Boundary Flux and Normalizability of Invariant Densities

From a Formal Density to an Invariant Probability

Solving the stationary Fokker–Planck equation produces a candidate density, not automatically a probability distribution for the diffusion. Two separate checks are essential: the candidate must have a finite integral, and its probability flux must be compatible with the process’s boundary behavior. A zero-current calculation can fail either check, and zero current is not the right boundary condition for every state space.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Consider a one-dimensional diffusion on an interval \(I=(l,r)\) with drift \(b(x)\) and diffusion coefficient \(\sigma(x)\), written in Ito form as \(dX_t=b(X_t)\,dt+\sigma(X_t)\,dB_t\). The Brownian integral means the Ito integral with progressively measurable integrand \(H_t=\sigma(X_t)\); on each finite time interval under consideration, assume \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). If this global integrability is unavailable, the integral can instead be defined locally by stopping, but the conclusions below require the stated regularity and boundary conditions for the diffusion being considered.

Write \(a(x)=\sigma(x)^2\), and suppose \(a(x)>0\) in the interior of \(I\). Its generator on twice continuously differentiable test functions is \(\mathcal{L}f=b f'+\tfrac12 a f''\). A probability measure \(\pi\) is invariant if starting the diffusion with distribution \(\pi\) gives distribution \(\pi\) at every later time. When it has a density \(p\), that density must be nonnegative and satisfy \(\int_I p(x)\,dx=1\).

Definition: For a sufficiently differentiable density \(p\), its probability current, or flux, is \(J(x)=b(x)p(x)-\tfrac12(ap)'(x)\). The Fokker–Planck equation is \(\partial_t p=-\partial_xJ\). A no-through boundary condition means that no probability crosses the boundary; on regular endpoints this requires the one-sided boundary flux to be zero.

What the Stationary Equation Actually Implies

The stationary Fokker–Planck equation says \(J'=0\), so the current is constant on each connected interior interval. It does not by itself say that this constant is zero. To reach the zero-current formula, one needs a suitable boundary condition, such as zero flux at an endpoint. On the real line, a zero-current calculation is also often natural when the boundary flux vanishes at infinity, but that limiting condition must be checked rather than presumed.

Theorem: Suppose \(p\) is a stationary density for a regular one-dimensional diffusion on \(I\), with \(a>0\), and that the stationary generator identity \(\int_I \mathcal{L}f(x)p(x)\,dx=0\) holds for every smooth test function \(f\) with compact support in \(I\). Then \(J\) is constant in the interior. If, in addition, the boundary condition gives zero current at an endpoint, then \(J=0\) throughout \(I\).

Proof. For each compactly supported \(f\), integration by parts has no boundary terms because \(f\) vanishes near both endpoints. Using \(J=bp-\tfrac12(ap)'\), we obtain

\int_I \mathcal{L}f\,p\,dx =\int_I bpf'\,dx+\frac12\int_I apf''\,dx =\int_I \left(bp-\frac12(ap)'\right)f'\,dx =\int_I J f'\,dx =-\int_I J'f\,dx.

The assumed generator identity therefore gives \(\int_I J'f\,dx=0\) for every such \(f\), which means \(J'=0\) in the distributional sense. Since the coefficients and density are sufficiently regular, \(J\) is constant in the ordinary sense. If its value at an endpoint is zero, that constant is zero, so \(J=0\) everywhere in the interior. \(\square\)

When the current is zero and \(p>0\), its defining equation is a first-order differential equation. Solving it gives the formal candidate below. This derivation assumes \(a>0\) and enough differentiability for the displayed derivatives and integral to make sense.

Zero-current formula: On an interval where \(a>0\), every positive solution of \(J=0\) has the form \(p(x)=C\,a(x)^{-1}\exp\!\left(\int_{x_0}^{x}\frac{2b(y)}{a(y)}\,dy\right)\), where \(x_0\in I\) is fixed and \(C>0\) is a constant.

Indeed, \(J=0\) is equivalent to \((ap)'=2bp\). Dividing by \(ap\) gives \((\log(ap))'=2b/a\). Integrating and then dividing by \(a\) yields the formula. To turn it into a probability density, the integral of the expression without \(C\) over the whole state space must be finite and positive; only then can \(C\) be chosen as the reciprocal of that integral.

Worked Checks: Normalization and Boundary Behavior

Worked Example: A Normalizable Candidate That Is Not Stationary When Killed

On \(I=(0,\infty)\), take \(a(x)=1\) and \(b(x)=-\tfrac12\). Compute the zero-current candidate, normalize it, and test whether it is an invariant density for the process killed when it reaches \(0\).

Here \(2b/a=-1\), so the zero-current formula gives \(p(x)=C e^{-x}\). Its integral is \(\int_0^\infty e^{-x}\,dx=[-e^{-x}]_0^\infty=1\), hence \(C=1\). The current is \(J(x)=-\tfrac12 e^{-x}-\tfrac12(-e^{-x})=0\) at every interior point.

