Chapter 4: Higher-Order Differential Equations Part 3 Department - PowerPoint PPT Presentation

Solving Systems of Linear Differential Equations Nonlinear Differential Equations Summary Chapter 4: Higher-Order Differential Equations – Part 3 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw October 29, 2013 1 / 24 DE Lecture 7 王奕翔 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations Summary 1 Solving Systems of Linear Differential Equations 2 Nonlinear Differential Equations 3 Summary 2 / 24 DE Lecture 7 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations 3 / 24 see how to solve it DE Lecture 7 So far we focus on solving an ODE with only one dependent variable. Systems of Linear Differential Equations Summary ⇒ 1 independent variable. Ordinary Differential Equation = ⇒ > 1 independent variables. Partial Differential Equation = Here we look at a system of linear ODE ( > 1 dependent variables), and Example : Let t be the independent variable, x , y be dependent variables. { x ′′ + 2 x ′ + y ′′ = x + 3 y + sin t . x ′ + y ′ = − 4 x + 2 y + e − t 王奕翔

Solving Systems of Linear Differential Equations Let us use the differential operator to rewrite it as follows: 4 / 24 Idea : Nonlinear Differential Equations = DE Lecture 7 How to Solve? Summary { x ′′ + 2 x ′ + y ′′ = x + 3 y + sin t . x ′ + y ′ = − 4 x + 2 y + e − t { D 2 { x } + 2 D { x } + D 2 { y } = x + 3 y + sin t D { x } + D { y } = − 4 x + 2 y + e − t {( ) ( ) D 2 + 2 D − 1 D 2 − 3 { x } + { y } = sin t ⇒ ( D + 4) { x } + ( D − 2) { y } = e − t Eliminate y and get a linear DE of x to solve x ( t ) . Eliminate x and get a linear DE of y to solve y ( t ) . 王奕翔

Solving Systems of Linear Differential Equations (1b) 5 / 24 To eliminate x : Nonlinear Differential Equations To eliminate y : Elimination (1a) Summary DE Lecture 7 { L 11 { x } + L 12 { y } = g 1 ( t ) L 21 { x } + L 22 { y } = g 2 ( t ) where L 11 = D 2 + 2 D − 1 , L 12 = D 2 − 3 , L 21 = D + 4 , L 22 = D − 2 . (1a) × L 22 − (1b) × L 12 = ⇒ ( L 11 L 22 − L 21 L 12 ) { x } = L 22 { g 1 }− L 12 { g 2 } (1a) × L 21 − (1b) × L 11 = ⇒ ( L 12 L 21 − L 22 L 11 ) { y } = L 21 { g 1 }− L 11 { g 2 } 王奕翔

Solving Systems of Linear Differential Equations (2a) 6 / 24 (1) and find out all additional constraints. ̸ Nonlinear Differential Equations (2b) DE Lecture 7 Summary Similar to the Cramer’s Rule, after the elimination we get (1b) (1a) { L 11 { x } + L 12 { y } = g 1 ( t ) L 21 { x } + L 22 { y } = g 2 ( t ) { L { x } = � g 1 ( t ) L { y } = � g 2 ( t ) � � � � � � � � � � � � g 1 ( t ) g 1 ( t ) � � � � � � L 11 L 12 L 12 L 11 where L = g 1 ( t ) = g 2 ( t ) = � � � , � , � � � � . g 2 ( t ) g 2 ( t ) L 21 L 22 L 22 L 21 ⇒ Solutions of (1) = ⇐ = Solutions of (2) Note : We should always plug the solutions found in solving (2) back to 王奕翔

Solving Systems of Linear Differential Equations . . . . . . . . . . . Nonlinear Differential Equations . . . . . . . . . L kk j -th column replaced by g k 4 Plug into the initial system, find additional constraints on the coefficients 7 / 24 . DE Lecture 7 Summary Solving a System of Linear DE with Constant Coefficients 1 Convert it into the following form:  L 11 { y 1 } + L 12 { y 2 } + · · · + L 1 k { y k } = g 1 ( t )     L 21 { y 1 } + L 22 { y 2 } + · · · + L 2 k { y k } = g 2 ( t )     L k 1 { y 1 } + L k 2 { y 2 } + · · · + L kk { y k } = g k ( t ) 2 Use Cramer’s rule to get L { y j } = � g j ( t ) , j = 1 , . . . , k , where � � � � · · · L 11 L 12 L 1 k � � � � · · · L 21 L 22 L 2 k � � L = � g j ( t ) = L | � � , [ ] T � � · · · g 1 � � � � · · · L k 1 L k 2 3 Solve each y j ( t ) , j = 1 , . . . , k . in the complimentary solutions { y 1 c , y 2 c , . . . , y kc } , and finalize. 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations 8 / 24 find the solution of the other dependent variable (see example). plug the general solution we found back to the original system and after solving one dependent variable, it may save some time if we operators (see example). DE Lecture 7 additional constraints on the coefficients in the complimentary It is very important to plug the general solutions found for Notes and Tips Summary { y 1 , y 2 , . . . , y k } back to the original system of equations to find solutions { y 1 c , y 2 c , . . . , y kc } (see example). When solving L { y j } = � g j ( t ) , sometimes we can eliminate redundant When k = 2 , that is, only two dependent variables to be solved, 王奕翔

