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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?

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1 Answer 1

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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.

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  • $\begingroup$ Just to add something more.... (and to catch the bonus). You cannot define the time domain as you did. From an intuitive point of view, it doesn't make sense to remove isolated points from the time interval. There are no time holes! From a mathematical point of view, the time domain must be a compact domain to guarantee that F(t) is bounded. Your domain is an open interval and it is not valid. $\endgroup$
    – Cantaro
    Jul 2, 2022 at 12:52
  • $\begingroup$ Quite disappointed I didn't get the bonus. But I repeat, $[0,T] \cap \mathcal{A}$ is not compact. In this domain the function F(t) explodes. And this condition does not satisfy the Theorem assumptions. $\endgroup$
    – Cantaro
    Jul 4, 2022 at 8:04
  • $\begingroup$ and $K$ is $\infty$.... not very useful. $\endgroup$
    – Cantaro
    Jul 4, 2022 at 8:10

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