Stationary Densities for Gradient Diffusions

From Gradient Drift to a Gibbs Density

A diffusion with gradient drift is pulled toward lower values of a potential, while its noise spreads probability mass. A stationary density balances these two effects. When the diffusion has constant isotropic noise and the stationary probability current is zero, that balance gives a density proportional to an exponential of the negative potential. The noise strength sets how concentrated the density is: stronger noise produces a flatter distribution.

Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space carrying a standard \(d\)-dimensional Brownian motion \(\mathbf{B}_t\). Consider the diffusion on \(\mathbb{R}^d\)

d\mathbf{X}_t=-\nabla U(\mathbf{X}_t)\,dt+\sigma\,d\mathbf{B}_t,\qquad \sigma>0,

where \(U:\mathbb{R}^d\to\mathbb{R}\) is twice continuously differentiable. The drift is \(-\nabla U\), and the diffusion coefficient is the constant scalar \(\sigma\), multiplying the identity matrix in \(d\) dimensions. In integral form the noise term is \(\int_0^t \sigma I_d\,d\mathbf{B}_s\); its integrand is progressively measurable and has finite squared integral expectation on every finite time interval, since \(\mathbb{E}\int_0^T\|\sigma I_d\|_F^2\,ds=d\sigma^2T<\infty\). Assume the solution is nonexplosive. When referring to a density solving the forward equation, assume it is sufficiently smooth for the derivatives below to exist.

The forward Kolmogorov equation for a time-dependent density \(\rho(t,x)\) is a standard result from the earlier material on forward equations. In this case its stationary form is \(\nabla\cdot(\rho\nabla U)+\tfrac{\sigma^2}{2}\Delta\rho=0\). It is useful to express this equation through the probability current.

Definition: For a density \(\rho(t,x)\), its probability current is \(J(t,x)=-\rho(t,x)\nabla U(x)-\tfrac{\sigma^2}{2}\nabla\rho(t,x)\). The forward equation is \(\partial_t\rho=-\nabla\cdot J\). A zero-current stationary density satisfies \(J(x)=0\) at every \(x\).

Stationarity requires only that the current have zero divergence. The stronger condition \(J=0\) specifies an equilibrium with no probability flow at any point. The Gibbs-form calculation below is for this zero-current condition; a stationary equation with nonzero divergence-free current is not ruled out merely by setting \(\partial_t\rho=0\).

The Gibbs-Form Characterization

Theorem: Let \(U\in C^2(\mathbb{R}^d)\). A strictly positive \(C^1\) density \(\rho\) has zero current for the diffusion above if and only if \(\rho(x)=C\exp\!\left(-2U(x)/\sigma^2\right)\) for some constant \(C>0\). Such a density is a probability density precisely when \(Z=\int_{\mathbb{R}^d}\exp\!\left(-2U(x)/\sigma^2\right)\,dx\) is finite; in that case \(C=Z^{-1}\).

Proof. Suppose first that \(\rho\) has zero current. By the definition of \(J\), for every \(x\in\mathbb{R}^d\),

0=-\rho(x)\nabla U(x)-\frac{\sigma^2}{2}\nabla\rho(x).

Since \(\rho(x)>0\) and \(\sigma>0\), divide by \(\rho(x)\) and use \(\nabla\log\rho=\nabla\rho/\rho\). This gives \(\nabla(\log\rho+2U/\sigma^2)=0\). A continuously differentiable function with zero gradient on the connected set \(\mathbb{R}^d\) is constant: along the line segment from \(x\) to \(y\), its derivative is the dot product of that gradient with \(y-x\), hence is zero, so its values at \(x\) and \(y\) agree. Therefore \(\log\rho+2U/\sigma^2\) is constant, and exponentiating gives \(\rho=C\exp(-2U/\sigma^2)\) with \(C>0\). Conversely, differentiating this expression gives \(\nabla\rho=-(2/\sigma^2)\rho\nabla U\); substituting into the formula for \(J\) gives \(J=0\). Finally, the integral of this positive function is \(CZ\). It equals one for some finite positive \(C\) exactly when \(0<Z<\infty\); since \(U\) is finite and continuous, its exponential is positive on every bounded region, so \(Z>0\). The required constant is then \(C=Z^{-1}\). \(\square\)

This theorem gives both a differential test and an integrability test. It does not say that every potential produces a stationary probability density: the exponential may fail to be integrable. Nor does the zero-current calculation alone establish all conditions needed for a diffusion to possess an invariant law. Under the standard forward-equation/semigroup correspondence, a normalized stationary solution gives an invariant probability law when the relevant well-posedness and boundary conditions hold.

Worked Densities and Normalizability

Worked Example: A Two-Dimensional Quadratic Potential

Take \(d=2\), \(\sigma=2\), and \(U(x_1,x_2)=x_1^2+4x_2^2\). Find the normalized zero-current density and check its covariance.

Here \(2U/\sigma^2=U/2=x_1^2/2+2x_2^2\), so the unnormalized density is \(\exp(-x_1^2/2-2x_2^2)\). The Gaussian integral in the first coordinate is \(\int_{\mathbb{R}}e^{-x_1^2/2}\,dx_1=\sqrt{2\pi}\). In the second coordinate, \(\int_{\mathbb{R}}e^{-2x_2^2}\,dx_2=\sqrt{\pi/2}\). Thus \(Z=\sqrt{2\pi}\sqrt{\pi/2}=\pi\), and

\rho(x_1,x_2)=\frac{1}{\pi}\exp\left(-\frac{x_1^2}{2}-2x_2^2\right).

