Verifying Theorem Hypotheses in Worked SDE Examples

What a Hypothesis Check Must Establish

For an SDE, recognizing familiar-looking coefficients is not enough: one must check the bounds in the theorem, including their uniformity in time, and then state only the conclusion those bounds support. The global Lipschitz and linear growth conditions are sufficient hypotheses for a strong existence and uniqueness theorem. They are not necessary conditions for a solution to exist.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to the filtration. Consider the scalar Ito SDE, with deterministic initial value $x_0$, on a finite interval $[0,T]$:

dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t,\qquad X_0=x_0.

Its integral form is $X_t=x_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s$. The stochastic integral here is the Ito integral: its integrand must be progressively measurable and belong to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For vector-valued integrands, use the squared Euclidean or Frobenius norm inside this condition. The drift is $\mu$; the diffusion coefficient is $\sigma$. No Stratonovich integral or Ito-to-Stratonovich correction is involved in these checks.

Definition: The scalar coefficients are globally Lipschitz in space, uniformly in time if there is a finite constant $L$ such that, for every $t\in[0,T]$ and $x,y\in\mathbb{R}$, $$ |\mu(t,x)-\mu(t,y)|^2+|\sigma(t,x)-\sigma(t,y)|^2\le L^2|x-y|^2. $$ They satisfy linear growth if there is a finite constant $C$, uniform in $t$, such that $$ |\mu(t,x)|^2+|\sigma(t,x)|^2\le C^2(1+|x|^2) $$ for all $t\in[0,T]$ and $x\in\mathbb{R}$.

For an $n$-dimensional state and an $m$-dimensional Brownian motion, $\mu(t,x)\in\mathbb{R}^n$ and $\sigma(t,x)\in\mathbb{R}^{n\times m}$. Replace absolute values by Euclidean norms for vectors and the Frobenius norm for matrices. The same conditions then read $|\mu(t,x)-\mu(t,y)|^2+\|\sigma(t,x)-\sigma(t,y)\|_F^2\le L^2|x-y|^2$ and $|\mu(t,x)|^2+\|\sigma(t,x)\|_F^2\le C^2(1+|x|^2)$.

How the Two Conditions Relate

Proposition: Suppose the paired coefficient $F(t,x)=(\mu(t,x),\sigma(t,x))$, with its product norm, is globally Lipschitz in $x$ with constant $L$, uniformly in $t$. If $\sup_{t\in[0,T]}|F(t,0)|\le C_0<\infty$, then the coefficients satisfy linear growth.

Proof. For every $t$ and $x$, the triangle inequality and the Lipschitz condition give $|F(t,x)|\le |F(t,x)-F(t,0)|+|F(t,0)|\le L|x|+C_0$. Since $|x|\le 1+|x|$, this is at most $(L+C_0)(1+|x|)$. Squaring and using $(1+|x|)^2\le2(1+|x|^2)$ yields $|F(t,x)|^2\le2(L+C_0)^2(1+|x|^2)$. This is the required growth bound with $C=\sqrt{2}(L+C_0)$. $\square$

This implication explains why a separate growth check is often short when global Lipschitz and a uniform bound at $x=0$ are already known. The uniform bound matters: a coefficient can be Lipschitz in $x$ at every time while its value at $x=0$ grows without bound over the time interval.

Standard result — global Lipschitz existence and uniqueness theorem: If the coefficients are measurable in time, globally Lipschitz in space uniformly on $[0,T]$, and satisfy linear growth uniformly in time, then for each deterministic $x_0$ there is a strong solution with continuous adapted paths and $\mathbb{E}[\sup_{0\le t\le T}|X_t|^2]<\infty$. It is pathwise unique: two such solutions on the same filtered probability space, driven by the same Brownian motion and with the same initial value, are equal for all $t\in[0,T]$ almost surely. This is the standard theorem established in the earlier material on global Lipschitz existence and uniqueness; its proof is not repeated here.

