Stationary Densities from the Zero-Current Equation

From Stationarity to Zero Current

For a one-dimensional diffusion, a stationary density must balance the probability entering and leaving each region. The forward equation expresses that balance through a probability flux. When the flux vanishes, the stationary equation becomes a first-order ordinary differential equation, which can be solved explicitly. The resulting formula is a candidate density; it represents a probability law only when it is nonnegative and integrable over the state space.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Consider a time-homogeneous diffusion on an interval $I\subseteq\mathbb{R}$ satisfying $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$. Assume the SDE has a solution and, on each finite time interval under consideration, its progressively measurable integrand $\sigma(X_t)$ belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T\sigma(X_s)^2\,ds<\infty$. Write $a(x)=\sigma(x)^2$. For the density calculations below, assume $a(x)>0$ on $I$, $a$ is continuously differentiable, and $\mu$ is continuous.

Definition: If $p(t,x)$ is a sufficiently differentiable probability density for $X_t$, its forward equation is $\partial_t p=-\partial_x J$, where $J(t,x)=\mu(x)p(t,x)-\tfrac12\partial_x(a(x)p(t,x))$ is the probability flux. A stationary density is a nonnegative density $p(x)$, with integral one over $I$, that is time-independent and satisfies the stationary forward equation. The zero-current equation is the stronger condition $J(x)=0$ at every interior point.

Stationarity gives $\partial_xJ=0$, so the flux is constant on a connected interval. The constant need not be zero: imposing zero current is a separate condition. For example, reflecting boundaries with zero flux at an endpoint force the constant to be zero. On a circle, a stationary density can instead carry a nonzero constant flux. Thus the formula derived next addresses zero-current stationary densities, not every possible stationary law.

Solving the Zero-Current Equation

Setting $J=0$ gives $(ap)'=2\mu p$. Because $a>0$, define $y=ap$, so $p=y/a$. The equation becomes $y'=(2\mu/a)y$, a linear first-order equation. Its integrating factor yields the following complete characterization.

Theorem: Suppose $I$ is a nonempty interval, $a$ is positive and continuously differentiable on $I$, and $\mu/a$ is continuous there. Fix $x_0\in I$ and define $$ \rho(x)=\frac{1}{a(x)}\exp\left(\int_{x_0}^{x}\frac{2\mu(z)}{a(z)}\,dz\right). $$ Every continuously differentiable solution $p$ of the zero-current equation on $I$ has the form $p(x)=C\rho(x)$ for a constant $C\in\mathbb{R}$. Among these solutions, every nonzero nonnegative solution has $C>0$.

Proof. Set $y=ap$. The zero-current equation says $y'=2\mu p=(2\mu/a)y$. Let $A(x)=\int_{x_0}^{x}2\mu(z)/a(z)\,dz$. Differentiation gives $(e^{-A}y)'=e^{-A}(y'-(2\mu/a)y)=0$, so $e^{-A(x)}y(x)$ is a constant $C\in\mathbb{R}$. Therefore $p(x)=C e^{A(x)}/a(x)=C\rho(x)$. Conversely, differentiating $y=C e^A$ gives $y'=(2\mu/a)y$, which is the zero-current equation after substituting $y=ap$. Since $\rho(x)>0$ at every $x\in I$, $C\rho$ is nonnegative everywhere exactly when $C\ge0$; it is nonzero exactly when $C\ne0$. Thus a nonzero nonnegative solution has $C>0$, while signed solutions may have any real $C$. $\square$

The reference point $x_0$ affects the displayed function $\rho$ only by a positive multiplicative constant, which is absorbed into $C$. A useful equivalent expression is obtained by expanding the derivative in the flux equation: $a p'+a'p=2\mu p$, and hence $p'/p=2\mu/a-a'/a$ wherever $p>0$. Integrating this logarithmic equation gives the same formula.

Normalization criterion: Let $Z=\int_I\rho(x)\,dx$, interpreted as an improper integral when $I$ has an infinite or excluded endpoint. Since $\rho>0$, the zero-current solution can be normalized to a probability density if and only if $0<Z<\infty$. In that case the unique normalized nonnegative solution is $p(x)=\rho(x)/Z$.

Proof. Every nonzero nonnegative solution is $C\rho$ with $C>0$, so its total mass is $CZ$. If $Z=\infty$, this mass is infinite for every $C>0$; if $Z=0$, its mass is zero. Conversely, if $0<Z<\infty$, choosing $C=1/Z$ gives a nonnegative solution with integral one. Any other nonnegative solution of mass one must satisfy $CZ=1$, so it has this same constant. The zero solution is not a probability density. $\square$

Worked Densities and Integrability Checks

Worked Example: A Quartic Exponential Density

On $I=\mathbb{R}$, take $a(x)=1$ and $\mu(x)=-x^3$. Find and normalize the zero-current solution.

The formula gives $\rho(x)=\exp(\int_0^x -2z^3\,dz)=e^{-x^4/2}$. To check integrability and compute the normalizer, use the substitution $u=x^4/2$ on the positive half-line. Then $x=(2u)^{1/4}$ and $dx=\tfrac14 2^{1/4}u^{-3/4}\,du$, so $\int_0^\infty e^{-x^4/2}\,dx=\tfrac14 2^{1/4}\Gamma(1/4)$, where $\Gamma(q)=\int_0^\infty u^{q-1}e^{-u}\,du$. The integrand is even, therefore $Z=\tfrac12 2^{1/4}\Gamma(1/4)=2^{-3/4}\Gamma(1/4)$, which is finite and positive.

