# Non-Holonomic Fokker–Planck Equation and Shaping Invariant Distributions

# The Non-Skew Adjoint Generator Case

Lecture 2 showed how density shaping carries over to systems that can move only along specific vector fields. By introducing operators that capture divergence and directional derivatives in this constrained geometry, the lecture derived the non-holonomic Fokker–Planck equation and showed how to select feedback so a chosen density becomes invariant. Normalizing the evolving density revealed a Lyapunov functional that always decreases, and controllability of the vector fields ensured that this decay stops only at the target distribution. Under bracket-generating conditions, a non-holonomic Poincaré inequality further guarantees exponential convergence. Simulations with a unicycle model illustrated how ensembles of particles still flow toward complex target densities despite restricted motion.

A key assumption that we made is we could choose control $v_i(x)$ such that the PDE
$$\partial_t p_t= \sum_{i=1}^m \mathcal{X}_i^2 p_t - \mathcal{X}_i(v_i p_t).
$$
could be transformed to the PDE:
$$
\partial_t p_t = -\sum_{i=1}^m \big(\mathcal{Y}_i^{}\big)^*\,\mathcal{Y}_i p_t + \big(\mathcal{Y}_i\big)^*(U_i\, p_t),
$$
for some functions $U_i$. An advantage of this transformation was, the resulting PDE was easier to analyze for its long term behavor.

A natural followup question is. When is this possible? One instance is when $\mathcal{X}^*_i = - \mathcal{X}_i$. That is, the generators are *skew-adjoint*. Then one can see that this is trivially true.

Lets look at the general situation, when such a transformation can be achieved.

## Claim:
For the choice of control law
\begin{equation}
v_i = \sum_{j=1}^n\frac{\partial g_i^j}{\partial x_j} - U_i
\end{equation}
for some functions $U_i$,
then the resulting PDE for the evolution of the probability density is
$$
\partial_t p_t = -\sum_{i=1}^m \big(\mathcal{Y}_i^{}\big)^*\,\mathcal{Y}_i p_t + \big(\mathcal{Y}_i\big)^*(U_i\, p_t),
$$
Hence, if $U_i = -\mathcal{Y}_i\log p_{\infty}$. Then
$$\lim_{t \rightarrow \infty} p_t = p_{\infty}$$
if the underlying control system $\{ g_i\}$ is controllable and the domain $\Omega$ is bounded and regular enough.
## Proof
The proof is just algebra. Ignore if you have better things to do.
We want the PDE
$$
\partial_t p
= \sum_{i=1}^m (\mathcal{Y}_i^*)^2 p - \mathcal{Y}_i^*(v_i p)
$$
to equal
$$
\partial_t p=-\sum_{i=1}^m \mathcal{Y}_i^* \mathcal{Y}_i\, p
\;+\;
\sum_{i=1}^m \mathcal{Y}_i^*(U_i p).
$$

Thus, for each \(i\), we require:

$$(\mathcal{Y}_i^*)^2 p - \mathcal{Y}_i^*(v_i p)=
-\,\mathcal{Y}_i^* \mathcal{Y}_i\, p + \mathcal{Y}_i^*(U_i p).
\tag{1}
$$

---

### Step 1 — Expand $\mathcal{Y}_i^*$

Fix an index $i$ and suppress it from the notation. Define
$$
d(x) := \sum_{j=1}^n \frac{\partial g^j}{\partial x_j}(x).
$$

Using the product rule,
$$
\mathcal{Y}^* f
= \sum_{j=1}^n \Big[ (\partial_{x_j} g^j) f + g^j\,\partial_{x_j} f \Big]
= d f - \mathcal{Y} f.
$$

Thus the operator identity is
$$
\mathcal{Y}^* = -\mathcal{Y} + d.
\tag{2}
$$

---

### Step 2 — Compute the LHS of

$$(\mathcal{Y}^*)^2 p -\mathcal{Y}^*(vp).$$

#### 2.1 Expand $(Y^*)^2 p$

Using $\mathcal{Y}^* = d - \mathcal{Y}$,
$$
(\mathcal{Y}^*)^2 p
= (d - \mathcal{Y})\big((d - \mathcal{Y})p\big)
= (d - \mathcal{Y})(dp - \mathcal{Y}p).
$$

Expanding each term,
$$
(\mathcal{Y}^*)^2 p
= d(dp) - \mathcal{Y}(dp) - d(\mathcal{Y}p) + \mathcal{Y}(\mathcal{Y}p).
$$

