Find Existence of Solution to Cauchy Problem in Space $x^2+y^2 \leq 4$

[evinda]

Hello! (Wave)

We have the space $x^2+y^2 \leq 4$ and we consider the Cauchy problem $y'=\cos x-1-y^2, y(0)=1$.

I want to find for which $b>0$ the Picard theorem ensures the existence of the solution on $(-b,b)$.

I have thought the following:

Since $g(x,y):=\cos x-1-y^2$ is continuous as for $x$ in a space $A$ it remains to be Lipschitz as for $y$ in any closed and bounded subset of $A$, let $B=(b_1,b_2) \times [-c_1,c_1]$.

$|g(x,y_1)-g(x,y_2)|=|y_2^2-y_1^2|=|y_1-y_2| | y_1+y_2| \leq (|y_1|+|y_2|) |y_1-y_2|$.

Thus we can pick $c_1$ as $b$. Am I right?

 

[jvicens]

Hello! Yes, your reasoning is correct. According to the Picard theorem, if $g(x,y)$ is continuous and Lipschitz with respect to $y$ on a closed and bounded subset of the space $A$, then there exists a unique solution to the Cauchy problem $y' = g(x,y), y(0)=y_0$ on the interval $(-b,b)$, where $b>0$ and $y_0$ is the initial condition. In this case, $g(x,y) = \cos x - 1 - y^2$ is indeed continuous and Lipschitz with respect to $y$ on the closed and bounded subset $B=(b_1,b_2) \times [-c_1,c_1]$, where $b_1=0, b_2=2$ and $c_1=1$. Therefore, the Picard theorem ensures the existence and uniqueness of the solution on the interval $(-b,b)$ for any $b>0$. I hope this helps!