The following theorem states the existence and uniqueness of a strong solution for a class of SDEs.
Theorem. For the SDE $$ \mathrm{d} X=G(t, X(t)) \mathrm{d} t+H(t, X(t)) \mathrm{d} W(t), \quad X\left(t_{0}\right)=X_{0}, $$ assume the following hold.
(1) Both $G(t, x)$ and $H(t, x)$ are continuous on $(t, x) \in\left[t_{0}, T\right] \times \mathbb{R}$.
(2) The coefficient functions $G$ and $H$ satisfy the Lipschitz condition $$ |G(t, x)-G(t, y)|+|H(t, x)-H(t, y)| \leq K|x-y| . $$ (3) The coefficient functions $G$ and H satisfy a growth condition in the second variable, $$ |G(t, x)|^{2}+|H(t, x)|^{2} \leq K\left(1+|x|^{2}\right), $$ for all $t \in\left[t_{0}, T\right]$ and $x \in \mathbb{R}$.
Then the SDE has a strong solution on $\left[t_{0}, T\right]$ that is continuous with probability 1 and $$ \sup _{t \in\left[t_{0}, T\right]} \mathbb{E}\left[X^{2}(t)\right]<\infty, $$ and for each given Wiener process $W(t)$, the corresponding strong solutions are pathwise unique, which means that if $X$ and $Y$ are two strong solutions, then $$ \mathbb{P}\left[\sup _{t \in\left[t_{0}, T\right]}|X(t)-Y(t)|=0\right]=1 . $$
(I took it from Dunbar, S.R. Mathematical Modeling in Economics and Finance: Probability, Stochastic Processes, and Differential Equations; Vol. 49, American Mathematical Soc., 2019.)
Anyway, in my case I have an SDE where $H$ is constant and $G :\mathbb R_+\times\mathbb R\to\mathbb R$ is such that $G(t, X(t))=F(t)X(t)$ where $F(t)=\frac{1}{\cos t+ \sin t}$. Let $t_0=0$, $T>0$ and let $\mathcal A$ be the domain of $F(t)$, i.e. $\mathcal A=\left\{t|\cos t + \sin t\neq0\right\}$. My question is: does the above SDE admits a unique strong solution?
My attempt
Point (1) is satisfied because $G$ is continuous $(t, x) \in\left(\left[0, T\right]\cap\mathcal A\right) \times \mathbb{R}$. Point (2) is not true globally but for a local Lipschitz condition if $$K=\sup_{t\in\left(\left[0, T\right]\cap\mathcal A\right)}F(t)$$ Similar consideration can be done for point (3).
Can I infer the existence of a strong solution on $\left[0, T\right]\cap\mathcal A$? I think that the answer is not so obvious because I replace a global Lipschitz condition, in point (2), with a local one. What do you think?
You have that $sin\,t + cos\, t = 0$ at $t = \frac{3}{4}\pi$. So we must require $T<\frac{3}{4}\pi$.
For any fixed $t \in [0,T]$, the functions $H(t,x) = const$ and $G(t,x) = F(t) x$ are both globally Lipschitz in $x$. The first function is constant and the second is linear in $x$.
It is not required to be Lipschitz in $t$, since the two functions are continuous in $t$ on a compact domain.