TOC Chapter 4. The Laplace Transform [part 1] 4.1 Preliminaries - PowerPoint PPT Presentation

TOC Chapter 4. The Laplace Transform [part 1] 4.1 Preliminaries 4.2 Laplace Transform 4.3 Basic Properties of the Laplace Transform 4.4 Inverse Laplace Transforms and Initial-Value Problems [part 2] 4.5 Piecewise Continuous Functions, Part I: Laplace Transforms 4.6 Piecewise Continuous Functions, Part II: Inverse Laplace Transforms 4.7 Initial-Value Problems with Piecewise Continuous Nonhomogeneous Terms

Recall: A method for solving initial-value problems for linear differential equations with constant coefficients. Can be also used if RHS is only piecewise continuous (so not continuous). � ∞ e − sx f ( x ) dx F ( s ) = L [ f ( x )] := 0 Main properties: I. L is linear and one-to-one II. L [ f ′ ( x )] = s L [ f ( x )] − f (0) III. L [ x n f ( x )] = ( − 1) n d n F ds n ⇒ L [ e rx f ( x )] = F ( s − r ) IV. L [ f ( x )] = F ( s ) = V. (Translated Function) Let L [ f ( x )] := F ( s ) and c > 0 : L [ f ( x − c ) u ( x − c )] = e − cs F ( s ) . L − 1 [ e − cs F ( s )] = f ( x − c ) u ( x − c ) . where u is the Heaviside step function � 0 x < c u c ( x ) = u ( x − c ) = 1 x ≥ c 1

f ( x ) F ( s ) = L [ f ( x )] c c s, s > 0 1 e αx s − α, s > α s cos βx s > 0 s 2 + β 2 , β sin βx s 2 + β 2 , s > 0 s − α e αx cos βx ( s − α ) 2 + β 2 , s > α β e αx sin βx ( s − α ) 2 + β 2 , s > α n ! x n , n = 1 , 2 , . . . s n +1 , s > 0 n ! x n e αx , n = 1 , 2 , . . . ( s − α ) n +1 , s > α s 2 − β 2 x cos βx ( s 2 + β 2 ) 2 , s > 0 2 βs x sin βx ( s 2 + β 2 ) 2 , s > 0 2

Section 4.5. Laplace Transform: Discontinuous Functions Types of discontinuities: 1 Infinite: f ( x ) = ( x − 1) 2 3

 2 x, 0 ≤ x < 3  Jump: f ( x ) = 4 , 3 ≤ x < ∞   x 2 − 4 x − 2 , x � = 2   Removable: f ( x ) = 1 , x = 2   4

Def. Let the function f = f ( x ) be defined on an interval I and continuous except at a point c ∈ I . If x → c − f ( x ) lim and x → c + f ( x ) lim exist but x → c − f ( x ) � = lim x → c + f ( x ), lim then f is said to have a jump (or finite ) discontinuity at c . 5

Def. A function f defined on an interval I is piecewise continuous on if it is continuous I on I except for at most a finite number of points c 1 , c 2 , . . . , c n of I at which it has jump discontinu- ities. 6

Theorem: If the function f is piecewise continuous on [0 , ∞ ), and of exponential order λ , then the Laplace transform L [ f ( x )] exists for s > λ . 7

Unit Step (Heaviside) Functions: Let c > 0. The function  0 x < c  u c ( x ) = u ( x − c ) = 1 x ≥ c  is called a unit step function . 8

Laplace Transform of a Unit Step Function: � ∞ e − sx u ( x − c ) dx L [ u ( x − c )] = 0 L [ u ( x − c )] = e − cs 1 s > 0. s, 9

Translation of a Function: if f is defined on [0 , ∞ ) and c > 0, then the function  0 , x < c   f ( x − c ) u ( x − c ) = f ( x − c ) u ( x − c ) , x ≥ c   is the translation of f to c . 10

f ( x ) f ( x − c ) 11

f ( x − c ) u ( x − c ) 12

Translations: Express f ( x ) in terms of ( x − c ) EXAMPLES: 1. Express f ( x ) = 5 x + 3 in terms of ( x − 3) Express f ( x ) = x 2 − 3 x + 7 in terms of ( x − 2) 2. 13

3. Express f ( x ) = sin 2 x in terms of ( x − π/ 2) 4. Express f ( x ) = cos πx in terms of ( x − 3) 14

Property V. Laplace Transform of a Translated Function: Suppose that L [ f ( x )] = F ( s ). Then, for any positive number c , L [ f ( x − c ) u ( x − c )] = e − cs F ( s ). 1. L − 1 [ e − cs F ( s )] = f ( x − c ) u ( x − c ). 2. 15

Proof 16

Examples: Given f ( x ), find F ( s ).  2 x, 0 ≤ x < 3  1. f ( x ) = 3 , 3 ≤ x < ∞  17

Step 1. Re-write f in terms of u ( x − 3): f ( x ) = 2 x − 2 x u ( x − 3) + 3 u ( x − 3) 18

Graphs: 2 x − 2 x u ( x − 3) 3 u ( x − 3) 19

Step 2. Write the coefficients in terms of x − 3 f ( x ) = 2 x − 2( x − 3) u ( x − 3) − 3 u ( x − 3) 20

Step 3. Determine L [ f ]: F ( s ) = 2 s 2 − 2 e − 3 s 1 s 2 − 3 e − 3 s 1 s 21

x 2 ,  0 ≤ x < 2  2. f ( x ) = 3 x, x ≥ 2  Step 1. Re-write f in terms of u ( x − 2): f ( x ) = x 2 − x 2 u ( x − 2) + 3 x u ( x − 2) 22

