1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you for your solution. May I ask how you computed $M(t)$? $\endgroup$ – Justin Le May 1 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/… $\endgroup$ – Carlo Beenakker May 1 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.