Lipschitz Conditions and Linear Growth¶

The existence and uniqueness theory for SDEs rests on two analytical conditions: the Lipschitz condition (controlling uniqueness) and the linear growth condition (preventing finite-time explosion). Together they guarantee a unique, globally-defined strong solution. The constructive proof is given in Picard Iteration; the distinction between strong and weak solutions is developed in Strong vs Weak Solutions.

Setting¶

Consider the $d$-dimensional SDE driven by an $m$-dimensional Brownian motion $W_t = (W_t^1, \ldots, W_t^m)$:

\[ dX_t = b(t, X_t)\,dt + \sigma(t, X_t)\,dW_t, \qquad X_0 = x_0 \in \mathbb{R}^d \]

where $b : [0,T] \times \mathbb{R}^d \to \mathbb{R}^d$ and $\sigma : [0,T] \times \mathbb{R}^d \to \mathbb{R}^{d \times m}$.

Norms used throughout: $|\cdot|$ is the Euclidean norm on $\mathbb{R}^d$; for matrices $|\sigma|^2 = \sum_{i,\alpha}(\sigma^{i\alpha})^2$ (Frobenius norm).

The Lipschitz Condition¶

Definition. The coefficients $(b, \sigma)$ satisfy the global Lipschitz condition if there exists $K > 0$ such that for all $t \in [0,T]$ and $x, y \in \mathbb{R}^d$:

\[ \boxed{|b(t,x) - b(t,y)| + |\sigma(t,x) - \sigma(t,y)| \leq K|x - y|} \]

Interpretation. Lipschitz continuity bounds how rapidly the coefficients can vary with the state $x$. The drift vector field has a uniformly bounded spatial slope, and each column of $\sigma$ varies at most proportionally to displacement.

Why Lipschitz implies uniqueness. Suppose $X_t$ and $Y_t$ are two solutions with $X_0 = Y_0 = x_0$. Set $Z_t = X_t - Y_t$. Then:

\[ Z_t = \int_0^t \bigl[b(s,X_s) - b(s,Y_s)\bigr]\,ds + \int_0^t \bigl[\sigma(s,X_s) - \sigma(s,Y_s)\bigr]\,dW_s \]

Using the Lipschitz bound, the Itô isometry, and Doob's maximal inequality one obtains:

\[ \mathbb{E}\Bigl[\sup_{s \leq t}|Z_s|^2\Bigr] \leq C \int_0^t \mathbb{E}|Z_s|^2\,ds \]

Gronwall's inequality with $Z_0 = 0$ then gives $\mathbb{E}[\sup_{s \leq t}|Z_s|^2] = 0$ for all $t$, hence $X_t = Y_t$ a.s. The full estimates are carried out in Picard Iteration.

Examples of Lipschitz coefficients¶

Ornstein–Uhlenbeck (with mean-reversion speed $\kappa > 0$):

\[ dX_t = -\kappa X_t\,dt + \nu\,dW_t, \qquad K = \kappa. \]

Affine SDE (with scalar coefficients $\alpha, \beta, \gamma, \delta$):

\[ dX_t = (\alpha + \beta X_t)\,dt + (\gamma + \delta X_t)\,dW_t, \qquad K = |\beta| + |\delta|. \]

Smooth, bounded-derivative case. If $b$ and $\sigma$ are $C^1$ in $x$ with bounded partial derivatives, the mean value theorem gives global Lipschitz with $K = \sup_x(|\nabla_x b| + |\nabla_x \sigma|)$.

A non-Lipschitz example: the CIR model¶

\[ dX_t = \kappa(\theta - X_t)\,dt + \sigma\sqrt{X_t}\,dW_t \]

The diffusion coefficient $\sigma\sqrt{x}$ fails global Lipschitz at $x = 0$. Taking $y = 0$ and letting $x \to 0^+$:

\[ \frac{|\sqrt{x} - \sqrt{0}|}{|x - 0|} = \frac{1}{\sqrt{x}} \to +\infty. \]

So no single constant $K$ bounds the ratio $|\sigma(x)-\sigma(y)|/|x-y|$ near the origin. Uniqueness here requires the Yamada–Watanabe condition; see below.

