# Solve a linear matrix ODE involving symmetric blocks

Let $$P \in \mathbb R^{n \times n}$$ be a symmetric positive definite matrix with eigenvalues denoted by $$\lambda_i$$ and corresponding eigenvectors denoted by $$v_i$$. For each $$j \in \{1, 2, 3, 4\}$$, let $$\alpha_j$$ be a non-zero real number. Let $$x: [0, \infty) \rightarrow \mathbb R^{2n}$$ be a continuous, differentiable function satisfying \begin{align*} &\frac{d}{dt}x(t) = Ax(t), \\ &A = \left[\begin{array}{cc} \alpha_1P & \alpha_2I \\ \alpha_3P & \alpha_4I \end{array}\right] \in \mathbb R^{2n \times 2n}, \\ &x(0) = z \in \mathbb R^{2n}, \end{align*} where $$I \in \mathbb R^{n \times n}$$ is the identity matrix. What methods can be used to manually obtain the solution $$x(t)$$ to the above differential equation, expressed in terms of $$\lambda_i$$ and $$v_i$$? We may manipulate $$z$$ for convenience (for example, as done below).

If we don't require the $$\alpha_j$$ to be non-zero, the case $$A = \left[\begin{array}{cc} 0 & I \\ -P & 0 \end{array}\right]$$ gives rise to the following solution. Define $$x_1 \in \mathbb R^n$$ and $$x_2 \in \mathbb R^n$$ such that $$x = \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right],$$ and let $$x_1(0) = \sum_{i=1}^n c_iv_i,$$ where each $$c_i$$ is a real number. I've been informed that, in this case, $$x(t) = \left[\begin{array}{c} \sum_{i=1}^n c_iv_i\cos\left(\sqrt{\lambda_i} t\right) \\ -\sum_{i=1}^n c_iv_i\sqrt{\lambda_i}\sin\left(\sqrt{\lambda_i} t\right) \end{array}\right].$$

Diagonalize $$P=O\Lambda O^T$$ with an $$n\times n$$ orthogonal matrix $$O$$ (containing the eigenvectors $$v_i$$) and a diagonal matrix $$\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_n)$$. Define $$X={{O\; 0}\choose{0\; O}}x$$ and $$Z={{O\; 0}\choose{0\; O}}z$$. Then the differential equation becomes $$\frac{d}{dt}X(t) = \left(\begin{array}{cc} \alpha_1\Lambda & \alpha_2I \\ \alpha_3\Lambda & \alpha_4I \end{array}\right)X(t), \;\; X(0) = Z,$$ with solution $$X(t)=M(t)Z,\;\;M(t)=\exp\left(\begin{array}{cc} \alpha_1\Lambda\, t & \alpha_2I t \\ \alpha_3\Lambda\, t & \alpha_4I t \end{array}\right)$$ $$\Rightarrow M(t)=\left( \begin{array}{cc} e^{\Omega_+t}\cosh\Xi\, t+\Omega_-\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t }& {\alpha_2}\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t } \\ {\alpha_3} \Lambda\Xi^{-1}e^{\Omega_+t}{\sinh \Xi\, t } &e^{\Omega_+t}\cosh\Xi \,t- \Omega_-\Xi^{-1}e^{\Omega_+t}{\sinh\Xi\, t } \\ \end{array}\right),$$ where we have defined $$\Omega_\pm=\tfrac{1}{2}(\alpha_1\Lambda\pm\alpha_4 I),\;\;\Xi=\sqrt{\Omega_-^2+ \alpha_2\alpha_3\Lambda}.$$ From here you recover $$x(t)={{O^T\; 0}\choose{0\; O^T}} X(t)$$.
• Thank you for your solution. May I ask how you computed $M(t)$? – Justin Le May 1 at 16:25
• $M(t)$ is the exponent of a 2x2 matrix, because $\Lambda$ is diagonal, which is elementary, see for example en.wikipedia.org/wiki/… – Carlo Beenakker May 1 at 16:57