We have already provided a proof for the autonomous case in our ODE solver post, based on a fixed-point argument. This argument extends directly to the non-autonomous case, since all vector fields $\{f(\cdot, t)\}_t$ share the same Lipschitz constant $L$.

The proof studies a Cauchy problem with a different boundary condition. With $\tau\in[0,1]$ and $\bar{y}\in\mathbb{R}^d$: $$ \tag{$\star$} \frac{dy}{dt}(t) = f\big(y(t), t\big)\; \text{ for any } t\in[0,\tau] \text{ and } y(\tau)=\bar{y}\;. $$ Clearly, the Cauchy-Lipschitz theorem holds for (1) iff it holds for ($\star$).

The flow’s injectivity is proven as follows. Fix $\tau$ and assume that there exist $y_0, y_0^\prime$ such that $\psi(y_0, \tau) = \psi(y_0^\prime, \tau)=\bar{y}$. Both $\psi(y_0, \cdot)$ and $\psi(y_0^\prime, \cdot)$ are solutions of ($\star$); by uniqueness of the solution, they are therefore equal on $[0,\tau]$ and in particular at $t=0$. This proves that $y_0 = y_0^\prime$, and the injectivity of $\psi$.

Surjectivity follows a similar argument; since a solution for ($\star$) exists for any $\bar{y}$, we can find a predecessor for any point of $\mathbb{R}^d$ under ($\star$). One can easily check that this provides a starting condition $y_0$ for (1) that will check $\psi(y_0, \tau)=\bar{y}$. This is true for any $\bar{y}$ and proves that the flow is a surjective map.

