Setup
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I have recently come across an ODE of the form
$$
0 = \dot{A}(t)^TG(t) + A(t)^T\dot{G}(t) + C(t) + \lambda B(t)^TA(t)G(t) + \frac{\lambda}{2} (D(t)A(t)G(t))(D(t)A(t)G(t))^T\bar{1},
$$
where $\dot{A}(t)$ is the time derivative of $A(t)$, $A$ and $D$ are $\mathbb{R}$eal-Matrix-valued functions, $G, C$ and $B$ are vector-valued functions, $\lambda$ is a constant and $\bar{1}$ is the vector with only $1$s as entries. 


Question
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My question is: when is there a solution to this equation and what is it explicitly? 


What I've been thinking
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I've looked up nonsymmetric Riccati equations but I'm not certain this falls into that class of equations or if it is even solvable. 



Differential Riccati-type equation
I do not see any relation to matrix Riccati equations here, so may guess is that the literature on nonsymmetric Riccati equations will not help. Anyway, here is an approach to solving the system. 


If we simply assume that $A$ is symmetric, the equation is not one for $A$, but rather one for $z(t):=A(t)G(t)= A(t)^TG(t)$. If we use this notation then the differential equation is
$$\dot{z}(t) = -C(t) - \lambda B(t) z(t) - \frac{\lambda}{2} (D(t)z(t))(D(t)z(t))^T\mathbf{1}.$$
This is a quadratic equation in $z$, so locally Lipschitz in $z$. So unique solutions to an inital value problem $z(t\_0) = z\_0$ will exist locally, if you assume $B,C,D$ are continuous (or if you want to appeal to Caratheodory's theorem, locally $L^1$ in $t$ will be fine). Because of the quadratic part forward completeness will be a problem of course and depend on data and initial conditions. For instance $\dot z = - z^2$ with $z\in\mathbb{R}$ is an example that fits in the class.


Now if $G$ is also part of your data, and you are looking for $A$, then you can pick a symmetric matrix-valued function $A(t)$ such that
$$ z(t;t\_0,z\_0) = A(t)G(t)$$
for all $t$ where $z(t;t\_0,z\_0)$ is defined. If $G(t)\neq 0$ always, then this can always be done. If $G(t)$ has zeros, you run into problems. Then it would depend on what $z$ happens to be doing at a zero of $G$.


Unfortunately, even if $G$ has no zeros, there are many symmetric matrix valued functions which solve the equation. So the problem seems to be, not whether a solution exists, but that you have way too many solutions. (Depending on what you want of course, maybe it is OK to assume that all entries of $G$ have no zeros and to insist on diagonal solutions for $A$).


