LN_23_F19(6).pdf - LN_23_F19 Step functions[Section 6.3 Discontinuous forcing functions[Section 6.4 \u2022 \u2022 \u2022 Jump discontinuities do occur in

LN_23_F19(6).pdf - LN_23_F19 Step functions[Section 6.3...

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Unformatted text preview: LN_23_F19: Step functions [Section 6.3]. Discontinuous forcing functions [Section 6.4]. • • • Jump discontinuities do occur in physical problems such as electric circuits with on/off switches. How to solve differential equations that have functions with jump discontinuities? For this, first we need to see how Olivier Heaviside has introduced the following unit step function. Heaviside function The graph on the right is The unit step function or Heaviside function () is: 0, < the graph of 2 () where, () = ( − ) = ( − ) = { 0, < 2 1, ≥ 2 () = { 1, ≥ 2 Note: As we mentioned above, some books denote () as ( − ) or as ( − ) Since the Laplace transform involves the values of in [0, ∞), we are interested only in the non-negative value of . So, for the Laplace transform purpose, we can define the unit step function () only on the nonnegative − as () = ( − ) = { 0, 1, ≤ < ≥ So, the smallest () value that we will calculate is the 0 () value which means 0 () = 1 for ≥ 0. Example 1: Express the given piecewise-defined function () in terms of the unit step function. () • We know that in the interval < , () has the expression () = 1 . 2, ≤ < ➢ So, in the interval [0,4), we have, () = 2. 5, ≤ < • We know that in the interval [, ), = { −1, ≤ < ➢ () experiences a jump (2 − 1 ). 1, ≥ ➢ Also, we know that 4 () = 0 for < 4, but, () = ≥ . That is , we have ➢ So, for the interval [0,7), we have: () () = 2 + (2 − 1 )4 () = + () 1 () ≤ < • We know that in the interval [, ), 2 () ≤ < ➢ () experiences a jump (3 − 2 ). = ➢ Also, we know that 7 () = 0 for < 7, but, () = ≥ . 3 () ≤ < ➢ So, for the interval [0,9), we have: ≥ {4 () () = 2 + 3 4 () + (3 − 2 )7 () = 2 + 3 4 () − () Make a graph below: • We know that in the interval [, ∞), ➢ () experiences a jump (4 − 3 ). ➢ Also, we know that 9 () = 0 for < 9, but, () = ≥ . ➢ So, for the interval [0, ∞), we have: () = 2 + 3 4 () − 6 7 () + (4 − 3 )9 () = + () − () + () Now, apply above example to construct corresponding step function for a given piecewise-defined function. Piecewise-defined function Corresponding unit step function () 1 () 2 () = 3 () {4 () ≤ < ≤ < ≤ < ≥ () = + [ − ] + [ − ] + [ − ] ❖ Note #1: From definition of 0 (), we know 0 () = 1 ≥ 0. ❖ Note #2: If the 1st interval is given as < , it can be expressed as ≤ < . So, () = + [ − ] + [ − ] + [ − ] 1 0 ≤ < 1 < 1 or as () = { 1 2 ≥ 1 2 ≥ 1 ℎ : () = + ( − ) () So, for a piecewise defined function () () = { 1 Laplace Transform of discontinuous functions Below are the Laplace transform formulas needed for Heaviside function. The proofs are in the last page. e−cs ℒ[uc (t)f(t − c)] = e−cs ℒ[f(t)] = e−cs F(s), where, c > 0 ℒ[uc (t)] = where, c ≥ 0, s • Be careful. The formula ℒ[ () ( − )] = − () is very tricky to use. • Please follow my lecture very carefully while I will be doing the related examples. Example #2: Find the Laplace transform of the following function. ≤ < 1 0 ≤ < 1 , () = { , This function is in the form () = { 2 ≥ 1 + ( − ) ≥ For a piecewise defined function expressed in the Example #2, we can find the unit step function as below: 0 ≤ < 1 < 1 For () = { 1 , or, for () = { 1 we have: () = + ( − ) () 2 ≥ 1 2 ≥ 1 Then, () = sin + { cos ( − )} () 4 So, 1 ℒ[()] = ℒ [sin + { cos ( − 4 )} ()] = ℒ[sin )] + ℒ [ () { cos ( − 4 )}] = s2 +1 + ℒ [ () ( − 4 )] where, we let, ( − ) = cos ( − ) so that () = cos() , and then, 4 4 ℒ[()] = F(s) = s2 s +1 Then, ℒ[()] = 1 1 1 s − s − 4 F(s) = 4 { () ( − )] = + ℒ + e + e [ } s2 + 1 4 s2 + 1 s2 + 1 s2 + 1 Example 3: If () = − [ + +] find, () = − [()] • • • Note that − above suggests that we should use the formula given by converse of above theorem: − [− ()] = () ( − ) Then, () = − [()] = () = − {− [ + } = − {− ()} = () ( − ) + What is the expression ( − ) ? For that, first find the expression for () using () = − [()] () = − [()] = () = − [ + = 3− [ ] + − [ = {3 (1)} + cos 2 + + Thus, ( − ) = 3 + cos 2( − 2) So, () = () ( − ) = () [3 + cos 2( − 2) ] 2 Example 4: Solve ′′ + ′ + = (), () = , ′ () = , where, 1, 0≤<1 () = { 0, ≥1 Here, ′′ + ′ + = (), (0) = , ′ (0) = , we get = 1, = 3, = 2, = 0 , = 1, = 0, = 0, = 1 Now, we need to apply the formula () + + + [()] + () = − [()] = − { ℎ, () = [()] } = − { }, ( + + ) ( + + ) • We need to find [()]. For that, first express () in terms of unit step function like below: We know that a piecewise defined function (), 0 ≤ < < expressed as () = { 1 , or, expressed as () = { 1 2 ≥ 2 ≥ can be expressed in terms of the unit step function as () = + ( − ) () Then, here, in this problem, we have () = + ( − ) () ⟹ () = − () 1 −1 − − − [()] = [ − ()] = [] − [ ()] = − [, ℎ ℒ[ () ] == • Now that we just found the expression for [()], substitute it back to the above expression for (). − − +1 ℒ[()] + 1 − − + ( + ) − − () = ℒ −1 { 2 } = ℒ −1 { } = ℒ −1 { } = ℒ −1 { } ( + 3 + 2) ( + + ) ( + + ) ( + )( + ) Then, − ( + ) − () = ℒ −1 { } − ℒ −1 { } = ℒ −1 { } − ℒ −1 { } ( + )( + ) ( + )( + ) ( + ) ( + )( + ) So we get, 1 () = − {()} − − {− ()}, ℎ, () = , , () = ( + 2) ( + )( + ) − {()}: Finding First, find the partial fraction decomposition of (). We get: 1 1 1 1 1 () = = ( )− ( ) ( + 2) 2 2 +2 So, 1 1 1 1 1 1 1 1 1 1 − {()} = ℒ −1 ( ) − ℒ −1 ( ) = ℒ −1 ( ) − ℒ −1 ( ) = − −2 2 2 +2 2 2 − (−2) 2 2 − {− Finding ()} : − {− ()} = () ( − ) [: , ℎ − [− ()] = () ( − ) ] In order to find the expressions for ( − ), first, we will find (), and, then, we will perform the shifting. 