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Let $\mathbf H$ be a fixed Hilbert space. Let $\mathcal L(\mathbf H)$ be the collection of linear operators (that may not be bounded) on $H$. For every $A\in \mathcal L(\mathbf H)$, denote by $D(A)$ the domain of $A$. I am looking for the references on the following differential equation

$$\frac {{\rm d}A_t}{{\rm d}t} = LA_t + f(A_t)A_t,$$

where $L\in \mathcal L(\mathbf H)$ is some operator, $LA$ denotes the multiplication between operators, $f: \mathcal L(\mathbf H)\to\mathbb R$ (or $f: \mathcal L(\mathbf H)\to\mathbb C$) is some suitable functional, e.g. $f(A_t)=trace(A_t)$, $f(A_t)=trace(A_t^2)$, $f(A_t)=(trace(A_t))^2$, etc.

I am happy to know the existing literature on the study of wellposedness and solution properties of the above equation.

PS : Suggested by Willie Wong, we consider an example as below (which is the main motivation). Consider the partial differential equation on $\mathbb R_+\times \mathbb R^d \times\mathbb R^d$:

\begin{eqnarray*} \partial_t \rho(t,x,y) &=& -\mathrm{i}\Big(-\frac 12 (\Delta_{x}-\Delta_{y})\rho(t,x,y) +\big( V(x)-V(y)\big) \rho(t,x,y)\Big)\\ && - \frac{1}{2}\Big(K(x)- K(y) \Big)^2 \rho(t,x,y), \end{eqnarray*} where $V, K:\mathbb R^d\to\mathbb R$ are suitable functions. Here we pursue the solution $\rho$ in the trace class. When $K\equiv 0$, it is known that $\rho(t,x,y)=\varphi(t,x)\overline\varphi(t,y)$, where $\varphi$ solves some Schrödinger equation, see e.g. Energy estimation of density operator to von Neumann equation

For a general $K$, I wish to know how to study this wellposedness. A natural idea, as above, is to treat this PDE as a differential equation valued in operator space. Namely, $\rho_t:=\rho(t,\cdot)$ can be seen as an operator, and $V, K$ can be seen as multiplication operators, then

\begin{equation} \mathrm{d} \rho_t = -\mathrm{i}\left[-\frac 12 \Delta + V,\rho_t\right]\mathrm{d} t + \left(K\rho_tK - \frac{1}{2}K^2\rho_t-\frac{1}{2}\rho_tK^2 \right)\mathrm{d} t\quad (\ast) \end{equation}

Here $[\cdot,\cdot]$ denotes the commutator of two operators. Does $(\ast)$ admit a solution in the trace class?

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  • $\begingroup$ Your equation includes a lot of not-defined quantities that requires suitable interpretation. At a formal level the linear part of the equation is exactly resolved by $C^0$ semigroups (Hille-Yoshida basically tells you how to solve $\frac{d A_t}{dt} = L A_t$ with $A_0 = Id$). So maybe that's a place to start? $\endgroup$ Commented Jan 10 at 20:21
  • $\begingroup$ @WillieWong Thanks for the comment. Indeed I'm very unfamiliar with this literature and I try to be as general as possible. A concrete context is that $L=-\Delta$, $L=-i\Delta$, $L=-\Delta+V$, $L=-i\Delta+V$ for example $\endgroup$ Commented Jan 10 at 20:30
  • $\begingroup$ @WillieWong I agree that for $dA_t=LA_tdt$ can be solved (at least formally) by the so-called semi group, as I saw for the Schordinger equation when I was student. While for the added non linear term, how to deal with it? $\endgroup$ Commented Jan 10 at 20:32
  • $\begingroup$ I should note a few things: in the operator semi-groups case, the solution operator $A_t$ turns out to be bounded linear maps of $H$, since there's a good topology in which continuity makes sense. So if you have a situation where whenever $A_t$ is a continuous function from an interval to the space of bounded linear maps, then $f(A_t)$ is a continuous scalar function, then you have hope of using the formalism of mild solutions to develop what you are doing. But the whole story will depend on what sorts of $f$ you consider, what notion of "solution" for the linear problem you use, etc. $\endgroup$ Commented Jan 10 at 20:42
  • $\begingroup$ If you do not even assume $A_t$ are bounded, I don't see any good way to start. This is to say nothing of your examples of $f$ are assuming that $A_t$ are trace class and so on. Is there a way where you can pose your question in a less general way? Right now your differential equation is setting equal two things both of which are not well-defined in general, so it is hard to say anything meaningful. $\endgroup$ Commented Jan 10 at 20:45

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