Pathwise Uniqueness Under Global Lipschitz Conditions

Comparing Solutions on the Same Probability Space

Pathwise uniqueness asks whether two solutions of an SDE must agree when they are constructed on the same filtered probability space, with the same Brownian motion and the same initial condition. The key is to compare their difference, rather than to construct either solution. Global Lipschitz bounds turn differences in the drift and diffusion coefficients into bounds proportional to the difference between the solutions. Stochastic-integral estimates then give an integral inequality, and Gronwall’s inequality forces that difference to vanish.

Fix a finite horizon $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Consider the scalar SDE $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$. Assume the coefficients are measurable in time, continuous in the state variable, and globally Lipschitz in that variable: for constants $L_\mu,L_\sigma\ge0$, all $t\in[0,T]$, and all $x,y\in\mathbb{R}$,

|\mu(t,x)-\mu(t,y)|\le L_\mu|x-y|, \qquad |\sigma(t,x)-\sigma(t,y)|\le L_\sigma|x-y|.

Assume also that $\sup_{t\in[0,T]}(|\mu(t,0)|+|\sigma(t,0)|)<\infty$. Together with the Lipschitz bounds, this gives linear growth: for example, $|\mu(t,x)|\le \sup_{s\le T}|\mu(s,0)|+L_\mu|x|$. Take a common $\mathcal{F}_0$-measurable initial value $\xi$ with $\mathbb{E}|\xi|^2<\infty$. The solutions considered below are continuous, adapted processes satisfying the integral SDE and $\mathbb{E}\sup_{0\le t\le T}|X_t|^2<\infty$. Their stochastic integrands are progressively measurable and in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; their drift integrals are well-defined Lebesgue integrals.

Definition: The SDE has pathwise uniqueness on $[0,T]$ in this solution class if, whenever $X$ and $Y$ are two solutions on the same filtered probability space, driven by the same $B_t$, and satisfying $X_0=Y_0=\xi$ almost surely, then $\mathbb{P}(X_t=Y_t\text{ for every }t\in[0,T])=1$. This conclusion is indistinguishability: the paths agree simultaneously at all times, outside one null set.

This is a statement about solutions that already exist. It is distinct from uniqueness in law, which compares distributions, and from existence, which asks whether a solution can be constructed. In particular, a uniqueness proof alone does not establish that any solution exists.

The Difference Estimate

Let $X$ and $Y$ be two solutions with the same initial value and the same Brownian motion. Write $D_t=X_t-Y_t$, and define $g(t)=\mathbb{E}\sup_{0\le r\le t}|D_r|^2$. The finite second-moment assumption makes $g(t)$ finite. The following estimate is the main calculation.

Proposition: Under the assumptions above, for every $t\in[0,T]$, $$ g(t)\le \bigl(2tL_\mu^2+8L_\sigma^2\bigr)\int_0^t g(s)\,ds. $$

Proof. Subtract the two integral equations. Since the initial values agree, for every $u\le t$, $$ D_u=\int_0^u\bigl(\mu(s,X_s)-\mu(s,Y_s)\bigr)\,ds +\int_0^u\bigl(\sigma(s,X_s)-\sigma(s,Y_s)\bigr)\,dB_s. $$ Set $H_s=\sigma(s,X_s)-\sigma(s,Y_s)$. It is progressively measurable. The Lipschitz bound and the assumed moment bounds give $\mathbb{E}\int_0^T H_s^2\,ds\le L_\sigma^2\int_0^T\mathbb{E}|D_s|^2\,ds<\infty$, so $H\in\mathcal{L}^2$. The Ito integral $M_u=\int_0^u H_s\,dB_s$ is therefore square-integrable: the Ito isometry gives $\mathbb{E}|M_u|^2=\mathbb{E}\int_0^u H_s^2\,ds<\infty$. It is a martingale, with $\mathbb{E}[M_v\mid\mathcal{F}_u]=M_u$ for $u\le v$; this conditional-expectation identity follows first for adapted simple integrands from the mean-zero Brownian increments, and then for $\mathcal{L}^2$ integrands by the defining $L^2$ approximation of the Ito integral.

For the drift part, the Cauchy–Schwarz inequality gives, for every $u\le t$, $$ \left|\int_0^u(\mu(s,X_s)-\mu(s,Y_s))\,ds\right|^2 \le t\int_0^t|\mu(s,X_s)-\mu(s,Y_s)|^2\,ds. $$ Using the Lipschitz bound, taking the supremum over $u\le t$, and then expectation, the right-hand side is at most $tL_\mu^2\int_0^t\mathbb{E}|D_s|^2\,ds\le tL_\mu^2\int_0^t g(s)\,ds$.

For the stochastic part, Doob’s $L^2$ maximal inequality states that a square-integrable martingale starting at zero satisfies $\mathbb{E}\sup_{u\le t}|M_u|^2\le4\mathbb{E}|M_t|^2$. This standard maximal inequality is used here without proof. The Ito isometry and the Lipschitz bound then give $$ \mathbb{E}\sup_{u\le t}|M_u|^2 \le4\mathbb{E}\int_0^t H_s^2\,ds \le4L_\sigma^2\int_0^t g(s)\,ds. $$ Finally, $|a+b|^2\le2|a|^2+2|b|^2$. Apply this inequality to the two integrals, take the supremum and expectation, and combine the bounds. The drift contribution is at most $2tL_\mu^2\int_0^t g(s)\,ds$; the martingale contribution is at most $8L_\sigma^2\int_0^t g(s)\,ds$. Their sum is the asserted estimate. $\square$

Gronwall’s Inequality Forces Equality

Theorem: Under the stated global Lipschitz and solution assumptions, the SDE has pathwise uniqueness on $[0,T]$.

