There is a stochastic differential equations: $$dX_t=a(t,X_t)dB_t+b(t,X_t)dt\quad t\in[0,T]$$ where $B_t$ is the one-dimensional Brownian motion.The "a" and "b" are functions: $$a,b:[0,T]\times R \rightarrow R.$$ Of course,there are some restrictions on the "a" and "b": $$|a(t,x)-a(t,y)|\leqslant L|x-y|\\(x-y)(b(t,x)-b(t,y))\leqslant L(x-y)^2\\|b(t,x)-b(t,y)|\leqslant L(1+|x|^{q-1}+|y|^{q-1})|x-y|$$ In short,the "a" satisfies the global Lipschitz condition,but the "b" satisfies the "semi-Lipschitz condition":I mean that it satisfies half the Lipschitz condition,and is stronger than the local Lipschitz.
Now,my question is :Under the above conditions, how to prove that the global solution of this equation exists？Suppose the initial value of the solution is $X_0\in L^2$.
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Here's my idea, but I can't figure it out：
First of all,the equation must have a local solution$(X,\tau)$.Because "a" and "b" is local Lipschitz，it is obvious by the classical conclusion.
Then,I want to prove the conclusion :$$\mathbb{P}(\tau=\infty)=1$$ by the ito formula and Gronwall inequality.In more detail,I hope to get the following conclusion by the ito formula：$\forall N$
$$\mathbb{E}\sup\limits_{t\in[0,T]}X^2_{t\wedge \tau_{N}}\leqslant \mathbb{E}X_0^2+C\int_0^t\mathbb{E}\sup\limits_{s\in[0,t]}X^2_{s\wedge \tau_{N}}ds$$
where $C$ is a constant.After that, the following conclusions are obtained by Gronwall inequality:
$$\mathbb{E}\sup\limits_{t\in[0,T]}X^2_{t\wedge \tau_{N}}\leqslant Ce^{CT}$$
where $C$ is still a large enough constant.The conclusion "$\mathbb{P}(\tau=\infty)=1$" is obvious if we make $N\rightarrow \infty$.
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My difficulty is,when I try to utilize the ito formula and expand the expectation by the Lipschitz condition,there is an item that I can't handle:
$$\int_0^tX_sb(s,X_s)ds$$
With the help of known condition,we can only get:
$$\int_0^tX_sb(s,X_s)ds\leqslant \int_0^tX_sb(s,0)ds+L\int_0^tX_s^2ds$$
The first item on the right is causing me so much trouble, I have no clue what to do with it.I don't want it to be a power of 1, instead I want it to be 2 so that I can use Gronwall inequality.I hope someone can help me with this trouble or have other better ideas to prove this problem.Thank you very much!