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Phase Plane Analysis of Differential Equations

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Phase Plane Analysis of Differential Equations Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations Outline Phase-plane Analysis of Ordinary Differential Equations • Midterm exam two weeks from tonight covering ODEs and Laplace transforms Larry Caretto • Review last class Mechanical Engineering 501A • Introduction to phase-plane analysis Seminar in Engineering Analysis • Look at two simultaneous ODEs dy1/dt and dy2/dt plotted as y1 vs. y2 November 1, 2017 • Look at different “critical points” for different systems of equations 2 Review Laplace Transforms Review Partial Fractions • Use transform tables to transform terms • Method to convert fraction with several in differential equation for y(t) into an factors in denominator into sum of algebraic equation for Y(s) individual factors (in denominator) – Derivative transforms give initial conditions • Example is F(s) = 1/(s+a)(s+b) on y(t) and its derivatives • Write 1/(s+a)(s+b) = A/(s+a) + B/(s+b) • Manipulate Y(s) equation to sum of • Multiply by (s+a)(s+b) and equate individual terms, “Y(s) subterms”, that coefficients of like powers of s you can find in transform tables – 1 = A(s + b) + B(s + a) – Manipulation may require use of method of – A + B = 0 for s1 terms and 1 = bA + aB for partial fractions s0 terms 3 4 Review Partial Fractions II Review Partial Fraction Rules – A + B = 0 for s1 terms and 1 = bA + aB for • Repeated fractions for repeated factors 1 A A A A s0 terms n n1 2 1 n n n1 2 – Solving for A and B gives A = -B = 1/(b – a) (s a) (s a) (s a) (s a) s a • Complex factors (s + –i)(s + + i) • Result: 1/(s+a)(s+b) = 1 As B 1/[(b – a)(s + a)] – 1/[(b – a)(s + b)] 2 2 – So f(t) = [e-at –e-bt]/(b – a) (s i )(s i ) (s ) • This actually matches a table entry • Pure imaginary factor • Follow same basic process for more 1 1 As B 2 2 2 2 complex fractions s (s i )(s i ) s • Real squared factors • Special rules for repeated factors and 1 A B complex factors 2 5 s s s 6 ME 501A Seminar in Engineering Analysis Page 1 Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations Review Systems of ODEs Review Group Exercise • Apply Laplace transforms to systems of •Solve y’’ –9y= e-t with y(0) = 0 and equations by transforming all ODEs y’(0) = 2 by Laplace transforms – Transform ODE terms like yk to Yk(s), dyk/dt • Transform differential equation: to sY (s) – y (0), etc. k k s2Y(s) – sy(0) – y’(0) –9Y(s)= 1/(s +1) • Transform all ODEs in system then use • Substitute initial conditions and solve Gaussian elimination to get an equation result for Y(s) for only one Y (s) k s2Y(s) –0 –2 –9Y(s) = 1/(s +1) • Get inverse transform from Y (s) to y (t) k k (s2 – 9)Y(s) = 2 + 1/(s +1) • Repeat for all ODEs 7 8 Review Group Exercise II Review Group Exercise III (s2 – 9)Y(s) = 2 + 1/(s +1) 1 1 3 1 3 9 2 1 s2 terms: 0 9 9 1 s1 terms: 042 0 • Use partial fractions for last term s terms: 1 3 3 9 1 •s1 equation gives B = 2A 9 1 3 3 1 • Substituting B = 2A into s2 equation 1 1 3 1 3 9 gives A + 2A + C = 0 or C = –3A A = 1/24 0 • Set sums of like powers to zero • Substitute B = 2A and C = –3A into s equation to get 1 = 3A – 3(2A) – 9(–3A) 9 10 Review Group Exercise IV Basic Phase Plane Equations • From A = 1/24 and B = 2A: B = 2/24 • Look at solution of system of two first- • From A = 1/24 and C = -3A: C = -3/24 order autonomous (no t dependence) 2 homogenous equations 9 3 3 1 • Can be single second order equation 2 1 1 2 3 written as two first order equations 9 24 3 3 1 d 2 y c dy k dy • From transform table 2 y 0 v dt m dt m dt 2 1 dv c k sinh 3 2 3 v y 0 3 24 dt m m dy sinh ⁄2 v 0 11 dt 12 ME 501A Seminar in Engineering Analysis Page 2 Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations Phase Plane Analysis What is a Stable Solution? • Look at solutions of systems of • Plot trajectories (plot one dependent equations, here use two equations as variable against the other) an example – For 2 ODE’s dy/dt and dv/dt, plot y vs. v • Find certain points, called critical points, • If at one time the trajectory is within a that have particular behavior depending distance of a point P0 and for all future on the eigenvalues of the ODE’s times it remains within a distance of • This leads to a discussion of stability; P0, the solution is stable will a solution tend to zero or increase • The solution is unstable if it is not stable