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How do I solve problems of the form $$X''(t) = -(A^2)X(t),$$ where $X$ is a $2 \times 1$ matrix and $A$ is a $2 \times 2$ matrix? We're given $X(0)$ and $X'(0)$.

Did
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Jian Liang
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    Is $A$ a constant? – Jan Eerland Mar 28 '16 at 14:05
  • This is an initial value problem. Where A is a square matrix. – Jian Liang Mar 28 '16 at 14:32
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    With $B=-A^2$, the solution is $$X(t)=C(t)X(0)-S(t)X'(0)$$ with $$C(t)=\sum_{k=0}^\infty\frac{t^{2k}}{(2k)!}B^k\qquad S(t)=\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}B^k.$$ – Did Mar 28 '16 at 14:36
  • @Did Why not $X(t) = e^{iAt}$? Because $X'(t) = iAe^{iAt}$ so $X''(t) = (iA)^2e^{iAt} = -A^2X(t)$. – John Martin Mar 28 '16 at 14:41
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    @JohnMartin Perhaps because 1. the solution would not depend on $X(0)$ and $X'(0)$, 2. the solution would take imaginary not real values while $X(t)$ is real valued if $X(0)$ and $X'(0)$ are, 4. the solution you suggest might correspond to the initial conditions $X(0)=I$ and $X'(0)=iA$. – Did Mar 28 '16 at 14:45
  • *Correction: $X(t)=C(t)X(0)+S(t)X'(0)$. – Did Mar 28 '16 at 14:47

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