2
$\begingroup$

I want to get some advice from you about the existence (and the uniqueness if possible) of a strong solution on my SDE. In fact, due to the structure of the problem that I consider, both the drift (usually denoted by mu) and diffusion (usually denoted by sigma) coefficients are discontinuous. More specifically, both coefficients are positive, stationary (i.e. time-homogeneous), and some polynomials in the state of the system. And each of them has a jump only at one point. In fact, my SDE is similar to the Tanaka equation but it has a drift term additionally and they are linear or quadratic rather than a step function. More precisely, the SDE is

$$ dX_t= (a+b(X_t))^2\frac{X_t^2}{X_t+1} dt + (a+b(X_t))X_t dB_t $$

where $ a>0 $, and $ b(X_t)=b $ if $ X_t \leq k $ and $ b(X_t)=0 $ if $ X_t > k $ for some constant $k>0$. The domain of $X_t$ is $(0,\infty)$ and of course I know $X_0$.

Since I know that the Tanaka equation doesn't admit a strong solution and my SDE is similar to it, I am afraid that my SDE only has a weak solution and doesn't have a strong solution. Sometimes a weak solution is fine, but I think that it is not the case for my problem. I tried to get some ideas from some journal papers, and to the best of my knowledge, I found out some results when the drift term is discontinuous, but the diffusion coefficient is still required to satisfy some regularity conditions like Lipschitz. Do you guys have any idea about this issue? Is there any general theorem that a strong solution doesn't exist if the diffusion coefficient has a jump like the Tanaka equation? Or is there any way to show that my SDE has a strong solution? Thanks for your help in advance!

$\endgroup$
0
$\begingroup$

A partial but quick answer. Put the equation in the natural scale, namely take $\varphi(x)=2\frac{x-1}{x+1}$ and set $Y_t = \varphi(X_t)$, then $$ dY=\frac14(4-Y^2)f\Bigl(\frac{2+Y}{2-Y}\Bigr)\,dB, $$ where $f(x)=a+b$ if $x\leq k$ and $f(x)=a$ otherwise.

Feller's test tells us that any solution starting in $y\in(-2,2)$ (consider that in the natural scale $0$ is $-2$ and $\infty$ is $2$ will go to $-2$ with probability $\frac{2-y}4$ and to $2$ otherwise, but this will not happen in finite time. In particular, the solution with initial condition $-2$ has a unique strong solution.

Now, choose $a$, $b$ and $k$ so that $\frac{2+y}{2-y}$ is $\mathop{sgn}$ (this is arbitrary, i need to think about the general case), then you can proceed as in the Tanaka equation: assume you have a strong solution, apply Ito-Tanaka formula to $|Y|$ and check that the driving Brownian motion is measurable with respect to $|Y|$, a contradiction.

(More details on the scale function, Feller's test, and the proof of non-existence strong solution for the Tanaka equation can be found on standard references, such as this or this)

$\endgroup$
  • $\begingroup$ Thanks for your answer! It is really helpful. And I have a few following-up questions. How can I know if this (transformed) SDE has a strong solution if Y0=-2? Is this the result that comes from Feller's test? But what do you mean by Feller's test here is Feller's test for explosion? I can't easily understand the argument here. So, the existence of a strong solution depends on the initial condition of the process? Thank you again! $\endgroup$ – user48678 Mar 25 '14 at 22:37
  • $\begingroup$ Never mind my first question. If Y0 is either -2 or 2, then Y is simply constant, right? And regarding the second part of your comment, in fact, I can't choose the number a,b, and k. In fact, they are given parameters or decision variables. I am not sure if the same argument from Tanaka holds here, but since you make the problem closer to it, I will try it. Thanks again! $\endgroup$ – user48678 Mar 25 '14 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.