Lesson 31
  First Order, Higher Dimensional Difference
                   Equations

                          Math 20


 ...
Recap

Higher dimensional linear systems
   Examples
        Markov Chains
        Population Dynamics
   Solution

Qualit...
one-dimensional linear difference equations
   Fact
   The solution to the inhomogeneous difference equation

            ...
Nonlinear equations

                      sl
                           op
                             e
               ...
Recap

Higher dimensional linear systems
   Examples
        Markov Chains
        Population Dynamics
   Solution

Qualit...
Let’s kick it up a notch and look at the multivariable, linear,
homogeneous difference equation

                         ...
Skipping class



   Example
   This example was a Markov chain with transition matrix

                                 0...
Example
Female lobsters have more eggs each season the longer they
live. For this reason, it is illegal to keep a lobster ...
Mmmm. . . Lobster
Formal solution


                  y(1) = Ay(0)
                  y(2) = Ay(1) = A2 y(0)
                  y(3) = Ay(2) =...
Formal solution


                       y(1) = Ay(0)
                       y(2) = Ay(1) = A2 y(0)
                      ...
Flop count




      To multiply two n × n matrices takes n3 (n − 1) additions or
      multiplications (flop=floating point...
Flop count




      To multiply two n × n matrices takes n3 (n − 1) additions or
      multiplications (flop=floating point...
Now what?
  Suppose v is an eigenvector of A with eigenvalue λ . Then the
  solution to the problem

                   y(...
Now what?
  Suppose v is an eigenvector of A with eigenvalue λ . Then the
  solution to the problem

                   y(...
Now what?
  Suppose v is an eigenvector of A with eigenvalue λ . Then the
  solution to the problem

                   y(...
Now what?
  Suppose v is an eigenvector of A with eigenvalue λ . Then the
  solution to the problem

                    y...
Now what?
  Suppose v is an eigenvector of A with eigenvalue λ . Then the
  solution to the problem

                    y...
The big picture



   Fact
   Let A have a complete system of eigenvalues and eigenvectors
   λ1 , λ2 , . . . , λn and v1 ...
Recap

Higher dimensional linear systems
   Examples
        Markov Chains
        Population Dynamics
   Solution

Qualit...
Iterating diagonal systems




   Consider a 2 × 2 matrix of the form

                                   λ1 0
           ...
Picture in terms of eigenvalues




      λ1 > λ2 > 1: repulsion away from the origin
Picture in terms of eigenvalues




      λ1 > λ2 > 1: repulsion away from the origin
      1 > λ1 > λ2 > 0: attraction to...
Picture in terms of eigenvalues




      λ1 > λ2 > 1: repulsion away from the origin
      1 > λ1 > λ2 > 0: attraction to...
Picture in terms of eigenvalues




      λ1 > λ2 > 1: repulsion away from the origin
      1 > λ1 > λ2 > 0: attraction to...
Picture in terms of eigenvalues




       λ1 > λ2 > 1: repulsion away from the origin
       1 > λ1 > λ2 > 0: attraction ...
Back to skipping class



   Example
   If
                          0.7 0.8
                     A=
                     ...
Back to skipping class



   Example
   If
                                0.7 0.8
                           A=
         ...
Back to skipping class



   Example
   If
                               0.7 0.8
                         A=
            ...
Back to the lobsters



   We had                            
                        0 100 400 700
                    ...
Back to the lobsters



   We had                            
                        0 100 400 700
                    ...
Recap

Higher dimensional linear systems
   Examples
        Markov Chains
        Population Dynamics
   Solution

Qualit...
The nonlinear case


  Consider now the nonlinear system

                         y(k + 1) = g(y(k)).