1 1 1 1 1 1 1 () = − [ ()] = − [ = ℒ −1 [ ] + ℒ −1 [ − ℒ −1 [ = + −2 − − . ( + )( + ) 2 2 +2 +1 2 2 ( ℎ, , ℎ ℎ ) 1 1 1 1 1 = ( )+ ( )− ( ) 2 +2 +1 ( + )( + ) 2 Then, 1 1 − {− ()} = () ( − ) = () [ + −2(−1) − −(−1) ] 2 2 Thus, finally 1 1 1 1 () = − {()} − − {− ()} = [ − −2 ] − () [ + −2(−1) − −(−1) ] 2 2 2 2 3 Example 5: Solve ′′ + = (), () = , ′ () = , 0, 0 ≤ < 5, −5 () = { , 5 ≤ < 10, 5 1, ≥ 10, where, () Answer: Step 1. Comparing with ′′ + ′ + = (), () = , ′ () = , we get = , = , = , = , and = Step 2. Now, we need to apply the formula () + + + () + 0 + 0 + 0 ℒ[()] () = ℒ −1 [()] = ℒ −1 { } = ℒ −1 { } = ℒ −1 { 2 } . Here, () = ℒ[()] 2 2 ( + + ) ( + 0 + 4) ( + 4) • We need find [()]. For that, first, express () in terms of unit step function. We know () expressed in the form 1 () ≤ < , () ≤ < , () = { 2 3 () ≥ , can be expressed in terms of unit step function as: () = + [ − ] + [ − ] So, here we have −5 −5 −5 5−+5 () = + [ { } − {}] + [ − { }] = [ { }] + [ { }] 5 5 5 5 which, when further simplified, reduces to: () = [ ()( − ) − () ( − )] [{ () ( − )} − { () ( − )}] Using the formula: ℒ[ () ( − )] = − () = − [()] on each of the terms above, we get: − − [()] = [− {} − − {}] = [(− ) ( ) − (− ) ( )] = [ − ] Now that we just found the expression for [()] substitute it back to the above expression for (). 1 −5 −10 − 2 ] 1 −1 −5 −10 5 [ 2 () = ℒ −1 [ = ℒ [ − = − [− () − − ()] 2 2 2 2 2 ( + 4) 5 ( + 4) ( + 4) then, • • [()] = where, () = 1 2 ( 2 + 4) Then, using the formula, ()] = () ( − ), we get the solution as, − − () = [ { ()} − − {− ()}] = [ () ( − ) − () ( − )] • Now, all that is left is to find the expressions for ( − ) and ( − ). For that first, find (), and perform the shift. () = − [ ()] = − [ ( + ) The partial fraction decomposition gives: = ( ) − ( ) . Thus, ( + ) + 1 1 1 1 1 1 1 1 2 () = ℒ −1 [ 2 2 = ℒ −1 [ 2 ] − ℒ −1 [ 2 = ℒ −1 [ 2 ] − ℒ −1 [ 2 = − 2 ( + 4) 4 4 +4 4 8 +2 So, on shifting, we get, ( − ) = ( − ) − ( − ) and ( − ) = ( − ) − ( − ) • Now, plug the expression for ( − ) and ( − ) into the solution below. Then, you will be done ! () = [ () ( − ) − () ( − )] − [− 4 Theorem: If () = ℒ[()] exists for > ≥ 0, and if is a positive constant, then [ () ( − )] = − () Converse of the above “Theorem”: If () = ℒ −1 [()], then () ( − ) = − [− ()] • Derive the formula: ℒ[ () ( − )] = − () > ∞ ∞ ℒ[ () ( − )] = ∫0 − { () ( − )} = ∫0 − { () ( − )} + ∫ − { () ( − )} ∞ = ∫ − {(0) ( − )} + ∫ − {(1) ( − )} 0 ∞ = ∫ −() ( − ) • • • Note that the way the unit step function () is defined, () = 0 for < , and () = 1 for ≥ . Make change of variable = − ⟹ + = so = . If = ⟹ = 0. If = ∞ ⟹ = ∞. Then, the above result reduces to: ∞ ∞ ℒ[ () ( − )] = ∫ −(+) () = − ∫ − () = − () 0 • • 0 For a non-negative , that is for ≥ 0, we have the following important formula. − [ ()] = , >0 We can derive the above formula easily: ∞ ∞ ∞ [ ()] = ∫ − { () } = ∫ − { () } + ∫ − { () } = ∫ − {0 } + ∫ − {1 } 0 0 ∞ 0 = ∫ − {1 } = lim {∫ − →∞ =∞ − − } = lim { | > 0 }= →∞ − = Important note: Some books denote () as ( − ) or as ( − ). Please do not get confused if you see such notations. Important note: Please practice the Example #1 from the Text-Book Section 6.4 as well. There, we see ( + + ) Note that (2 2 + + 2) is a nonreducible quadratic factor (because if we let (2 2 + + 2) = 0, then we get complex conjugate roots) After the partial fraction decomposition and simplification, we get 1 1 1 + + + 1 1 1 1 2 ] = 1 (1) − 1 [ 4 4 = ( ) − [ (2 2 + + 2) 2 2 2 + + 1 2 2 2 + 2. . 