There are elegant proofs of this result; this is not one of them—we will rely on brute-force computations. Fix $t\in[0,1]$. The density $p(\cdot, t)$ of the push-forward of $p$ under $\psi(\cdot, t)$ is given by the change-of-variable formula; equivalently $p$ is the push-forward of $p(\cdot, t)$ under $\psi(\cdot, t)^{-1}$. For any $y\in\mathbb{R}^d$ this writes: $$ p(y) = \vert \det J\psi(y, t)\vert p(\psi(y, t), t) \;, $$ with $J\psi$ being the flow’s Jacobian matrix. The left-hand side does not depend on time; differentiating wrt. time therefore yields (with $\det J\psi > 0$, by continuity from $\det J\psi(y, 0) = 1$): $$ \tag{$\star$} \frac{\partial \det J\psi}{\partial t}(y, t)\,p(\psi(y, t), t) + \vert \det J\psi(y, t)\vert\left(\frac{\partial \psi}{\partial t}(y, t)^\top\frac{\partial p}{\partial y}(\psi(y, t), t) + \frac{\partial p}{\partial t}(\psi(y, t), t)\right) = 0\;. $$ Observe that by definition we have $\partial_t \psi(y, t) = f(\psi(y, t), t)$. Dividing by $\vert \det J\psi(y, t)\vert$ (never zero since $\psi(\cdot, t)$ is a diffeomorphism) yields: $$ \underbrace{\vert \det J\psi(y, t)\vert^{-1} \frac{\partial \det J\psi}{\partial t}(y, t)\,p(\psi(y, t), t)}_{(\text{i})} + f(\psi(y, t), t)^\top\frac{\partial p}{\partial y}(\psi(y, t), t) + \frac{\partial p}{\partial t}(\psi(y, t), t) = 0\;. $$ Let’s focus on (i) for now. Recall that for any invertible matrix $A(t)$ we have $\partial_t \det(A(t))= \det(A(t))\text{Tr}(A^{-1}(t)\partial_t A(t))$. Applying it to the first term of the above, we get: $$ (\text{i}) = \text{Tr}\left(J\psi(y, t)^{-1}\frac{\partial J\psi}{\partial t}(y, t)\right)p(\psi(y, t), t)\;. $$ We can simplify this one further; indeed, for any $i, j$: $$ \begin{aligned} \frac{\partial }{\partial t} \frac{\partial \psi_i}{\partial y_j} &= \frac{\partial}{\partial y_j}\frac{\partial \psi_i}{\partial t}\;, &(\text{smoothness}) \\ &= \frac{\partial}{\partial y_j}f_i(\psi(y, t), t)\;, &(\text{def of flow}) \\ &= \sum_{k=1}^d \frac{\partial f_i}{\partial y_k}(\psi(y, t), t) \frac{\partial \psi_k}{\partial y_j}(y, t)\;. \end{aligned} $$ This yields that $\frac{\partial J\psi}{\partial t}(y, t) = Jf(\psi(y, t), t) \cdot J\psi(y, t)$ and therefore that: $$ \begin{aligned} (\text{i}) &= \text{Tr}\left(J\psi(y, t)^{-1}Jf(\psi(y, t), t) J\psi(y, t)\right)p(\psi(y, t), t)\;,\\ &= \text{Tr}\left(J\psi(y, t)J\psi(y, t)^{-1}Jf(\psi(y, t), t)\right)p(\psi(y, t), t)\;, &(\text{Tr}(AB)=\text{Tr}(BA))\\ &= \text{Tr}\left(Jf(\psi(y, t), t)\right)p(\psi(y, t), t)\;. \end{aligned} $$ Putting everything back together, and denoting $z=\psi(y, t)$ as a shorthand, we get: $$ \text{Tr}\left(Jf(z, t)\right)p(z, t) + f(z, t)^\top\frac{\partial p}{\partial y}(z, t) + \frac{\partial p}{\partial t}(z, t) = 0\;, $$ which is exactly the continuity equation at $z$ (expand the divergence of $pf$); since $\psi(\cdot, t)$ is surjective, it holds for all $z\in\mathbb{R}^d$. We proved that $p(\cdot, t)$ is a solution to the continuity equation. To prove uniqueness, we walk back to ($\star$). Observe that ($\star$) characterises $p(\cdot, t)$; if it holds, then $\vert \det J\psi(y, t)\vert p(\psi(y, t), t)$ is constant wrt. $t$—in particular, it equals for all $t$ its value at $t=0$: $$ \begin{aligned} \vert \det J\psi(y, t)\vert p(\psi(y, t), t) &= \vert \det J\psi(y, 0)\vert p(\psi(y, 0), 0) \;, \\ &= \vert \det I_d\vert p(y, 0) \;, \\ &= p(y)\;, \end{aligned} $$ since $\psi(y, 0)=y$ for any $y\in\mathbb{R}^d$. This is the change-of-variable formula; hence $p(\cdot, t)$ is the push-forward of $p$ by $\psi(\cdot, t)$, which concludes the proof.

Like almost always, we will not be able to solve the ODE in closed-form. Sampling through this generative process will require the use of ODE solvers; often, the simplest ones (e.g., the explicit Euler method) will be enough for the applications we pursue here.

Strictly speaking no flow can reach a point-mass since $\psi(\cdot, 1)$ is a bijection. A rigorous treatment targets $\mathcal{N}(z, \sigma_{\min}^2\mathbf{I}_d)$ instead, for some small $\sigma_{\min}>0$, and assumes that $p_\star$ has bounded support. In practice, though, using $\sigma_{\min}=0$ works fine since neither training nor sampling evaluates the vector fields at $t=1$.

This is a proof by verification; it is enough to check that for the vector fields (4) the continuity equation for $p(\cdot\vert t)$ is satisfied. (Indeed, the continuity equation characterises the push-forward.)

Unrolling the computation, for any $y\in\mathbb{R}^d$: $$ \begin{aligned} \frac{\partial p}{\partial t}(y\vert t) &= \frac{\partial}{\partial t}\int_{z}p_\star(z)p(y\vert z, t)dz \;, \\ &= \int_{z}p_\star(z) \frac{\partial}{\partial t}p(y\vert z, t)dz\;, &(\text{dominated convergence}) \\ &= -\sum_{i=1}^d \int_{z} p_\star(z)\frac{\partial (pf_i)}{\partial y_i}(y\vert z, t)dz\;, &(\text{by (3)})\\ &= -\sum_{i=1}^d \frac{\partial}{\partial y_i} \int_{z}p_\star(z)p(y\vert z, t)f_i(y\vert z, t)dz \;, \\ &= -\sum_{i=1}^d \frac{\partial}{\partial y_i}\Big[ p(y\vert t)\int_{z}p_\star(z)\frac{p(y\vert z, t)f_i(y\vert z, t)}{p(y\vert t)}dz\Big ]\;,\\ &= -\sum_{i=1}^d \frac{\partial (pf_\star)}{\partial y_i}(y\vert t)\;. \end{aligned} $$ The continuity equation is verified, which concludes the proof.