without bound? • Want to find criteria for stable solutions 13 14 Undamped Vibrations Example Phase Plane Introduction 2 2 2 2 • Equation: d x/dt + x = 0 ( = k/m) • Usual plot shows solutions for x = y1 – Solution: x = (v0/)sin t + x0 cos t and v = dx/dt = y2 as a function of time – As system of equations dx/dt = v and dv/dt 3 2 2 2 y1 = d x/dt = – x 2 y2 – Define y1 = x and y2 = v to get system of 1 2 equations as dy1/dt = y2 and dy2/dt = – y1 y 0 ' -1 y1 a11y1 a12 y2 y2 -2 ' 2 y2 a21y1 a22 y2 y1 -3 2 0 2.5 5 7.5 10 12.5 15 a11 a22 0 a12 1 a21 t 15 16 Phase Plane Introduction II Phase Plane Introduction III • Phase plane plot 2.5 2.5 3 shows y as a 2 2 2 2 1.5 1.5 function of y1 with t 1 1 1 as a parameter y 0 0.5 0.5 -1 2 2 y 0 • Same as previous -2 y 0 -0.5 plot y is displace- -3 -0.5 1 0 2.5 5 7.5 10 12.5 15 -1 y1 -1 ment and y is y2 t 2 -1.5 velocity -1.5 -2 -2 Initial point (t = 0) – Time differs along -2.5 -2.5 plot -1.5 -1 -0.5 0 0.5 1 1.5 repeats periodically -1.5 -1 -0.5 0 0.5 1 1.5 y 17 18 1 y1 ME 501A Seminar in Engineering Analysis Page 3 Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations General Form Stable Solution Criteria • Write as matrix equation dy/dt = Ay • Look at the system of two equations • General form for two equations and dy/dt = Ay solutions in terms of Ax = x – Autonomous systems (no t dependence) eigenvalues and eigenvectors is • We will show that stability depends on ' the trace of A = a + a = + , and dy y1 a11y1 a12 y2 11 22 1 2 Ay the determinant a a –a a and the dt y' a y a y 11 22 12 21 2 21 1 22 2 discriminant, = (trace A)2 – 4det A y C x e1t C x e2t y C x e1t C x e2t 1 1 1(1) 2 1(2) • Review eigenvalues for 2 x 2 matrix 1 (1) 2 (2) 1t 2t y2 C1x2(1)e C2x2(2)e from September 12 lecture 19 20 Two-by-two Matrix Eigenvalues Return to Previous Example • Quadratic • For undamped vibrations we had a11 a12 ' equation with y1 a11y1 a12 y2 y2 two roots for a21 a22 ' 2 y2 a21y1 a22 y2 y1 eigenvalues (a )(a ) a a 2 11 22 21 12 a11 a22 0 a12 1 a21 2 (a11 a22 ) a11a22 a21a12 0 • This gives trajectory slope as dy a y a y 2 y 0y 2 y • Eigenvalue solutions Trace A 2 21 1 22 2 1 2 1 (a a ) (a a )2 4(a a a a ) dy1 a11y1 a12 y2 0y1 (1)y2 y2 11 22 11 22 11 22 21 12 2 Det A 21 22 Continue Undamped Vibration What Happens if y1 = y2 = 0? • Have separable solution giving equation • Autonomous system of two equations for ellipse dy2 2 dy a y a y dy2 y1 2 2 dt 21 1 22 2 y2dy2 y1dy1 dy dy y dy1 1 a11y1 a12 y2 1 2 dt 2 2 y2 2 y1 2 2 2 2 2 2 C y2 y1 y2,0 y1,0 • Trajectory slope, dy2/dy1, depends on 2 2 values of A and may be indeterminate y2 y2 v2 x2 at y = y = 0 2 1 1 1 2 2 y2 y2 y2 2x2 v2 v2 1,0 2,0 2 2,0 0 0 x2 0 •y1 = y2 = 0 is called a critical point y1,0 2 0 2 – For multidimensional systems y = 0 23 24 ME 501A Seminar in Engineering Analysis Page 4 Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations Types of Critical Points Our First Critical Point 2.5 • Critical points are points on a y1-y2 plot • The undamped that are classified depending on the vibrations 2 trajectory shapes at or near these points solution is an 1.5 1 • Centers ellipse that does 0.5 2 • Improper Nodes not go through y1 y 0 = y = 0 -0.5 • Proper Nodes 2 • This type of -1 • Saddle Points -1.5 critical point is -2 • Spiral Points called a center -2.5 -1.5 -1 -0.5 0 0.5 1 1.5 25 26 y1 Improper Node Proper Node Unstable Stable • All trajectories, except two of them have the 10 1 0 ′ x same limiting direction of the tangents 01 0 1 • The two exceptions will have a different direction https://upload.wikimedia.org/wikipedia/commons/4/49/P https://upload.wikimedia.org/wikipedia/commons/ hase_Portrait_Unstable_Proper_Node.svg 4/4a/Phase_Portrait_Stable_Proper_Node.svg http://tutorial.math.lamar.edu/Classes/DE/ 27 28 RepeatedEigenvalues_files/image003.gif Unstable Saddle Point Stable Saddle Point 1 0 1 0 ′ ′ 04 04 http://systems-sciences.uni-graz.at/etextbook/assets/img/img_phpl/s9.png29 http://systems-sciences.uni-graz.at/etextbook/assets/img/img_phpl/s9.png 30 ME 501A Seminar in Engineering Analysis Page 5 Phase-plane Analysis of Ordinary November 1, 2017 Differential Equations Spiral Points Unstable Spiral Source 12 ′ 2 1 Asymptotically Unstable Asymptotically Stable http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues_files/image002.gif 31 http://systems-sciences.uni-graz.at/etextbook/assets/img/img_phpl/s12.png 32 ME 501A Seminar in Engineering Analysis Page 6.
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