Nevertheless, this is not an invariant probability density for the killed process. Starting from any \(x>0\), before killing the process is \(X_t=x+B_t-\tfrac12t\). Almost surely \(B_t/t\to0\), so \(x+B_t-\tfrac12t\) eventually becomes negative if the un-killed path is continued. By path continuity, the path must have hit \(0\) first. Thus the killing time is finite almost surely, and the probability of still being in the state space tends to zero as time tends to infinity. An invariant probability on \((0,\infty)\) would have total mass one there at every time, which is impossible.

Check: the formal interior calculation gives \(J=0\), but the normalized candidate has \(p(0+)=1\), not the zero boundary density required by the absorbing, killed boundary behavior. Normalizability and the interior equation do not repair that mismatch.

Worked Example: Zero Current but No Finite Normalization

On \(I=\mathbb{R}\), take \(a(x)=1\) and \(b(x)=\tfrac12\). Find the zero-current expression and check its integral.

Now \(2b/a=1\), so the candidate is \(p(x)=C e^x\). Its integral on the whole real line diverges: already \(\int_0^R e^x\,dx=e^R-1\to\infty\) as \(R\to\infty\). No positive constant \(C\) can make this expression a probability density. The formal current is indeed zero, since \(J(x)=\tfrac12 C e^x-\tfrac12(Ce^x)'=0\).

Check: multiplying a positive function with infinite integral by any positive constant leaves its integral infinite. Thus the zero-current equation alone has not produced an invariant probability.

Boundary Conditions Can Permit Nonzero Current

The no-through condition is appropriate at an impermeable endpoint, but not at every boundary. On a circle, the two ends of a coordinate interval are identified: probability leaving one endpoint re-enters through the other. A constant nonzero current is then compatible with stationarity. This is one reason that “stationary means zero current” is not a general rule.

Worked Example: Uniform Stationarity with Nonzero Current on a Circle

Consider the circle of circumference \(2\pi\), represented by coordinates modulo \(2\pi\), and the diffusion \(dX_t=2\,dt+dB_t\) modulo \(2\pi\). The Brownian integral has integrand \(H_t=1\), which is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_t^2\,dt=T<\infty\). Check the uniform density and its current.

The uniform density is \(p(x)=1/(2\pi)\), since its integral over a full period is \((1/(2\pi))2\pi=1\). Here \(a=1\), \(b=2\), and \(p'=0\), so \(J=bp-\tfrac12(ap)'=2/(2\pi)=1/\pi\), which is nonzero and constant.

To verify invariance directly, take \(X_0\) uniform on the circle and independent of \(B\). At time \(t\), \(X_t\) is \(X_0\) translated by \(2t+B_t\), with the result taken modulo \(2\pi\). For every fixed translation, the uniform distribution is unchanged; averaging over the independent random translation leaves it uniform as well. Hence the density is invariant despite its nonzero current.

Check: the current computed from the formula is \(1/\pi\), while the total probability remains one. The flux represents circulation around the circle, not loss of probability through an exterior boundary.

How to Use the Candidate Responsibly

A useful test has distinct stages. First solve the stationary equation under the boundary condition appropriate to the state space. If zero current is justified, compute the candidate and test its integral over the entire state space, including behavior near finite endpoints and at infinity. Finally check that the boundary condition matches the actual diffusion. The integral check is necessary for a probability density; the flux check is necessary for compatibility with probability conservation. Neither check can substitute for the other.

Even after both checks pass, a formal differential calculation is not a universal proof of invariance. For example, the generator identity \(\int \mathcal{L}f\,d\pi=0\) is a standard infinitesimal invariance test when the diffusion is conservative and regular and the relevant generator-domain and boundary hypotheses hold. Its general semigroup justification is beyond this tutorial. In a concrete problem, those hypotheses must be verified; behavior at a boundary cannot be inferred solely from the interior differential equation.

For a bounded interval \((l,r)\), integrating the Fokker–Planck equation gives the mass balance \(\frac{d}{dt}\int_l^r p(t,x)\,dx=J(t,l+)-J(t,r-)\), when the one-sided limits exist. A stationary total mass therefore requires equal endpoint fluxes, not automatically that both are zero. For an interval with impermeable endpoints the no-through condition sets each to zero. For identified periodic endpoints, equal nonzero fluxes can describe circulation. For an absorbing boundary, the appropriate boundary behavior is different, as the killed-process example demonstrated.

Common mistake. A common wrong conclusion is: “The candidate is integrable and solves \(J=0\), so it must be invariant.” The killed half-line example refutes this: \(e^{-x}\) integrates to one and has zero interior current, but killing removes mass. The correction is to check the actual boundary behavior as well as normalization. A second mistake is to impose \(J=0\) on a periodic state space; the circle example shows that a nonzero constant current can coexist with an invariant probability.

Check Your Understanding

For each question, distinguish the interior stationary equation from the probability and boundary conditions.

Starting from \(J=bp-\tfrac12(ap)'\), derive the zero-current formula for \(p\) on an interval where \(a>0\).
For \(a=1\) and \(b=-1\) on \((0,\infty)\), find the zero-current candidate and determine whether it can be normalized.
Why does a normalized interior solution with \(J=0\) fail to establish invariance for a process killed at a boundary?
On a circle, why can a stationary density have nonzero current without losing total probability?
For a stationary density on a bounded interval, what does the mass-balance equation require of the endpoint fluxes, and what extra condition expresses no-through boundaries?