1. Dynamical Systems Analysis I: Fixed Points & Linearization By Peter Woolf (pwoolf@umich.edu) University of Michigan Michigan Chemical Process Dynamics and Controls Open Textbook version 1.0 Creative commons

2. Problem: Given a large and complex system of ODEs describing the dynamics and control of your process, you want to know: (1)Where will it go? (2)What will it do? Is there anything fundamental you can say about it? E.g. With my control architecture, this process will always ________. Solution: Stability Analysis

3. Reactions: A+A --> B B+ A --> C Example: CSTR with cooling jacket and multiple reactions. Controls PID on jacket cooling water What will happen? What is possible? What effect will my controller have?

4. Aside: Linear vs. Nonlinear Linear systems are significantly easier to work with, and are the basis of stability analysis Few physical systems are linear, but all can be locally approximated as linear. Linear part Nonlinear part Feature: Any nonlinear terms in your model will render the whole system nonlinear, and as such harder to analyze. Example: Goal: convert nonlinear system to a simpler linear system

5. Linear System Notation Nonlinear system Linear approximation Identical linear system in matrix form How do we do this??

6. Linearization 1.Choose a relevant point to make your linear approximation. 2.Calculate the Jacobian matrix at that point 3.Solve to find the unknown constants

7. Linearization 1.Choose a relevant point to make your linear approximation. 1X 10X 100X Full nonlinear function Local linear approximation

8. Linearization 1.Choose a relevant point to make your linear approximation. Possible relevant points: Steady state value: points where the system does not change Current location: given where I am now, where will I go next. Time Temp

9. Calculating Steady State Values Given a system of ODEs, steady state values can be found by setting all time derivatives equal to zero and solving. Kinetics example: Set derivatives equal to zero Solve for steady state values of A and B {A=0,B=0} {A=0,B=1} {A=3,B=0} {A=4,B=-1} Four solutions

10. Calculating Steady State Values Mathematica function Solve[ ]: solves a system of algebraic expressions analytically or numerically. Or with variables…

11. Linearization 1.Choose a relevant point to make your linear approximation. 2.Calculate the Jacobian matrix at that point 100X Jacobian matrix is essentially a Tailor series expansion around a point. higher order terms Possibly nonlinear Linear approximation

12. Calculating a Jacobian matrix Jacobian shows how every variable changes with each other variable at a point. Always a square matrix (rows = columns) Example:

13. Calculating a Jacobian matrix {A=0,B=0}

14. Calculating a Jacobian matrix {A=0,B=0} Jacobian can also be solved analytically in Mathematica: New mathematica function D[f(x,y,z}, z] = Example: D[A 2 +BA+C+CA, A] --> 2A+B+C

15. Calculating a Jacobian matrix Original expressions Solve for steady state Calculate Jacobian

16. Calculating a Jacobian matrix Calculate Jacobian Substitute steady state values of A and B Note this is the same result we found by hand.

17. Linearization 1.Choose a relevant point to make your linear approximation. 2.Calculate the Jacobian matrix at that point 3.Solve to find the unknown constants Nonlinear modelLinear approximation Jacobian ??

18. Nonlinear modelLinear approximation Jacobian ?? Approach: 1) solve both models at the point A=0, B=0 2) Force derivatives of the linear and nonlinear model to agree by setting unknown constants k 13 = 0 k 23 = 0

19. Nonlinear modelLinear approximation Jacobian ?? Therefore the full linear approximation at A=0, B=0 is: Or in a different format

20. Nonlinear modelLinear approximation Jacobian ?? Note: the unknown constants are not always 0. E.g. linear approximation at steady state A=0, B=1 By definition will always be zero for a fixed point. Jacobian At A=0, B=1

21. Linear algebra aside: How to solve this? 1) Convert to a more familiar algebraic expression using matrix multiplication Substitute in derivatives from nonlinear expression 2) Solve k 13 =0, k 23 =2

22. Nonlinear modelLinear approximation Jacobian ?? Therefore the full linear approximation at A=0, B=1 is: Or in a different format

23. Take Home Messages Nonlinear models are more realistic but harder to manipulate Any nonlinear model can be approximated as a linear one at a point The linear approximation is exactly correct at the point, but less accurate away from the point.