“Strong” means the solution is constructed as an adapted process on the given space with its given Brownian motion. A weak solution may instead use a probability space and Brownian motion chosen as part of the solution. Pathwise uniqueness compares solutions driven by the same Brownian motion; uniqueness in law compares distributions of solutions and is a different assertion. The theorem above gives strong existence and pathwise uniqueness. Its hypotheses are sufficient, not a test that decides whether a solution exists in every case.

Worked Checks for Scalar Coefficients

Worked Example: Time-Dependent Affine Coefficients

Take $\mu(t,x)=-3x+2\cos t$ and $\sigma(t,x)=2x+\sin t$, with any deterministic $x_0$. Check the conditions on any finite interval $[0,T]$.

For $x,y\in\mathbb{R}$, the coefficient differences are $\mu(t,x)-\mu(t,y)=-3(x-y)$ and $\sigma(t,x)-\sigma(t,y)=2(x-y)$. Therefore

|\mu(t,x)-\mu(t,y)|^2+|\sigma(t,x)-\sigma(t,y)|^2 =9|x-y|^2+4|x-y|^2 =13|x-y|^2.

Thus $L=\sqrt{13}$ works, uniformly in time. For growth, $|\cos t|\le1$ and $|\sin t|\le1$, so $|\mu(t,x)|\le3|x|+2$ and $|\sigma(t,x)|\le2|x|+1$. Hence $|\mu|+|\sigma|\le5|x|+3\le5(1+|x|)$. Squaring and using $(1+|x|)^2\le2(1+|x|^2)$ gives $|\mu|^2+|\sigma|^2\le50(1+|x|^2)$.

Check: the pair of coefficients is globally Lipschitz with constant $\sqrt{13}$, and it has linear growth with $C=5\sqrt{2}$. The standard theorem therefore gives a strong continuous adapted solution on every finite horizon and pathwise uniqueness for that initial value and Brownian motion.

For systems, a reliable method is to bound each output coordinate by the input-coordinate differences, square those bounds, and add them. This produces a valid, if sometimes nonoptimal, Lipschitz constant; an optimal constant is not required.

Worked Example: Bounded Nonlinear Vector Coefficients

Let the state be $X_t\in\mathbb{R}^2$, driven by the scalar Brownian motion $B_t$, and set $\mu(x)=(\sin x_1,\sin(x_1+x_2))$ and $\sigma(x)=(\cos x_1,\sin x_2)\in\mathbb{R}^2$. Check the conditions using the Euclidean norm.

Write $h=x-y$. The mean value theorem and the bounds $|\sin a-\sin b|\le|a-b|$ and $|\cos a-\cos b|\le|a-b|$ give $|\mu_1(x)-\mu_1(y)|\le|h_1|$, $|\mu_2(x)-\mu_2(y)|\le|h_1|+|h_2|$, $|\sigma_1(x)-\sigma_1(y)|\le|h_1|$, and $|\sigma_2(x)-\sigma_2(y)|\le|h_2|$. Consequently,

\begin{aligned} |\mu(x)-\mu(y)|^2+|\sigma(x)-\sigma(y)|^2 &\le h_1^2+(|h_1|+|h_2|)^2+h_1^2+h_2^2\\ &=3h_1^2+2h_2^2+2|h_1h_2|\\ &\le4h_1^2+3h_2^2\\ &\le4|h|^2. \end{aligned}

Check: substituting these bounds gives a left-hand growth bound of at most $4$, while the stated right-hand side is $4(1+|x|^2)\ge4$ for every $x$, including $x=0$. Both conditions hold, so the theorem guarantees strong existence and pathwise uniqueness for each deterministic initial vector.

When Global Lipschitz Fails

Worked Example: Bounded but Not Globally Lipschitz

Consider the scalar drift $\mu(x)=\sin(x^2)$ and diffusion $\sigma(x)=0$. The growth check is direct: $|\mu(x)|^2+|\sigma(x)|^2\le1\le1+|x|^2$, so $C=1$ works.

