Show that $\frac{dy}{dx} = 5y +28 \cos(y), y(0) = 54$ has a unique solution on $\mathbb{R}$.
This is a spin off of one of the problems in Berkeley Problems in Mathematics.
My solution (attempt) is quite alot shorter than the one presented by the authours ( they show that a unique solution exists on some neighbourhood of $(0,54)$ using a local version of Picard's theorem and then use IFT to find an explicit solution on this neighbourhood and prove that this solution is valid on $\mathbb{R}$) so I wanted to check that I hadn't missed something.
Here is my solution:
Let $f(x,y)= 5y +28\cos(y)$. Fix $h >0$. By basic properties of continuous functions $f$ is continuous on $[-h,h] \times \mathbb{R}$ and moreover Lipschitz in $y$ on this strip. This follows from,
$|f_y (x,y)|=|5-28\sin(y)| \leq 5+28|\sin(y)| \leq 5+28 = 33$ and the MVT.
Picard's theorem applies and we see that the IVP has a unique solution on $[-h,h]$.
But $h$ was arbitrary so the IVP has a solution on all of $\mathbb{R}$. $\blacksquare$
Is this correct? In general I am bit unsure about how to prove the uniqueness/existence of global solutions… analytic continuation or global Picard?!
Note the version of Picard's theorem I am using is
The IVP $y'(x) = f(x,y), y(a)=b$, has a unique solution on $\mathbb{R}$ provided, $\forall h:$
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$f$ is continuous on $[a-h, a+h] \times \mathbb{R}$
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$f$ is Lipschitz in y on $[a-h, a+h] \times \mathbb{R}$.
Best Answer
Your idea is correct. With a sub-linear right side you get a global solution. The proof idea is explored for instance in
The problem with your source might be that they did not give the effort to prove this more global version of the theorem after the standard localized one. Thus they have to assemble the solution from many local solutions.
Note that with the formulation of your condition, you only get a solution on $[a-h,a+h]$, which should be no surprise as this is the explored domain of the ODE.