Picard Iteration and Gronwall in SDE Uniqueness Proofs

From an SDE to a Picard Scheme

For an ordinary differential equation, Picard iteration repeatedly substitutes an approximate path into the integral equation. The same idea applies to an SDE, but a convergence proof must control both the ordinary time integral and the Ito integral. Global Lipschitz bounds make successive differences contract after integration; iterating the resulting estimate produces a factorial in the denominator. Gronwall’s lemma then gives a concise uniqueness argument.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ satisfying the usual conditions, carrying a Brownian motion $B_t$ relative to this filtration. Fix $T>0$ and a deterministic initial value $x_0\in\mathbb{R}$. Consider the SDE $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$, $X_0=x_0$.

Definition: A strong solution on $[0,T]$ here is a continuous, adapted process $X_t$ satisfying the corresponding integral equation almost surely for every $t\in[0,T]$, with $\mathbb{E}[\sup_{0\le t\le T}|X_t|^2]<\infty$. The stochastic integral is the Ito integral of a progressively measurable integrand in $\mathcal{L}^2$: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. A strong solution is built on the given space with the given Brownian motion. Pathwise uniqueness means that any two such solutions on that space, driven by the same Brownian motion and with the same initial value, are indistinguishable. This is different from uniqueness in law, which asserts equality of distributions and does not require two solutions to be driven by the same Brownian path.

The assumptions below are uniform over $t\in[0,T]$. Joint Borel measurability is included so that evaluating the coefficients along a progressively measurable process gives measurable integrands; bounds alone would not ensure that the time integrals exist.

Assumptions: The functions $\mu,\sigma:[0,T]\times\mathbb{R}\to\mathbb{R}$ are jointly Borel measurable. There are finite constants $L,C\ge0$ such that for all $t\in[0,T]$ and $x,y\in\mathbb{R}$, $$ |\mu(t,x)-\mu(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le L|x-y|, \qquad |\mu(t,x)|+|\sigma(t,x)|\le C(1+|x|). $$ The first inequality is the global Lipschitz condition; the second is linear growth.

Gronwall’s Lemma

We will use an integral inequality in both the Picard estimates and the uniqueness proof. Its zero-initial-value case is particularly important: it says that a nonnegative function bounded by a constant times its own accumulated past must vanish.

Lemma: Let $u:[0,T]\to[0,\infty)$ be continuous, and suppose $a,b\ge0$ and $u(t)\le a+b\int_0^t u(s)\,ds$ for every $t\in[0,T]$. Then $u(t)\le ae^{bt}$. In particular, if $a=0$, then $u(t)=0$ throughout $[0,T]$.

Proof. First suppose $a>0$, and set $v(t)=a+b\int_0^t u(s)\,ds$. Then $u(t)\le v(t)$, $v(0)=a$, and $v'(t)=bu(t)\le bv(t)$. Differentiating $e^{-bt}v(t)$ gives $(e^{-bt}v(t))'=e^{-bt}(v'(t)-bv(t))\le0$. Thus $e^{-bt}v(t)\le v(0)=a$, so $u(t)\le v(t)\le ae^{bt}$. If $a=0$, apply the result just proved with any $\varepsilon>0$ in place of $a$, since $u(t)\le\varepsilon+b\int_0^t u(s)\,ds$. This gives $u(t)\le\varepsilon e^{bt}$ for every $\varepsilon>0$; letting $\varepsilon$ decrease to zero proves $u(t)=0$. $\square$

Picard Iteration and the Factorial Estimate

Define $X^0_t=x_0$ and, for each integer $n\ge0$, define $$ X^{n+1}_t=x_0+\int_0^t\mu(s,X^n_s)\,ds+\int_0^t\sigma(s,X^n_s)\,dB_s. $$ The time integral is the ordinary Lebesgue integral. Each stochastic integral is an Ito integral. At every iteration, its integrand must be progressively measurable and square-integrable; the estimates below verify these requirements. The Ito isometry used here states that for $H\in\mathcal{L}^2$, $\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds$.

We also use Doob’s $L^2$ maximal inequality: if $M_t$ is a square-integrable martingale, then $\mathbb{E}[\sup_{r\le t}|M_r|^2]\le4\mathbb{E}[|M_t|^2]$. This is a standard maximal inequality; its proof is not given here. For an Ito integral with an $\mathcal{L}^2$ integrand, the integral is a square-integrable martingale: its second moment is finite by the Ito isometry, and its conditional expectation at a later time, given $\mathcal{F}_s$, equals its value at $s$. The latter follows first for adapted step integrands from the Brownian increment property and then for $\mathcal{L}^2$ integrands by the construction of the Ito integral.

Theorem: Under the stated measurability, global Lipschitz, and linear growth assumptions, the Picard iterates converge in $L^2(\Omega;C([0,T]))$ to a strong solution. This solution is pathwise unique among strong solutions as defined above.

