What Degeneracy Does—and Does Not—Tell You
A diffusion coefficient can be zero everywhere, vanish only at selected states, or be constant and independent of the state. These cases change the form of the noise in an SDE, but they do not by themselves answer every question about its solutions. In particular, a zero diffusion coefficient removes the Brownian integral; it does not guarantee that the resulting deterministic equation has a unique solution.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. Consider the scalar equation with deterministic initial value \(x_0\): \(dX_t=b(X_t)\,dt+\sigma(X_t)\,dB_t\). Its integral form is \(X_t=x_0+\int_0^t b(X_s)\,ds+\int_0^t \sigma(X_s)\,dB_s\). The Ito integrand \(H_s=\sigma(X_s)\) must be progressively measurable and satisfy \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on each horizon where the equation is considered. The drift integral must also be defined, for example by requiring \(\int_0^T|b(X_s)|\,ds<\infty\) almost surely.
A strong solution is an adapted continuous process satisfying this integral equation on the specified stochastic basis and with its specified Brownian motion. A weak solution is a solution on some stochastic basis carrying a Brownian motion relative to its filtration; that basis need not have been prescribed. Pathwise uniqueness means any two solutions on the same basis with the same Brownian motion and initial value are indistinguishable. Uniqueness in law means that all weak solutions have the same distribution as processes. These are separate properties: diagnosing the coefficient does not replace checking which of these assertions is justified.
Identically Zero Diffusion: The Integral Equation Becomes Deterministic
Proof. If \(X\) solves the SDE, its stochastic integrand is \(H_s=\sigma(X_s)=0\). This process is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds=0\), so its Ito integral is zero. Substitution into the integral SDE gives \(X_t=x_0+\int_0^t b(X_s)\,ds\). Conversely, if this integral equation holds, adding the Ito integral of the zero integrand changes neither side, and gives the SDE’s integral form. The equality is understood for all \(t\) on one event of probability one, as required for continuous process solutions. \(\square\)
This proposition identifies the dynamics, not the number of solutions. For a deterministic equation with a non-Lipschitz drift, there can be multiple paths from the same initial value. The next example makes the distinction explicit.
Worked Example: Zero Diffusion but Multiple Strong Solutions
On the horizon \([0,2]\), take \(x_0=0\), \(b(x)=2\sqrt{x^+}\), where \(x^+=\max(x,0)\), and \(\sigma(x)=0\). For each \(a\in[0,2]\), define \(X^a_t=(t-a)_+^2\).
For \(t\le a\), \(X^a_t=0\), and for \(t>a\), \(X^a_t=(t-a)^2\). Thus \(X^a\) is continuous, adapted, and \(X^a_0=0\). Its derivative is zero for \(t<a\), and \(2(t-a)\) for \(t>a\); at \(t=a\), both one-sided derivatives are zero. Also \(2\sqrt{(X^a_t)^+}=0\) for \(t\le a\), and \(2(t-a)\) for \(t>a\). Therefore, at every \(t\), integration of the derivative gives \(X^a_t=\int_0^t 2\sqrt{(X^a_s)^+}\,ds\). The stochastic integrand is identically zero, hence belongs to \(\mathcal{L}^2\).
In particular, \(X^0_t=t^2\) and \(X^1_t=(t-1)_+^2\) both solve the same equation. They differ, for example, at \(t=\tfrac12\): \(X^0_{1/2}=\tfrac14\), whereas \(X^1_{1/2}=0\). On any stochastic basis carrying \(B_t\), both are strong solutions, since they are deterministic and adapted. They also give weak solutions. The two process laws differ: one assigns probability one to the path \(t\mapsto t^2\), the other to \(t\mapsto(t-1)_+^2\). Hence pathwise uniqueness and uniqueness in law fail in this example, although strong and weak existence hold.
Check: for \(t\ge a\), \(\int_0^t2\sqrt{(X^a_s)^+}\,ds=\int_a^t2(s-a)\,ds=(t-a)^2=X^a_t\); for \(t<a\), the integral and \(X^a_t\) are both zero.
