Checking Lipschitz Conditions for a Time-Varying SDE
Summary
The document considers an SDE with a drift proportional to the state and a diffusion coefficient that varies with time but not with the state. It asks how to verify that the proposed process is a strong solution, using the standard linear-growth and Lipschitz conditions. For the Lipschitz check, the diffusion terms cancel when comparing two states, while the drift difference is proportional to their distance. The coefficient is bounded for nonnegative time, so a finite Lipschitz constant exists on that domain.
The response sketches these checks and concludes that the equation has a unique strong solution. It is informal and acknowledges that mathematical details may be missing. In particular, the explanation about the coefficient tending to zero as time grows is not by itself the reason a uniform bound exists; the coefficient must be bounded over the entire time domain. The result also relies on the stated time domain and usual existence and uniqueness assumptions.
Key ideas
- The drift is Lipschitz in the state when its time-dependent multiplier is bounded.
- The diffusion coefficient has no state dependence, so its difference across states is zero.
- Linear-growth and Lipschitz bounds support existence and uniqueness of a strong solution under standard conditions.
- A coefficient tending to zero at large times does not alone establish a bound over the full time domain.
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Full text
# Show that the solution to a SDE is strong
# Show that the solution to a SDE is strong
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 by Pleb (score 0, accepted)
https://quant.stackexchange.com/a/69772
#### 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.
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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