MATH 251. ALl The Working Done. 100% Score

MATH 251 Homework 10 Vishal Vasan Solving linear systems of differential equations Exercise 1 Solve the following linear systems of differential equations. To do this, you need to find eigenvalu es, eigenvectors and write down the general form of the solution with two arbitrary constants. Be sure that your final answer is a real-valued function and don’t forget to use the initial conditions! Solution: Eigenvalues are given by which has roots λ = −3; −3. Eigenvector for λ = −3 given by which is equivalent to and so is the eigenvector for every value of C1. Notice, this is the only possible eigenvector that we can obtain given λ = −3. If we had obtained 0p + 0q = 0 then we can find two eigenvectors, namely C1(1; 0) and C2(0; 1). This is not the case here. So we need a generalized eigenvector. So which is equivalent to for every C2. So we can choose representative generalized eigenvector as The general solution to the differential equation is given by Finally we appy the initial conditions: and so thus Hence the solution to the differential equation is d dt Solution: Eigenvalues are given by det which has roots λ = −4; −4. Eigenvector for λ = −4 given by which is equivalent to and so is the eigenvector for every value of C1. Notice, this is the only possible eigenvector that we can obtain given λ = −3. If we had obtained 0p + 0q = 0 then we can find two eigenvectors, namely C1(1; 0) and C2(0; 1). This is not the case here. So we need a generalized eigenvector. So which is equivalent to and so for every C2. So we can choose representative generalized eigenvector as rs = −10=2 . The general solution to the differential equation is given by and so Finally we appy the initial conditions: nd so thus Hence the solution to the differential equation is dt Solution: Eigenvalues are given by which has roots λ = 1 ± i. Eigenvector for λ = 1 + i given by which is equivalent to and so is the eigenvector for every value of C1. Notice, we have a complex conjugate pair of eigenvalues and so we also have a complex conjugate pair of eigenvectors. To find real-valued solutions we find the real and imaginary parts of Thus the general solution is given by Finally we appy the initial conditions: −0 1 and so 2C1 − C2 = 0 C1 = 2 thus C1 = 2 C4 = 0 Hence the solution to the differential equation is ~x = 2et 2 coscos t + sin t t + 4et 2 sinsin t −tcos t Classification of the origin Exercise 2 Based on the eigenvalues, classify the equilibrium point at the origin (0; 0) according to its type. Also state whether the origin is stable, asymptotically stable or unstable. which is equivalent to λ2 + 6λ + 9 = 0 ) λ = −3; −3 Hence we have repeated, real negative eigenvalues. We now find the eigenvector which is given by which is equivalent to is the eigenvector for every value of C1. Notice, this is the only possible eigenvector that we can obtain given λ = −3. If we had obtained 0p + 0q = 0 then we can find two eigenvectors, namely C1(1; 0) and C2(0; 1). This is not the case here. So we need a generalized eigenvector. So which is equivalent to for every C2. So we can choose representative generalized eigenvector as rs = −10=4 . Thus we have two, equal, real negative eigenvalues with one eigenvector and a generalized eigenvector. Hence we have an Asymptotically stable proper node.(b) d Solution: which is equivalent to Thus we have a complex conjugate pair of eigenvalues with zero real part. Hence we have a Stable center. Solution: Thus we have a complex conjugate pair of eigenvalues with positive real part. Hence we have an Unstable spiral.Autonomous systems and stability Exercise 3 Find all critical points for the folllowing systems: 1. dx Obtained by setting dx=dt = 0 and dy=dt = 0. Thus From the second equation either y = 0 or x = 1. Consider, y = 0 in the first equation. This implies So (0; 0) is a critical point since y = 0 ) x = 0. Now consider x = 1 in the first equation. This implies Thus x = 1 ) y = −2; −1 and so we have two additional critical points (1; −1) and (1; −2). Solution: Critical points are (0; 0); (0; 1); (1=2; 1=2); (−1=2; 1=2) Obtained by setting dx=dt = 0 and dy=dt = 0. Thus From the first equation either y = 1=2 or x = 0. Consider, x = 0 in the second equation. This implies So (0; 0); (0; 1) are critical points since x = 0 ) y = 0; 1. Now consider y = 1=2 in the second equation. This implies 2 and so we have two additional critical points (1=2; 1=2) and Solution: Critical points are (0; 0); (p6; 0); (−p6; 0) Obtained by setting dx=dt = 0 and dy=dt = 0. Thus y = 0 and x − x3 0From the first equation y = 0. This implies for second equation we have x − ±p6 So (0; 0); (p6; 0); (−p6; 0) are critical points since y = 0 ) x = 0; p6; −p6.Exercise 4 For each of the following system given below, a. Find all critical points for the system. b. Find the corresponding linear system near each critical point. c. Based on the eigenvalues and eigenvectors for each linear system, classify the critical point according to type and stability. d. Do your conclusions from (c) for the critical point hold for the nonlinear system? What is the type and stability for the nonlinear system that you would expect? 1. dx Solutions: a. Find critical points: x + x2 − 2y2 = 0 and y(1 − x) = 0 From second equation y = 0 or x = 1. Substituting y = 0 in first equation we have x + x2 = 0 which implies x = 0; −1. Thus we have (0; 0); (−1; 0) as critical points. Similarly, substituting x = 1 in first equation we have 2 − 2y2 = 0 which implies y = ±1 and so (1; 1); (1; −1) are also critical points. Totally we have four critical points (0; 0); (−1; 0); (1; 1); (1; −1) b. Linear system about each critical point: linear system about the critical point (x0; y0) is given by d where f(x; y) = x + x2 −2y2 and g(x; y) = y(1−x) are the right-hand sides of given nonlinear equation. We find the partial derivatives as follows fx (x; y) = 1 + 2x; fy(x; y) = −4y; gx(x; y) = −y; gy(x; y) = 1 − x Evaluating each of these partial derivatives at each critical point we obtain the following systems: Linear stability based on eigenvalues: For (0; 0) Eigenvalues are given by Thus we have real positive repeated eigenvalues. Since they are repeated, we would like to determine whether we have a proper or improper node. So we determine eigenvectors: which is satisfied for every p and q. So we can take (C1; 0) and (0; C2) as two eigenvectors or C1(1; 0) and C2(0; 1) as eigenvectors. Thus we have two eigenvectors. Hence we have a Unstable proper node. For (−1; 0) Eigenvalues are given by Thus we have two real distinct eigenvalues. One is positive and other is negative. Hence we have a Unstable Saddle. Note, saddle points are always unstable. For (1; 1) Eigenvalues are given by Thus once again we have an Unstable saddle point. For (1; −1) Eigenvalues are given by Once again we have an Unstable saddle point. d. For the critical points (−1; 0); (1; 1); (1; −1) the conclusions from the linear analysis based on eigenvalues is the same for the nonlinear system. For the critical point at (0; 0) the nonlinear system may have either a node or a spiral point, but will still be unstable (see pg 523 of book).2. Solutions: a. Find critical points: From first equation x = y3. Substituting x = y3 in second equation we have 1−y3y = 0 which implies y4 = 1 or y = ±1. We restrict ourselves to real values of y. This leads to x = y3 = 1 when y = 1 and x = y3 = −1 when y = −1. Totally we have two critical points b. Linear system about each critical point: linear system about the critical point (x0; y0) is given by d are the right-hand sides of given nonlinear equation. We find the partial derivatives as follows fx Evaluating each of these partial derivatives at each critical point we obtain the following systems: For (1; 1): d c. Linear stability based on eigenvalues: For (1; 1) Eigenvalues are given by Thus we have an Unstable saddle point. For (1; −1) Eigenvalues are given by Since eigenvalues are a complex conjugate pair, we have an Unstable spiral point. d. In both cases, the conclusions from the linear analysis based on eigenvalues is the same for the nonlinear system (see pg 523 of book).