Math AI - Expand Score Matching to Flow Matching

Flow Sampling (from $\mathbf{x}_0$ and $u_t$ to get $\mathbf{x}_t$)

最重要的是 $\mathbf{x}_t$不是直線！因為 $\mathbf{u}_t(\mathbf{x}_t)$ 是一個平均的結果，不是一個 constant vector! 但是 condition vector 是直線 (in the OT case).

理論上非常簡單，就是解一個 ODE with initial condition $\mathbf{x}_0$ is: $\frac{d\mathbf{x}}{dt} = \mathbf{u}_t(\mathbf{x})$ The exact solution for 例一 of the ODE $\mathbf{x}_t = (2t-1)\boldsymbol{\mu} + \sqrt{2t^2 - 2t + 1} \cdot (\mathbf{x}_0 + \boldsymbol{\mu})$ 通用的表示：（Two indepedent Gaussians, Appendix E and F) $\boxed{\mathbf{u}_t(\mathbf{x}) = (\boldsymbol{\mu}_1 - \boldsymbol{\mu}_0) + \frac{ {d\log\sigma}_t^{2}}{2 dt} (\mathbf{x} - \boldsymbol{\mu}_t)}$ $\begin{aligned} \mathbf{x}(t) &= \left[(1-t) \boldsymbol{\mu}_0 + t \boldsymbol{\mu}_1\right] + \dfrac{\sqrt{(1-t)^2 \sigma_0^2 + t^2 \sigma_1^2}}{\sigma_0} (\mathbf{x}_0 - \boldsymbol{\mu}_0)\\ \phi_t(\mathbf{x}_0) &= \mathbf{x}(t)=\boldsymbol{\mu}_t + \dfrac{\sigma_t}{\sigma_0} (\mathbf{x}_0 - \boldsymbol{\mu}_0)\\ \end{aligned}$

$\boldsymbol{\mu}_t = (1-t) \boldsymbol{\mu}_0 + t \boldsymbol{\mu}_1 = \boldsymbol{\mu}_0 + t(\boldsymbol{\mu}_1-\boldsymbol{\mu}_0)$ 是以 $\boldsymbol{\mu}_0$ 為起點，斜率為 $(\boldsymbol{\mu}_1-\boldsymbol{\mu}_0)$ 的直綫。只有在$\mathbf{x}_0 = \boldsymbol{\mu}_0$ 才會走這條直綫。當 $\mathbf{x}_0 \ne \boldsymbol{\mu}_0$ 偏離的部分就會照 standard deviation 比例 ($\frac{\sigma_t}{\sigma_0}$) 加到這條直綫。
$t =0, \phi_0(\mathbf{x}_0) = \mathbf{x}_0$
$t =1, \phi_1(\mathbf{x}_0) = \boldsymbol{\mu}_1 + \frac{\sigma_1}{\sigma_0}(\mathbf{x}_0-\boldsymbol{\mu}_0)$. 如果 $\sigma_1=\sigma_0=\sigma$ , $\phi_1(\mathbf{x}_0) = \mathbf{x}_0+(\boldsymbol{\mu}_1-\boldsymbol{\mu}_0)$. 即是所有的終點都是起點加上 mean difference.

$\boldsymbol{\mu}_t = (1 - t) \boldsymbol{\mu}_0 + t \boldsymbol{\mu}_1$ and $\boldsymbol{\Sigma}_t = (1 - t)^2 \boldsymbol{\Sigma}_0 + t^2 \boldsymbol{\Sigma}_1$ If $\boldsymbol{\Sigma}_0$ and $\boldsymbol{\Sigma}_1$ commute, the flow simplifies to: (Appendix H) $\boxed{\phi_t(\mathbf{x}_0)=\mathbf{x}(t) = \boldsymbol{\mu}_t + \boldsymbol{\Sigma}_t^{1/2} \boldsymbol{\Sigma}_0^{-1/2} \left( \mathbf{x}_0 - \boldsymbol{\mu}_0 \right)}$

Mean Flow 的 Idea 就是 Sampling 的時候走直綫！這樣只需要 1-step! 初中幾何的概念！

https://www.youtube.com/watch?v=swKdn-qT47Q&ab_channel=%E9%98%BF%E5%86%89

Typical flow model training

![[Pasted image 20250615091041.png]]

Difference between flow matching vs. mean flow matching: one-step generation ![[Pasted image 20250615091217.png]]

Mean flow model training

![[Pasted image 20250615081535.png]]

![[Pasted image 20250615081608.png]]

![[Pasted image 20250615081653.png]]

How about variable time step mean flow matching? on manifold, slow, off manifold - fast?

