Matrix Ordinary Differential Equation Incorrect Solution?
Posted: November 5th, 2024, 1:37 pm
I have been trying to derive expressions for dA/dt and dB/dt from the below equation(s) for some time now in order to try and replicate the results from a research paper. However, when I attempt to calculate A and B @ t = 0 using numerical methods my solution blows up wich seems to suggest that the equations I have derived for dA/dt and dB/dt are incorrect.
I have checked my math more times than I can count and sadly can't find an error. Can anyone help me validate the below solution?
\begin{equation}
0 = \frac{\partial \theta}{\partial t} - k_{E}(E - D)(q^{F} + \frac{\partial \theta}{\partial E}) - k_{D} (D - \bar{D}) + \frac{1}{2}Tr(\tilde{\Sigma}\nabla^{2}_{ED}) - \frac{\gamma}{2} \begin{pmatrix} q^{S} + q^{F} \\ q^{F} + \frac{\partial \theta}{\partial E} \\ \frac{\partial \theta}{\partial D} \end{pmatrix}^T \Sigma \begin{pmatrix} q^{S} + q^{F} \\ q^{F} + \frac{\partial \theta}{\partial E} \\ \frac{\partial \theta}{\partial D} \end{pmatrix} + \alpha(\frac{\partial^2 \theta}{\partial (q^{S})^{2}} + \frac{\partial \theta}{\partial q^{S}}) ^ 2 + \frac{1}{4\nu^{S}}(\frac{\partial \theta}{\partial q^{S}})^2 + \frac{1}{4\nu^{F}}(\frac{\partial \theta}{\partial q^{F}})^2
\end{equation}
Where:
\begin{equation}
\theta = - \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix}^{T} A(t) \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix} - \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix}^{T}B(t) - C(t)
\end{equation}
Given that A(t) is 4 by 4, B(t) is 4 by 1 and $\Sigma$ is a 3 by 3 positive symmetric (variance covariance matrix). I have come to the below solutions for dA/dt and dB/dt :
\begin{equation}
\frac{\partial A}{\partial t} = A(t)M^{A}A(t) + A(t)U^{A} + (U^{A})^{T}A(t) + R^{A}
\end{equation}
\begin{equation}
\frac{\partial B}{\partial t} = A(t)M^{A}B(t) + A(t)V^{B} + (U^{A})^{T}B(t)
\end{equation}
Where:
\begin{equation}
M^{A} = \begin{pmatrix} 4\alpha + \frac{1}{\nu^{S}} & 0 & 0 & 0 \\0 & \frac{1}{\nu^{F}} & 0 & 0 \\ 0 & 0 & -2\gamma\Sigma_{22} & -2\gamma\Sigma_{23} \\ 0 & 0 & -2\gamma\Sigma_{32} & -2\gamma\Sigma_{33} \end{pmatrix}
\end{equation}
\begin{equation}
U^{A} = \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\ \gamma\Sigma_{21} & \gamma(\Sigma_{21} + \Sigma_{22}) & k_{E} & -k_{E} \\ \gamma\Sigma_{31} & \gamma(\Sigma_{31} + \Sigma_{32}) & 0 & k_{D} \end{pmatrix}
\end{equation}
\begin{equation}
R^{A} = -\frac{\gamma}{2} \begin{pmatrix} \Sigma_{11} & \Sigma_{11} + \Sigma_{12} & 0 & 0 \\\Sigma_{11}+ \Sigma_{21} & \Sigma_{11} + \Sigma_{12} + \Sigma_{21} + \Sigma_{22} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} - \frac{k_{E}}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}
\end{equation}
\begin{equation}
V^{B} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ -2k_{D}\bar{D} \end{pmatrix}
\end{equation}
\begin{equation}
\Sigma = \begin{pmatrix} \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \\ \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \\ \Sigma_{31} & \Sigma_{32} & \Sigma_{33}\end{pmatrix}
\end{equation}
\begin{equation}
\tilde{\Sigma} = \begin{pmatrix} \Sigma_{22} & \Sigma_{23} \\ \Sigma_{32} & \Sigma_{33}\end{pmatrix}
\end{equation}
I have checked my math more times than I can count and sadly can't find an error. Can anyone help me validate the below solution?
