I am not very good at differential equations so I am struggling with this most likely very simple Cauchy problem:
$x' = f(x,t) = \sin(\cos(x)), x(0) = 1$.
I do not have to find a solution, but rather describe it.
Could you please verify my presumptions and give some hints for remaining questions.
Since $f$ and $f'$ are continuous, a unique solution exists. It is an analytic function and it is also continuous for all $t \in R$.
How could I find out if the solution is always positive?
Thanks in advance!
Best Answer
Your claims are right: a unique solution exists and it is analytic. This is an autonomous ODE (i.e., it is of the form $x'=f(x)$). Therefore, its $\textit{phase diagram}$, that is to say, the plot of $x'$ (in the Y axis) versus $x$ (in the X axis) will be the graph of the function f(x), which I will display here:
Now, let's observe that if $x(t_0)=x_0$ is exactly one of the roots of $f$ for some $t_0\in\mathbb{R}$, then $x(t)=x_0$ for all $t\in\mathbb{R}$, because of the uniqueness of the solution (it is a solution, and therefore it is unique). In particular, this means that if $x(0)$ is not a root of $f$, then $x(t)$ can never be a root of $f$ for any $t$.
In this particular case, $x(0)=1$, which is not a root of $f$ ($f(1)=\sin(\cos(1))\neq 0$). By the observation we made before, and since $x$ must be continuous, $x(t)\in(c_1,c_2)\ \forall t\in\mathbb{R}$, where $c_1$ and $c_2$ denote the two roots of $f$ that $1$ lies in between of. Finally, we observe that $f(x)$ is positive for $x\in(c_1,c_2)$, hence, $x'(t)>0\ \forall t\in\mathbb{R}$ and $x$ is strictly increasing with $t$.
Finally (this is a well-known fact in autonomous ODEs), we have that $$\lim_{t\to\infty}x(t)=c_2>0$$ and $$\lim_{t\to-\infty}x(t)=c_1<0$$ thus we can conclude that the solution is not always positive.