oxymoron
11-26-03, 09:43 AM
Just quickly before I continue...
Ed Lorenz had three differential equations...
dx/dt = o(y-x)
dy/dt = rx - y - xz
dz/dt = xy - bz
But what do they mean? Did they actually have something to do with the weather? And how did he work them out? (Note: the "o" should be a "sigma")
DIFFERENCE EQUATIONS
If you have a difference equation x_n+2 + ax_n+1 + bx_n = 0 is it always possible to write this as an ODE? Like x" + ax' + bx = 0.
Solving such an ODE would be hard!
You could substitute x = x_0 e^st. But where did the "s" come from? Is it a parameter that satisfies s^2 + as + b = 0; this coming from the auxillary equation from the ODE? If it is then the general solution is x = x_0 e^st.
From the auxillary equation this implies that there are two linearly independent solutions. With that in mind and knowing x = x_0 e^st how would you go about substituting this back into the difference equation???
I heard that you try x_n = x_0 L^n (where L is lambda).
But if you tried this then you would get x_0L^n+2 + ax_0L^n+1 + bx_0L^n = 0
The general solution is of the form x_n = Ar^2 + Br^2 where A and B are constants. I am getting the idea that you may treat difference equations like ODE's. Is this a correct assumption?
Say you wanted to solve x_n+2 - 3x_n+1 + 2x_n = 0 with initial conditions x_0 = 0 and x_1 = 1
The auxillary equation is L^2 - 3L + 2 = 0 which is a quadratic with roots L = 2 and L = 1.
So the general solution will be x_n = A.2^n + B.1^n
Using the initial conditions...
0 = A + B
1 = 2A + B } => A = 1 and B = -1
The solution is x_n = 2^n - 1
But finding the solution of a difference equation by the above method was also covered by the iterative method. Is both methods okay? Is the method outlined above more practical? I would think so...
THE LOGISTIC EQUATION
The general form of the logistic equation is x_n+1 = ax_n(1-x_n). I have seen this too, as a 1st order differential equation! But not as a difference equation.
Looking at this we would expect x to be positive since this equation can govern population and we cannot have negative population. This restricts x_n to [0,1] If x_n is greater than 1 then x_n+1 is negative and we do not want that!
I would like to know how you get the fact that it has a maximum value of x = 1/2 resulting in the fact that if a > 4 then x_n+1 is negative again.
FIXED POINTS
What is a fixed point? Is it that if we follow say the logistic equation above through each iterate x_n+1 -> x_n -> x_n-1 ...and they all equal each other then does the equation map to a fixed point X.
So if it tends to X under iteration we say that the fixed point X is attracting. and if it moves away from X then it is repelling.
So it would be efficient to find all the attractors of a nonlinear system?
Finally, could someone explain why for a logistic map, the fixed points satsify X = f(X) = aX(1-X) and that when you solve it there is one fixed point for a<1 and two for a>1 and how can you determine whether these are attractors or repellors?
Thanks.
Ed Lorenz had three differential equations...
dx/dt = o(y-x)
dy/dt = rx - y - xz
dz/dt = xy - bz
But what do they mean? Did they actually have something to do with the weather? And how did he work them out? (Note: the "o" should be a "sigma")
DIFFERENCE EQUATIONS
If you have a difference equation x_n+2 + ax_n+1 + bx_n = 0 is it always possible to write this as an ODE? Like x" + ax' + bx = 0.
Solving such an ODE would be hard!
You could substitute x = x_0 e^st. But where did the "s" come from? Is it a parameter that satisfies s^2 + as + b = 0; this coming from the auxillary equation from the ODE? If it is then the general solution is x = x_0 e^st.
From the auxillary equation this implies that there are two linearly independent solutions. With that in mind and knowing x = x_0 e^st how would you go about substituting this back into the difference equation???
I heard that you try x_n = x_0 L^n (where L is lambda).
But if you tried this then you would get x_0L^n+2 + ax_0L^n+1 + bx_0L^n = 0
The general solution is of the form x_n = Ar^2 + Br^2 where A and B are constants. I am getting the idea that you may treat difference equations like ODE's. Is this a correct assumption?
Say you wanted to solve x_n+2 - 3x_n+1 + 2x_n = 0 with initial conditions x_0 = 0 and x_1 = 1
The auxillary equation is L^2 - 3L + 2 = 0 which is a quadratic with roots L = 2 and L = 1.
So the general solution will be x_n = A.2^n + B.1^n
Using the initial conditions...
0 = A + B
1 = 2A + B } => A = 1 and B = -1
The solution is x_n = 2^n - 1
But finding the solution of a difference equation by the above method was also covered by the iterative method. Is both methods okay? Is the method outlined above more practical? I would think so...
THE LOGISTIC EQUATION
The general form of the logistic equation is x_n+1 = ax_n(1-x_n). I have seen this too, as a 1st order differential equation! But not as a difference equation.
Looking at this we would expect x to be positive since this equation can govern population and we cannot have negative population. This restricts x_n to [0,1] If x_n is greater than 1 then x_n+1 is negative and we do not want that!
I would like to know how you get the fact that it has a maximum value of x = 1/2 resulting in the fact that if a > 4 then x_n+1 is negative again.
FIXED POINTS
What is a fixed point? Is it that if we follow say the logistic equation above through each iterate x_n+1 -> x_n -> x_n-1 ...and they all equal each other then does the equation map to a fixed point X.
So if it tends to X under iteration we say that the fixed point X is attracting. and if it moves away from X then it is repelling.
So it would be efficient to find all the attractors of a nonlinear system?
Finally, could someone explain why for a logistic map, the fixed points satsify X = f(X) = aX(1-X) and that when you solve it there is one fixed point for a<1 and two for a>1 and how can you determine whether these are attractors or repellors?
Thanks.