The Linear Growth Condition¶

Definition. The coefficients satisfy the linear growth condition if there exists $K > 0$ such that for all $t \in [0,T]$ and $x \in \mathbb{R}^d$:

\[ \boxed{|b(t,x)|^2 + |\sigma(t,x)|^2 \leq K^2(1 + |x|^2)} \]

Equivalently, $|b(t,x)| + |\sigma(t,x)| \leq K(1 + |x|)$.

Why linear growth prevents explosion. Compare the ODE $\dot{x} = x^p$:

$p > 1$: solution explodes at finite time $t^* = \tfrac{1}{(p-1)x_0^{p-1}}$.
$p = 1$: solution $x(t) = x_0 e^t$ is finite for all $t < \infty$.

Linear growth keeps the SDE in the $p = 1$ regime, yielding the a-priori bound:

\[ \mathbb{E}\Bigl[\sup_{0 \leq t \leq T}|X_t|^2\Bigr] \leq C(T)(1 + |x_0|^2) \]

for a constant $C(T)$ that is finite for each finite $T$.

Examples¶

Coefficient	Linear growth?	Note
$b(x) = \sin x$	✓	bounded
$b(x) = \kappa(\theta - x)$	✓	affine
$\sigma(x) = \sqrt{x},\ x \geq 0$	✓	sublinear
$b(x) = x^2$	✗	may explode
$\sigma(x) = e^x$	✗	may explode

The Main Existence and Uniqueness Theorem¶

Theorem (Itô). Let $b$ and $\sigma$ be measurable and satisfy the global Lipschitz and linear growth conditions with constant $K$. Then for any $\mathcal{F}_0$-measurable initial condition $X_0 = x_0 \in \mathbb{R}^d$:

\[ \boxed{\text{There exists a unique strong solution } X_t \in C([0,T],\mathbb{R}^d)\ \text{a.s.}} \]

The solution satisfies the moment bound:

\[ \mathbb{E}\Bigl[\sup_{0 \leq t \leq T}|X_t|^2\Bigr] \leq C(1+|x_0|^2)e^{CT} \]

for a constant $C$ depending only on $K$ and dimension.

Lipschitz implies linear growth

Global Lipschitz implies linear growth with the same constant $K$. Indeed, $|b(t,x)| \leq |b(t,0)| + K|x|$, and the theorem assumes $\sup_{t \in [0,T]}|b(t,0)| < \infty$ (satisfied whenever $b$ is measurable and bounded at $x=0$, e.g. if $b$ is continuous). The two conditions are listed separately only to highlight their distinct roles: Lipschitz controls uniqueness, linear growth controls non-explosion.

The proof constructs the solution via Picard iteration; see Picard Iteration for the full argument.

Local Lipschitz Conditions¶

Definition. The coefficients are locally Lipschitz if for every $R > 0$ there exists $K_R > 0$ such that

\[ |b(t,x) - b(t,y)| + |\sigma(t,x) - \sigma(t,y)| \leq K_R|x-y| \quad \text{for all } |x|, |y| \leq R. \]

Under local Lipschitz alone, a unique continuous solution exists up to the explosion time (since $X_t$ is continuous, each $\tau_n$ is a stopping time):

\[ \tau_\infty = \lim_{n \to \infty} \tau_n, \qquad \tau_n = \inf\{t \geq 0 : |X_t| \geq n\}. \]

Adding the linear growth condition forces $\mathbb{P}(\tau_\infty = \infty) = 1$, recovering a global solution.

Beyond Lipschitz: Yamada–Watanabe Conditions¶

For one-dimensional SDEs, the diffusion coefficient need not be Lipschitz for pathwise uniqueness to hold.