It is enough to prove that: $$ \argmin_{\tilde f\in\mathcal{F}} \mathbb{E}_{z\sim p_\star}\mathbb{E}_{y\sim p(\cdot\vert z, t)}\Big[ \big\| \tilde f(y\vert t) - f(y\vert z, t)\big\|^2\Big] = \argmin_{\tilde f\in\mathcal{F}}\mathbb{E}_{y\sim p(\cdot\vert t)}\Big[ \big\| \tilde f(y\vert t) - f_\star(y\vert t)\big\|^2\Big]\;. $$ Let’s expand the left-hand side; the term inside the expectation decomposes as: $$ \begin{aligned} \| \tilde f(y\vert t) - f(y\vert z, t)\|^2 &=\| \tilde f(y\vert t) - f_\star(y\vert t)\|^2 + \|f_\star(y\vert t)- f(y\vert z, t)\|^2 + 2(\tilde f(y\vert t) - f_\star(y\vert t))^\top(f_\star(y\vert t)- f(y\vert z, t))\;.\\ \end{aligned} $$ Since $\tilde f(y\vert t) - f_\star(y\vert t)$ does not depend on $z$, the cross term writes: $$ \int_y \big(\tilde f(y\vert t) - f_\star(y\vert t)\big)^\top \underbrace{\Big[f_\star(y\vert t)p(y\vert t) - \int_z f(y\vert z, t)p_\star(z)p(y\vert z, t)dz\Big]}_{=0 \text{ by (4)}}dy = 0\;, $$ where we used $\int_z p_\star(z)p(y\vert z, t)dz = p(y\vert t)$. For the same reason, the first term integrates against $p(y\vert t)$. Hence: $$ \mathbb{E}_{z\sim p_\star}\mathbb{E}_{y\sim p(\cdot\vert z, t)}\Big[ \big\| \tilde f(y\vert t) - f(y\vert z, t)\big\|^2\Big] = \mathbb{E}_{y\sim p(\cdot\vert t)}\Big[ \big\| \tilde f(y\vert t) - f_\star(y\vert t)\big\|^2\Big] + \blacksquare\;, $$ where $\blacksquare$ is a constant wrt. $\tilde{f}$—which concludes the proof.

The methods described here naturally extend to modelling conditional distributions $p_\star(\cdot\vert c)$—typically, with $c$ some label describing some class to generate a sample from. The extension works well on paper; in practice, it needs some augmentation like classifier-free guidance [4] to produce satisfying results.

The ’typical’ flow-matching schedule writes $\mu(t)=t$ and $\sigma(t)=1-t$; it yields so-called straight interpolating paths. Also of interest are variance preserving paths, where $\mu(t) = \sqrt{\alpha(t)}$ and $\sigma(t) = \sqrt{1-\alpha(t)}$ for some well-chosen schedule $\alpha$. In this case, rewriting (6) accordingly: $$ p(y\vert z, t) = \mathcal{N}(y\vert \sqrt{\alpha(t)}z, (1-\alpha(t))\mathbf{I}_d)\;, $$ is reminiscent of denoising diffusion models—see eq. (3) in this post. The relationship between flow matching and denoising diffusion models goes deeper; see [1] [3] .

