I have a difference equation y(k+1)=y(k)+k*(a-y^2(k)) k>0 ,a>0 The equilibrium points are easy to find y* = +/- sqrt(a) How can I find the limit on k for stability? I am beginning to think there is not a closed-form solution. N.
Solution of deifference equation
Started by ●November 29, 2009
Reply by ●November 30, 20092009-11-30
On Sun, 29 Nov 2009 19:56:45 -0800, naebad wrote:> I have a difference equation > > y(k+1)=y(k)+k*(a-y^2(k)) > > k>0 ,a>0 > > The equilibrium points are easy to find > > y* = +/- sqrt(a) > > How can I find the limit on k for stability? I am beginning to think > there is not a closed-form solution. > > > N.Do you mean for your gain 'k' to be the same as your time-step variable 'k'? With that y^2 in there you'll always be able to make the system unstable whenever a and k are nonzero and your starting value of y is of sufficient magnitude and the correct sign. -- www.wescottdesign.com
Reply by ●November 30, 20092009-11-30
naebad <naebad@yahoo.com> writes:> I have a difference equation > > y(k+1)=y(k)+k*(a-y^2(k)) > > k>0 ,a>0 > > The equilibrium points are easy to find > > y* = +/- sqrt(a) > > How can I find the limit on k for stability? I am beginning to think > there is not a closed-form solution.If y(k) = p + epsilon, where p is one of the equilibrium points, y(k+1) = p + (1 - 2*k*p)*epsilon - k*epsilon^2 so small deviations from p are amplified if k*p < 0 or k*p > 1. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Reply by ●November 30, 20092009-11-30
On Nov 30, 5:54�pm, Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:> naebad <nae...@yahoo.com> writes: > > I have a difference equation > > > y(k+1)=y(k)+k*(a-y^2(k)) > > > k>0 ,a>0 > > > The equilibrium points are easy to find > > > y* = +/- sqrt(a) > > > How can I find the limit on k for stability? I am beginning to think > > there is not a closed-form solution. > > If y(k) = p + epsilon, where p is one of the equilibrium points, > y(k+1) = p + (1 - 2*k*p)*epsilon - k*epsilon^2 > so small deviations from p are amplified if k*p < 0 or k*p > 1. > -- > Robert Israel � � � � � � �isr...@math.MyUniversitysInitials.ca > Department of Mathematics � � � �http://www.math.ubc.ca/~israel > University of British Columbia � � � � � �Vancouver, BC, CanadaNow should I make that P+epsilon.exp(jtheta) where 0<theta<pi ?
Reply by ●November 30, 20092009-11-30
On Nov 30, 5:33�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On Sun, 29 Nov 2009 19:56:45 -0800, naebad wrote: > > I have a difference equation > > > y(k+1)=y(k)+k*(a-y^2(k)) > > > k>0 ,a>0 > > > The equilibrium points are easy to find > > > y* = +/- sqrt(a) > > > How can I find the limit on k for stability? I am beginning to think > > there is not a closed-form solution. > > > N. > > Do you mean for your gain 'k' to be the same as your time-step variable > 'k'? > > With that y^2 in there you'll always be able to make the system unstable > whenever a and k are nonzero and your starting value of y is of > sufficient magnitude and the correct sign. > > --www.wescottdesign.comyes k is the step size if you like.
Reply by ●November 30, 20092009-11-30
naebad <naebad@yahoo.com> writes:> On Nov 30, 5:54=A0pm, Robert Israel > <isr...@math.MyUniversitysInitials.ca> wrote: > > naebad <nae...@yahoo.com> writes: > > > I have a difference equation > > > > > y(k+1)=3Dy(k)+k*(a-y^2(k)) > > > > > k>0 ,a>0 > > > > > The equilibrium points are easy to find > > > > > y* =3D +/- sqrt(a) > > > > > How can I find the limit on k for stability? I am beginning to think > > > there is not a closed-form solution. > > > > If y(k) =3D p + epsilon, where p is one of the equilibrium points, > > y(k+1) =3D p + (1 - 2*k*p)*epsilon - k*epsilon^2 > > so small deviations from p are amplified if k*p < 0 or k*p > 1. > > -- > > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0 > > =A0isr...@math.MyUniversitysInitial= > s.ca > > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel > > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC, > > Cana= > da > > Now should I make that P+epsilon.exp(jtheta) where 0<theta<pi ?No. You should think about what stability and instability mean. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Reply by ●November 30, 20092009-11-30
