Chapter 4. Systems of ODEs. Phase plane. Qualitative methods C.O.S. Sorzano Biomedical Engineering September 13, 2014 4. Systems of ODEs September 13, 2014 1 / 83 Outline 1 Systems of ODEs. Phase plane. Qualitative methods Systems of ODEs as models Basic theory of systems of ODEs. Wronskian Constant-coefficient systems. Phase plane method Criteria for critical points. Stability Qualitative methods for nonlinear systems Nonhomogeneous linear systems of ODEs 4. Systems of ODEs September 13, 2014 2 / 83 References E. Kreyszig. Advanced Engineering Mathematics. John Wiley & sons. Chapter 4. 4. Systems of ODEs September 13, 2014 3 / 83 Outline 1 Systems of ODEs. Phase plane. Qualitative methods Systems of ODEs as models Basic theory of systems of ODEs. Wronskian Constant-coefficient systems. Phase plane method Criteria for critical points. Stability Qualitative methods for nonlinear systems Nonhomogeneous linear systems of ODEs 4. Systems of ODEs September 13, 2014 4 / 83 Systems of ODEs as models Mixing tanks Solution: y lb gal y lb gal y 0 = inflow-outflow = 2 2 − 1 2 1 100 gal min 100 gal min y lb gal y lb gal y 0 = inflow-outflow = 1 2 − 2 2 2 100 gal min 100 gal min 0 0 y1 = −0.02y1 + 0.02y2 y1 −0.02 0.02 y1 0 0 ⇒ 0 = ⇒ y = Ay y2 = 0.02y1 − 0.02y2 y2 0.02 −0.02 y2 4. Systems of ODEs September 13, 2014 5 / 83 Systems of ODEs as models Mixing tanks (continued) We try a solution of the form y = xeλt y0 = λxeλt Now we substitute into the ODE y0 = Ay λxeλt = Axeλt λx = Ax That is y = xeλt can be a solution of the ODE if x is an eigenvector of the matrix A and λ its associated eigenvalue. T λ1 = 0, x1 = (1, 1) A ⇒ T λ2 = −0.04, x2 = (1, −1) 4. Systems of ODEs September 13, 2014 6 / 83 Systems of ODEs as models Mixing tanks (continued) The general solution is λ1t λ2t y = c1x1e + c2x2e 1 1 y = c + c e−0.04t 1 1 2 −1 To determine the coefficients c1 and c2, we impose the initial conditions y1(0) = 0, y2(150) 0 1 1 c = c + c ⇒ c = 75, c = −75 1 150 1 1 2 −1 1 2 The solution to the problem is −0.04t 1 1 −0.04t y1 = 75 − 75e y = 75 − 75 e ⇒ −0.04t 1 −1 y2 = 75 + 75e 4. Systems of ODEs September 13, 2014 7 / 83 Systems of ODEs as models Mixing tanks (continued) −0.04t 1 1 −0.04t y1 = 75 − 75e y = 75 − 75 e ⇒ −0.04t 1 −1 y2 = 75 + 75e 4. Systems of ODEs September 13, 2014 8 / 83 Systems of ODEs as models Electrical network Solution: 0 0 LI1 + R1(I1 − I2) = E ⇒ I1 = −R1I1 + R1I2 + E 1 Z 1 R (I − I ) + I dt + R I = 0 ⇒ −R I0 + (R + R )I0 = − I 1 2 1 C 2 2 2 1 1 1 2 2 C 2 4. Systems of ODEs September 13, 2014 9 / 83 Systems of ODEs as models Electrical network (continued) 0 I1 = −R1I1 + R1I2 + E 1 −R I0 + (R + R )I0 = − I 1 1 1 2 2 C 2 0 0 I1 = −4I1 + 4I2 + 12 1 0 I1 −4 4 I1 12 0 0 ⇒ 0 = + −4I1 + 10I2 = −4I2 −4 10 I2 0 −4 I2 0 0 I1 −4 4 I1 12 0 = + I2 −1.6 1.2 I2 4.8 The eigenvalues and eigenvectors of the matrix A are T λ1 = −2, x1 = (2, 1) A ⇒ T λ2 = −0.8, x2 = (1, 0.8) 4. Systems of ODEs September 13, 2014 10 / 83 Systems of ODEs as models Electrical network (continued) T λ1 = −2, x1 = (2, 1) A ⇒ T λ2 = −0.8, x2 = (1, 0.8) The general solution of the H problem is 2 1 I = c e−2t + c e−0.8t 1 1 2 0.8 For a particular solution of the NH problem we try a constant vector I = a 0 −4 4 a1 12 = + ⇒ a1 = 3, a2 = 0 0 −1.6 1.2 a2 4.8 So the general solution of the NH problem is 2 1 3 I = c e−2t + c e−0.8t + 1 1 2 0.8 0 4. Systems of ODEs September 13, 2014 11 / 83 Systems of ODEs as models Electrical network (continued) 2 1 3 I = c e−2t + c e−0.8t + 1 1 2 0.8 0 To determine the unknown coefficients we impose the initial condition I(0) = 0 2 1 3 0 = c e−2t + c e−0.8t + ⇒ c = −4, c = 5 1 1 2 0.8 0 1 2 The solution to the problem is 2 1 3 I = −4 e−2t + 5 e−0.8t + 1 0.8 0 4. Systems of ODEs September 13, 2014 12 / 83 Systems of ODEs as models Electrical network (continued) 4. Systems of ODEs September 13, 2014 13 / 83 Conversion of an n-th order ODE to a system Conversion of an ODE An n-th order ODE y (n) = F (t, y, y 0, ..., y (n−1)) can be converted to a system of n first-order ODEs by setting y1 = y and 0 y1 = y2 0 y2 = y3 ... 