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I look for examples of SDEs (stochastic differential equations) s.t. the uniqueness of the solution fails, i.e.

$$dX_t = B(t,X_t)dt + \Sigma(t,X_t)dW_t,\quad \forall t\ge 0.$$

More precisely, the coefficients $B,\Sigma$ satisfy :

  1. $B: \mathbb R_+\times \mathbb R\to\mathbb R$ is Lipschitz and bounded;
  2. $\Sigma:\mathbb R_+\times \mathbb R\to\mathbb R_+$ is of form $\Sigma(t,x)=(1+{\bf 1}_{\{B(t,x)>0\}})\sigma(t,x)$, where $\sigma$ is Lipschitz, bounded and $\inf_{(t,x)}\sigma(t,x)>0$.

The set of discontinuities of $\Sigma$ is clearly included in $E:=\{(t,x)\in\mathbb R_+\times\mathbb R: B(t,x)=0\}$. Under which conditions the above SDE has a unique solution? Under which conditions the above SDE has more than one solution?

Any answer, comments and references are highly appreciated.

PS : Here we the uniqueness can be considered for either strong solution or weak solution.

Two cases are trivial. If $B$ does not change sign, i.e. $B>0$ or $B\le 0$, there is a unique solution. If $B\equiv B(t)$, and the set $E^o:=\{t\ge 0: B(t)=0\}$ is negligible.

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  • $\begingroup$ It's probably a good idea to specify the notion of uniqueness (pathwise?) and I think there is an omission in the definition of $E$. $\endgroup$ Commented Jul 22, 2022 at 16:40
  • $\begingroup$ @NawafBou-Rabee Thanks for pointing out this. Edits have been made $\endgroup$
    – GJC20
    Commented Jul 22, 2022 at 19:13
  • $\begingroup$ Thanks, the definition of $E$ still seems to be omitting a condition on $B(t.x)$. $\endgroup$ Commented Jul 22, 2022 at 19:39
  • $\begingroup$ @NawafBou-Rabee You are absolutely right! Corrected $\endgroup$
    – GJC20
    Commented Jul 24, 2022 at 8:21

1 Answer 1

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If the diffusion coefficient is not continuous but uniformly positive, there is a general pathwise uniqueness result for homogenous SDE coefficients:

Theorem (Pathwise Uniqueness). Suppose the SDE coefficients are homogeneous, measurable and bounded; and, the noise coefficient is uniformly positive and of bounded variation on any compact interval. Then pathwise uniqueness holds.

(To be sure, pathwise uniqueness implies strong and hence weak uniqueness.) A reference for this pathwise uniqueness result is:

Nakao, Shintaro, On the pathwise uniqueness of solutions of one-dimensional stochastic differential equations, Osaka J. Math. 9, 513-518 (1972). ZBL0255.60039.

Remark 1. Generalizations of this result to time-dependent coefficients, but with several fairly technical requirements on the coefficients, can be found in:

Nakao, Shintaro, On pathwise uniqueness and comparison of solutions of one-dimensional stochastic differential equations, Osaka J. Math. 20, 197-204 (1983). ZBL0517.60061.

Veretennikov, A. Yu., On the strong solutions of stochastic differential equations, Theory Probab. Appl. 24, 354-366 (1980). ZBL0434.60064.

Remark 2. If one only requires weak uniqueness, then the assumption of bounded variation of the noise coefficient can be dropped (and the result holds for planar SDEs too) but again one requires homogeneity of the coefficients, see:

Krylov, N. V., On Ito's stochastic differential equations, Theory of Probability and Its Applications, 14, No. 2, 330-336 (1969).

Remark 3. Even for the homogeneous SDE $dX_t = \Sigma(X_t) dW_t$, continuity and positivity of the noise coefficient do not guarantee pathwise uniqueness; see

Barlow, M. T., One dimensional stochastic differential equations with no strong solution, J. Lond. Math. Soc., II. Ser. 26, 335-347 (1982). ZBL0456.60062.

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    $\begingroup$ Thanks for the answer and apologies for the delayed reply due to family reasons. Indeed, for my case the drift term $B$ has the form $B(t,x)=b(x)+c(t)$ with $c\ge 0$. Thus, even $/sigma/equiv /sigma(x)$, the coefficients can not be homogeneous. I will read carefully the references that you listed and accept your answer as soon as I finish $\endgroup$
    – GJC20
    Commented Jul 28, 2022 at 11:53
  • $\begingroup$ For a weak solution of the SDE you describe, it seems Girsanov might do the trick as implemented, e.g., mathoverflow.net/questions/424364/…. The idea is to relate the inhomogeneous SDE to a homogeneous one. See also mathoverflow.net/questions/426905/… $\endgroup$ Commented Aug 26, 2022 at 14:02

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