Probability Flow Through an Interval
The forward equation describes how a diffusion’s probability density changes with time. Its conservation-law form makes the role of a boundary explicit: probability can move through the interval’s interior, enter or leave at an endpoint, or be prevented from crossing that endpoint. The sign of the flux matters. Positive flux points toward increasing \(x\); at the left endpoint it points into the interval, while at the right endpoint it points out.
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 \((a,b)\), with drift \(\mu(x)\) and diffusion coefficient \(\sigma(x)\), described in the interior by the Itô equation \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\). The integral \(\int_0^t \sigma(X_s)\,dB_s\) is an Itô integral; on each finite time interval its integrand is assumed progressively measurable and in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T\sigma(X_s)^2\,ds<\infty\). For the calculations below, assume the coefficients and density are regular enough for the displayed derivatives, integrations by parts, and endpoint values to exist.
Let \(p(t,x)\) be the density of the probability that the process is in the interior at time \(t\). The one-dimensional forward equation is \(\partial_t p=-\partial_x(\mu p)+\tfrac12\partial_{xx}(\sigma^2p)\). Define the probability flux, or current, by
The derivative in the second term acts on the product \(\sigma^2p\). Thus, when \(\sigma\) varies with position, the flux is not in general \(\mu p-\tfrac12\sigma^2\partial_xp\). The equivalent expansion is \(J=\mu p-\tfrac12(\sigma^2)'p-\tfrac12\sigma^2\partial_xp\), whenever \(\sigma^2\) is differentiable.
Mass Balance and Endpoint Signs
Write \(M(t)=\int_a^b p(t,x)\,dx\) for the probability mass in the interior. The flux at \(a\), \(J(t,a)\), points to the right and therefore into the interval. The flux at \(b\), \(J(t,b)\), points to the right and therefore out of the interval. Integrating the conservation law determines the net change in interior mass.
Proof. Using \(\partial_t p=-\partial_xJ\) and the fundamental theorem of calculus, \[ \frac{d}{dt}\int_a^b p(t,x)\,dx =\int_a^b\partial_t p(t,x)\,dx =-\int_a^b\partial_xJ(t,x)\,dx =J(t,a)-J(t,b). \] The endpoint signs follow directly from the orientation: outward flow is \(-J(t,a)\) at the left endpoint and \(J(t,b)\) at the right endpoint. Adding them gives \(J(t,b)-J(t,a)\), the negative of the rate of change of interior mass. \(\square\)
This calculation is the mass-balance statement, not an assumption that mass is always conserved inside \((a,b)\). Whether it is conserved depends on the boundary behavior. At a reflecting endpoint, probability is turned back into the interval, so the normal flux there is zero. At an absorbing endpoint, probability leaves the interior when it reaches the endpoint; the interior density has a zero boundary value for a regular, nondegenerate absorbing boundary.
Reflecting and Absorbing Boundary Conditions
Zero flux, rather than zero density, is the reflecting condition. For constant nonzero \(\sigma\), it reads \(\mu p-\tfrac12\sigma^2\partial_xp=0\) at each endpoint. Drift and diffusion can both contribute to the current, so setting \(\partial_xp=0\) alone is not generally the correct reflecting condition. In a reflected stochastic model, the process is constrained to the interval by a finite-variation boundary regulator in addition to its drift and Itô Brownian integral. The zero-current condition expresses that constraint in the forward equation.
For absorption, the interior density vanishes at the absorbing endpoint under the stated regularity and nondegeneracy assumptions. This does not mean that the outgoing flux vanishes: its nonzero value records probability lost from the interior. If instead one defines a process that stays at the endpoint after absorption, its full probability law can include atoms at the endpoints. The density \(p\) here is only the density of the part not yet absorbed; the missing mass is accounted for by the accumulated absorption probabilities.