0 yn−1 = yn 0 yn = F (t, y1, y2, ..., yn) 4. Systems of ODEs September 13, 2014 14 / 83 Conversion of an n-th order ODE to a system Example my 00 + cy 0 + ky = 0 Solution: c k y 00 = − y 0 − y m m Now we write it as 0 y1 = y2 k c y 0 = − y − y 2 m 1 m 2 0 y1 0 1 y1 0 = k c y2 − m − m y2 Its characteristic polynomial is −λ 1 2 c k k c = λ + λ + = 0 − m − m − λ m m 4. Systems of ODEs September 13, 2014 15 / 83 Conversion of an n-th order ODE to a system Example (continued) c k λ2 + λ + = 0 m m Let us now give values, m = 1, c = 2, k = 0.75 2 λ + 2λ + 0.75 = 0 ⇒ λ1 = −0.5, λ2 = −1.5 With eigenvectors T T v1 = (2, −1) , v2 = (1, −1.5) So, the general solution is 2 1 y = c e−0.5t + c e−1.5t 1 −1 2 −1.5 The first component is −0.5t −1.5t y1 = y = 2c1e + c2e The second component is its derivative. 4. Systems of ODEs September 13, 2014 16 / 83 Exercises Exercises From Kreyszig (10th ed.), Chapter 4, Section 1: 4.1.1 4.1.12 4. Systems of ODEs September 13, 2014 17 / 83 Outline 1 Systems of ODEs. Phase plane. Qualitative methods Systems of ODEs as models Basic theory of systems of ODEs. Wronskian Constant-coefficient systems. Phase plane method Criteria for critical points. Stability Qualitative methods for nonlinear systems Nonhomogeneous linear systems of ODEs 4. Systems of ODEs September 13, 2014 18 / 83 Basic theory Basic theory In general, an ODE system is of the form 0 y1 = f1(t, y1, ..., yn) y 0 = f (t, y , ..., y ) 2 2 1 n ⇒ y0 = f(t, y) ... 0 yn = fn(t, y1, ..., yn) An Initial Value Problem needs n initial conditions y1(t0) = K1, y2(t0) = K2, ..., yn(t0) = Kn ⇒ y(t0) = K Existence and uniqueness ∂f1 Let f1, f2, ..., fn be continuous functions with continuous partial derivatives , ∂y1 ∂f1 ∂fn , ..., in some domain R of the ty1y2...yn-space containing the point ∂y2 ∂yn (t0, K1, ..., Kn). Then the ODE system has a solution on some interval t0 − α < t < t0 + α satisfying the initial conditions, and this solution is unique. 4. Systems of ODEs September 13, 2014 19 / 83 Basic theory Linear systems 0 y1 = a11(t)y1 + a12(t)y2 + ... + a1n(t)yn + g1(t) y 0 = a (t)y + a (t)y + ... + a (t)y + g (t) 2 21 1 22 2 2n n 2 ⇒ y0 = Ay + g ... 0 yn = an1(t)y1 + an2(t)y2 + ... + ann(t)yn + gn(t) with a11 a12 ... a1n a21 a22 ... a2n A = ... ... ... ... an1 an2 ... ann If g = 0, the system is homogeneous. Existence and uniqueness Let the aij ’s andgi ’s functions be continuous functions of t in an open interval I containing t0, then there exists a solution satisfying the initial conditions, and this solution is unique. 4. Systems of ODEs September 13, 2014 20 / 83 Basic theory Superposition principle The linear combination of any two solutions, y1 and y2, of the H problem is also a solution of the H problem. Proof: y = c1y1 + c2y2 0 0 0 y = c1y1 + c2y2 = c1(Ay1) + c2(Ay2) = A(c1y1 + c2y2) = Ay 4. Systems of ODEs September 13, 2014 21 / 83 Basic theory General solution. Wronskian If the aij ’s functions are continuous, then the general solution of the H problem can be written as y = c1y1 + c2y2 + ... + cnyn where y1, y2, ..., yn constitute a basis or fundamental system of solutions, and there is no singular solution. We can write the n basis functions as the columns of a matrix Y Y = y1 y2 ... yn and write the general solution as y = Y c The Wronskian is the determinant of Y W = |Y | 4. Systems of ODEs September 13, 2014 22 / 83 Outline 1 Systems of ODEs. Phase plane. Qualitative methods Systems of ODEs as models Basic theory of systems of ODEs. Wronskian Constant-coefficient systems. Phase plane method Criteria for critical points. Stability Qualitative methods for nonlinear systems Nonhomogeneous linear systems of ODEs 4. Systems of ODEs September 13, 2014 23 / 83 Constant coefficients systems Constant coefficients systems y0 = Ay We try a function of the form y = xeλt y0 = λxeλt And substitute it in the ODE λxeλt = Axeλt λx = Ax That is y is a solution if x is an eigenvector of A.