Gaussian Flow Mean Flow Field

Correct Closed-Form Vector Fields

Given:

$p_0 = \mathcal{N}(\boldsymbol{\mu}_0, \sigma_0^2 I)$
$p_1 = \mathcal{N}(\boldsymbol{\mu}_1, \sigma_1^2 I)$
Given $\mathbf{x}_1$, conditional straight-line trajectory: $\mathbf{x}_t = (1-t)\mathbf{x}_0 + t \mathbf{x}_1$
Marginal distribution: $\mathbf{x}_t \sim \mathcal{N}(\boldsymbol{\mu}_t, \sigma_t^2 I)$
where $\boldsymbol{\mu}_t = (1-t)\boldsymbol{\mu}_0 + t\boldsymbol{\mu}_1$
and $\sigma_t^2 = (1-t)^2\sigma_0^2 + t^2\sigma_1^2$

Example 1: Two Gaussians with opposite mean and same variance. (Appendix A)

先看一般 flow matching, 也就是 instantaneous vector field。

$\boxed{v_t(x) = \dot{\boldsymbol{\mu}}_t + \dfrac{\dot{\sigma_t^2}}{2\sigma_t^2}(\mathbf{x} - \boldsymbol{\mu}_t)=\boldsymbol{\mu}_1-\boldsymbol{\mu}_0 + \dfrac{-(1-t)\sigma_0^2 + t\sigma_1^2}{(1-t)^2\sigma_0^2 + t^2\sigma_1^2}(\mathbf{x} - \boldsymbol{\mu}_t)}$ 假設 $\mu_1=-\mu$ and $\mu_2=+\mu$, $\sigma_0=\sigma_1=\sigma$ $v_t(\mathbf{x}) = 2\boldsymbol{\mu} + \dfrac{2t-1}{2t^2-2t+1}(\mathbf{x} - (2t-1)\boldsymbol{\mu})$ $v_t(\mathbf{x}) = 2\boldsymbol{\mu} + \dfrac{\dot{\sigma^2_t}}{2\sigma^2_t}(\mathbf{x} - \boldsymbol{\mu}_t)$ 平均向量場是 $2\boldsymbol{\mu}$，在 $t<0.5$, $\sigma_t$ 變小，所以微分是負值。大於 mean $\boldsymbol{\mu}_t$ 會被壓縮，小於 mean 則被膨脹。

實際 unconditional trajectory (given $x_0$ and $v_t(x)$): 兩個重點

不是直線，而是曲線
$t = 0, v_t = 2 \mu - (x_0 + \mu) = \mu-x_0$, 其實是 $E(x_1) - x_0$
$t = 0.5, v_t = 2 \mu$, 其實是 $E(x_1) - E(x_0)$
$t = 1, v_t = 2 \mu + (x_1 - \mu) = x_1-\mu$,
$t=1$, $\mathbf{x}_1 = \mathbf{x}_0 + 2\boldsymbol{\mu}$, 也就是雖然是曲線，最後的終端相當於平均平移 $2\boldsymbol{\mu}$ $\boxed{ \mathbf{x}_t = \boldsymbol{\mu}_t + \frac{ {\sigma}_t} { {\sigma}_0} \left( \mathbf{x}_0 - \boldsymbol{\mu}_0 \right)=(2t-1) \boldsymbol{\mu}+\sqrt{2t^2-2t+1}\left( \mathbf{x}_0 + \boldsymbol{\mu} \right)}$