\begin{equation}
0 = \frac{\partial \theta}{\partial t} - k_{E}(E - D)(q^{F} + \frac{\partial \theta}{\partial E}) - k_{D} (D - \bar{D}) + \frac{1}{2}Tr(\tilde{\Sigma}\nabla^{2}_{ED}) - \frac{\gamma}{2} \begin{pmatrix} q^{S} + q^{F} \\ q^{F} + \frac{\partial \theta}{\partial E} \\ \frac{\partial \theta}{\partial D} \end{pmatrix}^T \Sigma \begin{pmatrix} q^{S} + q^{F} \\ q^{F} + \frac{\partial \theta}{\partial E} \\ \frac{\partial \theta}{\partial D} \end{pmatrix} + \alpha(\frac{\partial^2 \theta}{\partial (q^{S})^{2}} + \frac{\partial \theta}{\partial q^{S}}) ^ 2 + \frac{1}{4\nu^{S}}(\frac{\partial \theta}{\partial q^{S}})^2 + \frac{1}{4\nu^{F}}(\frac{\partial \theta}{\partial q^{F}})^2
\end{equation}
Where:
\begin{equation}
\theta = - \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix}^{T} A(t) \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix} - \begin{pmatrix} q^{S} \\ q^{F} \\ E \\ D \end{pmatrix}^{T}B(t) - C(t)
\end{equation}
Given that A(t) is 4 by 4, B(t) is 4 by 1 and $\Sigma$ is a 3 by 3 positive symmetric (variance covariance matrix). I have come to the below solutions for dA/dt and dB/dt :
\begin{equation}
\frac{\partial A}{\partial t} = A(t)M^{A}A(t) + A(t)U^{A} + (U^{A})^{T}A(t) + R^{A}
\end{equation}
\begin{equation}
\frac{\partial B}{\partial t} = A(t)M^{A}B(t) + A(t)V^{B} + (U^{A})^{T}B(t)
\end{equation}
Where:
\begin{equation}
M^{A} = \begin{pmatrix} 4\alpha + \frac{1}{\nu^{S}} & 0 & 0 & 0 \\0 & \frac{1}{\nu^{F}} & 0 & 0 \\ 0 & 0 & -2\gamma\Sigma_{22} & -2\gamma\Sigma_{23} \\ 0 & 0 & -2\gamma\Sigma_{32} & -2\gamma\Sigma_{33} \end{pmatrix}
\end{equation}
\begin{equation}
U^{A} = \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\ \gamma\Sigma_{21} & \gamma(\Sigma_{21} + \Sigma_{22}) & k_{E} & -k_{E} \\ \gamma\Sigma_{31} & \gamma(\Sigma_{31} + \Sigma_{32}) & 0 & k_{D} \end{pmatrix}
\end{equation}
\begin{equation}
R^{A} = -\frac{\gamma}{2} \begin{pmatrix} \Sigma_{11} & \Sigma_{11} + \Sigma_{12} & 0 & 0 \\\Sigma_{11}+ \Sigma_{21} & \Sigma_{11} + \Sigma_{12} + \Sigma_{21} + \Sigma_{22} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} - \frac{k_{E}}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}
\end{equation}
\begin{equation}
V^{B} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ -2k_{D}\bar{D} \end{pmatrix}
\end{equation}
\begin{equation}
\Sigma = \begin{pmatrix} \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \\ \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \\ \Sigma_{31} & \Sigma_{32} & \Sigma_{33}\end{pmatrix}
\end{equation}
\begin{equation}
\tilde{\Sigma} = \begin{pmatrix} \Sigma_{22} & \Sigma_{23} \\ \Sigma_{32} & \Sigma_{33}\end{pmatrix}
\end{equation}