3.2 Systems of Two First Order Linear DE s
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1 Agenda Section 3.2 Reminders Lab 1 write-up due 9/26 or 9/28 Lab 2 prelab due 9/26 or 9/28 WebHW due 9/29 Office hours Tues, Thurs 1-2 pm (5852 East Hall) MathLab office hour Sun 7-8 pm (MathLab)
2 3.2 Systems of Two First Order Linear DE s Objectives Be able to write a linear system of DE s using matrices Be able to recognize component plots, direction fields, and phase portraits for linear systems Be able to decide when a linear system is guaranteed to have unique solutions Be able to find the critical points of an autonomous linear system Be able to rewrite a 2nd order DE as a 1st order system
3 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
4 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
5 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
6 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
7 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
8 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
9 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
10 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
11 Consider two interconnected tanks. Tank 1 initially contains 55 oz of salt, and Tank 2 initially contains 26 oz of salt. Write a system of DE s for the amount of salt in each tank. dq 1 dt dq 2 dt 1.5 gal 1 oz gal Q 2 oz gal Q 1 oz = min gal min 20 gal min 30 gal, = 1 gal 3 oz gal Q 1 oz gal Q 2 oz +3 4 min gal min 30 gal min 20 gal, Q 1 (0) = 55 oz, Q 2 (0) = 26 oz.
12 Simplifying our expression gives Let q = [ Q1 Q 2 dq 1 = 0.1Q Q , dt dq 2 = 0.1Q 1 0.2Q 2 + 3, dt Q 1 (0) = 55, Q 2 (0) = 26. [ 1.5, b = 3 [ , A = We can express the system, using matrix notation, as [ dq 55 = Aq + b, q(0) =. dt 26
13 The solution of the system turns out to be [ [ [ [ Q1 (t) q = = 7 e t/4 + 2 e t/20 + Q 2 (t) Combining terms gives us equations for Q 1 (t) and Q 2 (t). Q 1 (t) = 7e t/4 + 6e t/ , Q 2 (t) = 14e t/4 + 4e t/ We can visualize Q 1 and Q 2 by plotting both functions on the same graph. Plots of Q 1 and Q 2 versus t are called component plots. (Note that this solution was given. We have not yet discussed how to solve such a system.).
14 Component Plot for Q 1 (t) and Q 2 (t) Q 1, Q 2 t
15 In our example, the right-hand side of the DE only involved the dependent variables Q 1 and Q 2. Such a system is called autonomous, just like it was when we had individual DE s. Another way to visualize solutions of 2 2 autonomous systems is to consider plots with Q 1 on the x-axis and Q 2 on the y-axis. The two dependent variables Q 1 and Q 2 are sometimes referred to as state variables since the state of the system at any time depends on their values. In addition, the Q 1 Q 2 -plane is sometimes referred to as state space, the state plane, or the phase plane. There are two common plots in state space. The first one, called a vector field, is similar to a slope field, except each line segment is a normalized vector (an arrow) pointing in the direction of the vector dq 1 /dt, dq 2 /dt, the slope of which can be easily computed using the chain rule. That is, dq 2 dq 1 = dq 2 dq 1 dt dt. The second plot, called a phase portrait, is the 2-dimensional version of a phase line.
16 Vector Field showing Q 2 vs Q 1 Q 2 Q 1 The solid line corresponds to the solution with initial condition Q 1 (0) = 55, Q 2 (0) = 26.
17 Phase Portrait showing Q 2 vs Q 1
18 Example Find the equilibrium solutions of the linear system q = Aq + b, where q = [ Q1 Q 2 [ , A = [ 1.5, b = 3 Remember that equilibrium solutions are constant solutions. Therefore, we need to set q = 0. 0 = Aq eq + b Aq eq = b (1) We now have a linear system of equations. We can write the system as an augmented matrix. That is,
19 Example Find the equilibrium solutions of the linear system q = Aq + b, where q = [ Q1 Q 2 [ , A = [ 1.5, b = 3 Remember that equilibrium solutions are constant solutions. Therefore, we need to set q = 0. 0 = Aq eq + b Aq eq = b (2) We now have a linear system of equations. We can write the system as an augmented matrix. That is,
20 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
21 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
22 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
23 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
24 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
25 [ Add R 1 to R 2 [ The matrix is now in row echelon form. We can see that columns 1 and 2 are pivot columns, so Q 1 and Q 2 are basic variables. Therefore, we will try to solve for both variables. Converting back to a linear system gives us 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. We can use the last equation to solve for Q 2. That is, Q 2 = 4.5/0.125 = 36.
26 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. Plugging Q 2 = 36 into the first equation and solving for Q 1 gives Q 1 = (1.5/0.1) + ( )/0.1 = 42. Therefore, q eq = [
27 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. Plugging Q 2 = 36 into the first equation and solving for Q 1 gives Q 1 = (1.5/0.1) + ( )/0.1 = 42. Therefore, q eq = [
28 0.1Q Q 2 = 1.5, 0Q Q 2 = 4.5. Plugging Q 2 = 36 into the first equation and solving for Q 1 gives Q 1 = (1.5/0.1) + ( )/0.1 = 42. Therefore, q eq = [
29 Example Find the critical points of x = Ax + b, where x = [ x1 x 2 [ 2 3, A = 4 6 [ 1, b = 2. Recall that a critical point is the same thing as an equilibrium solution. Therefore, we ll set x = 0. 0 = Ax eq + b Ax eq = b We now have a linear system we can express as an augmented matrix.