$\boxed{u(z_t, t_{\text{end}}, t) = 2\boldsymbol{\mu} + \dfrac{\sqrt{2t_{\text{end}}^2 - 2t_{\text{end}} + 1} - \sqrt{2t^2 - 2t + 1}}{\sqrt{2t^2 - 2t + 1} \cdot (t_{\text{end}} - t)} \left( z_t - (2t - 1) \boldsymbol{\mu} \right)}$ Assume $t_{end}=1$

$u(z_t, t_{\text{end}}, t) = 2\boldsymbol{\mu} + \dfrac{1 - \sqrt{2t^2 - 2t + 1}}{\sqrt{2t^2 - 2t + 1} \cdot (1 - t)} \left( z_t - (2t - 1) \boldsymbol{\mu} \right)=2\boldsymbol{\mu} + \dfrac{1 - \sqrt{2t^2 - 2t + 1}}{ {2t^2 - 2t + 1} \cdot (1 - t)} \left( z_t - (2t - 1) \boldsymbol{\mu} \right)$

所以 $u(\mathbf{x}_0, 1, 0) = 2\boldsymbol{\mu}$, 也就是 $\mathbf{x}_1= \mathbf{x}_0+2\boldsymbol{\mu}$, 和之前一樣。

1. Instantaneous Flow Matching Vector Field

Given:

$p_0 = \mathcal{N}(\boldsymbol{\mu}_0, \sigma_0^2 I)$
$p_1 = \mathcal{N}(\boldsymbol{\mu}_1, \sigma_1^2 I)$
Marginal distribution: $\mathbf{x}_t \sim \mathcal{N}(\boldsymbol{\mu}_t, \sigma_t^2 I)$
where $\boldsymbol{\mu}_t = (1-t)\boldsymbol{\mu}_0 + t\boldsymbol{\mu}_1$
and $\sigma_t^2 = (1-t)^2\sigma_0^2 + t^2\sigma_1^2$

Example 1: Two Gaussians with opposite mean and same variance. (Appendix A)

先看一般 flow matching, 也就是 instantaneous vector field。

\[\boxed{\mathbf{v}_t(\mathbf{x}) = \dot{\boldsymbol{\mu}}_t + \dfrac{\dot{\sigma_t^2}}{2\sigma_t^2}(\mathbf{x} - \boldsymbol{\mu}_t)=\dot{\boldsymbol{\mu}}_t + \dfrac{\dot{\sigma_t}}{\sigma_t}(\mathbf{x} - \boldsymbol{\mu}_t)=\boldsymbol{\mu}_1-\boldsymbol{\mu}_0 + \dfrac{-(1-t)\sigma_0^2 + t\sigma_1^2}{(1-t)^2\sigma_0^2 + t^2\sigma_1^2}(\mathbf{x} - \boldsymbol{\mu}_t)}\]

對應的 trajectory 是: $\boxed{ \mathbf{x}_t = \boldsymbol{\mu}_t + \frac{ {\sigma}_t} { {\sigma}_0} \left( \mathbf{x}_0 - \boldsymbol{\mu}_0 \right)}$ prove $\frac{d\mathbf{x}_t}{dt} = \mathbf{v}_t(\mathbf{x})$

$\frac{\mathbf{x}_t - \boldsymbol{\mu}_t}{ {\sigma}_t} = \frac{\mathbf{x}_0 - \boldsymbol{\mu}_0} { {\sigma}_0}$

Core Intuition: Constant Normalized Offset or z-score

Left side: $x_t = \mu_t + \frac{\sigma_t}{\sigma_0}(x_0 - \mu_0)$ is the z-score of xtxt at time tt.
It measures how many standard deviations (σtσt) the point xtxt is from the current mean μtμt.
Right side: x0−μ0σ0σ0x0−μ0 is the z-score of x0x0 at time t=0t=0.

The equality states that:
The z-score (normalized position relative to the current distribution) remains constant along the entire trajectory.

