MA 282-Assignment #4: Firstβorder Linear Systems We restrict our discussion to systems of the form ππ₯ = ππ₯ + ππ¦ ππ‘ ππ¦ = ππ₯ + ππ¦ ππ‘ where π, π, π, π are constants. We let = ( π π π₯(π‘) ππ π ) , π(π‘) = ( ), and we write the system in the form ππ‘ = π΄π. π¦(π‘) π 1. Strait βLine(Eigen-line) Solutions ππ Recall, if the matrix π΄ has a real eigenvalue π with associated eigenvector π£, then the linear system ππ‘ = π΄π has the straight line solution π(π‘) = π ππ‘ π£. Example ππ Consider the system ππ‘ = ( 2 2 ) π. 1 3 We use the MATLAB command eig to calculate the eigenvalues and their associated eigenvectors as follows. The eigenpairs of the system are: π1 = 1, π£1 = [ β2 1 ] and π2 = 4, π£2 = [ ]. 1 1 So the system has two straight- line solutions namely, π1 (π‘) = π π‘ [ We can plot the direction field of the system. π‘ 4π‘ β2 ] = [β2π ] and π2 (π‘) = [π 4π‘ ]. π‘ 1 π π To plot half of the straight line of eigenvectors, we type Any multiple of the vector π£1 is also an eigeinvector of π1 . In particular β π£1 is an eigenvector and the lower half of the line is obtained by plotting the solution β π π‘ π£1 . Similarly we can plot the second straight line of eigenvectors as follows. Please consult the textbook for detailed analysis of the importance of straight line solutions. Next we use MATLAB to solve linear systems in two different ways. We will focus on the technical parts; students should consult their textbooks to learn about the valuable information that eigenpairs contain. 2. Solving Systems in terms of Eigenpairs Case I: two distinct real eigenvalues Example As an example we solve the system ππ ππ‘ β3 1 2 =( ) π, π(0) = [ ]. 0 2 5 β3 1 We enter the coefficient matrix π΄ = ( ) in MATLAB: 0 2 Next we use the command eig to calculate the eigenpairs of π΄: 1 1/5 ) provides the eigenvectors associated to the distinct 0 1 eigenvalues π1 = β3 and π2 = 2 found along the diagonal of the matrix Evl. Note that the columns of the matrix Evr = ( Therefore the general solution is 1 1/5 π(π‘) = π1 π β3π‘ [ ] + π2 π 2π‘ [ ]. 0 1 2 Substituting for the initial condition π(0) = [ ]: 5 1 1/5 2 π1 [ ] + π2 [ ]=[ ] 0 5 1 which can be written in the form: 1 1/5 π1 2 ( ) [π ] = [ ] 5 2 0 1 1 1/5 ), hence we obtain: 0 1 Observe that Evr = ( π1 2 Evr [π ] = [ ]. 5 2 Now we use MATLAB to solve for π1 and π2 as follows: that is π1=1, π2 = 5, and the solution to our initial value problem is 1 1/5 π(π‘) = π β3π‘ [ ] + 5π 2π‘ [ ] 0 1 or 1 1 π(π‘) = [ ] π β3π‘ + [ ] π 2π‘ 0 5 Case II: two complex eigenvalues As an example we find the general solution of the system So the eigenpairs are ππ ππ‘ =( β2 β3 ) π. 3 β2 βπ π1 = β2 β 3π, π£1 = [ ] 1 and π2 = β2 + 3π, π π£2 = [ ]. 1 Consider the second pair and let us calculate the real and imaginary parts of the complex solution: π π(π‘) = [ ] π (β2+3π)π‘ 1 π π(π‘) = [ ] π 1 β2π‘ (πππ 3π‘ ππ β2π‘ πππ 3π‘ β π β2π‘ π ππ3π‘ + ππ ππ3π‘) = [ βπ β2π‘ π ππ3π‘ ]=[ π β2π‘ πππ 3π‘ + ππ β2π‘ π ππ3π‘ π β2π‘ πππ 3π‘ ]+[ π β2π‘ πππ 3π‘ Consequently the general solution of this system is βπ β2π‘ π ππ3π‘ π(π‘) = πΆ1 [ π β2π‘ πππ 3π‘ ] + πΆ2 [ π β2π‘ πππ 3π‘ ] π β2π‘ π ππ3π‘ Case III: The eigenvalue π has multiplicity 2 i. ππ 3 0 )π 0 3 There are two independent eigenvectors associated with π : ππ‘ = ( 1 0 In this case the general solution is π(π‘) = π1 π 3π‘ [ ] + π2 π 3π‘ [ ] 0 1 ii. ππ 3 1 There is only one linearly independent vector associated with π: ππ‘ = ( 0 )π 3 ]π π β2π‘ π ππ3π‘ 0 So there is one vector, π£ = [ ], associated with π = 3. 1 0 A particular solution is given by π1 = π π π‘ π£ = π 3π‘ [ ]. 1 A second solution is π2 = π‘π π π‘ π£ + π€π ππ‘ , where (π΄ β ππΌ)π€ = π£. Note that Evr is just the vector π£ hence we can easily solve for π€: 1 0 Therefore, π2 = π‘π π π‘ π£ + π€π ππ‘ = π‘π 3 π‘ [ ] + [ ] π 3π‘ and the general solution is π(π‘) = πΆ1 π1 (π‘) + πΆ2 π2 (π‘). 0 1 3. Solving Systems using the command dsolve Let us consider the system discussed in Case III (ii): ππ ππ‘ That is the system ππ₯ = 3π₯ ππ‘ ππ¦ = π₯ + 3π¦ ππ‘ To find the general solution, we type 3 1 =( 0 ) π. 3 To solve the same system but with the initial conditions π₯(0) = 2, π¦(0) = 5: MA 282- MatLab Project #4: Firstβorder Linear Systems Due Thursday, April 16 @ 9:30AM Total = 100 points Important Reminder: ο· Project without Honesty Statement & Signature will NOT be acceptable . ο· Follow the format on the instruction file. ο· Submit at the beginning of the class. Late submission will not be accepted. ************************************************************************************* Problem #1. Choose either (a) or (b) and do the following for the system: 1. Use the MATLAB to find the eigenvalues and associated eigenvectors. 2. Plot the direction field of the system and the straight-lines of eigenvectors(eigenlines) on the same axes. 3. Observe the behavior of trajectories and determine the type(stable node, unstable node, saddle point, degenerate stable node or degenerate unstable node) of the equilibrium point at the origin. (a) Consider the system β4 2 ππ =( 5 )π β 2 ππ‘ 2 (b) Consider the system 5 β 2 ππ 2 =( )π 3 ππ‘ β2 4 Problem #2. Let π be a real number different from 0 and consider the system ππ₯ = ππ₯ + 4π¦ ππ‘ ππ¦ = ππ¦ ππ‘ 1. 2. 3. 4. Use MATLAB to verify that the system has a repeated eigenvalue π Use MATLAB to verify that there is only one eigenvector associated with π Find the general solution of the system. Verify the result in 3 by using the command dsolve to solve the system