Proof. Write $D_n=X^n-X^{n-1}$ for $n\ge1$, and put $A_n(t)=\mathbb{E}[\sup_{0\le r\le t}|D_n(r)|^2]$. For the first difference, $D_1$ is the sum of the drift and stochastic integrals evaluated at $x_0$. Using $|p+q|^2\le2|p|^2+2|q|^2$, Cauchy–Schwarz for the time integral, the growth bound, and Doob’s inequality with the Ito isometry gives, for $t\le T$, $$ A_1(t)\le 2t\int_0^t|\mu(s,x_0)|^2\,ds +8\int_0^t|\sigma(s,x_0)|^2\,ds \le 2C^2(1+|x_0|)^2(T^2+4T)=:K. $$ This is finite. In particular, the first stochastic integrand is square-integrable. Inductively, the Lipschitz condition and the bound on $A_n$ imply square-integrability of each next stochastic integrand; the time integrals are also well-defined.

For $n\ge1$, subtract the two consecutive iteration equations. For $r\le t$, apply the same square-of-a-sum inequality, Cauchy–Schwarz to the drift, Doob’s inequality to the stochastic integral, and Ito isometry. Since the coefficient differences are at most $L|D_n(s)|$, this yields $$ A_{n+1}(t) \le 2L^2t\int_0^t\mathbb{E}|D_n(s)|^2\,ds +8L^2\int_0^t\mathbb{E}|D_n(s)|^2\,ds \le 2L^2(T+4)\int_0^t A_n(s)\,ds. $$ Set $c=2L^2(T+4)$. Starting from $A_1(t)\le K$, induction gives $A_n(t)\le Kc^{n-1}t^{n-1}/(n-1)!$: the claim holds for $n=1$, and integrating its bound in the recurrence gives the claim for $n+1$.

At $t=T$, the series $\sum_{n\ge1}\sqrt{A_n(T)}$ converges. If $cKT=0$, this follows directly from the bounds (all terms after the first vanish when $c=0$); otherwise the ratio of consecutive bounds is $\sqrt{cT/n}$, which tends to zero. By the triangle inequality in $L^2(\Omega;C([0,T]))$, the partial sums of $D_n$ are Cauchy in that space. Since $X^N=x_0+\sum_{n=1}^N D_n$, the iterates converge there to a continuous process $X$. A subsequence converges almost surely uniformly, so $X$ is adapted, using the usual completeness conditions on the filtration.

It remains to pass to the limit in the equation. The Lipschitz bound gives $\mathbb{E}\int_0^T|\mu(s,X^n_s)-\mu(s,X_s)|^2\,ds\le L^2\int_0^T\mathbb{E}|X^n_s-X_s|^2\,ds\to0$; the same bound holds for $\sigma$. Cauchy–Schwarz makes the drift integrals converge in $L^2$, and Ito isometry makes the stochastic integrals converge in $L^2$ at each fixed time. Thus the limit satisfies the integral equation. It belongs to $L^2(\Omega;C([0,T]))$ by convergence in that space, so it is a strong solution.

For uniqueness, let $X,Y$ be two strong solutions driven by the same $B_t$, with the same $x_0$, and set $u(t)=\mathbb{E}|X_t-Y_t|^2$. This function is continuous: both processes have continuous paths and finite expected squared supremum, so dominated convergence applies to their squared difference. Subtract their integral equations and apply the same square, Cauchy–Schwarz, and Ito isometry estimates to obtain $$ u(t)\le 2L^2t\int_0^t u(s)\,ds+2L^2\int_0^t u(s)\,ds \le 2L^2(T+1)\int_0^t u(s)\,ds. $$ Gronwall’s lemma with $a=0$ implies $u(t)=0$ for every $t$. For each rational $q\in[0,T]$, $X_q=Y_q$ almost surely; intersecting these countably many probability-one events and using continuity shows $X_t=Y_t$ for all $t\in[0,T]$ on one probability-one event. Hence the solutions are indistinguishable. $\square$

Worked Examples

Worked Example: Constant Drift and Diffusion

Take $dX_t=dt+dB_t$, $X_0=0$, on $[0,2]$. The coefficients are jointly Borel, have Lipschitz constant $L=0$, and satisfy linear growth, for example with $C=1$. The first iterate is $X^1_t=0+\int_0^t1\,ds+\int_0^t1\,dB_s=t+B_t$. Because neither coefficient depends on the state, substituting $X^1$, or any other iterate, leaves both integrands unchanged. Thus $X^2_t=t+B_t$, and every subsequent iterate has that same value.

Check: the candidate has drift integral $\int_0^t1\,ds=t$ and Ito integral $\int_0^t1\,dB_s=B_t$, so it satisfies the SDE. Its mean is $t$, since $\mathbb{E}[B_t]=0$, and its variance is $t$, since $B_t\sim N(0,t)$. At $t=2$, these are $2$ and $2$, respectively.