Theorem (Yamada–Watanabe). Consider $dX_t = b(X_t)\,dt + \sigma(X_t)\,dW_t$ on $\mathbb{R}$. Pathwise uniqueness holds if:

$b$ is globally Lipschitz: $|b(x) - b(y)| \leq K|x-y|$.
$\sigma$ satisfies $|\sigma(x) - \sigma(y)| \leq \rho(|x-y|)$ for a function $\rho : (0,\infty) \to (0,\infty)$ with

\[ \int_0^\epsilon \frac{du}{\rho^2(u)} = +\infty \quad \text{for every } \epsilon > 0. \]

The divergence is required at $0$: for any fixed $\epsilon > 0$ the tail integral $\int_\epsilon^1 \rho^{-2}(u)\,du$ is finite whenever $\rho$ is bounded away from zero on $[\epsilon, 1]$, so the condition is purely a constraint on the behaviour of $\rho$ near $u = 0$.

CIR application. Take $\sigma(x) = \sigma_0\sqrt{x}$ for $x \geq 0$. To bound $|\sigma(x) - \sigma(y)|$, use the elementary inequality $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x-y|}$ (which follows from $(\sqrt{x}-\sqrt{y})^2 \leq |x-y|$ when $x, y \geq 0$), giving $|\sigma(x) - \sigma(y)| \leq \sigma_0|x - y|^{1/2}$, so $\rho(u) = \sigma_0 u^{1/2}$. Then:

\[ \int_0^\epsilon \frac{du}{\sigma_0^2\,u} = +\infty. \quad \checkmark \]

Hence the CIR model has a pathwise unique solution despite non-Lipschitz diffusion. The relationship between pathwise uniqueness and strong existence is detailed in Strong vs Weak Solutions.

Summary¶

\[ \boxed{\text{Lipschitz} + \text{Linear Growth} \implies \text{Unique Global Strong Solution}} \]

Condition	Role	Formula
Global Lipschitz	Uniqueness	$
Linear growth	No explosion	$
Local Lipschitz	Local uniqueness	Lipschitz on each ball $
Yamada–Watanabe	Uniqueness for $	\sigma

Standard models at a glance:

GBM, Vasicek, OU: global Lipschitz + linear growth ✓
CIR: linear growth + Yamada–Watanabe ✓
$dX = X^2\,dt + dW$: linear growth violated, may explode ✗

Exercises¶

Exercise 1. Consider the SDE

\[ dX_t = (3 + 2X_t)\,dt + (1 - X_t)\,dW_t \]

Verify that the coefficients satisfy the global Lipschitz condition and identify the Lipschitz constant $K$.

Solution to Exercise 1

The coefficients are $b(t,x) = 3 + 2x$ and $\sigma(t,x) = 1 - x$. For any $x, y \in \mathbb{R}$:

\[ |b(t,x) - b(t,y)| = |2x - 2y| = 2|x - y| \]

\[ |\sigma(t,x) - \sigma(t,y)| = |(1-x) - (1-y)| = |x - y| \]

Therefore:

\[ |b(t,x) - b(t,y)| + |\sigma(t,x) - \sigma(t,y)| = 2|x-y| + |x-y| = 3|x-y| \]

The global Lipschitz condition is satisfied with Lipschitz constant $K = 3$.

Exercise 2. Determine which of the following diffusion coefficients $\sigma(x)$ satisfy the global Lipschitz condition on $\mathbb{R}$. For those that do, give the Lipschitz constant; for those that do not, explain why.

(a) $\sigma(x) = \sin(x)$

(b) $\sigma(x) = |x|^{2/3}$

(c) $\sigma(x) = \dfrac{x}{1 + |x|}$

(d) $\sigma(x) = x^2$

Solution to Exercise 2

(a) $\sigma(x) = \sin(x)$. By the mean value theorem, $|\sin(x) - \sin(y)| \leq |x - y|$ for all $x, y \in \mathbb{R}$, since $|\cos(\xi)| \leq 1$ for all $\xi$. So $\sigma$ is globally Lipschitz with $K = 1$.

(b) $\sigma(x) = |x|^{2/3}$. This is not globally Lipschitz. Near the origin, consider $x > 0$ and $y = 0$:

\[ \frac{|x^{2/3} - 0|}{|x - 0|} = x^{-1/3} \to +\infty \quad \text{as } x \to 0^+ \]

No finite constant $K$ can bound the ratio uniformly.