The failure here is not caused by the stochastic integral. Once \(\sigma\equiv0\), the solution question is the deterministic integral-equation question. The drift \(b(x)=2\sqrt{x^+}\) is not locally Lipschitz at zero, which is compatible with the branching family of solutions just computed.
State-Independent Diffusion: Additive Noise
If \(\sigma(x)=c\), where \(c\) is constant, the stochastic term is \(cB_t\), since \(B_0=0\). For a constant drift \(b(x)=\mu\), the solution is obtained directly from the integral equation. This is additive noise: the diffusion coefficient does not depend on the current state. When \(c=0\), this reduces to the identically zero diffusion case; when \(c\ne0\), the noise remains present at every state.
Proof. The constant drift integrates to \(\mu t\), and the Ito integral of the constant integrand \(c\) is \(cB_t\). This gives the displayed process and verifies the equation. Any solution must equal this expression by the integral equation itself, which proves uniqueness on the given basis. For the quadratic variation, the finite-variation path \(x_0+\mu t\) has zero quadratic variation and zero covariation with \(B_t\); Brownian motion has \([B]_t=t\). Therefore \([x_0+\mu t+cB]_t=c^2[B]_t=c^2t\). This uses the quadratic-variation rules, not the ordinary chain rule. \(\square\)
Worked Example: Constant Drift and State-Independent Noise
Let \(x_0=1\), \(\mu=-1\), and \(c=2\). Then \(X_t=1-t+2B_t\). The stochastic integrand is the constant \(H_s=2\), which is progressively measurable, and \(\mathbb{E}\int_0^T H_s^2\,ds=4T<\infty\). Thus the Ito integral is defined on each finite horizon and equals \(2B_t\).
At \(T=2\), the mean is \(\mathbb{E}[X_2]=1-2+2\mathbb{E}[B_2]=-1\). Since \(B_2\sim N(0,2)\), the variance is \(\operatorname{Var}(X_2)=4\operatorname{Var}(B_2)=4\cdot2=8\). The quadratic variation at that time is \([X]_2=2^2\cdot2=8\). The random fluctuation comes from the diffusion coefficient, while the drift changes the mean but not the quadratic variation.
Check: the integral equation gives \(X_2=1+(-1)\cdot2+2B_2=-1+2B_2\). Its mean is \(-1+2\cdot0=-1\), and its variance is \(2^2\cdot2=8\), agreeing with the computations above.
Vanishing at One State Is Not Vanishing Everywhere
If \(\sigma(x_*)=0\), the Brownian coefficient is zero at the instant the process is at \(x_*\). That fact alone does not imply that the process stays there. The drift may move the process away from \(x_*\), after which the diffusion coefficient may be nonzero. Conversely, if both drift and diffusion vanish at a state, the constant path at that state satisfies the equation; that observation by itself does not establish uniqueness.
Worked Example: Diffusion Vanishes Initially, Then Becomes Active
Consider \(dX_t=dt+X_t\,dB_t\), with \(X_0=0\). The coefficient \(\sigma(x)=x\) vanishes at the initial state. Define \(Y_t=\exp(B_t-t/2)\) and \(A_t=\int_0^t\exp(-B_s+s/2)\,ds\), and set \(X_t=Y_tA_t\).
To verify the equation, apply Ito’s formula to \(f(t,b)=\exp(b-t/2)\). Its derivatives are \(f_t=-\tfrac12f\), \(f_b=f\), and \(f_{bb}=f\). Including the second-order correction gives \(dY_t=f_t\,dt+f_b\,dB_t+\tfrac12f_{bb}\,dt=Y_t\,dB_t\). The process \(A_t\) has finite variation and derivative \(A'_t=Y_t^{-1}\). The product rule has no quadratic-covariation term between a continuous finite-variation process and \(Y\), so \(dX_t=A_t\,dY_t+Y_t\,dA_t=X_t\,dB_t+dt\). Also \(X_0=0\). Hence \(X\) satisfies the stated SDE.