Worked Example: Deterministic Picard Series

Take $dX_t=X_t\,dt$, $X_0=1$, on $[0,1]$, with $\sigma=0$. Here $\mu(t,x)=x$, so $L=C=1$. The iterates are $X^0_t=1$ and $X^{n+1}_t=1+\int_0^tX^n_s\,ds$. Direct substitution gives $X^1_t=1+t$, $X^2_t=1+t+\tfrac{t^2}{2}$, and $X^3_t=1+t+\tfrac{t^2}{2}+\tfrac{t^3}{6}$. In general $X^n_t=\sum_{k=0}^n t^k/k!$: integrating each term $s^k/k!$ adds $t^{k+1}/(k+1)!$, which proves the formula by induction.

At $t=1$, the third iterate is $1+1+\tfrac12+\tfrac16=\tfrac83$. The series converges to $e^t$, which satisfies $e^t=1+\int_0^t e^s\,ds$, since the integral is $e^t-1$. Check: differentiating the candidate gives $(e^t)'=e^t$ and $e^0=1$, matching the SDE and initial value. The finite iterate at $t=1$ is below the limit, as its omitted series terms are positive.

Worked Example: A Stochastic Difference in the First Iterations

Take $dX_t=X_t\,dB_t$, $X_0=1$, on $[0,2]$, with zero drift. The coefficients are jointly Borel, globally Lipschitz with $L=1$, and of linear growth with $C=1$. The first two iterates are $X^1_t=1+B_t$ and $X^2_t=1+\int_0^t(1+B_s)\,dB_s$. The difference is $D_2(t)=\int_0^tB_s\,dB_s$. The integrand is in $\mathcal{L}^2$, because $\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2$. By Ito isometry, $$ \mathbb{E}|D_2(2)|^2=\int_0^2\mathbb{E}[B_s^2]\,ds =\int_0^2s\,ds=\left.\frac{s^2}{2}\right|_0^2=2. $$ This is a terminal-time second moment, not the maximal-in-time bound used in the theorem.

Check: Ito’s formula for $f(x)=x^2$, with $f'(x)=2x$, $f''(x)=2$, and $dB_t\,dB_t=dt$, gives $d(B_t^2)=2B_t\,dB_t+dt$. Integrating shows $D_2(t)=\tfrac12(B_t^2-t)$. At $t=2$, its second moment is $\tfrac14\mathbb{E}[(B_2^2-2)^2]=\tfrac14(3\cdot2^2-4\cdot2+4)=\tfrac14(12-8+4)=2$, agreeing with the independent Ito-isometry calculation.

What the Estimates Do—and a Common Error

The factorial estimate is stronger than a one-step contraction: the iteration need not reduce each difference by the same fixed factor. Instead, repeated time integration contributes $t^{n-1}/(n-1)!$, which makes the sum of the $L^2$ norms of the differences finite on every fixed bounded time interval. The proof establishes strong existence and pathwise uniqueness in the stated square-integrable class. It does not by itself establish uniqueness in law for arbitrary weak solutions; those are different assertions.

Common mistake: omitting time measurability. Lipschitz and growth bounds concern values of the coefficients, but do not imply that the drift can be integrated in time. For instance, a coefficient of the form $\mu(t,x)=\mathbf{1}_A(t)$, with a non-Lebesgue-measurable set $A\subset[0,T]$, is bounded and independent of $x$, but its time integral is not defined as a Lebesgue integral. The iteration would therefore not even be defined. Requiring joint Borel measurability (as above) prevents this failure and ensures that the coefficient compositions used in the construction are measurable.

Another error is to use only the terminal-time Ito isometry when proving convergence of continuous paths. That controls $\mathbb{E}|D_n(T)|^2$, not $\mathbb{E}\sup_{t\le T}|D_n(t)|^2$. The maximal inequality supplies the latter control for the stochastic integral; Cauchy–Schwarz supplies it for the drift. Both are needed for convergence in $L^2(\Omega;C([0,T]))$.

Check Your Understanding

For each question, identify which assumption or estimate in the proof applies.

Why is joint Borel measurability needed in addition to the Lipschitz and linear growth bounds?
For $A_n(t)=\mathbb{E}[\sup_{r\le t}|X^n_r-X^{n-1}_r|^2]$, derive the factorial bound from the recurrence $A_{n+1}(t)\le c\int_0^tA_n(s)\,ds$.
Which inequality controls the supremum of the stochastic integral, and which controls the supremum of the drift integral?
Explain why $u(t)\le b\int_0^t u(s)\,ds$, with $u$ continuous and nonnegative, forces $u$ to vanish.
In the equation $dX_t=X_t\,dB_t$, calculate $\mathbb{E}|\int_0^tB_s\,dB_s|^2$ directly using Ito isometry.