30 Example Find the critical points of x = Ax + b, where x = [ x1 x 2 [ 2 3, A = 4 6 [ 1, b = 2. Recall that a critical point is the same thing as an equilibrium solution. Therefore, we ll set x = 0. 0 = Ax eq + b Ax eq = b We now have a linear system we can express as an augmented matrix.
31 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
32 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
33 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
34 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
35 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
36 [ Add 2R 1 to R 2 [ The matrix is now in row echelon form. In this case, only the first column is a pivot column. Therefore, x 1 is a basic variable and x 2 is a free variable. We ll start by setting x 2 equal to a parameter. That is, x 2 = c. We can write the row reduced matrix as a linear system. That is, 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0.
37 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0 If we look at the first equation, plug c in for x 2, and solve for x 1, we get x 1 = c. Therefore, x eq = [ x1 x 2 = [ c c = [ 1/2 0 + c [ 3/2 1. In other words, every point on a line in the x 1 x 2 -plane is a critical point. Such a situation arises because the two DE s are exactly the same, just off by a constant.
38 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0 If we look at the first equation, plug c in for x 2, and solve for x 1, we get x 1 = c. Therefore, x eq = [ x1 x 2 = [ c c = [ 1/2 0 + c [ 3/2 1. In other words, every point on a line in the x 1 x 2 -plane is a critical point. Such a situation arises because the two DE s are exactly the same, just off by a constant.
39 2x 1 + 3x 2 = 1, 0x 1 + 0x 2 = 0 If we look at the first equation, plug c in for x 2, and solve for x 1, we get x 1 = c. Therefore, x eq = [ x1 x 2 = [ c c = [ 1/2 0 + c [ 3/2 1. In other words, every point on a line in the x 1 x 2 -plane is a critical point. Such a situation arises because the two DE s are exactly the same, just off by a constant.
40 Example Rewrite the second order DE as a first order system. 2 d 2 y dt 0.5dy 2 dt + 8y = 6 sin (2t) If we let x = dy/dt, then dx/dt = d 2 y/dt 2 and dx dt = 1 x 4y + 3 sin (2t) 4 dy dt = x We can express this system of DE s in the form x = Ax + b(t). That is, [ [ [ [ d x 1/4 4 x 3 sin (2t) = + dt y 1 0 y 0.
41 Example Rewrite the second order DE as a first order system. 2 d 2 y dt 0.5dy 2 dt + 8y = 6 sin (2t) If we let x = dy/dt, then dx/dt = d 2 y/dt 2 and dx dt = 1 x 4y + 3 sin (2t) 4 dy dt = x We can express this system of DE s in the form x = Ax + b(t). That is, [ [ [ [ d x 1/4 4 x 3 sin (2t) = + dt y 1 0 y 0.
42 Example Rewrite the second order DE as a first order system. 2 d 2 y dt 0.5dy 2 dt + 8y = 6 sin (2t) If we let x = dy/dt, then dx/dt = d 2 y/dt 2 and dx dt = 1 x 4y + 3 sin (2t) 4 dy dt = x We can express this system of DE s in the form x = Ax + b(t). That is, [ [ [ [ d x 1/4 4 x 3 sin (2t) = + dt y 1 0 y 0.
43 Example Rewrite the second order DE as a first order system. t 2 u + tu + (t )u = 0 If we let v = du/dt. Then, dv/dt = d 2 u/dt 2 and u = v v = 1 t v t u t 2 We can express the system in the form u = A(t)u. That is, [ [ [ d u 0 1 u = dt v (0.25 t 2 )/t 2. 1/t v
44 Example Rewrite the second order DE as a first order system. t 2 u + tu + (t )u = 0 If we let v = du/dt. Then, dv/dt = d 2 u/dt 2 and u = v v = 1 t v t u t 2 We can express the system in the form u = A(t)u. That is, [ [ [ d u 0 1 u = dt v (0.25 t 2 )/t 2. 1/t v
45 Example Rewrite the second order DE as a first order system. t 2 u + tu + (t )u = 0 If we let v = du/dt. Then, dv/dt = d 2 u/dt 2 and u = v v = 1 t v t u t 2 We can express the system in the form u = A(t)u. That is, [ [ [ d u 0 1 u = dt v (0.25 t 2 )/t 2. 1/t v
46 Theorem: Existence and Uniqueness of Solutions Let each of the functions p 11,..., p 22, g 1, and g 2 be continuous on an open interval I = α < t < β. Let t 0 be any point in I, and let x 0 and y 0 be any given numbers. Then, there exists a unique solution to the IVP dx dt = P(t)x + g(t), x(t 0) = [ x0 y 0, where [ x x = y [ p11 (t) p, P(t) = 12 (t) p 21 (t) p 22 (t) [ g1 (t), g(t) = g 2 (t). Furthermore, the solution exists throughout the interval I.
47 Example Transform the given IVP into an IVP with two first order equations. Then write the system in matrix form. tu + u + tu = 0, u(1) = 1, u (1) = 0
48
49 Example Find the critical points of the system of DE s. x = x + y + 1, y = x + y 3
50
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