On Sun, 29 Nov 2009 23:48:09 -0800, naebad wrote:> On Nov 30, 5:33 pm, Tim Wescott <t...@seemywebsite.com> wrote: >> On Sun, 29 Nov 2009 19:56:45 -0800, naebad wrote: >> > I have a difference equation >> >> > y(k+1)=y(k)+k*(a-y^2(k)) >> >> > k>0 ,a>0 >> >> > The equilibrium points are easy to find >> >> > y* = +/- sqrt(a) >> >> > How can I find the limit on k for stability? I am beginning to think >> > there is not a closed-form solution. >> >> > N. >> >> Do you mean for your gain 'k' to be the same as your time-step variable >> 'k'? >> >> With that y^2 in there you'll always be able to make the system >> unstable whenever a and k are nonzero and your starting value of y is >> of sufficient magnitude and the correct sign. >> >> --www.wescottdesign.com > > yes k is the step size if you like.OK, now you've introduced a _third_ possibility. In your equation, you clearly use 'k' to mean the step index -- i.e. y(k) is the k'th value of y. But you also use 'k' as a gain -- in the expression k*(a-y^2(k)). Do you mean that y(1) = y(0) + 0 * (a - y^2(0)), y(2) = y(1) + 1 * (a - y^2(1)), y(3) = y(2) + 2 * (a - y^2(2)), etc.? Or did you inadvertently use 'k' for two different things? -- www.wescottdesign.com
Reply by ●November 30, 20092009-11-30
On Dec 1, 9:15�am, Tim Wescott <t...@seemywebsite.com> wrote:> On Sun, 29 Nov 2009 23:48:09 -0800, naebad wrote: > > On Nov 30, 5:33�pm, Tim Wescott <t...@seemywebsite.com> wrote: > >> On Sun, 29 Nov 2009 19:56:45 -0800, naebad wrote: > >> > I have a difference equation > > >> > y(k+1)=y(k)+k*(a-y^2(k)) > > >> > k>0 ,a>0 > > >> > The equilibrium points are easy to find > > >> > y* = +/- sqrt(a) > > >> > How can I find the limit on k for stability? I am beginning to think > >> > there is not a closed-form solution. > > >> > N. > > >> Do you mean for your gain 'k' to be the same as your time-step variable > >> 'k'? > > >> With that y^2 in there you'll always be able to make the system > >> unstable whenever a and k are nonzero and your starting value of y is > >> of sufficient magnitude and the correct sign. > > >> --www.wescottdesign.com > > > yes k is the step size if you like. > > OK, now you've introduced a _third_ possibility. > > In your equation, you clearly use 'k' to mean the step index -- i.e. y(k) > is the k'th value of y. > > But you also use 'k' as a gain -- in the expression k*(a-y^2(k)). > > Do you mean that > > y(1) = y(0) + 0 * (a - y^2(0)), > y(2) = y(1) + 1 * (a - y^2(1)), > y(3) = y(2) + 2 * (a - y^2(2)), > etc.? > > Or did you inadvertently use 'k' for two different things? > > --www.wescottdesign.comyes!
Reply by ●November 30, 20092009-11-30
On Mon, 30 Nov 2009 16:15:00 -0800, naebad wrote:> On Dec 1, 9:15 am, Tim Wescott <t...@seemywebsite.com> wrote: >> On Sun, 29 Nov 2009 23:48:09 -0800, naebad wrote: >> > On Nov 30, 5:33 pm, Tim Wescott <t...@seemywebsite.com> wrote: >> >> On Sun, 29 Nov 2009 19:56:45 -0800, naebad wrote: >> >> > I have a difference equation >> >> >> > y(k+1)=y(k)+k*(a-y^2(k)) >> >> >> > k>0 ,a>0 >> >> >> > The equilibrium points are easy to find >> >> >> > y* = +/- sqrt(a) >> >> >> > How can I find the limit on k for stability? I am beginning to >> >> > think there is not a closed-form solution. >> >> >> > N. >> >> >> Do you mean for your gain 'k' to be the same as your time-step >> >> variable 'k'? >> >> >> With that y^2 in there you'll always be able to make the system >> >> unstable whenever a and k are nonzero and your starting value of y >> >> is of sufficient magnitude and the correct sign. >> >> >> --www.wescottdesign.com >> >> > yes k is the step size if you like. >> >> OK, now you've introduced a _third_ possibility. >> >> In your equation, you clearly use 'k' to mean the step index -- i.e. >> y(k) is the k'th value of y. >> >> But you also use 'k' as a gain -- in the expression k*(a-y^2(k)). >> >> Do you mean that >> >> y(1) = y(0) + 0 * (a - y^2(0)), >> y(2) = y(1) + 1 * (a - y^2(1)), >> y(3) = y(2) + 2 * (a - y^2(2)), >> etc.? >> >> Or did you inadvertently use 'k' for two different things? >> >> --www.wescottdesign.com > > yes!Which one????????? -- www.wescottdesign.com
Reply by ●November 30, 20092009-11-30
Tim Wescott wrote: > On Mon, 30 Nov 2009 16:15:00 -0800, naebad wrote: ...>>> Or did you inadvertently use 'k' for two different things?>> yes! > > Which one?????????Huh? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
