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Say I have the following second order 7 x 7 system of equations:

  • $x_1'' = 10(x_2- x_1- 1)$
  • $x_2'' = 10(x_3- 2x_2+ x_1)$
  • $x_3'' = 10(x_4- 2x_3+ x_2)$
  • $x_4'' = 10(x_5- 2x_4+ x_3)$
  • $x_5'' = 10(x_6- 2x_5+ x_4)$
  • $x_6'' = 10(x_7- 2x_6+ x_5)$
  • $x_7'' = 10(x_6- x_7)$.

How would I convert this second order 7 x 7 system into a first order 14 x 14 system using the additional equations $v_j = x'_j$, where $j = 1, 2, 3, ..., 7$?

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  • $\begingroup$ By using $x'=v$, $v'=x''=f(x)$. $\endgroup$ – Dr. Lutz Lehmann May 3 '15 at 21:02
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You would define variables $x_8,x_9\dots x_{14}$, add equations $x_i'=x_{i+7}$ and rewrite your existing equations to look like $x_1''=x_8'=10(x_2-x_1-1)$

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  • $\begingroup$ So there would basically be two of each equation? $\endgroup$ – Daniel Derfus May 3 '15 at 21:25
  • $\begingroup$ Could you elaborate on defining $x_8, x_9, ..., x_{14}$ $\endgroup$ – Daniel Derfus May 3 '15 at 21:42
  • $\begingroup$ Not two of each equation, but rather you have that $x_1' = x_8$. $\endgroup$ – Emily May 4 '15 at 0:37
  • $\begingroup$ As you now have fourteen variables, you need fourteen equations. The seven new ones are the ones that define the new variables as I have shown. The old ones are then rewritten to remove the second derivative, using the new variables. There is nothing special about the fact that you have seven variables to start with. The same process could take a single second order equation and make two first order equations. $\endgroup$ – Ross Millikan May 4 '15 at 4:14
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Just have the 7 equations $v_j = x'_j$ and the seven equations with $v'_j$ equal to the right hand sides of the ones you have there.

If the $x_j$ were positions, all this would do is to say, instead of giving 7 differential equations for relating the accelerations and positions, you get 7 equations relating the velocities with the rate of change of the positions $x' = v$ (which are indeed identical) and the 7 equations relating the change in velocities with the position.

You can see this is a bit artificial but absolutely true.

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  • $\begingroup$ I get the second part, where $v'_j$ equals the right hand side of the equations I have, but for the 7 equations $v_j = x'_j$, $v_j$ is of order 0 and there is no $x'_j$ in the present system. $\endgroup$ – Daniel Derfus May 3 '15 at 21:37
  • $\begingroup$ $v'_j$ is the time derivative of $v_j$ which itself equals $x'_j$, which is the time derivative of $x_j$. That's why $x'_j = v_j$ is a first order differential equation. $\endgroup$ – Andrea May 4 '15 at 0:12

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