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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.

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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))

Prints

0.15722869700957792 0.15722869700957764
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