To test global Lipschitz continuity, suppose $\mu$ had a finite Lipschitz constant $L$. Since $\mu$ is differentiable, taking limits of its difference quotients would give $|\mu'(x)|\le L$ for every $x$. But $\mu'(x)=2x\cos(x^2)$. At $x_n=\sqrt{2\pi n}$, for positive integers $n$, $\cos(x_n^2)=\cos(2\pi n)=1$, so $|\mu'(x_n)|=2\sqrt{2\pi n}$, which is unbounded as $n$ increases. No finite global Lipschitz constant exists.

Check: the global Lipschitz theorem cannot be invoked because one of its hypotheses fails. However, this drift is smooth and therefore locally Lipschitz; it also satisfies linear growth. The earlier result on locally Lipschitz coefficients with linear growth applies, and gives a strong global solution with pathwise uniqueness. Failure of the global Lipschitz check does not, by itself, mean that a solution fails to exist.

Worked Example: Superlinear Drift and Loss of a Hypothesis

Take $\mu(x)=x^2$, $\sigma(x)=0$, and $x_0=1$. The drift is not globally Lipschitz: for $x\ne y$, its difference quotient has magnitude $|x^2-y^2|/|x-y|=|x+y|$, which is unbounded as $x,y$ vary. It also fails linear growth. For $x\ge1$, the ratio $x^2/(1+x)$ is unbounded, so there can be no finite $C$ with $x^4\le C^2(1+x^2)$ for all $x$.

In this example the SDE has no diffusion term, so its integral equation is the ordinary differential equation $X_t=1+\int_0^tX_s^2\,ds$. For $0\le t<1$, $X_t=1/(1-t)$ satisfies it: its derivative is $1/(1-t)^2=X_t^2$, and $X_0=1$. This solution diverges as $t$ increases to $1$; it cannot be extended as a finite continuous solution on an interval containing $1$. Indeed, any differentiable solution while it is positive satisfies $(1/X_t)'=-1$, hence $1/X_t=1-t$.

Check: the example shows why the theorem’s conclusion cannot be asserted after its hypotheses fail: here there is no global finite solution. It does not show that failure of either condition always causes nonexistence; the preceding example already showed otherwise.

Common Mistakes in Applying the Conditions

Checking only a derivative at one point. A finite derivative at $x=0$ does not establish a global Lipschitz bound. For $\sin(x^2)$, the derivative at zero is $0$, but the derivatives at $x_n=\sqrt{2\pi n}$ are unbounded. The correction is to prove a bound valid for every pair $x,y$, or to find a sequence of difference quotients ruling out any such bound.

Checking growth separately for each fixed time. The theorem requires constants uniform over the whole time interval. Bounds whose constants depend on $t$ are not enough unless those constants are bounded on $[0,T]$. Also, satisfying the hypotheses permits an existence-and-uniqueness conclusion; it does not identify an explicit solution or its distribution.

Coefficient check	What it establishes	What it does not establish alone
Global Lipschitz and linear growth	The standard theorem’s strong existence and pathwise uniqueness conclusion	An explicit formula for the solution
Linear growth without global Lipschitz	A moment-growth bound on the coefficients	The global Lipschitz theorem’s conclusion
Failure of a stated hypothesis	Only that this theorem cannot be invoked from those checks	Nonexistence of a solution

Check Your Understanding

For each question, distinguish a bound that holds for every state from one that has only been checked at a single state.

For $\mu(x)=4x-1$ and $\sigma(x)=3x$, find a global Lipschitz constant using the paired-coefficient definition.
For the coefficients in question 1, give a finite linear growth constant $C$.
Why does boundedness of $\sin(x^2)$ establish linear growth but not global Lipschitz continuity?
State the difference between pathwise uniqueness and uniqueness in law.
If an SDE coefficient fails global Lipschitz continuity, what additional check would allow the earlier locally Lipschitz and linear growth result to apply?