Discrete dynamical systems

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  1. Discrete Dynamical Systems Overview:
  2. The Differential Equation xk+1=Axkx_{k+1}=Ax_k
    • Linear Transformation
    • Multiplying with A k times
    • Generalized formula
    • Why is this formula useful?
  3. Doing an Example
    • Calculate xkx_k
    • Long-term behaviour (k(k ) \infty)
  4. Application: Predator and Prey Model
    • Analyzing the predator and prey equations
    • Converting to the equation xk+1=Axkx_{k+1}=Ax_k
    • Calculating the general solution
    • Long term behaviour (k(k ) \infty)
  1. Finding the General Solution
    Let AA be a 3×33 \times 3 matrix with eigenvalues 3,2,3,2, and 12\frac{1}{2}, and corresponding eigenvectors corresponding eigenvectors 1, 2 and corresponding eigenvectors 3. if initial vector x_0 , find the general solution of the equation xk+1=Axkx_{k+1}=Ax_k.
    1. Analyzing the Long Term Behaviour
      Explain the long term behaviour (k(k ) \infty) of the equation xk+1=Axkx_{k+1}=Ax_k, where
      Analyzing the Long Term Behaviour, Matrix A

      And Analyzing the Long Term Behaviour, initial vector x_0 where c1c_1 > 00 and c2c_2 > 00.
      1. Explain the long term behaviour (k(k ) \infty) of the equation xk+1=Axkx_{k+1}=Ax_k, where
        Analyzing the Long Term Behaviour, Matrix A

        And Analyzing the Long Term Behaviour, initial vector x_0 where c1c_1 > 00 and c2c_2 > 00
        1. Predator and Prey Model
          Let the eagle and rabbit population at time kk be denoted as initial vector k, where kk is the time in years, EkE_k is the number of eagles at time kk, and RkR_k is the number of rabbits at time kk (all measured in thousands). Suppose there are two equations describing the relationship between these two species:

          Ek+1=(.4)Ek+(.5)Rk E_{k+1}=(.4) E_k+(.5)R_k
          Rk+1=(.207)Ek+(1.2)Rk R_{k+1}=(-.207) E_k+(1.2) R_k
          1. What happens to the rabbits if there are no eagles? What happens to the eagles if there are no rabbits?
          2. Find the matrix AA for the equation xk+1=Axkx_{k+1}=Ax_k.
          3. Suppose the eigenvalues of AA is λ1=1.0377\lambda_1=1.0377 and λ2=0.562303\lambda_2=0.562303 and the corresponding eigenvectors are corresponding eigenvector v_1 and corresponding eigenvector v_2 (assume a,b,c,da,b,c,d > 00). Find the general solution.
          4. Assuming that c1,  c2c_1,\;c_2 > 00, explain the population of rabbits and eagles as kk \infty.
        Topic Notes

        Discrete dynamical systems

        Dynamic system definition

        We define dynamic system as a body pertaining to a geometrical space which depends on time. This time dependence means that such a system changes as time progresses and there is a mathematical function which describes such change.

        In simple words, a dynamical system refers to any system that changes through time and so, its state can be explained and observed at any particular moment or lapse in time, but this state will not be the same as time progresses. Dynamical systems can be classified as discrete and continuous dynamical systems depending on the size of the periods of time required for the system to change, or transform. We will talk about the differences of the two in the next section, for now let us put this concept into context:
        Since everything in our world changes as time passes by, the idea of a dynamical system can be applied to an infinite amount of scenarios to be studied mathematically. For example: the water flowing through a river is a system in constant change. Another example is a building being built. Notice that these two cases are very different and they change in steps of time in different scales, but both of them are dynamical, they move, they change state, they are not the same as they were in an earlier stage of their history.

        Discrete and continuous dynamical systems

        Once we have learnt the dynamic system definition, let us have an introduction to dynamical systems continuous and discrete.

        Take again the examples mentioned in the last section: the river water flowing and the construction of a building, as we all know, they are both changing but notice how they do it: the water in the river is continuously flowing through and so, every infinitesimally small moment in time, the particles of water that passed through a certain point of the length of the river are gone onto another location by the next moment; on the other hand, a building in construction will change every certain periods of time that are palpable, very definable by conditions of the system, in other words, changes in construction of that building will be noticeable by the day: workers come in the morning, work their 8-hour shifts and by the end of the day they finished a certain part of the project and so they have a recount of what was done for that day. At the end of that day the state of the building changed, but you cannot go back into minutes or seconds during the day and see palpable difference from the day before, until the project for the day has been finished, you can now say oh today this and that was done.