The correct general form for isotropic Gaussians is: $\boxed{v_t(x) = \underbrace{\dot{\boldsymbol{\mu}}_t}_{\text{mean velocity}} + \underbrace{\frac{\dot\sigma_t}{\sigma_t}}_{\text{radial scale}} (x - \boldsymbol{\mu}_t)}$

where we compute the derivatives:

$\dot{\boldsymbol{\mu}}_t = \frac{d}{dt}\boldsymbol{\mu}_t = \boldsymbol{\mu}_1 - \boldsymbol{\mu}_0$
$\dot\sigma_t^2 = \frac{d}{dt}\sigma_t^2 = 2(1-t)\sigma_0^2(-1) + 2t\sigma_1^2 = -2(1-t)\sigma_0^2 + 2t\sigma_1^2$
→ $\dot\sigma_t = \frac{\dot\sigma_t^2}{2\sigma_t} = \dfrac{-(1-t)\sigma_0^2 + t\sigma_1^2}{\sigma_t}$

Final form: $\boxed{v_t(x) = (\boldsymbol{\mu}_1 - \boldsymbol{\mu}_0) + \left(\dfrac{-(1-t)\sigma_0^2 + t\sigma_1^2}{(1-t)^2\sigma_0^2 + t^2\sigma_1^2}\right)(x - \boldsymbol{\mu}_t)}$

\[\boxed{\mathbf{v}_t(\mathbf{x}_t) = \dot{\boldsymbol{\mu}}_t + \dfrac{\dot{\sigma_t^2}}{2\sigma_t^2}(\mathbf{x}_t - \boldsymbol{\mu}_t)=\dot{\boldsymbol{\mu}}_t + \dfrac{\dot{\sigma_t}}{\sigma_t}(\mathbf{x}_t - \boldsymbol{\mu}_t)}\]

對應的 trajectory 是: $\boxed{ \mathbf{x}_t = \boldsymbol{\mu}_t + \frac{ {\sigma}_t} { {\sigma}_0} \left( \mathbf{x}_0 - \boldsymbol{\mu}_0 \right)}$

2. Mean Flow Matching Vector Field (Average Velocity)

For an interval $[t, t_{\text{end}}]$, the average velocity is: $u(t, t_{\text{end}}, x) = \frac{1}{t_{\text{end}} - t} \int_t^{t_{\text{end}}} v_\tau(x) d\tau$

Key simplification:
Because the vector field $v_t(x)$ is linear in $x$ (affine transformation), the average velocity retains the same functional form: $\boxed{u(t, t_{\text{end}}, x) = v_t(x)}$

Why Are They Identical?

Path linearity:
$\mathbf{x}_t$ evolves linearly → instantaneous velocity $v_t(x)$ is constant along each path. $\frac{d\mathbf{x}_t}{dt} = \mathbf{x}_1 - \mathbf{x}_0 \quad \text{(constant for fixed } \mathbf{x}_0, \mathbf{x}_1)$
Averaging constant velocity:
For any interval $[t, t_{\text{end}}]$: $u = \frac{1}{t_{\text{end}}-t} \int_t^{t_{end}} (\mathbf{x}_1 - \mathbf{x}_0) d\tau = \mathbf{x}_1- \mathbf{x}_0$
Conditional expectation:
Both fields reduce to: $\mathbb{E}[\mathbf{x}_1 - \mathbf{x}_0 \mid \mathbf{x}_t] = (\boldsymbol{\mu}_1 - \boldsymbol{\mu}_0) + \left(\dfrac{-(1-t)\sigma_0^2 + t\sigma_1^2}{(1-t)^2\sigma_0^2 + t^2\sigma_1^2}\right)(\mathbf{x}_t - \boldsymbol{\mu}_t)$

Special Case Verification

Variance-preserving flow ($\sigma_0 = \sigma_1 = \sigma$):

$\dot\sigma_t = 0$
$v_t(x) = u(t, t_{\text{end}}, x) = \boldsymbol{\mu}_1 - \boldsymbol{\mu}_0$
Matches expectation: $\mathbb{E}[\mathbf{x}_1 - \mathbf{x}_0] = \boldsymbol{\mu}_1 - \boldsymbol{\mu}_0$

Data-to-noise ($\sigma_0 = \sigma > 0, \sigma_1 = 0$):

$\boldsymbol{\mu}_t = (1-t)\boldsymbol{\mu}_0$
$\sigma_t^2 = (1-t)^2\sigma^2$
$\dot\sigma_t = -\sigma(1-t)$
$v_t(x) = u(t, t_{\text{end}}, x) = \dfrac{\boldsymbol{\mu}_0 - x}{1-t}$
Matches the OU process limit.