Solving Systems of Linear Differential Equations (3a) 9 / 24 D Nonlinear Differential Equations D Based on the method mentioned above, we compute (3b) DE Lecture 7 Example Solve Summary A: Rewrite it as { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 {( ) D 2 − 4 { x } + D 2 { y } = t 2 ( D + 1) { x } + D { y } = 0 � � D 2 − 4 � � D 2 � = − ( D 2 + 4 D ) = − D ( D + 4) � � L = � D + 1 � � � � D 2 − 4 � � { t 2 } � � { t 2 } t 2 D 2 t 2 � � � � � g 1 ( t ) = � = D � g 2 ( t ) = � = − ( D + 1) , � � 0 D + 1 0 王奕翔

Solving Systems of Linear Differential Equations Hence, solutions of the original system of equations must be solutions of 10 / 24 Question : why can we cancel the repeated operators on both sides? D Nonlinear Differential Equations the the following: DE Lecture 7 Summary Example: Convert into two separate linear equations Solve { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( ) { t 2 } { x } = � − � L { x } = � g 1 ( t ) ⇐ ⇒ D ( D + 4) { t 2 } L { y } = � g 2 ( t ) ⇐ ⇒ ( − D ( D + 4)) { y } = − ( D + 1) Ans : because instead of eliminating y by D { (3a) } − D 2 { (3b) } , we can simply eliminate y by (3a) − D { (3b) } . 王奕翔

Solving Systems of Linear Differential Equations We convert the above into 11 / 24 = Nonlinear Differential Equations (4b) (4a) DE Lecture 7 Solution 1: Solving x and y separately Solve Summary { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( D + 4) { x } = − t 2 ( D ( D + 4)) { y } = t 2 + 2 t Step 1. Solve (5a): x c = c 1 e − 4 t , x p = A + Bt + Ct 2 , and − t 2 = ( D + 4) { x p } = (4 A + B ) + (4 B + 2 C ) t + 4 Ct 2 ⇒ C = − 1 4 , B = − 1 2 C = 1 8 , A = − 1 4 B = − 1 32 . Hence x ( t ) = c 1 e − 4 t − 1 32 + 1 8 t − 1 4 t 2 . 王奕翔

Solving Systems of Linear Differential Equations We convert the above into 12 / 24 = Nonlinear Differential Equations (5b) (5a) DE Lecture 7 Solution 1: Solving x and y separately Solve Summary { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( D + 4) { x } = − t 2 ( D ( D + 4)) { y } = t 2 + 2 t Step 2. Solve (5b): y c = c 2 + c 3 e − 4 t , y p = At + Bt 2 + Ct 3 , and t 2 + 2 t = ( D 2 + 4 D ) { y p } = (4 A + 2 B ) + (8 B + 6 C ) t + 12 Ct 2 ⇒ C = 1 12 , B = 1 4 − 3 4 C = 3 16 , A = − 1 2 B = − 3 32 . Hence y ( t ) = c 2 + c 3 e − 4 t − 3 32 t + 3 16 t 2 + 1 12 t 3 . 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations 13 / 24 DE Lecture 7 Solve Solution 1: Solving x and y separately Summary { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 We find that for some c 1 , c 2 , c 3 , { c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 x ( t ) = c 2 + c 3 e − 4 t − 16 t 2 + 3 3 12 t 3 1 y ( t ) = 32 t + Final Step. Plug them back to find the constraints on { c 1 , c 2 , c 3 } . { (12 c 1 + 16 c 3 ) e − 4 t + t 2 = t 2 − (3 c 1 + 4 c 3 ) e − 4 t = 0 which implies c 3 = − 3/4 c 1 . Hence, the final solution is { c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 x ( t ) = 4 c 1 e − 4 t − 16 t 2 + c 2 − 3 3 3 12 t 3 1 32 t + y ( t ) = 王奕翔

Solving Systems of Linear Differential Equations Plug this back to the second equation, we get 14 / 24 Plug it back to the first equation, it is also satisfied. Done! = Nonlinear Differential Equations DE Lecture 7 Summary Solve Solution 2: Solving x first and then plugging in to find y { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 After Step 1., we find that for some c 1 , x ( t ) = c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 . y ′ = − x ′ − x = 3 c 1 e − 4 t − 3 32 + 3 8 t + 1 4 t 2 ⇒ y = c 2 − 3 4 c 1 e − 4 t − 3 32 t + 3 16 t 2 + 1 12 t 3 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations Summary 1 Solving Systems of Linear Differential Equations 2 Nonlinear Differential Equations 3 Summary 15 / 24 DE Lecture 7 王奕翔

Solving Systems of Linear Differential Equations Nonlinear Differential Equations Summary Linear vs. Nonlinear Differential Equations Differences : 1 Superposition principle does not hold for nonlinear DE. 2 Nonlinear DE can have singular solutions, while linear DE will not have singular solutions. 3 Usually there is no analytical tool to solve a nonlinear DE of higher order. We will present some special kinds of nonlinear second order DE that can be solved analytically. 16 / 24 DE Lecture 7 王奕翔