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We have a linear difference equation with constant coefficient $$\begin{cases}x(t+1)=ax(t)+by(t)\\y(t+1)=cx(t)+dy(t)\end{cases}$$

What is the differential equation associated with the above difference equation?And what is the reason of such association? What about if the coefficient are not constant?May you give a reference for more materials on such topics?Thank you.

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  • $\begingroup$ What do you mean by 'the differential equation associated with the above difference equation'? $\endgroup$ – copper.hat Dec 7 '15 at 18:59
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    $\begingroup$ @copper.hat I mean some things like this math.stackexchange.com/questions/145523/… $\endgroup$ – Ali Taghavi Dec 7 '15 at 19:04
  • $\begingroup$ but I need more explanation for this particular 2 dim system in my question. $\endgroup$ – Ali Taghavi Dec 7 '15 at 19:04
  • $\begingroup$ You can always create some sort of correspondence, as the other answer shows, but presumably you have some intent in mind? One possible interpretation is to find an equivalent ODE whose solution is given by the above if the input is constant on the intervals $[n,n+1)$. $\endgroup$ – copper.hat Dec 7 '15 at 19:10
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To do this, you need to construct a derivative from the pieces you have $$\frac{dx}{dt} \approx \frac{x(t+\Delta t)-x(t)}{\Delta t}$$ For the difference equation, $\Delta t=1$, so this translates to $$\frac{x(t+1)-x(t)}{1}=\Delta x$$ So for the first equation, you need to move a $1x$ over the left side, yielding $$\frac{dx}{dt} \approx x(t+1)-x(t)=(a-1)x(t)+by(t)$$

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