The normalized density is consequently $p(x)=2^{3/4}e^{-x^4/2}/\Gamma(1/4)$. Check the flux directly: $p'(x)=-2x^3p(x)$, so $J(x)=-x^3p(x)-\tfrac12p'(x)=-x^3p(x)+x^3p(x)=0$. Its integral is $2^{3/4}Z/\Gamma(1/4)=2^{3/4}2^{-3/4}=1$, as required.

Worked Example: A Zero-Flux Density on a Bounded Interval

On $I=(0,1)$, take $a(x)=1$ and constant drift $\mu(x)=\tfrac12\ln 2$. Find the zero-current density and normalize it. Interpret the endpoints as reflecting boundaries, so the zero-flux condition is appropriate.

Here $2\mu=\ln 2$, and the formula with $x_0=0$ gives $\rho(x)=e^{x\ln2}=2^x$. Its integral is $Z=\int_0^1 2^x\,dx=(2^1-2^0)/\ln2=(2-1)/\ln2=1/\ln2$. Thus $p(x)=(\ln2)2^x$, which is positive and has integral $(\ln2)(1/\ln2)=1$.

For the flux check, $p'(x)=(\ln2)^2 2^x$, and $J(x)=\tfrac12(\ln2)(\ln2)2^x-\tfrac12(\ln2)^2 2^x=0$. The values at the endpoints are understood as one-sided limits: the same expression gives zero flux at both $x=0$ and $x=1$. The bounded interval makes the candidate integrable; the reflecting boundary condition supplies the zero-current interpretation.

Worked Example: A Density with a Degenerate Diffusion Coefficient

On $I=(0,\infty)$, take $a(x)=2x$ and $\mu(x)=2-x$. Solve and normalize the zero-current equation. These coefficients are smooth and $a$ is strictly positive in the open state space, though it tends to zero at the excluded endpoint.

We have $2\mu/a=(2-x)/x=2/x-1$, so integrating from $1$ to $x$ gives $\int_1^x(2/z-1)\,dz=2\ln x-(x-1)$. Therefore $\rho(x)=(2x)^{-1}e^{2\ln x-x+1}=(e/2)x e^{-x}$. The factor $e/2$ is a positive constant and can be absorbed into $C$, so use $\widetilde\rho(x)=xe^{-x}$. Its integral is $\int_0^\infty xe^{-x}\,dx=[-(x+1)e^{-x}]_0^\infty=1$. The normalized answer is $p(x)=xe^{-x}$.

Substitution checks the equation without relying on the integrating-factor derivation: $ap=2x^2e^{-x}$, and $\tfrac12(ap)'=(2x-x^2)e^{-x}$. Also $\mu p=(2-x)xe^{-x}=(2x-x^2)e^{-x}$. Hence $J=\mu p-\tfrac12(ap)'=0$, and the computed integral confirms unit mass.

When the Formula Applies—and When It Does Not

The criterion $Z<\infty$ must be checked over the entire state space, especially near endpoints where either $a$ or the exponential factor changes rapidly. A formal solution of the ODE may fail to define a probability density because its integral diverges. Even when it is integrable, identifying it as an invariant distribution for a particular diffusion can require checking that the process respects the stated boundary conditions and that the forward equation describes its transition laws. The calculation here establishes the normalized zero-current stationary solution under the stated regularity assumptions; it does not replace those boundary and process checks.

Common pitfall: confusing stationary flux with zero flux. Stationarity implies that $J$ is constant in space, not that it is zero. For example, on a circle of circumference $L>0$, take constant drift $\mu$ and diffusion coefficient $a=1$. The uniform density $p(x)=1/L$ is stationary because it is constant and periodic, but its flux is $J=\mu/L$. If $\mu\ne0$, this is nonzero. Setting the flux to zero would therefore omit this stationary density. On an interval with reflecting boundaries, by contrast, the boundary condition sets the constant flux to zero.

A second common pitfall is to claim that every nonzero solution has a positive constant. The differential equation is linear: if $\rho$ solves it, so does $-\rho$. The full solution family uses $C\in\mathbb{R}$; positivity of $C$ follows only after restricting attention to nonzero nonnegative solutions, such as probability densities.

Question	What to check
Does the candidate solve the stationary equation?	Differentiate $ap$ and verify $J$ is constant.
Is it a zero-current solution?	Verify $J=0$, including any imposed boundary condition.
Can it be a probability density?	Check nonnegativity and $0<\int_I\rho(x)\,dx<\infty$.
Is it an invariant law of the diffusion?	Also check the process's boundary behavior and the applicable forward-equation conditions.

Check Your Understanding

Use the flux equation and normalization criterion to organize each calculation.

For $a(x)=1$ and $\mu(x)=-x$ on $\mathbb{R}$, write the zero-current solution up to a constant and determine whether it is integrable.
Explain why the zero-current solution family includes $C=0$ and negative values of $C$, even though a probability density must be nonnegative.
For $a(x)=1$ and $\mu(x)=c$ on $(0,1)$, derive the zero-current solution when $c\ne0$, and state its limit as $c$ tends to zero.
Show directly that a uniform density on a circle with constant nonzero drift has constant nonzero flux but still satisfies the stationary forward equation.
State the endpoint-integrability condition that must hold for a positive zero-current solution to be normalized.

Question	What to check
Does the candidate solve the stationary equation?	Differentiate \(ap\) and verify \(J\) is constant.
Is it a zero-current solution?	Verify \(J=0\), including any imposed boundary condition.
Can it be a probability density?	Check nonnegativity and \(0<\int_I\rho(x)\,dx<\infty\).
Is it an invariant law of the diffusion?	Also check the process's boundary behavior and the applicable forward-equation conditions.

Stationary Densities from the Zero-Current Equation

What You'll Learn