I have the differential equation $\frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$\frac{d^2 x}{dt^2}=\left(\frac{d^2 x_1}{dt^2},\frac{d^2 x_2}{dt^2},\ldots,\frac{d^2 x_n}{dt^2}\right)^T\text{ for }x=(x_1,x_2,\ldots,x_n).$$
I know the solution is $x=\cos(\sqrt{A}(t-t_0))x_0+(\sqrt{A})^{-1}\sin(\sqrt{A}(t-t_0))\dot{x}_0$, but I have no idea why this is.
How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.
Best Answer
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $\vec{x}''+A\vec{x}=0$ can be rewritten as $P\vec{x}''+A_dP\vec{x}=0$. Denote $\vec{y}=P\vec{x}$, then $$\vec{y}''+A_d\vec{y}=0.$$ But this is equivalent to $n$ independent scalar ODEs $$ y_k''+\lambda_k y_k=0,$$ where $\lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are $$y_k(t)=\cos\sqrt{\lambda_k}(t-t_0)\,y_k(t_0)+\frac{1}{\sqrt{\lambda_k}}\sin\sqrt{\lambda_k}(t-t_0)\,y'_k(t_0).$$ Or, in matrix form $$ \vec{y}(t)=\cos \sqrt{A_d}(t-t_0)\,\vec{y}(t_0)+\sqrt{A_d}^{-1}\sin \sqrt{A_d}(t-t_0)\,\vec{y}\,'(t_0). $$ Now using that $\vec{x}=P^{-1}\vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.