Proof. Substituting \(J(t,a)=J(t,b)=0\) into the mass-balance identity gives \(dM/dt=0\), so the interior mass is constant. With absorbing boundaries, the same identity applies; no boundary condition in this calculation sets the flux to zero. The mass lost over a time interval \([0,t]\) is therefore the integral of the outward flux over time: \[ M(0)-M(t)=\int_0^t\big(J(s,b)-J(s,a)\big)\,ds. \] At \(b\), outward flow is \(J(s,b)\); at \(a\), it is \(-J(s,a)\). Thus absorption removes probability at precisely these outward rates. Nonnegativity of each outward rate is the probabilistic requirement that probability leaves, rather than enters, through an absorbing endpoint. \(\square\)
Worked Flux Calculations
Worked Example: A Reflecting Brownian Density
Consider standard Brownian motion on \([0,1]\) with reflecting boundaries. Its drift is \(\mu=0\), its diffusion coefficient is \(\sigma=1\), and its forward equation is \(\partial_t p=\tfrac12\partial_{xx}p\). Check the density \(p(t,x)=1+\tfrac12e^{-\pi^2t/2}\cos(\pi x)\).
First, it is nonnegative: since \(-1\leq\cos(\pi x)\leq1\) and \(0<e^{-\pi^2t/2}\leq1\), we have \(p(t,x)\geq1-\tfrac12=\tfrac12\). Its total mass is \(\int_0^1p(t,x)\,dx=1+\tfrac12e^{-\pi^2t/2}[\sin(\pi x)/\pi]_0^1=1\). For these coefficients \(J=-\tfrac12\partial_xp\), and \(\partial_xp=-\tfrac{\pi}{2}e^{-\pi^2t/2}\sin(\pi x)\). Hence \(J(t,0)=J(t,1)=0\).
To check the equation, differentiating in time gives \(\partial_tp=-\tfrac{\pi^2}{4}e^{-\pi^2t/2}\cos(\pi x)\). Differentiating twice in space gives \(\partial_{xx}p=-\tfrac{\pi^2}{2}e^{-\pi^2t/2}\cos(\pi x)\), so \(\tfrac12\partial_{xx}p=\partial_tp\). Thus the stated density satisfies the heat equation and the reflecting conditions. Its mass remains one, also as predicted by zero endpoint flux.
Worked Example: Absorption and Outgoing Flux
Take standard Brownian motion killed at the endpoints of \([0,L]\), where \(L>0\). Check the interior density \(p(t,x)=\dfrac{\pi}{2L}e^{-\pi^2t/(2L^2)}\sin(\pi x/L)\), for \(0<x<L\) and \(t\geq0\).
The initial density is nonnegative and has mass \(\dfrac{\pi}{2L}\int_0^L\sin(\pi x/L)\,dx =\dfrac{\pi}{2L}\cdot\dfrac{2L}{\pi}=1\). At both endpoints \(p=0\). With \(\lambda=\pi^2/(2L^2)\), direct differentiation gives \(\partial_tp=-\lambda p\) and \(\partial_{xx}p=-(\pi/L)^2p\). Therefore \(\partial_tp=\tfrac12\partial_{xx}p\), since \(\tfrac12(\pi/L)^2=\lambda\).
The flux is \(J=-\tfrac12\partial_xp\). The derivative is \(\partial_xp=\dfrac{\pi^2}{2L^2}e^{-\lambda t}\cos(\pi x/L)\), so \(J(t,0)=-\dfrac{\pi^2}{4L^2}e^{-\lambda t}\) and \(J(t,L)=\dfrac{\pi^2}{4L^2}e^{-\lambda t}\). Consequently the outward flux is \(J(t,L)-J(t,0)=\dfrac{\pi^2}{2L^2}e^{-\lambda t}=\lambda e^{-\lambda t}\).