Summary

| Property | Flow Matching | Mean Flow Matching | | ——————- | ——————————————————————————————– | ———————————- | | Form | $v_t(x) = \dot{\boldsymbol{\mu}}t + \dfrac{\dot\sigma_t}{\sigma_t}(x - \boldsymbol{\mu}_t)$ | Identical to $v_t(x)$ | | Time Dependence | Explicit $t$ dependence | Independent of $t{\text{end}}$ | | Sampling | Requires ODE integration | One-step: $\mathbf{x}_1 = \mathbf{x}_0 + u(0,1,\mathbf{x}_0)$ | | Complexity | Same closed form | Same closed form |

Key Insight:
For Gaussian distributions with linear interpolation paths, the average velocity $u$ is path-constant and identical to the instantaneous velocity $v_t(x)$. Mean flow matching achieves one-step sampling by exploiting this constancy, but the functional form of the vector field is identical to flow matching.

Thank you for the correction—this is an important nuance in the Gaussian case! Would you like me to derive the conditional expectation $\mathbb{E}[\mathbf{x}_1 - \mathbf{x}_0 \mid \mathbf{x}_t]$ step-by-step?

Reference

MIT 6.S184: Flow Matching and Diffusion Models https://www.youtube.com/watch?v=GCoP2w-Cqtg&t=28s&ab_channel=PeterHolderrieth

Yaron Meta paper: [2210.02747] Flow Matching for Generative Modeling

An Introduction to Flow Matching: https://mlg.eng.cam.ac.uk/blog/2024/01/20/flow-matching.html

Appendix C

Compute Expectation** (Appendix C)

The conditional vector field should be: ![[Pasted image 20250603200549.png]]

Since $u_t(x \mid \mathbf{x}_1) = \frac{\mathbf{x}_1 - x}{1-t}$ is linear in $\mathbf{x}_1$, and the posterior is Gaussian: $u_t(x) = \mathbb{E}_{\mathbf{x}_1 \sim p_{1\mid t}(\mathbf{x}_1|x)}\left[\frac{\mathbf{x}_1 - x}{1-t} \right] = \frac{\overbrace{\mathbb{E}_{p_{1\mid t}}[\mathbf{x}_1 ]}^{\text{posterior mean}} - x}{1-t}$ $\boxed{\mathbf{u}_t(\mathbf{x}) = \frac{(2t-1)\mathbf{x} + \boldsymbol{\mu}}{2t^2-2t+1} = \frac{(2t-1)\mathbf{x} + \boldsymbol{\mu}}{\sigma^2_t}}$

Geometric Interpretation

看下圖比較清楚：想像是一團 centered at (-10, 0) 的點，隨著時間往 (+10, 0) 移動的過程。綠色是 trace, 是每個點經過 vector field 被改變之後的 trace. 可以想像是微分方程的解。

$\frac{d\boldsymbol{x}_t}{dt} = \boldsymbol{u}_t(\boldsymbol{\mathbf{x}_t})$

For $t < 0.5$:
The term $(2t-1)$ is negative, 所以是一個壓縮的流。
For $t > 0.5$:
The term $(2t-1)$ is positive, 所以是一個膨脹的流。

![[Pasted image 20250604104124.png]]

The posterior mean is: $\mathbb{E}[\mathbf{x}_1 \mid \mathbf{x}_t = x] = \frac{tx + (1-t)\boldsymbol{\mu}}{2t^2-2t+1}$