(c) $\sigma(x) = \dfrac{x}{1 + |x|}$. For any $x, y \in \mathbb{R}$, write $f(x) = x/(1+|x|)$. Since $f'(x) = 1/(1+|x|)^2$ for $x \neq 0$ and $|f'(x)| \leq 1$ everywhere, the mean value theorem gives $|f(x) - f(y)| \leq |x - y|$. So $\sigma$ is globally Lipschitz with $K = 1$.

(d) $\sigma(x) = x^2$. This is not globally Lipschitz. For $y = 0$:

\[ \frac{|x^2|}{|x|} = |x| \to +\infty \quad \text{as } |x| \to \infty \]

The ratio grows without bound, so no finite Lipschitz constant exists.

Exercise 3. Let $b(x) = -x^3$ and $\sigma(x) = 1$. Show that $b$ is locally Lipschitz but not globally Lipschitz. Then verify that the linear growth condition fails for $b$. What does this imply about the solution to $dX_t = -X_t^3\,dt + dW_t$?

Solution to Exercise 3

Locally Lipschitz: For any $R > 0$ and $|x|, |y| \leq R$:

\[ |b(x) - b(y)| = |{-x^3 + y^3}| = |x - y| \cdot |x^2 + xy + y^2| \leq |x-y| \cdot 3R^2 \]

so $b$ is Lipschitz on the ball of radius $R$ with constant $K_R = 3R^2$.

Not globally Lipschitz: Taking $y = 0$, the ratio $|b(x) - b(0)|/|x| = x^2 \to \infty$ as $|x| \to \infty$, so no single $K$ works for all $x \in \mathbb{R}$.

Linear growth fails: We need $|b(x)|^2 + |\sigma(x)|^2 \leq K^2(1 + |x|^2)$. Since $|b(x)|^2 = x^6$ and $|\sigma(x)|^2 = 1$:

\[ x^6 + 1 \leq K^2(1 + x^2) \]

is impossible for large $|x|$, since the left side grows as $|x|^6$ while the right side grows as $|x|^2$.

Implication: Despite the failure of linear growth, the drift $b(x) = -x^3$ is dissipative (it pushes the solution toward the origin for large $|x|$). Local Lipschitz guarantees a unique local solution, and dissipativity prevents explosion. The solution to $dX_t = -X_t^3\,dt + dW_t$ exists globally and is unique, but this cannot be concluded from the standard existence-uniqueness theorem — a Lyapunov function argument is needed instead.

Exercise 4. Prove that the global Lipschitz condition implies the linear growth condition. Specifically, show that if

\[ |b(t,x) - b(t,y)| + |\sigma(t,x) - \sigma(t,y)| \leq K|x - y| \]

for all $x, y \in \mathbb{R}^d$, and $M := \sup_{t \in [0,T]}\bigl(|b(t,0)| + |\sigma(t,0)|\bigr) < \infty$, then

\[ |b(t,x)| + |\sigma(t,x)| \leq (M + K)(1 + |x|) \]

Solution to Exercise 4

Let $b$ and $\sigma$ satisfy the global Lipschitz condition. For any $t \in [0,T]$ and $x \in \mathbb{R}^d$, apply the Lipschitz bound with $y = 0$:

\[ |b(t,x) - b(t,0)| + |\sigma(t,x) - \sigma(t,0)| \leq K|x - 0| = K|x| \]

By the triangle inequality:

\[ |b(t,x)| \leq |b(t,0)| + |b(t,x) - b(t,0)| \leq |b(t,0)| + K|x| \]

\[ |\sigma(t,x)| \leq |\sigma(t,0)| + |\sigma(t,x) - \sigma(t,0)| \leq |\sigma(t,0)| + K|x| \]