The integrand \(X\) is progressively measurable. Its square-integrability on a finite horizon follows from the explicit expression and Cauchy–Schwarz: \(\mathbb{E}[X_t^2]\le t\int_0^t\mathbb{E}[(Y_t/Y_s)^2]\,ds=t\int_0^t e^{t-s}\,ds\le t^2e^t\). Here \(Y_t/Y_s=\exp(B_t-B_s-(t-s)/2)\), and the normal moment formula gives \(\mathbb{E}[(Y_t/Y_s)^2]=e^{t-s}\). Consequently \(\mathbb{E}\int_0^T X_s^2\,ds\le\int_0^T s^2e^s\,ds<\infty\), as required for the Ito integral.
For every \(t>0\), both \(Y_t\) and the integrand defining \(A_t\) are strictly positive, so \(X_t>0\). The diffusion coefficient starts at zero but becomes positive immediately along this solution.
Check: substituting the differentials gives \(dt+X_t\,dB_t\), exactly the original drift and diffusion terms; at \(t=0\), \(X_0=0\), so the initial diffusion coefficient is indeed zero.
Coefficient Tests and Their Limits
Two familiar coefficient conditions are useful diagnostics. Global Lipschitz continuity means that for some \(L<\infty\), \(|b(x)-b(y)|+|\sigma(x)-\sigma(y)|\le L|x-y|\) for all states \(x,y\). Linear growth means that for some \(C<\infty\), \(|b(x)|^2+|\sigma(x)|^2\le C(1+|x|^2)\) for every state \(x\). These are sufficient hypotheses in the standard global Lipschitz existence-and-uniqueness theorem; its proof belongs to the later treatment of existence and uniqueness. They are not definitions of strong or weak solution, and checking that a coefficient vanishes at a point does not check either condition.
For the additive example, constant coefficients satisfy both bounds: the Lipschitz differences are zero, and \(|\mu|^2+|c|^2\) is bounded by a suitable constant \(C(1+|x|^2)\). In the zero-diffusion example with \(b(x)=2\sqrt{x^+}\), the drift is not globally Lipschitz: for \(x>0\), \(|b(x)-b(0)|/|x-0|=2/\sqrt{x}\), which is unbounded as \(x\) decreases to zero. The explicit solutions, rather than a coefficient label alone, establish the nonuniqueness claim.
A useful diagnostic order is therefore:
- Check whether \(\sigma\) is identically zero, state-independent, or merely zero at selected states.
- Write the integral equation and identify whether the stochastic integral disappears or remains active.
- Check coefficient regularity separately from the definitions of solution and uniqueness.
- When asserting uniqueness or nonuniqueness, specify whether the comparison is on one common noise basis or between process laws.
A common mistake is to infer “deterministic and unique” from \(\sigma\equiv0\). The incorrect argument says that the Brownian term vanishes, so there is no randomness and therefore only one solution. The first implication is valid: the equation reduces to an integral equation with no Brownian term. The second is not. In the example above, \(X^0_t=t^2\) and \(X^1_t=(t-1)_+^2\) are two different deterministic solutions from the same initial value. Zero diffusion removes noise from the equation; uniqueness still depends on the drift equation.
Check Your Understanding
For each question, distinguish what follows directly from the coefficient structure from what requires a separate solution argument.
- If \(\sigma\equiv0\), what integral equation must every solution satisfy?
- For \(b(x)=\mu\) and \(\sigma(x)=c\), what are the explicit solution and its quadratic variation?
- Why does \(\sigma(x_0)=0\) not imply that the diffusion term remains zero for all later times?
- For \(b(x)=2\sqrt{x^+}\), \(\sigma=0\), and \(x_0=0\), verify the family \(X^a_t=(t-a)_+^2\) and state which uniqueness properties fail.
- State the global Lipschitz and linear growth conditions, and explain why they are not definitions of strong or weak solution.