Nonlinear (quadratic) matrix differential equation Let $X_0, A \in \mathbf R^{n \times n}$ be symmetric positive semidefinite matrices, and consider the equation
$$
\dot X = - X A X, \qquad X(0) = X_0.
$$
Does this equation admit an explicit solution (possibly in terms of the eigenvalues and eigenvectors of $A$ or $X_0$)? If not, is there a standard way of approaching it, in order to for example deduce convergence rates to equilibrium?
Edit: diagonalizing $A$ as $V D V^T$ and introducing $Y = V^TXV$, we can write the following equation for $Y$:
$$ \dot Y = - YDY, \qquad Y(0) = Y_0:= V^T X_0V.$$
 A: There was a silly typo in my original comment but after a discussion with a friend I think there's a way to arrive at an answer for some cases which includes the original comment.
Given $\dot{X}=-XAX$, we can try to write the solution $X$ as a power series. $X(0)=X_0$,
$$\dot{X}(0)=-X_0AX_0=(-1)^1 (X_0A)^1X_0$$
$$\ddot{X}(0)=+2X_0AX_0AX_0=(-1)^2\cdot 2\cdot (X_0A)^2X_0$$
$$\dddot{X}(0)=(-1)^3\cdot 6\cdot (X_0A)^3X_0$$
$$X^{(n)}(0)=(-1)^n\cdot n!\cdot (X_0A)^nX_0 $$
So without taking into account questions of convergence or invertibility for the moment, it seems like
$$X=\sum_{n=0}^{\infty}(-1)^n (X_0A)^nX_0t^n=(\mathbb{1}+X_0At)^{-1}X_0$$
You could also choose to factor out each of the derivatives as $X^{(n)}(0)=(-1)^n\cdot n!\cdot X_0 (AX_0)^n$ so that 
$$X=\sum_{n=0}^{\infty}(-1)^n X_0(AX_0)^nt^n=X_0(\mathbb{1}+AX_0t)^{-1}$$
This answer by Robert Israel points out something to be wary of for non-hermitian positive semi-definite matrices, but it looks like from the last part of the answer $AX_0$ or $X_0A$ should have non negative eigenvalues and so it looks like it could be ok for something like $0\leq t \leq \rho(AX_0)^{-1}$ where $\rho(AX_0)$ is its spectral radius. I'm not very familiar with these things so maybe someone can correct me on this.
Although I think by using $\frac{d}{dt}U^{-1}=-U^{-1}\frac{dU}{dt}U^{-1}$, then it could be true in general that the solution is $X(t)=(\mathbb{1}+X_0At)^{-1}X_0$ or $X(t)=X_0(\mathbb{1}+AX_0t)^{-1}$.
