When a Theorem’s Hypotheses Fail
Existence and uniqueness theorems give sufficient conditions: if their hypotheses hold, the theorem guarantees a solution with specified properties. If a hypothesis fails, the guarantee is unavailable, but that alone says nothing about whether a solution exists or is unique. To diagnose a failed proof, check the coefficients directly and then analyze the equation itself. The distinction is important: some equations really do explode or have multiple solutions, while others are well posed despite failing a global Lipschitz bound.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider the time-homogeneous scalar SDE \(dX_t=b(X_t)\,dt+\sigma(X_t)\,dB_t\), with \(X_0=x_0\). A solution is a continuous adapted process satisfying the corresponding integral equation. The Ito integral is defined for a progressively measurable integrand \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). In particular, a solution in the usual square-integrable class requires this condition for \(H_s=\sigma(X_s)\).
The global Lipschitz existence and uniqueness theorem, established in the earlier material on SDE existence and uniqueness, gives a global strong solution and pathwise uniqueness when both conditions hold. A strong solution is constructed on the given filtered space and driven by its given Brownian motion. Pathwise uniqueness means any two solutions on that space with the same initial value and Brownian motion agree at all times almost surely. By contrast, uniqueness in law means that all weak solutions (which may be defined on different probability spaces) have the same distribution as processes. These are different claims: a theorem’s failure to guarantee pathwise uniqueness does not itself establish nonuniqueness in law, or vice versa.
Worked Example: Superlinear Drift Can Cause Explosion
Worked Example: The Equation \(dX_t=X_t^3\,dt\)
Take \(x_0=1\), \(b(x)=x^3\), and \(\sigma(x)=0\). Although the probability space carries Brownian motion, the stochastic integral is zero: its integrand \(H_s=0\) is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds=0\). The integral equation is simply \(X_t=1+\int_0^t X_s^3\,ds\).
For \(x\ne y\), the Lipschitz ratio is \(|b(x)-b(y)|/|x-y|=|x^2+xy+y^2|\). Setting \(y=0\) gives the ratio \(x^2\), which is unbounded. Linear growth also fails: \(b(x)^2/(1+x^2)=x^6/(1+x^2)\) tends to infinity as \(|x|\) tends to infinity.
To determine what happens to the solution, any continuous solution of the integral equation is absolutely continuous and satisfies \(X'_t=X_t^3\) almost everywhere. Since \(X_0=1\) and \(X'_t\ge0\) whenever \(X_t\ge0\), it stays at least \(1\) for as long as it exists. The chain rule for absolutely continuous functions therefore gives \((X_t^{-2})'=-2X_t^{-3}X'_t=-2\) almost everywhere. Integrating from \(0\) to \(t\) yields \(X_t^{-2}=1-2t\), and hence \(X_t=(1-2t)^{-1/2}\) for \(0\le t<\tfrac12\).
Check: differentiating this candidate gives \(X'_t=(1-2t)^{-3/2}=X_t^3\), and \(X_0=1\), so it satisfies the equation before \(t=\tfrac12\). But \(X_t\) tends to infinity as \(t\) increases to \(\tfrac12\). A finite continuous solution on an interval containing \(\tfrac12\) is therefore impossible: the formula was forced for every such solution before that time, and it has no finite continuous extension to \(\tfrac12\). Here the theorem’s hypotheses fail and there is actual finite-time explosion.
Worked Example: Linear Growth Does Not Prevent Nonuniqueness
Worked Example: Waiting-Time Solutions
Let \(b(x)=2\sqrt{x^+}\), where \(x^+=\max(x,0)\), let \(\sigma(x)=0\), and set \(X_0=0\). Again the Ito integrand is \(H_s=0\), which belongs to \(\mathcal{L}^2\) on every finite interval. The equation becomes \(X_t=\int_0^t2\sqrt{X_s^+}\,ds\).
This drift satisfies linear growth. Indeed, for every \(x\), \(2\sqrt{x^+}\le2(1+|x|)\), so \(|b(x)|^2\le4(1+|x|)^2\le8(1+|x|^2)\). It is not locally Lipschitz at \(0\): for \(h>0\), \(|b(h)-b(0)|/|h-0|=2/\sqrt h\), which is unbounded as \(h\) decreases to zero.
For every waiting time \(a\ge0\), define \(X^a_t=(t-a)_+^2\). If \(t<a\), its derivative is \(0\), and \(2\sqrt{(X^a_t)^+}=0\). If \(t>a\), its derivative is \(2(t-a)\), while \(2\sqrt{(X^a_t)^+}=2\sqrt{(t-a)^2}=2(t-a)\). At \(t=a\), both sides are \(0\). Thus \(X^a\) satisfies the differential equation at every time, and integration from \(0\) shows that it satisfies the integral equation with initial value zero.