Adding these two inequalities:

\[ |b(t,x)| + |\sigma(t,x)| \leq \bigl(|b(t,0)| + |\sigma(t,0)|\bigr) + 2K|x| \leq M + 2K|x| \]

where $M = \sup_{t \in [0,T]}(|b(t,0)| + |\sigma(t,0)|)$. To get the stated form, note that $M \leq (M + K)$ and $2K|x| \leq (M+K) \cdot 2|x|$. More directly, since $1 + |x| \geq 1$ and $1 + |x| \geq |x|$:

\[ M + 2K|x| \leq M(1 + |x|) + 2K(1+|x|) = (M + 2K)(1 + |x|) \]

A tighter bound matching the stated form uses $|x| \leq 1 + |x|$ and $1 \leq 1 + |x|$:

\[ |b(t,x)| + |\sigma(t,x)| \leq M \cdot 1 + K|x| + K|x| \leq M(1+|x|) + K(1+|x|) = (M+K)(1+|x|) \]

This holds because $|b(t,x)| \leq |b(t,0)| + K|x| \leq M + K|x|$ and similarly for $\sigma$, and the Lipschitz bound gives $K|x|$ for each separately (not $2K|x|$ when combined). Specifically:

\[ |b(t,x)| + |\sigma(t,x)| \leq M + K|x| \leq (M + K)(1 + |x|) \]

where the last step uses $M \leq (M+K)$ and $K|x| \leq (M+K)|x|$.

Exercise 5. Consider the diffusion coefficient $\sigma(x) = |x|^\alpha$ for $x \in \mathbb{R}$ and $\alpha \in (0,1)$. Show that $\sigma$ satisfies the Yamada--Watanabe condition by verifying that $\rho(u) = u^\alpha$ gives

\[ \int_0^\epsilon \frac{du}{\rho^2(u)} = \int_0^\epsilon u^{-2\alpha}\,du = +\infty \]

if and only if $\alpha \geq \tfrac{1}{2}$. What happens when $\alpha < \tfrac{1}{2}$?

Solution to Exercise 5

Take $\rho(u) = u^\alpha$ for $\alpha \in (0,1)$. The Yamada--Watanabe integral becomes:

\[ \int_0^\epsilon \frac{du}{\rho^2(u)} = \int_0^\epsilon u^{-2\alpha}\,du \]

The integrand $u^{-2\alpha}$ is integrable near $0$ if and only if $-2\alpha > -1$, i.e., $\alpha < 1/2$. Conversely, the integral diverges if and only if $-2\alpha \leq -1$, i.e., $\alpha \geq 1/2$. Explicitly:

If $2\alpha < 1$ (i.e., $\alpha < 1/2$):

\[ \int_0^\epsilon u^{-2\alpha}\,du = \frac{\epsilon^{1-2\alpha}}{1 - 2\alpha} < \infty \]

If $2\alpha = 1$ (i.e., $\alpha = 1/2$):

\[ \int_0^\epsilon u^{-1}\,du = \ln\epsilon - \ln 0 = +\infty \]

If $2\alpha > 1$ (i.e., $\alpha > 1/2$):

\[ \int_0^\epsilon u^{-2\alpha}\,du = \frac{u^{1-2\alpha}}{1-2\alpha}\Bigg|_0^\epsilon = +\infty \]

since $1 - 2\alpha < 0$ and $u^{1-2\alpha} \to +\infty$ as $u \to 0^+$.

Therefore the Yamada--Watanabe condition holds if and only if $\alpha \geq 1/2$.

When $\alpha < 1/2$, the integral converges, so the Yamada--Watanabe theorem does not apply. Pathwise uniqueness may fail in this regime; indeed, for $dX_t = |X_t|^\alpha\,dW_t$ with $\alpha < 1/2$ and $X_0 = 0$, pathwise uniqueness is known to fail.

Exercise 6. The Vasicek model is $dX_t = \kappa(\theta - X_t)\,dt + \sigma\,dW_t$ with $\kappa, \sigma > 0$. Compute the Lipschitz constant $K$ for the pair $(b, \sigma)$ and verify the linear growth condition. Then state what the main existence and uniqueness theorem guarantees about the solution.