Step 2. Write the coefficients in terms of x − 2: f ( x ) = x 2 − ( x − 2) 2 u ( x − 2) − ( x − 2) u ( x − 2)+2 u ( x − 2) 23

Step 3. Determine L [ f ]: f ( x ) = x 2 − ( x − 2) 2 u ( x − 2) − ( x − 2) u ( x − 2)+2 u ( x − 2) F ( s ) = 2 s 3 − e − 2 s 2 s 3 − e − 2 s 1 s 2 + 2 e − 2 s 1 s 24

 x + 1 , 0 ≤ x < 3  3. f ( x ) = sin πx, x ≥ 3  Step 1. Re-write f in terms of u ( x − 3): f ( x ) = x + 1 − ( x + 1) u ( x − 3) + sin( πx ) u ( x − 3) 25

Step 2. Write the coefficients in terms of x − 3: f ( x ) = x + 1 − 4 u ( x − 3) − ( x − 3) u ( x − 3) − sin[ π ( x − 3)] u ( x − 3) 26

Step 3. Determine L [ f ]: F ( s ) = 1 s 2 + 1 s − 4 e − 3 s 1 s − e − 3 s 1 π s 2 − e − 3 s s 2 + π 2 27

 1 0 ≤ x < π/ 2  4. f ( x ) = sin x π/ 2 ≤ x < π 2 cos x x ≥ π  Step 1. Re-write in terms of u ( x − π/ 2) and u ( x − π ) : f 28

x − π Step 2. Write coefficients in terms of and x − π : 2 29

(continued) f ( x ) = 1 − u ( x − π 2 ) + cos( x − π 2 ) u ( x − π 2 ) + sin( x − π ) u ( x − π ) − 2 cos( x − π ) u ( x − π ) 30

Step 3. Determine L [ f ]: F ( s ) = 1 s − e − πs/ 2 1 s 1 s s + e − πs/ 2 s 2 + 1 + e − πs s 2 + 1 − 2 e − πs s 2 + 1 31

Section 4.6. Inverse Transforms & Piecewise Continuous Functions: Recall Property V : Suppose that L [ f ( x )] = F ( s ). Then, for any positive number c , L [ f ( x − c ) u ( x − c )] = e − cs F ( s ). 1. L − 1 [ e − cs F ( s )] = f ( x − c ) u ( x − c ). 2. 32

Examples: Given F ( s ), find f ( x ): F ( s ) = 3 s − 2 e − 2 s 1 1 s + e − 2 s 1. s − 2 33

Answer: f ( x ) = 3 − 2 u ( x − 2) + e 2( x − 2) u ( x − 2)  3 , 0 ≤ x < 2  = 1 + e 2( x − 2) , x ≥ 2  34

F ( s ) = 2 1 s + 4 e − 3 s 2. s ( s + 2) 35

Answer: f ( x ) = 2 + 2 u ( x − 3) − 2 e − 2( x − 3) u ( x − 3)  2 , 0 ≤ x < 3   = 4 − 2 e 6 e − 2 x , x ≥ 3   36

F ( s ) = 1 − e − πs 3. s ( s 2 + 4) 37

Answer: f ( x ) = 1 4 − 1 4 cos 2 x − 1 4 u ( x − π )+ 1 4 cos(2[ x − π ]) u ( x − π ) = 1 4 − 1 4 cos 2 x − 1 4 u ( x − π ) + 1 4 cos 2 x u ( x − π )  4 − 1 1 4 cos 2 x, 0 ≤ x < π   = . 0 , x ≥ π   38

4. F ( s ) = 4 s + 2 s 2 +3 e − 2 s 1 s − 2 e − 2 s 1 s 2 − 5 e − 4 s 1 1 s 2 + e − 4 s s − 2 39

 4 + 2 x, 0 ≤ x < 2      3 + 4 x, 2 ≤ x < 4 Answer: f ( x ) =   23 − x + e 2( x − 4) ,  x ≥ 4   40

Section 4.7. Application to Initial-Value Prob- lems with piecewise continuous RHS Examples: 1. Use the Laplace transform method to solve the initial-value problem y ′ + 2 y = f ( x ) , y (0) = 1 .  x 0 ≤ x < 3  where f ( x ) = 1 x ≥ 3  41

Laplace transform of the solution: e − 3 s e − 3 s 1 1 Y ( s ) = s 2 ( s + 2) − 2 s ( s + 2) + s 2 ( s + 2) − s + 2 The solution: − 1 4 + 1 2 x + 5  4 e − 2 x , 0 ≤ x < 3       y = 1 2 + 5 4 e − 2 x + 3  4 e − 2( x − 3) ,  x ≥ 3 .     42

y 2 1 x 1 2 3 4 5 6 38

2. Use the Laplace transform method to solve the initial-value problem y ′′ + 2 y ′ + y = f ( x ) , y (0) = y ′ (0) = 0 .  1 0 ≤ x < 2  where f ( x ) = x + 1 x ≥ 2  Laplace transform of the solution: 2 e − 2 s e − 2 s 1 Y ( s ) = s ( s + 1) 2 + s ( s + 1) 2 + s 2 ( s + 1) 2 44

The solution: y = 1 − ( x + 1) e − x ,  0 ≤ x < 2   x − 1 − ( x + 1) e − x − ( x − 2) e − ( x − 2) , x ≥ 2   45

y 3 2 1 x 2 41