Check: On any interval \([0,T]\) with \(T>0\), the choices \(a=0\) and \(a=T/2\) give different solutions: at \(t=T/4\), \(X^0_t=T^2/16\), whereas \(X^{T/2}_t=0\). Both are deterministic, adapted processes. They are solutions driven by the same Brownian motion (which has no effect because \(\sigma=0\)), so pathwise uniqueness fails. Their distinct deterministic paths also give distinct laws. In this example, failure of local Lipschitz regularity corresponds to actual nonuniqueness, even though linear growth holds.
Worked Example: Failure of Global Lipschitz Does Not Decide the Answer
Worked Example: Locally Lipschitz Coefficients with Linear Growth
Consider \(dX_t=X_t\sin(X_t^2)\,dt+dB_t\), with a fixed finite initial value \(x_0\). Set \(b(x)=x\sin(x^2)\) and \(\sigma(x)=1\). For every \(x\), \(|b(x)|\le|x|\), so \(|b(x)|^2+|\sigma(x)|^2\le x^2+1=1+|x|^2\). Thus the linear growth condition holds with \(C=1\).
The derivative is \(b'(x)=\sin(x^2)+2x^2\cos(x^2)\). On \([-R,R]\), its absolute value is at most \(1+2R^2\). The mean value theorem therefore gives \(|b(x)-b(y)|\le(1+2R^2)|x-y|\) whenever \(x,y\in[-R,R]\), so \(b\) is locally Lipschitz. The constant diffusion coefficient is Lipschitz as well.
Nevertheless, \(b\) is not globally Lipschitz. At \(x_n=\sqrt{2\pi n}\), for positive integers \(n\), \(b'(x_n)=4\pi n\). These derivative values are unbounded, whereas a differentiable globally Lipschitz function must have derivative bounded by its global Lipschitz constant wherever the derivative exists. The locally Lipschitz and linear-growth existence and uniqueness theorem therefore applies, despite the failure of global Lipschitz. It gives a global strong solution and pathwise uniqueness on every finite time interval. The diffusion integrand \(H_s=1\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_s^2\,ds=T<\infty\).
Check: the coefficient tests give two distinct conclusions: the global Lipschitz test fails, but the local Lipschitz and linear-growth tests both pass. It is those latter hypotheses—not the global Lipschitz test alone—that support the applicable existence and uniqueness result.
A Reliable Diagnostic
These examples illustrate the logical direction of an existence theorem. A sufficient-condition theorem has the form “if the hypotheses hold, then the conclusion holds.” If a hypothesis fails, one cannot infer that the conclusion fails: the conclusion might still be true by another argument or by a theorem with weaker assumptions. Conversely, simply writing down coefficients that are not globally Lipschitz does not prove that an SDE has no solution or that it has multiple solutions.
Check the stated inequalities on their full domain. To disprove a global bound, exhibit a sequence or ratio that becomes unbounded; do not rely only on a graph or a derivative that appears large.
A failed global Lipschitz condition means that theorem cannot be applied in its global-Lipschitz form. It does not by itself settle existence, uniqueness, or explosion.
Look for an explicit solution, a contradiction to finite-time existence, distinct solutions, or a weaker theorem whose hypotheses do hold. State whether the result concerns strong solutions, pathwise uniqueness, or uniqueness in law.
A common error is to treat the failed estimate itself as a proof of failure. For \(b(x)=x^3\), one might write, “The global Lipschitz condition fails, so no solution exists.” The displayed ratio \( |b(x)-b(0)|/|x|=x^2\) proves only that no finite global Lipschitz constant works; it does not prove nonexistence. The correction is to solve or otherwise analyze the equation: with initial value \(1\), the solution is \((1-2t)^{-1/2}\) until time \(\tfrac12\), and that divergence proves the specific claim of no finite continuous solution extending through that time. For the drift \(x\sin(x^2)\), the global Lipschitz condition also fails, but local Lipschitz regularity and linear growth still give global existence and pathwise uniqueness.
The practical lesson is to keep coefficient diagnostics separate from conclusions about the process. Global Lipschitz and linear-growth assumptions are powerful sufficient conditions, not definitions of well-posedness. When one fails, establish what happens by checking other hypotheses or by studying solutions directly.
Check Your Understanding
For each question, distinguish a failed theorem hypothesis from a conclusion about the SDE.
- For \(b(x)=x^3\), use the ratio with \(y=0\) to show why no global Lipschitz constant exists. What additional calculation establishes explosion for initial value \(1\)?
- Verify the integral equation for \(X^a_t=(t-a)_+^2\), including the cases \(t<a\), \(t=a\), and \(t>a\).
- Why does linear growth of \(b(x)=2\sqrt{x^+}\) not rule out the distinct solutions in the waiting-time example?
- For \(b(x)=x\sin(x^2)\), give the sequence showing that the derivative is unbounded, then state which coefficient conditions still allow the existence and uniqueness theorem to apply.
- Explain why failure of a sufficient theorem’s hypothesis alone cannot prove nonexistence or nonuniqueness.