1 + 1 + 15 2 4 16 16 which on further simplification gives () = 1 1 1 ( )− 2 2 1 ( + ) 4 2 1 2 √15 ( + ) + ( ) 4 4 [{ } () = + 1 4 2 1 2 √15 ( + ) + ( ) 4 4 { }] 1 1 1 = ( )− ( + + ) 2 2 1 1 1 = ( )− 2 2 1 ( − (− )) 4 2 so that 1 ( − (− )) 4 2 2 1 √15 ( − (− )) + ( ) 4 4 [{ } 2 + √ + √ 4 1 √ 2 2 1 √15 ( − (− )) + ( ) 4 4 { }] √ 4 2 2 1 1 √15 √15 ( − (− )) + ( ) ( − (− )) + ( ) 4 4 4 4 [{ } { }] To get () apply the inverse Laplace Transform formulas related to formula #7 and #9 from Lecture Note 22. 5 Example #6: Find the Laplace transform of the following function. 0 0 ≤ < 1, () = { 2 1 ≤ < 3, 0 ≥ 3, , 1 () = 0, 2 () = t 2 , and 3 () = 0 • • First, express () in terms of unit step function as () = 1 + [ 2 − 1 ]1 + [ 3 − 2 ]2 so, here we have () = 0 + [ { 2 } − {0}]1 () + [ 0 − { 2 }]3 () ⟹ () = 1 (){ 2 } − 3 (){ 2 } Then, apply the Laplace transform: 2 2 ℒ[()] = ℒ [1 (){ }] − ℒ [3 (){ }] ⟹ [()] = [ () ( − )] − [ () ( − )] where, we make the expression ( − ) = , and ( − ) = [Why did we have to such expressions? Please think about it.] The answer is: After expressing in such a manner, we will be able to make use of the formula • [ () ( − )] = − [()] Then, ℒ[()] = ℒ[1 ()( − 1)] − ℒ[3 ()ℎ( − 3)] ⟹ [()] = − [()] − − [()] • • Before we attempt to calculate [()] and [()] , we must figure out the expressions for () and (). Although we do not have expression s for () and (), we can get such expressions by shifting the already available expressions for ( − ) and ( − ). Look below: Add 1 to t here After applying shifting on ( − 1) = 2 , , () = ( + 1)2 . Therefore, () = + + Add 1 to t here • • • • • • • • Similarly, after the shifting on ℎ( − 3) = 2 , we get, () = ( + 3)2 . Therefore, () = + + Add 3 Then, [()] = e−s ℒ[()] − e−3s ℒ[()] gives 2 1 1 2 6 9 [()] = e−s ℒ[ + + ] − e−3s ℒ[ + + ] = e−s [ 3 + 2 + ] − e−3s [ 3 + 2 + ] Notes: Alternatively, to find [ () ], you could express it as [ ()( − )], where, ( − ) = . Then, let, ( − ) = so that = + so we get ( − ) = converted to () = ( + ) so that () = + + . Since is a dummy variable so you can now have () = + + also to be expressed as () = + + . Now, we can make use of the formula ℒ[ () ( − )] = − () = − ℒ[()] to get 2 1 1 [ () ] = [ ()( − )] = − ℒ[()] = − ℒ[ + + ] = e−s [ 3 + 2 + ] Using the similar procedure, you can find [ () ]. Look below: Alternatively, to find [ () ], you could express it as [ ()( − )], where, ( − ) = . Then, let, ( − ) = so that = + so we get ( − ) = converted to () = ( + ) so that () = + + . Since is a dummy variable so you can now have () = + + also to be expressed as () = + + . Now, we can make use of the formula ℒ[ () ( − )] = − () = − ℒ[()] to get 2 6 9 [ () ] = [ ()( − )] = −3 ℒ[()] = −3 ℒ[ + + ] = e−3s [ 3 + 2 + ] 6 ...
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