I am interested in a system with state-dependent diffusion coefficients. This paper by Lau and Lubensky derives the correct Forward FPE in this case:

$$\partial_tP(x,t) = \frac{\partial}{\partial x} D(x) \Bigg(\beta \frac{\partial \mathcal{H}}{\partial x} + \frac{\partial}{\partial x}\Bigg) P(x,t)$$

I want to find the corresponding backwards FPE, however since this is not the standard form, I have not been able to find a derivation which applies.


Define the inner product $\langle f, g\rangle = \int fg \,\mathrm{d}x$. Define also the adjoint $\mathcal{A}^\dagger$ of $\mathcal{A}$ as one satisfying $\langle \mathcal{A}f,g\rangle = \langle f, \mathcal{A}^\dagger g\rangle$ for all $f, g$. Obviously $(\mathcal{A}^\dagger)^\dagger = \mathcal{A}$.

Given the forward FPE: $$\partial_t p = \mathcal{L}_t p, \quad \text{with} \quad p(x, 0) = \delta(x - x_0)$$ And the backward FPE: $$\partial_t u = \mathcal{B}_t u, \quad \text{with} \quad u(x, T) = f(x)$$

Now for these to be the forward and the backward PDE, they have to be connected.

Now clearly $\mathbb{E}(f(X_T) | X_0 = x_0) = \int p(x, T)f(x)\,\mathrm{d}x = \langle p(\bullet, T), f\rangle$ where $p$ is a solution to the forward FPE starting from $x_0$. This has to equal the backward solution $u(x_0, 0)$, as it is just diffusing $f$ from $T$ to the starting point.

We thus have $\langle p(\bullet, T), u(\bullet, T) \rangle = \langle p(\bullet, T), f\rangle = u(x_0, 0) = \langle u(\bullet, 0), p(\bullet, 0) \rangle$, and so: \begin{align} 0 &= \langle p(\bullet, T), u(\bullet, T) \rangle - \langle u(\bullet, 0), p(\bullet, 0) \rangle \\ &= \int_0^T \left(\langle \partial_t p, u \rangle + \langle p, \partial_t u\rangle\right)\,\mathrm{d}t \\ &= \int_0^T \left(\langle \mathcal{L}_t p, u \rangle + \langle p, \mathcal{B}_t u \rangle\right)\,\mathrm{d}t \end{align}

This is always true if we define, as one does, the backwards operator as $\mathcal{B}_t = -\mathcal{L}_t^\dagger$.

Ok, so that was a long reminder (+ some intuition) of the definition of the backwards operator as the (minus) adjoint of the forward. So taking the adjoint of the forward operator we're done:

Given arbitrary $f, g$ and defining $\mathcal{L} = \partial_x D (\beta \mathcal{H}' + \partial_x)$: \begin{align} \langle \mathcal{L} f, g\rangle &= \int \left(\partial_x D (\beta \mathcal{H}' + \partial_x) f\right) g\,\mathrm{d}x \\ &= -\int \left(D (\beta \mathcal{H}' + \partial_x) f\right) \partial_x g\,\mathrm{d}x \\ &= -\int f D \beta \mathcal{H}'\partial_x g + (\partial_x f) D \partial_x g\,\mathrm{d}x \\ &= -\int f D \beta \mathcal{H}'\partial_x g - f \partial_x D \partial_x g\,\mathrm{d}x \\ &= \langle f, \mathcal{L}^\dagger g\rangle \end{align} where we have defined $\mathcal{L}^\dagger = -(D \beta \mathcal{H}' - \partial_x D )\partial_x$, which is to say that the backwards FPE comes out as: $$\frac{\partial u(x, t)}{\partial t} = \left(D(x) \beta \frac{\partial \mathcal{H}}{\partial x} - \frac{\partial}{\partial x} D(x) \right)\frac{\partial}{\partial x} u(x, t)$$

In closing, let's do a simple and crude numerical experiment of the example B in the paper, $\frac{\partial \mathcal{H}}{\partial x} = 0$, $D(x) = 1 - x^2$ (for $L=1$). We compute the PDF at time 0.1 using the forward equation given the the initial condition $x_0 = 0$, and then we solve the backwards equation $f(x) = \max(x, 0)$ and confirm that the expectations match:

import numpy as np
TT = np.linspace(0, .1, 8192)
XX = np.linspace(-1, 1, 51)
dt = np.diff(TT)[0]
dx = np.diff(XX)[0]
D = 1 - XX**2

p  = np.zeros(len(XX))
p[(len(XX)-1)//2] = 1./dx
u  = np.array([max(x, 0) for x in XX])

for t in TT:
    p[1:-1] += dt * (np.diff(D[1:] * np.diff(p)/dx)/dx)
    u[1:-1] += -dt * (-np.diff(D[1:] * np.diff(u)/dx)/dx)

print(u[(len(XX)-1)//2], np.sum(np.array([max(x, 0) for x in XX]) * p * dx))


0.15722869700957792 0.15722869700957764

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.