The density factors into centered Gaussian densities with variances \(1\) and \(1/4\): their normalizing constants multiply to \((2\pi)^{-1/2}(2/\sqrt{2\pi})=1/\pi\). Hence its covariance matrix is \(\operatorname{diag}(1,1/4)\). To check the current directly, \(\nabla U=(2x_1,8x_2)\), while \(\nabla\rho=(-x_1\rho,-4x_2\rho)\). Since \(\sigma^2/2=2\), the two current terms are \(-\rho(2x_1,8x_2)\) and \(-2(-x_1\rho,-4x_2\rho)\); their sum is \((0,0)\). The covariance entries also match the one-dimensional quadratic formula \(\sigma^2/(2\lambda)\), with potential \(\lambda x^2/2\): the coordinate curvatures are \(\lambda=2,8\), giving \(4/4=1\) and \(4/16=1/4\).

Worked Example: A Quartic Potential

In one dimension, take \(U(x)=x^4\) and \(\sigma=\sqrt{2}\). Compute the normalized density and its second moment, then verify zero current.

Now \(2U/\sigma^2=x^4\). The integral on the positive half-line is evaluated by \(u=x^4\): \(dx=\tfrac14u^{-3/4}du\), so \(\int_0^\infty e^{-x^4}\,dx=\tfrac14\Gamma(1/4)\), where \(\Gamma(a)=\int_0^\infty u^{a-1}e^{-u}\,du\). Doubling for the negative half-line gives \(Z=\tfrac12\Gamma(1/4)\). Therefore

\rho(x)=\frac{2}{\Gamma(1/4)}e^{-x^4}.

The density is even, so its mean is zero. For its second moment, the same substitution gives \(\int_{\mathbb{R}}x^2e^{-x^4}\,dx=\tfrac12\Gamma(3/4)\). Dividing by \(Z=\tfrac12\Gamma(1/4)\) yields \(\mathbb{E}[X^2]=\Gamma(3/4)/\Gamma(1/4)\). To check the current, \(U'(x)=4x^3\) and \(\rho'(x)=-4x^3\rho(x)\). Since \(\sigma^2/2=1\),

J(x)=-4x^3\rho(x)-\rho'(x)=-4x^3\rho(x)+4x^3\rho(x)=0.

The normalizer is finite because the exact integral \(Z=\Gamma(1/4)/2\) is finite. The rapidly increasing quartic potential suppresses the tails sufficiently for a probability density.

Worked Example: A Logarithmic Potential and a Tail Test

Take \(U(x)=\log(1+x^2)\) on \(\mathbb{R}\) and \(\sigma=1\). Test normalizability and verify the resulting density.

The unnormalized Gibbs factor is \(\exp(-2U(x))=(1+x^2)^{-2}\). With \(x=\tan\theta\), \(dx=\sec^2\theta\,d\theta\) and \(1+x^2=\sec^2\theta\); as \(x\) runs over \(\mathbb{R}\), \(\theta\) runs from \(-\pi/2\) to \(\pi/2\). Consequently

Z=\int_{\mathbb{R}}\frac{dx}{(1+x^2)^2} =\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta =\frac{\pi}{2}, \qquad \rho(x)=\frac{2}{\pi(1+x^2)^2}.

For the current, \(U'(x)=2x/(1+x^2)\) and \(\rho'(x)=-8x/(\pi(1+x^2)^3)\). The drift contribution is \(-\rho U'=-4x/(\pi(1+x^2)^3)\), and the diffusion contribution is \(-\tfrac12\rho'=4x/(\pi(1+x^2)^3)\). They cancel. As a tail check, the unnormalized factor is asymptotic to \(|x|^{-4}\), whose integral at infinity is finite; the exact substitution above also confirms the normalization.

When the Gibbs Formula Does Not Give a Density

The criterion is the integral \(Z\), not merely whether \(U\) tends to infinity. Growth of the potential is a useful guide, but different growth rates have different consequences. For example, with \(U(x)=-x^2\) and \(\sigma=1\), the Gibbs factor is \(e^{2x^2}\). Its integral over \(\mathbb{R}\) is infinite, so no positive constant can normalize it. The formula still solves the pointwise zero-current equation, but it is not a probability density.

By contrast, the logarithmic potential in the preceding example tends to infinity slowly and nevertheless gives an integrable factor because its tail decays like \(|x|^{-4}\). In higher dimensions, radial volume also matters: a tail that is integrable on the line may fail to be integrable in \(\mathbb{R}^d\). Always test the full integral over the state space.

Common Pitfall: Confusing Stationarity with Normalization

A frequent error is to solve the zero-current differential equation and stop as soon as an exponential expression is obtained. For \(U(x)=-x^2\), that calculation gives \(\rho(x)=C e^{2x^2}\). Differentiating confirms \(J=0\), but \(\int_{\mathbb{R}} e^{2x^2}\,dx=\infty\); no \(C>0\) makes its integral one. The correction is to compute or bound \(Z\) before calling the expression a stationary probability density.

A second distinction is also important: a stationary solution satisfies \(\nabla\cdot J=0\), whereas the Gibbs derivation imposes the stronger condition \(J=0\). The latter gives a particularly direct equilibrium with no current. Do not infer the zero-current condition from stationarity alone without additional assumptions.

Check Your Understanding

Use the zero-current equation and the normalizability criterion to organize your answers.

For \(U(x)=\tfrac{3}{2}x^2\) and \(\sigma=3\), find the normalizing constant and the variance.
For \(U(x)=a x^4\), with \(a>0\) and \(\sigma>0\), write the normalizing integral and use a change of variables to express it using the Gamma function.
For \(U(x)=\log(1+x^2)\) and \(\sigma=2\), determine whether the Gibbs factor is integrable on \(\mathbb{R}\).
Explain why \(\nabla\cdot J=0\) does not by itself imply \(J=0\).
For a proposed Gibbs density, state the two checks needed before interpreting it as a zero-current stationary probability density.