Proof. From the proposition and $t\le T$, set $C=2TL_\mu^2+8L_\sigma^2$ to obtain $g(t)\le C\int_0^t g(s)\,ds$ for every $t\in[0,T]$. Define $G(t)=\int_0^t g(s)\,ds$. Since $g$ is nonnegative and integrable, $G(0)=0$, $G(t)\ge0$, and $G'(t)=g(t)$ for almost every $t$. Thus $G'(t)\le CG(t)$ almost everywhere. The absolutely continuous function $e^{-Ct}G(t)$ has derivative $e^{-Ct}(G'(t)-CG(t))\le0$ almost everywhere, so it is nonincreasing. It starts at zero and is nonnegative; hence it is identically zero. Therefore $G(t)=0$, and the inequality for $g$ gives $g(t)=0$ for every $t$.

In particular, $\mathbb{E}\sup_{0\le r\le T}|X_r-Y_r|^2=0$. A nonnegative random variable with expectation zero is zero almost surely, so $\sup_{0\le r\le T}|X_r-Y_r|=0$ almost surely. This is precisely equality of the two paths at every time, outside one null set. The argument also covers $C=0$: the same differential inequality yields $G'=0$ almost everywhere and $G(0)=0$. $\square$

Worked Estimates

Worked Example: Linear Drift and Diffusion

Suppose the drift is $\mu(t,x)=-2x$, the diffusion coefficient is $\sigma(t,x)=3x$, and $T=1$. Let $X$ and $Y$ be solutions with the same square-integrable initial value and the same $B_t$. Compute the difference estimate.

For any $x,y$, $|-2x-(-2y)|=2|x-y|$, so $L_\mu=2$. Also $|3x-3y|=3|x-y|$, so $L_\sigma=3$. The proposition gives, for $0\le t\le1$, $$ g(t)\le(2t\cdot 2^2+8\cdot3^2)\int_0^t g(s)\,ds =(8t+72)\int_0^t g(s)\,ds \le80\int_0^t g(s)\,ds. $$ Gronwall’s argument implies $g(t)=0$, so the solutions agree almost surely at all times up to $1$.

Check: substituting $t=1$, $L_\mu=2$, and $L_\sigma=3$ into the coefficient gives $2\cdot1\cdot4+8\cdot9=8+72=80$, as in the bound. The estimate does not require an explicit formula for either solution.

Worked Example: Bounded Nonlinear Coefficients

Take $\mu(t,x)=\sin x$, $\sigma(t,x)=\cos x$, and $T=1$. Find Lipschitz constants and the resulting bound for two solutions with the same initial value and driver.

Check: the uniform constant on $[0,1]$ is $2\cdot1\cdot1^2+8\cdot1^2=10$. It bounds the time-dependent coefficient $2t+8$, whose largest value on that interval is $10$.

Worked Example: Why the Initial Values Must Match

Consider $dX_t=4\,dB_t$ and $dY_t=4\,dB_t$ on $[0,2]$, but set $X_0=2$ and $Y_0=-1$. Compute both solutions and compare them.

Integrating each equation gives $X_t=2+4B_t$ and $Y_t=-1+4B_t$. Subtracting, $$ X_t-Y_t=(2+4B_t)-(-1+4B_t)=3 $$ for every $t$. Thus the paths do not agree, even though the Brownian motion and coefficients match. The theorem does not apply because the initial values differ.

Check: at $t=0$, the difference is $2-(-1)=3$, and the calculation shows it remains $3$. The diffusion difference is $4-4=0$, and the drift difference is also zero, so neither changes the initial gap.

What the Estimate Does—and Does Not—Say

The two solutions must be driven by the same Brownian motion for the subtraction in the proof to produce the diffusion difference $\sigma(s,X_s)-\sigma(s,Y_s)$ integrated against one common noise. Equality of the coefficients or equality in distribution alone does not give this comparison. Likewise, matching initial values are essential: as the last example shows, the difference can preserve a nonzero initial gap.

A common mistake is to say that the stochastic integrals cancel because the driving Brownian motion is the same. For $dX_t=X_t\,dB_t$ and $dY_t=Y_t\,dB_t$, subtraction gives $$ X_t-Y_t=(X_0-Y_0)+\int_0^t(X_s-Y_s)\,dB_s, $$ not $(X_0-Y_0)$ alone. The integrals have the same driver, but their integrands need not be equal. The correction is to subtract the integrands and then use $|X_s-Y_s|$ to bound their difference. The Lipschitz condition supplies exactly that bound; the Ito isometry and maximal inequality control the resulting stochastic integral.

The estimate is a uniqueness argument, not an existence argument. It proves that any two solutions in the stated class with the same initial value and driver must coincide. It does not construct a solution, and it does not assert that solutions driven by different Brownian motions have identical paths.

Check Your Understanding

Use the difference estimate and its hypotheses to answer the following questions.

For $T=2$, $L_\mu=1$, and $L_\sigma=2$, what uniform value of $C$ gives $g(t)\le C\int_0^t g(s)\,ds$ on $[0,2]$?
Why does the proof require the stochastic integrand $\sigma(s,X_s)-\sigma(s,Y_s)$ to be in $\mathcal{L}^2$?
Which martingale inequality bounds the expected squared supremum of the stochastic-integral term, and what integrability condition allows its use?
Explain why the same Brownian motion does not, by itself, make the stochastic-integral terms cancel.
Does pathwise uniqueness establish existence of a solution? State what it establishes instead.