Step 3: Substitute and Simplify

$u_t(x) = \frac{1}{1-t} \left( \frac{tx + (1-t)\boldsymbol{\mu}}{2t^2-2t+1} - x \right)$

\[= \frac{1}{1-t} \left( \frac{t x + (1-t)\boldsymbol{\mu} - x(2t^2-2t+1)}{2t^2-2t+1} \right)\] \[= \frac{1}{1-t} \cdot \frac{(t - 2t^2 + 2t - 1)x + (1-t)\boldsymbol{\mu}}{2t^2-2t+1}\]

$= \frac{1}{1-t} \cdot \frac{(3t - 2t^2 - 1)x + (1-t)\boldsymbol{\mu}}{2t^2-2t+1}$ $\boxed{\mathbf{u}_t(\mathbf{x}) = \frac{(2t-1)\mathbf{x} + \boldsymbol{\mu}}{2t^2-2t+1}}$

Verification.

At $t = 0$: $\mathbf{u}_0(\mathbf{x}) = \frac{(-1)\mathbf{x} + \boldsymbol{\mu}}{1} = -\mathbf{x}+\boldsymbol{\mu}$
- At prior mean $\mathbf{x} = -\boldsymbol{\mu}$: $\mathbf{u}_0(-\boldsymbol{\mu}) = -(-\boldsymbol{\mu})+\boldsymbol{\mu} = 2\boldsymbol{\mu}$
  (Points toward $+2\boldsymbol{\mu}$, correct)
At $t = 1$: $\mathbf{u}_1(\mathbf{x}) = \frac{(2-1)\mathbf{x} + \boldsymbol{\mu}}{2-2+1} = \mathbf{x} + \boldsymbol{\mu}$
- At target mean $\mathbf{x} = \boldsymbol{\mu}$: $\mathbf{u}_1(\boldsymbol{\mu}) = \boldsymbol{\mu} + \boldsymbol{\mu} = 2\boldsymbol{\mu}$
  (Consistent with linear interpolation)

Implementation

import numpy as np

def vector_field(x: np.ndarray, t: float, mu: np.ndarray) -> np.ndarray:
    """
    Computes CFM vector field for p0 = N(-μ, I) → p1 = N(+μ, I).
    
    Args:
        x: Current position (n-dimensional vector)
        t: Time in [0, 1]
        mu: Target mean vector (+μ)
    
    Returns:
        u_t(x): Vector field direction
    """
    numerator = (2*t - 1) * x + mu
    denominator = 2*t**2 - 2*t + 1
    return numerator / (denominator + 1e-8)  # Avoid division by zero

One-sided flow matching? Two sided flow matching?

Reference

youtube video: next generation model https://www.youtube.com/watch?v=swKdn-qT47Q&ab_channel=%E9%98%BF%E5%86%89

MIT Kaiming He https://arxiv.org/abs/2505.13447

Appendix A: Two Gaussian Case

Given the specific conditions $\sigma_0^2 = \sigma_1^2 = \sigma^2$, $\boldsymbol{\mu}_0 = -\boldsymbol{\mu}$, and $\boldsymbol{\mu}_1 = +\boldsymbol{\mu}$, the instantaneous vector field is:

\[\mathbf{v}_t(\mathbf{x}) = \frac{(2t-1)\mathbf{x} + \boldsymbol{\mu}}{2t^2 - 2t + 1}\]

The mean flow matching field $u(z_t, t_{\text{end}}, t)$ is defined as the time-average of $\mathbf{v}$ along the trajectory from $t$ to $t_{\text{end}}$:

\[u(z_t, t_{\text{end}}, t) = \frac{1}{t_{\text{end}} - t} \int_t^{t_{\text{end}}} \mathbf{v}_s(\mathbf{x}_s) ds\]

where the trajectory $\mathbf{x}_s$ starting from $z_t$ at time $t$ is given by:

\[\mathbf{x}_s = \boldsymbol{\mu}_s + \frac{\sigma_s}{\sigma_t} (z_t - \boldsymbol{\mu}_t)\]

with:

$\boldsymbol{\mu}_s = (2s - 1) \boldsymbol{\mu}$
$\sigma_s = \sigma \sqrt{2s^2 - 2s + 1}$
$\boldsymbol{\mu}_t = (2t - 1) \boldsymbol{\mu}$
$\sigma_t = \sigma \sqrt{2t^2 - 2t + 1}$

