Solution of deifference equation
naebad
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.
Tim Wescott
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.
Robert Israel
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, Canada
naebad
Now should I make that P+epsilon.exp(jtheta) where 0<theta<pi ?
naebad
yes k is the step size if you like.
Robert Israel
> 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.
Tim Wescott
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?
naebad
yes!
Tim Wescott
Jerry Avins
> 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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Tim Wescott
> 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
Exactly!
I asked him "A or B", and he said "yes!".
It left me a bit confused.
Jerry Avins
> On Mon, 30 Nov 2009 21:03:30 -0500, Jerry Avins wrote:
>
>> 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
>
> Exactly!
>
> I asked him "A or B", and he said "yes!".
>
> It left me a bit confused.
He meant "Yes! I used k to represent two different things."
Greg Heath
Note that
y(t) = a*tanh( arctanh(y0/a) + a*K*(t-t0) )
satisfies
y'(t) = K*(a^2-y^2).
Hope this helps.
Greg
Tim Wescott
Assuming that Jerry is right and you really meant
y(n+1)=y(n)+k*(a-y^2(n)),
What happens when k = -1, a = 1 and y(0) > 1? Or k = -1, a = 1, and
-sqrt(2) > y > 1?
How about k = 1, a = 1, and y(0) < -sqrt(2)?
See a pattern?
Greg Heath
> 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.
>
> Assuming that Jerry is right and you really meant
>
> y(n+1)=y(n)+k*(a-y^2(n)),
>
> What happens when k = -1, a = 1 and y(0) > 1? Or k = -1, a = 1,
The Op specified k > 0. Do you want to consider k = -1 anyway?
> and -sqrt(2) > y > 1?
Impossible. A sign typo?
> How about k = 1, a = 1, and y(0) < -sqrt(2)?
>
> See a pattern?
Waiting for your reply.
Hope this helps.
Greg
Tim Wescott
> On Dec 1, 2:04 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.
>>
>> Assuming that Jerry is right and you really meant
>>
>> y(n+1)=y(n)+k*(a-y^2(n)),
>>
>> What happens when k = -1, a = 1 and y(0) > 1? Or k = -1, a = 1,
>
> The Op specified k > 0. Do you want to consider k = -1 anyway?
>
I missed that, but it doesn't make much difference.
>
>> and -sqrt(2) > y > 1?
>
> Impossible. A sign typo?
Yes; I thought that looked weird. -sqrt(2) < y < 1.
>> How about k = 1, a = 1, and y(0) < -sqrt(2)?
>>
>> See a pattern?
There's still a pattern in there -- is there _any_ strictly positive k
for which you can find a y(0) that guarantees stability? For that
matter, is there any strictly positive k for which you _cannot_ find a y
(0) that guarantees instability?
Jon Slaughter
This is called the linearized riccati equation(analogous to the DE) and the
closed form solution to your equation is known. That is, your equation is of
the form y(k+1) = a_k + b_k*y(k) + c_k*y^2(k)
Look up on google for the solution.
naebad
Yes but for a white-noise input? This is unforced - from initial
conditions.
Greg Heath
> On Nov 29, 10:56 pm, naebad <nae...@yahoo.com> 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.
>
> Note that
>
>
> satisfies
>
> y'(t) = k*(a^2-y^2).
Rescaling with t0 = 1 yields
z'(t) = K*( 1 - z^2), z(t=1) = z1.
satisfied by
z(t) = tanh( arctanh(z1) + K*(t-1) ).
Regardless of z1, the solution is
continuous and bounded with the limit
z(t) --> tanh(K*t) --> sgn(K)
as t --> inf.
Notice, however, that arctanh(z1) and
z(t) become complex when abs(z1) > 1.
For a numerical solution consider the FDE
z(n) = z(n-1) + Kdt*(1-z(n-1)^2), z(1) = z1
with Kdt == K*dt , dt > 0 and z1 real. It is
sufficient to consider K > 0 (K < 0 just
changes the sign of z).
A good approximation to the ODE should occur
if Kdt is sufficiently small. However, the
continuous time solution to the ODE is complex
when abs(z1) > 1 whereas, if z1 is real, the
solution to the FDE will always be real.
The stationary solutions are zs = +/- 1
with 1st order perturbations
z1 = zs + eps1
z2 ~ zs + eps1 - 2Kdt*zs*eps1
When zs = +1, the small perturbation
will decrease in magnitude regardless of
it's polarity and, similar to the ODE solution,
z(n) --> +1 as n --> inf
is a reasonable expectation.
However, when zs = -1, the small perturbation
will increase in magnitude.
When eps1 > 0, abs(z1) < 1 and, similar to the
ODE solution, it is expected that
z(n) --> 1 as n --> inf
i.e z will tend toward the other stationary point!
However, when eps1 < 0, abs(z1) > 1 and it
is expected that
z(n) --> -inf as n --> +inf.
In summary, it is predicted that
K > 0, zs = -1, eps1 < 0
and
K < 0, zs = +1, eps1 > 0
both yield instability.
Hope this helps.
Greg
Greg Heath
In general, stability is expected for K > 0 when
z(2) = z1+K*(1-z1^2) > -1
K*z1^2 - z1- (K+1) < 0
(1-sqrt(1+4*K*(K+1)))/(2K) < z1 < (1+sqrt(1+4*K*(K+1)))/(2K)
Therefore,
-1 < z1 < 2 for K = 1
Hope this helps.
Greg