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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, $$x(t) = \left[\begin{array}{c} \sum_{i=1}^n c_iv_i\cos\left(\sqrt{\lambda_i} t\right) \\ -\sum_{i=1}^n c_iv_i\sqrt{\lambda_i}\sin\left(\sqrt{\lambda_i} t\right) \end{array}\right].$$

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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 $x(t)={{O^T\; 0}\choose{0\; O^T}} X(t)$.

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  • $\begingroup$ Thank you for your solution. May I ask how you computed $M(t)$? $\endgroup$
    – Justin Le
    May 1, 2019 at 16:25
  • $\begingroup$ $M(t)$ is the exponent of a 2x2 matrix, because $\Lambda$ is diagonal, which is elementary, see for example en.wikipedia.org/wiki/… $\endgroup$ May 1, 2019 at 16:57

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