        That is the difference between continuous and discrete dynamical systems. A continuous dynamical system, as its name mentions, changes CONTINUOUSLY, meaning that it changes by every infinitesimally small moment in time. The water at certain point in the river is gone to another location in fractions of a second, the same as our aging process that every moment in time that passes is changing progressively and there is no way to stop the process until is done.
        Then, a discrete dynamical system is that system which requires of specific periods of time to produce a change, in this case, the construction of a building will have certain changes for a day, then the workers go home and the state of the building remains unchanged until they come back next day and achieve a new change. The state of the system varies by separate points of time, but in between these points there is no change.

        The names of the dynamical systems then come from how time as a variable is taken on them, time is a continuous variable when studying continuous dynamical systems, while time is a discrete variable (looked as in measurable steps) when studying discrete dynamical systems. Continuous dynamical systems require differential equations for their study, and so, today we will focus on discrete dynamical systems only.

        Discrete dynamical system

        It is time for us to focus on the mathematics of the topic for today, and so, it is time for us to learn about the differential equation of a discrete dynamical system:

        xk+1=Axk x_{k+1} = Ax_k
        Equation 1: Differential equation for a discrete dynamical system

        We have said already that a discrete dynamical system is a system which is changing through time and for this system, time is seen as a discrete variable, in other words, the system is changing by lapses of time, every particular lapse an specific change (not a continuum of progression). Thus, to define this mathematically, we will have to derive what is called the differential equation for the discrete dynamical system (notice differential means is changing through a variable, and for this case is through the variable of time).

        And so, in order to produce the mathematical expression for discrete event dynamics systems, let us assume we have a matrix A, which is diagonalizable, and which has n linearly independent eigenvectors v1,v2,...,vnv_1, v_2, ... , v_n and their corresponding eigenvalues λ1,λ2,...,λn\lambda_1, \lambda_2, ... , \lambda_n. Then we can write an initial vector x0x_0 to be:

        x0=c1v1+c2v2+...+cnvnx_0 = c_1 v_1 + c_2v_2 + ... +c_nv_n
        Equation 2: Initial vector x0

        And no, we perform a transformation of x0x_0 with matrix AA.
        Remember from our lesson on image and range of linear transformations, that a linear transformation is defined as: T(x)=AxT(x)=Ax where AA is a matrix and xx is a vector. Then, from our lesson on eigenvalues and eigenvectors , we learnt that multiplying AA by an eigenvector xx just rescales the vector with a corresponding scalar which is called and eigenvalue following the condition Ax=λxAx= \lambda x. Therefore, if we say that we are to transform x0x_0 with matrix AA, we will have that T(x0)=Ax0=λx0T(x_0)=Ax_0= \lambda x_0 (for each eigenvalue of AA). We call the new re-scaled vector coming from this transformation x1x_1 and so, we have that is defined as:

        x1=Ax0=c1Av1+c2Av2+...+cnAvn=c1λ1v1+c2λ2v2+...+cnλnvnx_1=Ax_0=c_1Av_1 \,+ \, c_2Av_2 \, +...+ \, c_nAv_n = c_1 \lambda_1 v_1 \, + \, c_2 \lambda_2 v_2 \, +...+ \, c_n \lambda_n v_n
        Equation 3: First transformation of the initial vector

        Equation 3 is what we call the first transformation of the initial vector x0x_0. We could continuing transforming the initial vector with the matrix, lets say we do it a second time and we call this new resulting vector x2x_2, therefore, this second transformation would have to satisfy the condition: x2=Ax1=λx1x_2 = Ax_1= \lambda x_1\, for each corresponding eigenvalue of AA, and would be defined as:

        x2=Ax1=c1λ1Av1+c2λ2Av2+...+cnλnAvn x_2 = Ax_1 = c_1 \lambda_1 Av_1 \, + \, c_2 \lambda_2 Av_2 \,+...+ \, c_n \lambda_n Av_n

        x2=c1λ1λ1v1+c2λ2λ2v1+...+cnλnλnvnx_2 = c_1 \lambda_1 \lambda_1 v_1 \, + \, c_2 \lambda_2 \lambda_2 v_1 \, +...+ \, c_n \lambda_n \lambda_n v_n

        x2=c1λ12v1+c2λ22v2+...+cnλn2vn x_2 = c_1 \lambda^{2}_{1} v_1 \, + \, c_2 \lambda^{2}_{2} v_2 \,+...+ \, c_n \lambda^{2}_{n} v_n
        Equation 4: Second transformation of the initial vector

        So in the same way as the initial vector x0x_0 was transformed here a second time, we could continue transforming it as many times as we want. Each of this transformations is the way how the system described in the initial vector is evolving through time, specifically, through particular lapses of time, and each of those lapses is represented by the transformations.