Solution to Exercise 6

The Vasicek model has $b(t,x) = \kappa(\theta - x) = \kappa\theta - \kappa x$ and $\sigma(t,x) = \sigma$ (constant).

Lipschitz condition: For any $x, y \in \mathbb{R}$:

\[ |b(t,x) - b(t,y)| = |\kappa\theta - \kappa x - \kappa\theta + \kappa y| = \kappa|x - y| \]

\[ |\sigma(t,x) - \sigma(t,y)| = |\sigma - \sigma| = 0 \]

So the Lipschitz constant is $K = \kappa$.

Linear growth condition: We have $|b(t,0)| = \kappa\theta$ and $|\sigma(t,0)| = \sigma$, so $M = \kappa\theta + \sigma$. By the result of Exercise 4:

\[ |b(t,x)| + |\sigma(t,x)| \leq (M + K)(1 + |x|) = (\kappa\theta + \sigma + \kappa)(1 + |x|) \]

confirming linear growth.

Conclusion from the main theorem: Since the coefficients are globally Lipschitz and satisfy linear growth, the main existence and uniqueness theorem guarantees that for any initial condition $X_0 = x_0$, there exists a unique strong solution $X_t \in C([0,T], \mathbb{R})$ a.s., with the moment bound:

\[ \mathbb{E}\!\left[\sup_{0 \leq t \leq T}|X_t|^2\right] \leq C(1 + |x_0|^2)e^{CT} \]

Exercise 7. Let $X_t$ and $Y_t$ be two strong solutions of $dZ_t = b(t,Z_t)\,dt + \sigma(t,Z_t)\,dW_t$ with $X_0 = Y_0$, where the coefficients are globally Lipschitz with constant $K$. Define $\varphi(t) = \mathbb{E}[\sup_{s \leq t}|X_s - Y_s|^2]$. Starting from the estimate

\[ \varphi(t) \leq C \int_0^t \varphi(s)\,ds \]

apply Gronwall's inequality to conclude that $\varphi(t) = 0$ for all $t \in [0,T]$. Then explain why this proves pathwise uniqueness, not merely uniqueness in law.

Solution to Exercise 7

Applying Gronwall's inequality: The integral form of Gronwall's inequality states: if $\varphi : [0,T] \to [0,\infty)$ is a continuous function satisfying

\[ \varphi(t) \leq \alpha + \beta \int_0^t \varphi(s)\,ds \]

for constants $\alpha \geq 0$ and $\beta > 0$, then $\varphi(t) \leq \alpha e^{\beta t}$ for all $t \in [0,T]$.

In our case, $\alpha = 0$ and $\beta = C$, so:

\[ \varphi(t) \leq 0 \cdot e^{Ct} = 0 \]

Since $\varphi(t) = \mathbb{E}[\sup_{s \leq t}|X_s - Y_s|^2] \geq 0$ by definition, we conclude $\varphi(t) = 0$ for all $t \in [0,T]$.

Pathwise uniqueness, not merely uniqueness in law: The conclusion $\mathbb{E}[\sup_{s \leq t}|X_s - Y_s|^2] = 0$ means that $\sup_{s \leq t}|X_s - Y_s|^2 = 0$ a.s. for each $t$. Taking $t = T$ gives $X_s = Y_s$ for all $s \in [0,T]$ a.s. (the sup over a compact interval of a continuous function being zero forces pointwise equality everywhere).

This is pathwise uniqueness because $X$ and $Y$ are defined on the same probability space with the same Brownian motion $W_t$, and we proved $X_t(\omega) = Y_t(\omega)$ for a.e. $\omega$. Uniqueness in law would only assert $\mathrm{Law}(X) = \mathrm{Law}(Y)$, which is a weaker statement — two processes can have the same distribution while taking different values on a common probability space.

Coefficient	Linear growth?	Note
\(b(x) = \sin x\)	✓	bounded
\(b(x) = \kappa(\theta - x)\)	✓	affine
\(\sigma(x) = \sqrt{x},\ x \geq 0\)	✓	sublinear
\(b(x) = x^2\)	✗	may explode
\(\sigma(x) = e^x\)	✗	may explode