Since $d(dp) = d^2 p$ and
$$
\mathcal{Y}(dp) = (\mathcal{Y}d)p + d(\mathcal{Y}p),
$$
we obtain
$$
(\mathcal{Y}^*)^2 p
= \mathcal{Y}^2 p - (\mathcal{Y}d)p - 2 d\,\mathcal{Y}p + d^2 p.
$$

#### 2.2 Expand $Y^*(vp)$

Using Leibniz,
$$
\mathcal{Y}^*(vp) = (d - \mathcal{Y})(vp) = d(vp) - \mathcal{Y}(vp),
$$
and
$$
\mathcal{Y}(vp) = (\mathcal{Y}v)p + v(\mathcal{Y}p),
$$
we obtain
$$
\mathcal{Y}^*(vp)
= (dv)p - (\mathcal{Y}v)p - v\,\mathcal{Y}p.
$$

#### 2.3 Combine terms

The LHS becomes

$$\boxed{ \mathrm{LHS} = \mathcal{Y}^2 p + (-2d + v)\,\mathcal{Y}p + \big[-\mathcal{Y}d + \mathcal{Y}v + d^2 - dv\big]p } \tag{LHS}
$$

---

### Step 3 — Compute the RHS of
$$
-\,\mathcal{Y}^*\mathcal{Y} p + \mathcal{Y}^*(U p).
$$

#### 3.1 Expand $-\mathcal{Y}^*\mathcal{Y} p$

Since
$$
\mathcal{Y}^*\mathcal{Y} p = d(\mathcal{Y}p) - \mathcal{Y}(\mathcal{Y}p),
$$
we have
$$
-\,\mathcal{Y}^*\mathcal{Y} p = -d\,\mathcal{Y}p + \mathcal{Y}^2 p.
$$

#### 3.2 Expand $\mathcal{Y}^*(Up)$

Using Leibniz,
$$
\mathcal{Y}(Up) = (\mathcal{Y}U)p + U(\mathcal{Y}p),
$$
so
$$
\mathcal{Y}^*(Up)
= d(Up) - \mathcal{Y}(Up)
= (dU)p - (\mathcal{Y}U)p - U\,\mathcal{Y}p.
$$

#### 3.3 Combine terms

Thus the RHS is

$$ \boxed{\mathrm{RHS}= \mathcal{Y}^2 p+ (-d - U)\,\mathcal{Y}p + \big[dU - \mathcal{Y}U\big]p}
\tag{RHS}
$$

---

### Step 4 — Equate LHS and RHS

We require
$$
\mathrm{LHS} = \mathrm{RHS}
\quad\text{for all smooth }p.
$$

---

#### 4.1 Second-order term

Coefficients of $\mathcal{Y}^2 p$:
$$
1 = 1,
$$
so no condition arises.

---

#### 4.2 First-order term

Equating coefficients of $Yp$:

$$ -2d + v = -d - U \quad\Longrightarrow\quad \boxed{v = d - U.} \tag{*} $$

---

#### 4.3 Zeroth-order term

Insert $v = d - U$ into
$$- \mathcal{Y}d + \mathcal{Y}v + d^2 - dv = dU - YU. $$

Compute:
$$
Yv = Y(d - U) = Yd - YU,
\qquad
dv = d(d - U) = d^2 - dU.
$$

Then the LHS becomes
$$
-\,\mathcal{Y}d + (\mathcal{Y}d - \mathcal{Y}U) + d^2 - (d^2 - dU)
= dU - \mathcal{Y}U,
$$
which matches the RHS automatically.

---

### Final Result

Restoring the index \(i\), define
$$
d_i(x) := \sum_{j=1}^n \frac{\partial g_i^j}{\partial x_j}(x).
$$

The required control law is therefore
$$ \boxed{v_i(x) = d_i(x) - U_i(x) =
\sum_{j=1}^n \frac{\partial g_i^j}{\partial x_j}(x) \;-\; U_i(x).}
$$

With this choice,

$$ (\mathcal{Y}_i^*)^2 p - \mathcal{Y}_i^*(v_i p) = -\,\mathcal{Y}_i^* \mathcal{Y}_i p + \mathcal{Y}_i^*(U_i p),
$$
and summing over $i = 1,\dots,m$ yields the desired transformed PDE.