Finally, integrating the density gives the interior mass \(M(t)=e^{-\lambda t}\), whose derivative is \(-\lambda e^{-\lambda t}\). This equals \(J(t,0)-J(t,L)\), as required. Unlike the reflecting example, the interior mass decreases: the outgoing flux accounts for the probability absorbed at the endpoints.
Worked Example: Endpoint Flux for a Nonuniform Density
At a fixed time, let the interior density on \([0,1]\) be \(p(x)=6x(1-x)\), and let \(\mu=1\), \(\sigma=1\). This is a normalized nonnegative density because \(\int_0^1 6x(1-x)\,dx=6(\tfrac12-\tfrac13)=1\). Compute its instantaneous boundary fluxes and the resulting rate of change of interior mass under absorbing conditions.
Here \(J=p-\tfrac12p'\), and \(p'(x)=6-12x\). The endpoint values are \(p(0)=p(1)=0\), \(p'(0)=6\), and \(p'(1)=-6\). Therefore \(J(0)=-3\) and \(J(1)=3\). The outward flux is \(J(1)-J(0)=3-(-3)=6\), so the mass-balance identity gives \(dM/dt=-6\) at this instant.
Check: the derivative of the mass can also be computed from the forward equation at this instant. Since \(\partial_t p=-\partial_xJ\), its integral is \(-[J(1)-J(0)]=-6\). The negative sign is consistent with loss of interior probability. This calculation is an instantaneous flux check; the chosen density alone is not asserted to be a solution at later times.
A Weak Formulation and a Common Mistake
The flux also identifies the boundary terms in the forward equation’s weak formulation. Let \(f\) be twice continuously differentiable on \([a,b]\), and suppose the integrations below are valid. Integrating by parts once gives \[ \frac{d}{dt}\int_a^b f(x)p(t,x)\,dx =-[f(x)J(t,x)]_a^b+\int_a^b f'(x)J(t,x)\,dx. \] Writing \(J=\mu p-\tfrac12\partial_x(\sigma^2p)\) and integrating its last term by parts yields \[ \frac{d}{dt}\int_a^b fp\,dx =\int_a^b\left(\mu f'+\tfrac12\sigma^2f''\right)p\,dx -[fJ]_a^b-\frac12[f'\sigma^2p]_a^b. \] The integral contains the diffusion generator \(\mathcal{L}f=\mu f'+\tfrac12\sigma^2f''\). Boundary conditions determine whether the displayed endpoint terms vanish; they must not be discarded without justification.
A common error is to call \(J=0\) the absorbing condition. For the killed Brownian density in the preceding example, \(p(t,0)=p(t,L)=0\), but for \(t\geq0\) the computed fluxes are \(J(t,0)=-\dfrac{\pi^2}{4L^2}e^{-\lambda t}\) and \(J(t,L)=\dfrac{\pi^2}{4L^2}e^{-\lambda t}\), which are not zero. Setting both fluxes to zero would instead force the mass-balance formula to give \(dM/dt=0\), contradicting the actual mass \(M(t)=e^{-\lambda t}\). The correction is to use zero flux for reflection and zero density, with generally nonzero outgoing flux, for absorption.
Check Your Understanding
Use the flux definition and endpoint orientation in each question. When discussing absorption, distinguish the interior mass from any probability retained at a boundary.
- For \(J=\mu p-\tfrac12\partial_x(\sigma^2p)\), derive the interior-mass balance on \([a,b]\) and identify the outward flux at each endpoint.
- For constant drift \(\mu=c\) and constant diffusion coefficient \(\sigma\), write the reflecting condition at \(a\) explicitly in terms of \(p\) and \(\partial_xp\).
- A killed diffusion has \(J(a)=-2\) and \(J(b)=5\) at a given time. What is the outward flux, and what is \(dM/dt\) at that time?
- Why does a reflecting boundary condition generally constrain the flux rather than require \(\partial_xp=0\)?
- For the absorbing Brownian density in the second worked example, explain why nonzero endpoint flux is consistent with zero endpoint density.