As we said, this is about verification; let’s show that (6) and (7) check the continuity equation. Observe that $$ \begin{aligned} \frac{\partial p}{\partial t}(y\vert z, t) &= p(y\vert z, t)\frac{\partial\log p}{\partial t}(y\vert z, t) \;, \\ &= p(y\vert z, t)\frac{\partial}{\partial t}\left(-\frac{d}{2}\log(2\pi\sigma(t)^2) - \| y - \mu(t)z\|^2/(2\sigma(t)^2)\right) \;, \\ &= p(y\vert z, t)\left(-d\frac{\dot\sigma(t)}{\sigma(t)}+ \frac{\dot\mu(t)z^\top(y-\mu(t)z)}{\sigma(t)^2}+\frac{\dot\sigma(t)\| y - \mu(t)z\|^2}{\sigma(t)^3}\right)\;. \end{aligned} $$ On the other hand: $$ \begin{aligned} \frac{\partial (pf_i)}{\partial y_i}(y\vert z, t) &= f_i(y\vert z, t)\frac{\partial p}{\partial y_i}(y\vert z, t) + p(y\vert z, t)\frac{\partial f_i}{\partial y_i}(y\vert z, t)\;,\\ &= f_i(y\vert z, t)p(y\vert z, t)\frac{\partial \log p}{\partial y_i}(y\vert z, t) + p(y\vert z, t)\frac{\dot\sigma(t)}{\sigma(t)}\;, \\ &= -p(y\vert z, t)\big( \dot\mu(t) z_i + \dot\sigma(t)(y_i-\mu(t)z_i)/\sigma(t)\big)(y_i - \mu(t)z_i)/\sigma(t)^2 + p(y\vert z, t)\frac{\dot\sigma(t)}{\sigma(t)}\;, \\ &= p(y\vert z, t)\left(-\frac{\dot\mu(t) z_i(y_i - \mu(t)z_i)}{\sigma(t)^2} - \frac{\dot\sigma(t)(y_i - \mu(t)z_i)^2}{\sigma(t)^3} + \frac{\dot\sigma(t)}{\sigma(t)}\right)\;. \end{aligned} $$ Clearly, the continuity equation is verified.

With the ‘straight’ schedule $\mu(t)=t$ and $\sigma(t)=1-t$, the vector fields of (7) become $$f(y\vert z, t) = (z-y)/(1-t)\;.$$ Along the conditional flow started at $y_0$, we have $y(t) = tz + (1-t)y_0$ and $f(y(t)\vert z, t) = z - y_0$: each conditional trajectory is a straight line, travelled at constant speed. That is often understood to back the intuition behind ‘flow matching requires fewer integration steps’.

This is however misleading. ‘Straightness’ does not carry over to the marginal trajectories, which will mechanically bend if $p_\star$ has finite support (which, through the lense of the training process, is true). To get truly straight marginal trajectories, additional work is needed [2] .

Observe that: $$ \begin{aligned} p(y(t)\vert t)\frac{dg}{dt}(t) &= \frac{d}{dt}p(y(t)\vert t) \;, \\ &= \frac{\partial p}{\partial t} (y(t)\vert t) + \sum_{i=1}^d \frac{\partial p}{\partial y_i}(y(t)\vert t) \frac{d y_i}{dt}(t) \;,\\ &= \frac{\partial p}{\partial t} (y(t)\vert t) + \sum_{i=1}^d \frac{\partial p}{\partial y_i}(y(t)\vert t)f_i(y(t)\vert t) \;. \end{aligned} $$ since $y(t)$ solves (1). From there, using the continuity equation at $y(t)$ yields the announced result: $$ \begin{aligned} p(y(t)\vert t)\frac{dg}{dt}(t) &= -\sum_{i=1}^d \left(\frac{\partial p}{\partial y_i}(y(t)\vert t)f_i(y(t)\vert t) + p(y(t)\vert t)\frac{\partial f_i}{\partial y_i}(y(t)\vert t)\right) + \sum_{i=1}^d \frac{\partial p}{\partial y_i}(y(t)\vert t)f_i(y(t)\vert t) \;, \\ &= -p(y(t)\vert t)\sum_{i=1}^d \frac{\partial f_i}{\partial y_i}(y(t)\vert t)\;. \end{aligned} $$ Dividing both sides by $p(y(t)\vert t)>0$ yields the announced result.