  The process is a...
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Lesson31 Higher Dimensional First Order Difference Equations Slides

  1. 1. Lesson 31 First Order, Higher Dimensional Difference Equations Math 20 April 30, 2007 Announcements PS 12 due Wednesday, May 2 MT III Friday, May 4 in SC Hall A Final Exam: Friday, May 25 at 9:15am, Boylston 110 (Fong Auditorium)
  2. 2. Recap Higher dimensional linear systems Examples Markov Chains Population Dynamics Solution Qualitative Analysis Diagonal systems Examples Higher dimensional nonlinear
  3. 3. one-dimensional linear difference equations Fact The solution to the inhomogeneous difference equation yk+1 = ayk + b (with a = 1) has solution b b yk = ak y0 − + 1−a 1−a Please try not to memorize this. When a and b have actual values, it’s either to follow this process: 1. Start with ak times an undetermined parameter c (this satisfies the homogenized equation) 2. Find the equilibrium value y∗ . 3. Add the two and pick c to match y0 when k = 0.
  4. 4. Nonlinear equations sl op e = − Fact 1 slo The equilibrium pe =g y0 point y∗ of the (y ∗) nonlinear y2 difference equation yk+1 = g(yk ) is y1 stable if 1 |g (yk )| < 1. = e op sl
  5. 5. Recap Higher dimensional linear systems Examples Markov Chains Population Dynamics Solution Qualitative Analysis Diagonal systems Examples Higher dimensional nonlinear
  6. 6. Let’s kick it up a notch and look at the multivariable, linear, homogeneous difference equation y(k + 1) = Ay(k ) (we move the index into parentheses to allow y(k ) to have coordinates and to avoid writing yk,i .)
  7. 7. Skipping class Example This example was a Markov chain with transition matrix 0.7 0.8 A= 0.3 0.2 Then the probability of going or skipping on day k satisfies the equation p(k + 1) = Ap(k )
  8. 8. Example Female lobsters have more eggs each season the longer they live. For this reason, it is illegal to keep a lobster that has laid eggs. Let yi be the number of lobsters in a fishery which are i years alive. Then the difference equation might have the simplified form   0 100 400 700 0.1 0 0 0 y(k + 1) =   y(k)  0 0.3 0 0 0 0 0.9 0
  9. 9. Mmmm. . . Lobster
  10. 10. Formal solution y(1) = Ay(0) y(2) = Ay(1) = A2 y(0) y(3) = Ay(2) = A3 y(0) So
  11. 11. Formal solution y(1) = Ay(0) y(2) = Ay(1) = A2 y(0) y(3) = Ay(2) = A3 y(0) So Fact The solution to the homogeneous system of linear difference equations y(k + 1) = Ay(k) is y(k) = Ak y(0)
  12. 12. Flop count To multiply two n × n matrices takes n3 (n − 1) additions or multiplications (flop=floating point operation)
  13. 13. Flop count To multiply two n × n matrices takes n3 (n − 1) additions or multiplications (flop=floating point operation) So finding Ak takes about n4k flops!
  14. 14. Now what? Suppose v is an eigenvector of A with eigenvalue λ . Then the solution to the problem y(k + 1) = Ay(k), y(0) = v is
  15. 15. Now what? Suppose v is an eigenvector of A with eigenvalue λ . Then the solution to the problem y(k + 1) = Ay(k), y(0) = v is y(k ) = λ k v
  16. 16. Now what? Suppose v is an eigenvector of A with eigenvalue λ . Then the solution to the problem y(k + 1) = Ay(k), y(0) = v is y(k ) = λ k v Suppose y(0) = c1 v1 + c2 v2 + · · · + cm vm
  17. 17. Now what? Suppose v is an eigenvector of A with eigenvalue λ . Then the solution to the problem y(k + 1) = Ay(k), y(0) = v is y(k ) = λ k v Suppose y(0) = c1 v1 + c2 v2 + · · · + cm vm Then Ay(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm A2 y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm 2 2 2
