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Nonlinear (quadratic) matrix differential equation

Nonlinear Ordinary differential equationNonlinear differential equation typeNonlinear differential equationNonlinear differential equationNonlinear Differential Equation questionNonlinear matrix differential equationDifferential Equation with a Singular point/singularityNonlinear ODE differential equationnonlinear nonhomogeneous differential equationCheaper way of calculating this (very) multidimensional derivative

2

$begingroup$

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 $Y$:
$$ dot Y = - YDY, qquad Y(0) = Y_0:= V^T X_0V.$$

ordinary-differential-equations

share|cite|improve this question

edited 2 days ago

Roberto Rastapopoulos

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it something like $X=(At+C)^-1$, where $dotx=fracdxdt$?
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe that's only the case when $x_0$ is invertible
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I was thinking about this too, but it's already a good partial solution. Thank you very much!
  $endgroup$
  – Roberto Rastapopoulos
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just thinking about the case where $X=f(At)$, and $f$ can be expanded in a power series, so that $A$ commutes with $X$, and then you can just solve the ode $dotX=-AX^2$ which is separable.
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $Y$:
$$ dot Y = - YDY, qquad Y(0) = Y_0:= V^T X_0V.$$

ordinary-differential-equations

share|cite|improve this question

edited 2 days ago

Roberto Rastapopoulos

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it something like $X=(At+C)^-1$, where $dotx=fracdxdt$?
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe that's only the case when $x_0$ is invertible
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I was thinking about this too, but it's already a good partial solution. Thank you very much!
  $endgroup$
  – Roberto Rastapopoulos
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just thinking about the case where $X=f(At)$, and $f$ can be expanded in a power series, so that $A$ commutes with $X$, and then you can just solve the ode $dotX=-AX^2$ which is separable.
  $endgroup$
  – snulty
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

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 $Y$:
$$ dot Y = - YDY, qquad Y(0) = Y_0:= V^T X_0V.$$

ordinary-differential-equations

share|cite|improve this question

edited 2 days ago

Roberto Rastapopoulos

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

$endgroup$

share|cite|improve this question

edited 2 days ago

Roberto Rastapopoulos

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

share|cite|improve this question

edited 2 days ago

Roberto Rastapopoulos

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425

asked 2 days ago

Roberto RastapopoulosRoberto Rastapopoulos

953425