left[\dfrac{3}{2},\dfrac{5}{2}\right]$. Notice that, using the notation of your enunciation of the theorem, \begin{align*} &\bar{B}_{\mathbb{R}}(x_0, a) = \bar{B}_{\mathbb{R}}(0, 1) = [-1,1]\\ &\bar{B}_{\mathbb{R^n}}(y_0, b) = \bar{B}_{\mathbb{R}}(2, 1/2) = \left[\dfrac{3}{2},\dfrac{5}{2}\right], \end{align*} where here your dimension $n$ of the second ball is $n=1$ (you only have one dependent variable, $y$), $a$ and $b$ are the radious of the balls (in your case simply intervals) $1$ and $1/2$, respectively.

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