        In other words, if we are to transform a vector kk amount of times, all of these transformations will define the state of the system from the initial vector at any particular lapse after that particular transformation, in other words, the change for the system in that particular amount of time. The system will not change until another transformation is applied (so, the system is not continuously changing through every infinitely small moment of time time, is just changing through every specific lapses of time, or pieces of time, defined by the transformations.
        The general formula to define the change a system went through for kk number of transformations is:

        xk=c1λ1kv1+c2λ2kv2+...+cnλnkvnx_k = c_1 \lambda^{k}_{1} v_1 \, + \, c_2 \lambda^{k}_{2} v_2 \, +...+ \, c_n \lambda^{k}_{n} v_n
        Equation 5: General formula for k transformations of the initial vector

        Which gives sense to equation 1: xk+1=Axkx_{k+1} = Ax_{k} because all equation 1 is saying is that one more transformation ( one more than k) has been done to the system. Or in other words, after passing through k transformations, the system has gone through another one because another defined lapse of time has passed and the state of the system is now defined by equation 1: xk+1x_{k+1}.

        Equation 5 is very useful because it will allow you to see the behaviour or a system, and the changes of state that it goes through discrete pieces of time. Such equation is then useful not only for mathematical studies but also for physics, biology and chemistry (actually, this is good for any branch of science, since it allows you to study the changes in behaviour or stage evolution of anything, object or body, that changes through time, which is EVERYTHING!).

        Let us work through an example exercise where we will be finding the general solution to the discrete dynamical system as defined:

        Example 1

        Let A be a 3x3 matrix with eigenvalues 3, 2 and 1/2 , and the corresponding eigenvectors:

        Discrete dynamical systems
        Equation 6: Eigenvectors of A

        Find the general solution of the equation xk+1=Axkx_{k+1}=Ax_k if the initial vector x0x_0 is defined as:

        Discrete dynamical systems
        Equation 7: Initial vector x0

        Since we have 3 eigenvalues and their corresponding 3 eigenvectors, we can automatically rewrite equation 5 for the general solution to this problem:

        xk=c1λ1kv1+c2λ2kv2+c3λ3kv3x_k=c_1 \lambda^{k}_{1}v_1 \, + \, c_2 \lambda^{k}_{2} v_2 \, + \, c_3 \lambda^{k}_{3} v3

        Discrete dynamical systems
        Equation 8: General formula for k transformations of the initial vector

        But a problem arises, what are the values of c1,c2,c3c_1, c_2, c_3? In order to obtain those constant values we have to write the initial vector expression following equation 2:

        Discrete dynamical systems
        Equation 9: Initial vector x0

        Notice that we have transformed the expression for the initial vector x0x_0 into a matrix equation that we can use to solve for the values of c1,c2,c3c_1, c_2, c_3. Thus, we rewrite this matrix equation as an augmented matrix and row-reduce it into its reduced echelon form in order to obtain the constant values:

        Discrete dynamical systems

        Discrete dynamical systems

        Discrete dynamical systems

        Discrete dynamical systems

        Discrete dynamical systems
        Equation 10: Obtaining an augmented matrix and finding its reduced echelon form

        And so the values of the constants are: c1=3,c2=1,c3=2c_1=3, c_2=1, c_3=2 . Using this we can write our final answer:

        Discrete dynamical systems
        Equation 11: General formula for k transformations of the initial vector

        Example 2

        Explain the long term behaviour (k  k \;    \; \infty )of the equation xk+1=Axkx_{k+1}=Ax_k, where the matrix AA is defined as:

        Discrete dynamical systems
        Equation 12: Matrix A

        And the initial vector contains the two components c1>0,c2>0c_1 > 0, c_2 > 0, and is defined as:

        Discrete dynamical systems
        Equation 13: Initial vector

        Since we want to explain long term behaviour, then it means we want to find the general solution of the way the system is changing through time, therefore, we are being asked to find the expression for xkx_k as shown in equation 5. The problem here is that we are not given neither the eigenvalues or eigenvectors of the provided matrix AA, therefore, our first step to work through this problem in to find them:
        Using the characteristic equation det(AIλ)=0det (A-I\lambda )=0 we solve for the eigenvalues:

        Discrete dynamical systems
        Equation 14: Finding the eigenvalues

        So the eigenvalues are: λ1=1.32  \lambda 1 = 1.32 \; and  λ2=1.1 \; \lambda 2 = 1.1
        In order to find the eigenvectors associated with it, we use the relationship (AIλ)v=0(A-I\lambda )v=0 where vv is the eigenvalue associated to the eigenvector used. And so, for λ1=1.32\lambda 1 = 1.32 we have:

        Discrete dynamical systems
        Equation 15: Matrix equation (A-I\lambda )v=0

        With this, we can form an augmented matrix and solve for the components of the eigenvector associated to the eigenvalue λ1=1.32\lambda 1=1.32.

        Discrete dynamical systems
        Equation 16: Augmented matrix to find components of the eigenvector

        And so, we see that v1v_1 is a free variable (that we will set equal to 1 ) and v2=0 \, v_2=0, then our eigenvector for this case is:

        Discrete dynamical systems
        Equation 17: Eigenvector associated to λ\lambda1 = 1.32

        Now forλ2\lambda_2 =1.1 we have:

        Discrete dynamical systems
        Equation 18: Matrix equation (A-Iλ\lambda)v=0

        And the augmented matrix to solve for the components of the eigenvector associated to the eigenvalue λ2 \lambda_2 =1.1

        Discrete dynamical systems
        Equation 19: Augmented matrix to find components of the eigenvector

        And so, we see that v1=0v_1=0 \, and v2\, v_2 is a free variable that we will set equal to 1, then our eigenvector for this case is:

        Discrete dynamical systems
        Equation 20: Eigenvector associated to λ\lambda2 = 1.1

        Therefore, for this problem we have that the eigenvalues are: λ1=1.32\lambda_1=1.32 \, and λ2=1.1 \, \lambda_2=1.1 \, and the corresponding eigenvectors are:

        Discrete dynamical systems
        Equation 21: Eigenvectors

        Having this, we can finally write the expression for xk=c1λ1kv1+c2λ2kv2x_k = c_1 \lambda^{k}_{1} v_1 \, + \, c_2 \lambda^{k}_{2} v_2

        Discrete dynamical systems
        Equation 22: General solution for the transformations of the initial vector

        With the expression found in equation 22 we can finally talk about the long term behaviour of the system as k   k \;    \; \infty . Look what happens to equation 22 when kk goes to infinity:

        Discrete dynamical systems
        Equation 23: Infinity transformations of the initial vector

        This means that as the transformations progress through time on this system, and they continue to infinity, then the final value of the system will go to infinity too since it just continues to increase every time there is a new transformation.


        We recommend you to watch all of the videos from this lesson, since the last set of videos talks about a special case called the predator and prey model. Make sure you understand this and remember, you can always asks us if in doubt!.

        To finalize this lesson we would like to recommend you to visit the next lesson on an introduction to discrete dynamical systems, which could complement what you have learned today on discrete dynamics with example models.

        So, this is it for our lesson on discrete dynamical systems linear algebra, we hope you enjoyed it. See you in the next one!
        Assume that AA is diagonalizable, with nn linearly independent eigenvectors v1,v2,,vnv_1, v_2, \cdots , v_n, and corresponding eigenvalues λ1,λ2,λn\lambda _1, \lambda _2, \cdots \lambda _n. Then we can write an initial vector x0x_0 to be:
        x0=c1v1+c2v2++cnvnx_0=c_1 v_1+c_2 v_2+ \cdots +c_n v_n

        Let's say we want to transform x0x_0 with matrix AA. Let's call the transformed vector to be x1x_1. Then,
        x1=Ax0=c1Av1+c2Av2++cnAvnx_1=Ax_0=c_1 Av_1+c_2 Av_2+\cdots+c_n Av_n
        =c1λ1v1+c2λ2v2++cnλnvn=c_1 \lambda_1 v_1+c_2 \lambda_2 v_2+\cdots+c_n \lambda_n v_n

        Let's say we want to keep transforming it with matrix A  kA\; k times. Then we can generalize this to be:
        xk=c1(λ1)kv1+c2(λ2)kv2++cn(λn)kvnx_k=c_1 (\lambda_1 )^k v_1+c_2 (\lambda_2 )^k v_2+\cdots+c_n (\lambda_n )^k v_n

        This is useful because we get to know the behaviour of this equation when kk \infty.