Substitute $\mathbf{v}_s(\mathbf{x}_s)$ and simplify:

\[\mathbf{v}_s(\mathbf{x}_s) = \frac{(2s-1)\mathbf{x}_s + \boldsymbol{\mu}}{2s^2 - 2s + 1}\]

Using $\mathbf{x}_s = \boldsymbol{\mu}_s + \frac{\sigma_s}{\sigma_t} (z_t - \boldsymbol{\mu}_t)$ and $\boldsymbol{\mu}_s = (2s-1)\boldsymbol{\mu}$:

\[\mathbf{v}_s(\mathbf{x}_s) = 2\boldsymbol{\mu} + \frac{(2s-1)}{\sqrt{2s^2 - 2s + 1} \sqrt{2t^2 - 2t + 1}} (z_t - \boldsymbol{\mu}_t)\]

The integral becomes:

\[\int_t^{t_{\text{end}}} \mathbf{v}_s(\mathbf{x}_s) ds = 2\boldsymbol{\mu} (t_{\text{end}} - t) + \frac{1}{\sqrt{2t^2 - 2t + 1}} (z_t - \boldsymbol{\mu}_t) \left[ \sqrt{2s^2 - 2s + 1} \right]_t^{t_{\text{end}}}\]

Evaluating the definite integral:

\[\int_t^{t_{\text{end}}} \mathbf{v}_s(\mathbf{x}_s) ds = 2\boldsymbol{\mu} (t_{\text{end}} - t) + \frac{\sqrt{2t_{\text{end}}^2 - 2t_{\text{end}} + 1} - \sqrt{2t^2 - 2t + 1}}{\sqrt{2t^2 - 2t + 1}} (z_t - \boldsymbol{\mu}_t)\]

Divide by $t_{\text{end}} - t$:

\[u(z_t, t_{\text{end}}, t) = 2\boldsymbol{\mu} + \frac{\sqrt{2t_{\text{end}}^2 - 2t_{\text{end}} + 1} - \sqrt{2t^2 - 2t + 1}}{\sqrt{2t^2 - 2t + 1} \cdot (t_{\text{end}} - t)} (z_t - \boldsymbol{\mu}_t)\]

Substituting $\boldsymbol{\mu}_t = (2t - 1) \boldsymbol{\mu}$:

\[\boxed{u(z_t, t_{\text{end}}, t) = 2\boldsymbol{\mu} + \dfrac{\sqrt{2t_{\text{end}}^2 - 2t_{\text{end}} + 1} - \sqrt{2t^2 - 2t + 1}}{\sqrt{2t^2 - 2t + 1} \cdot (t_{\text{end}} - t)} \left( z_t - (2t - 1) \boldsymbol{\mu} \right)}\]

where:

$z_t$ is the position at time $t$,
$t_{\text{end}}$ is the end time,
$t$ is the current time,
$\boldsymbol{\mu}$ is a constant vector,
$\sigma$ is a constant scalar (though it cancels out in the expression).

This expression simplifies to the general form derived earlier when $\sigma_0^2 = \sigma_1^2$, but is presented here explicitly for the given parameters.

Flow Sampling (from $\mathbf{x}_0$ and $u_t$ to get $\mathbf{x}_t$)

Mean Flow 的 Idea 就是 Sampling 的時候走直綫！ 這樣只需要 1-step! 初中幾何的概念！

How about variable time step mean flow matching? on manifold, slow, off manifold - fast?

Gaussian Flow Mean Flow Field

Correct Closed-Form Vector Fields

1. Instantaneous Flow Matching Vector Field

Core Intuition: Constant Normalized Offset or z-score

2. Mean Flow Matching Vector Field (Average Velocity)

Why Are They Identical?

Special Case Verification

Summary

Reference

Appendix C

Geometric Interpretation

Step 3: Substitute and Simplify

Verification.

Implementation

Reference

Appendix A: Two Gaussian Case

Mean Flow 的 Idea 就是 Sampling 的時候走直綫！這樣只需要 1-step! 初中幾何的概念！