I am working on variational analysis for the Schrodinger bridge problem (SBP), defined as follows.
The SBP is equivalent to seeking a flow $\rho(t, x)$ and a velocity field $v(t, x)$, that are a solution to the following constrained minimization problem: \begin{align} \inf_{(\rho, v)} \int_{\mathbb{R}^{n}} \int_{0}^{1} &\left[ \frac{1}{2} \lVert v(t,x) \rVert^{2} + \frac{\gamma^{2}}{8} \lVert \nabla \log{ \rho(t,x) } \rVert^{2} \right] \rho(t, x) \mathrm{d} t \mathrm{d} x\\ &\frac{ \partial \rho }{ \partial t } + \nabla \cdot (v \rho) = 0 \end{align}
The analysis starts by writing out the Lagrangian for the above problem, as: \begin{align*} \mathcal{L}(\rho, v, \lambda) = \int_{\mathbb{R}^{n}} \int_{0}^{1} \left[ \frac{1}{2} \lVert v(t,x) \rVert^{2} \rho(t,x) + \frac{\gamma^{2}}{8} \lVert \nabla \log{ \rho(t,x) } \rVert^{2} \rho(t, x) + \underbrace{\lambda(t, x) \left( \frac{ \partial \rho }{ \partial t } + \nabla \cdot (v \rho) \right)}_{\Lambda} \right] \mathrm{d} t \mathrm{d} x \end{align*}
Then, the proof transforms the $\Lambda$ term, via integration by parts and assuming limits of $x \rightarrow \infty$. It shows that the above $\mathcal{L}$ is equivalent to the following expression: \begin{align*} \mathcal{L}(\rho, v, \lambda) = \int_{\mathbb{R}^{n}} \int_{0}^{1} \left[ \frac{1}{2} \lVert v(t, x) \rVert^{2} + \frac{\gamma^{2}}{8} \lVert \nabla \log{ \rho(t, x) } \rVert^{2} + \left( -\frac{\partial \lambda}{\partial t} - \nabla \lambda \cdot v \right) \right] \rho(t, x) \mathrm{d} t \mathrm{d} x + \int_{\mathbb{R}^{n}} \left[ \lambda(1, x) \rho(1, x) - \lambda(0, x) \rho(0, x) \right] \mathrm{d} x \end{align*}
I am uncertain about whether there is an equality here. I am confused how to transform from the first $\mathcal{L}$ to the second form. Can anyone help me show the derivation process?
Since the total derivative of any scalar function $\lambda$ with respect to $t$ is: $$ d_t\lambda(t,\mathbf{x}) = \partial_t\lambda(t,\mathbf{x}) + \frac{d\mathbf{x}}{dt}\cdot\nabla\lambda(t,\mathbf{x}) = \partial_t\lambda(t,\mathbf{x}) + v(t,\mathbf{x})\cdot\nabla\lambda(t,\mathbf{x}) $$
and $\nabla\cdot(\rho(t,\mathbf{x}) v(t,\mathbf{x})) = \rho(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x}) + v(t,\mathbf{x})\cdot\nabla\rho(t,\mathbf{x})$, you can rewrite your $\Lambda$ term as : $$ \lambda(t,\mathbf{x})\left(d_t\rho(t,\mathbf{x}) + \rho(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x})\right) $$ where $v(t,\mathbf{x})$ is the same velocity field.
It is necessary to write the total derivative since you can not integrate in the following integration by parts the $\partial_t\rho(t,\mathbf{x})$ term, with the derivative being $\partial$ instead of $d$. The first symbol is called partial derivative, because it is a partial derivation relative to one variable.
The integration by parts gives : $$ \int_{\mathbb{R}^N}d\mathbf{x}\int_0^1dt\lambda(t,\mathbf{x})\left(d_t\rho(t,\mathbf{x}) + \rho(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x})\right) = \\ \int_{\mathbb{R}^N}d\mathbf{x}[\lambda(t,\mathbf{x})\rho(t,\mathbf{x})]_0^1 - \int_{\mathbb{R}^N}d\mathbf{x}\int_0^1dt\rho(t,\mathbf{x})\left(d_t\lambda(t,\mathbf{x}) - \lambda(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x})\right) = \\ \int_{\mathbb{R}^N}d\mathbf{x}[\lambda(t,\mathbf{x})\rho(t,\mathbf{x})]_0^1 + \\ \int_{\mathbb{R}^N}d\mathbf{x}\int_0^1dt\rho(t,\mathbf{x})(-\partial_t\lambda(t,\mathbf{x}) - v(t,\mathbf{x})\cdot\nabla\lambda(t,\mathbf{x})) + \\ \int_{\mathbb{R}^N}d\mathbf{x}\int_0^1dt\rho(t,\mathbf{x})\lambda(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x}) $$ As you can see for the last equality, the first and second lines are expected, but the third is not. Therefore the equality between the two Lagrangians should hold if $$ \int_{\mathbb{R}^N}d\mathbf{x}\int_0^1dt\lambda(t,\mathbf{x})\rho(t,\mathbf{x})\nabla\cdot v(t,\mathbf{x}) = 0 $$
Do you have any condition on the velocity field, such as incompressibility ($\nabla\cdot v(t,\mathbf{x}) = 0)$ ?
Details: