Show that the solution to a SDE is strong

Question

I have the following SDE

\begin{equation} dX_t = - \frac{1}{1+t}X_t dt + \frac{1}{1+t}dB_t \end{equation}

that has the solution:

\begin{equation} \begin{aligned} X_t = \frac{X_0 + B_t}{1+t} = \frac{B_t}{1+t} \;\;\; X_0 = 0 \end{aligned} \end{equation}

Now how can I show that this is a strong solution?

I have found online that I should show that this 2 conditions are met:

\begin{align} |\mu(t,x)| + |\sigma(t,x)| \leq c(1+|x|) \\ |\mu(t,x) - \mu(t,y)|+|\sigma(t,x) - \sigma(t,y)| \leq D|x-y| \end{align}

I know that $\mu(t,x) = \frac{-1}{1+t}X_t$ and $\sigma(t,x) = \frac{1}{1+t}$ so the first inequality would be

\begin{equation} \begin{aligned} \frac{1}{1+t}(|X_t| + 1) \leq c(1 + |X_t|) \\ \frac{1}{1+t} \leq c \end{aligned} \end{equation}

which should be fulfilled as when $t \rightarrow \infty$ the LHS goes to 0. For the second one I'm just not sure how to advance.

Answer 1

This was too long for a comment:

On the basis of the above, $\sigma(t,x) = \sigma(t)$ does not depend on $x$. Therefore, the calculations looks similar to your derivations done with the first inequality.

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Let $D<\infty$ be a constant and define,

$$\sigma(t,\cdot)=\sigma(t)=\frac{1}{1+t}, \qquad \mu(t,X_t) = \frac{-X_t}{1+t} \qquad \text{and} \qquad \mu(t,Y_t) = \frac{-Y_t}{1+t}$$

Then see that:

\begin{align} |\mu(t,x) - \mu(t,y)|+|\sigma(t,x) - \sigma(t,y)| &= \bigg|\frac{-X_t}{1+t} - \left(\frac{-Y_t}{1+t}\right)\bigg| + \bigg|\frac{1}{1+t}-\frac{1}{1+t}\bigg|\\ &=\bigg|\frac{1}{1+t} \cdot \left(Y_t - X_t\right)\bigg|\\ &=\frac{1}{1+t} |Y_t - X_t|\\ &\leq D |X_t - Y_t|, \end{align} is satisfied when $\frac{1}{1+t}\leq D$ which is true for $t \rightarrow \infty$. In conclusion, the SDE satisfy the Lipschitz condition and has a (unique) strong solution. I might have missed some mathematical formalities. Nevertheless, this is how I would approach the second inequality.