  18. 18. Now what? Suppose v is an eigenvector of A with eigenvalue λ . Then the solution to the problem y(k + 1) = Ay(k), y(0) = v is y(k ) = λ k v Suppose y(0) = c1 v1 + c2 v2 + · · · + cm vm Then Ay(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm A2 y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm 2 2 2 If A is diagonalizable, we can take m = n and write any initial vector as a linear combination of eigenvalues.
  19. 19. The big picture Fact Let A have a complete system of eigenvalues and eigenvectors λ1 , λ2 , . . . , λn and v1 , v2 , . . . , vn . Then the solution to the difference equation y(k + 1) = Ay(k) is y(k ) = Ak y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cn λn vn k k k where c1 , c2 , . . . , cn are chosen to make y(0) = c1 v1 + c2 v2 + · · · + cn vn
  20. 20. Recap Higher dimensional linear systems Examples Markov Chains Population Dynamics Solution Qualitative Analysis Diagonal systems Examples Higher dimensional nonlinear
  21. 21. Iterating diagonal systems Consider a 2 × 2 matrix of the form λ1 0 D= 0 λ2 Then the λ ’s tell the behavior of the system.
  22. 22. Picture in terms of eigenvalues λ1 > λ2 > 1: repulsion away from the origin
  23. 23. Picture in terms of eigenvalues λ1 > λ2 > 1: repulsion away from the origin 1 > λ1 > λ2 > 0: attraction to the origin
  24. 24. Picture in terms of eigenvalues λ1 > λ2 > 1: repulsion away from the origin 1 > λ1 > λ2 > 0: attraction to the origin λ1 > 1 > λ2 : saddle point
  25. 25. Picture in terms of eigenvalues λ1 > λ2 > 1: repulsion away from the origin 1 > λ1 > λ2 > 0: attraction to the origin λ1 > 1 > λ2 : saddle point
  26. 26. Picture in terms of eigenvalues λ1 > λ2 > 1: repulsion away from the origin 1 > λ1 > λ2 > 0: attraction to the origin λ1 > 1 > λ2 : saddle point For negative eigenvalues just square them and use the above results.
  27. 27. Back to skipping class Example If 0.7 0.8 A= 0.3 0.2
  28. 28. Back to skipping class Example If 0.7 0.8 A= 0.3 0.2 The eigenvectors (in decreasing order of absolute value) are −1 8/11 1 with eigenvalue 1 and 12 with eigenvalue − 10 . 3/11 2
  29. 29. Back to skipping class Example If 0.7 0.8 A= 0.3 0.2 The eigenvectors (in decreasing order of absolute value) are −1 8/11 1 with eigenvalue 1 and 12 with eigenvalue − 10 . So the 3/11 2 8/11 system converges to a multiple of 3 . /11
  30. 30. Back to the lobsters We had   0 100 400 700 0.1 0 0 0 A=   0 0.3 0 0 0 0 0.9 0 The eigenvalues are 3.80293, −2.84895, −0.476993 + 1.23164i, −0.476993 − 1.23164i and the first eigenvector is T 0.999716 0.0233099 0.00489153
  31. 31. Back to the lobsters We had   0 100 400 700 0.1 0 0 0 A=   0 0.3 0 0 0 0 0.9 0 The eigenvalues are 3.80293, −2.84895, −0.476993 + 1.23164i, −0.476993 − 1.23164i and the first eigenvector is T 0.999716 0.0233099 0.00489153 The population will grow despite the increased harvesting!
  32. 32. Recap Higher dimensional linear systems Examples Markov Chains Population Dynamics Solution Qualitative Analysis Diagonal systems Examples Higher dimensional nonlinear
  33. 33. The nonlinear case Consider now the nonlinear system y(k + 1) = g(y(k)). The process is as it was with the one-dimensional nonlinear: 1. Look for equilibria y∗ with g(y∗ ) = y∗ 2. Linearize about the equilibrium using the matrix ∂ gi A = Dg(y∗ ) = ∂ yj 3. The eigenvalues of A determine the stability of y∗ .

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