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Let $$\begin{cases}p'' = -q\\q'' = p \end{cases}$$ Goal: convert above system into first order equations.

I expressed the system in matrix form. Next, did some calculation and ended up getting $$p' = e^t\quad\text{and}\quad q' = e^{-t}$$

But I doubt what I did is correct. Is there a proper method of doing this?

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3 Answers 3

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You are starting with 2 second order equations in two unknowns. In effect, you have a single vector ODE of the form v'' = some function of v, where v = (p,q). What you want is something of the form w' = a function of w. Try letting w = (p, q, u, s), where the new variables u and s are u = p' and s = q'.

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  • $\begingroup$ so (p q) ' = [(0 -1), (1,0)] (p q) ?? and then basically I would get exponential answers, correct? $\endgroup$
    – user147037
    Sep 4, 2015 at 20:34
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Often I make mistakes in calculations but the idea will hopefully be clear.

We have \begin{eqnarray*} \partial _{t}^{2}p &=&-q \\ \partial _{t}^{2}q &=&p \end{eqnarray*} Define \begin{eqnarray*} p_{1} &=&p,\;p_{2}=\partial _{t}p \\ q_{1} &=&q,\;q_{2}=\partial _{t}q \end{eqnarray*} Then \begin{eqnarray*} \partial _{t}p_{1} &=&p_{2} \\ \partial _{t}p_{2} &=&-q_{1} \\ \partial _{t}q_{1} &=&q_{2} \\ \partial _{t}q_{2} &=&p_{1} \end{eqnarray*} or $$ \partial _{t}\left( \begin{array}{c} q_{2} \\ p_{1} \\ p_{2} \\ q_{1} \end{array} \right) =\mathsf{B}\left( \begin{array}{c} q_{2} \\ p_{1} \\ p_{2} \\ q_{1} \end{array} \right) $$ where \begin{eqnarray*} \mathsf{B} &=&\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1% \end{array}% \right) ,\;\mathsf{B}^{2}=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1% \end{array}% \right) =\mathsf{I} \\ \mathsf{I}+\mathsf{B} &=&2\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1% \end{array}% \right) ,\;\mathsf{I}-\mathsf{B}=-2\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) \end{eqnarray*} With $$ \mathbf{x}(t)=\left( \begin{array}{c} q_{2} \\ p_{1} \\ p_{2} \\ q_{1} \end{array} \right) $$ we have \begin{eqnarray*} \partial _{t}\mathbf{x} &=&\mathsf{B}\mathbf{\cdot x} \\ \mathbf{x}(t) &=&\exp [\mathsf{B}t]\mathbf{x}(0)=\{\mathsf{I}+\mathsf{B}t+% \frac{1}{2}\mathsf{B}^{2}t^{2}+\frac{1}{3!}\mathsf{B}^{3}t^{3}+\cdots \}\cdot \mathbf{x}(0) \\ &=&\{\mathsf{I}+\mathsf{B}t+\frac{1}{2}\mathsf{I}t^{2}+\frac{1}{3!}\mathsf{B}% t^{3}+\cdots \}\cdot \mathbf{x}(0) \\ &=&\{\mathsf{I}\cosh t+\mathsf{B}\sinh (t)\}\cdot \mathbf{x}(0) \\ &=&\{\mathsf{I}\frac{e^{t}+e^{-t}}{2}+\mathsf{B}\frac{e^{t}-e^{-t}}{2}% \}\cdot \mathbf{x}(0) \\ &=&\{\frac{e^{t}}{2}(\mathsf{I}+\mathsf{B})+\frac{e^{-t}}{2}(\mathsf{I}-% \mathsf{B})\}\cdot \mathbf{x}(0) \\ &=&\{e^{t}\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1% \end{array}% \right) -e^{-t}\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0% \end{array}% \right) \}\cdot \mathbf{x}(0) \end{eqnarray*} from which you can read off the expressions for $q(t)$ and $p(t)$.

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I find laplace tranforms to be the cleanest way for such problems. for example http://tutorial.math.lamar.edu/Classes/DE/SystemsLaplace.aspx

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