Chapter 4 Laplace Tranform 4 Introduction Reading aignment: In thi chapter we will cover Section 4 45 4 Definition and the Laplace tranform of imple function Given f, a function of time, with value f(t at time t, the Laplace tranform of f which i denoted by L(f (or F i defined by L(f( = F ( = e t f(t dt > ( Example 4 The Laplace tranform i linear: L(αf(t+βg(t = αl(f(t+βl(g(t e t (αf(t + βg(t dt = α e t f(t dt + β e t g(t dt 2 f(t = F ( = e t dt = e t = 3 f(t = e at for > a F ( = e t e at dt = e ( at dt = ( a e ( at = ( a 4 For a poitive integer n, f(t = t n ( F ( = L(t n = e t t n dt = e t t n dt ( ( n = e t t n e t t n dt ( n ( n = e t t n dt = L(t n
( n So we find that L(t n = L(t n We can now ue thi formula over and over to ucceively reduce the power of t to obtain L(t n = ( n L(t n = ( n(n 2 L(t n 2 = = ( n! n ( n! L( = n+ 5 f(t = co(at To compute the Laplace tranform we will ue the Euler formula decribed in the note for Chapter 3 e iθ = co(θ + i in(θ (2 which implie that co(θ = eiθ + e iθ 2 Alo, uing i 2 = we can write ( + ib( ib = 2 (ib 2 = 2 + b 2 Combing the above we can write ( e ibt + e ibt L(co(bt =L 2 = 2 = 2 ( ib + + ib ( ( + ib ( ib + ( 2 + b 2 ( 2 + b 2 = ( ( + ib + ( ib = 2 ( 2 + b 2 ( 2 + b 2 So we arrive at L(co(bt = ( 2 + b 2 2
6 Given a function f(t we can find L(f (t by applying integration by part a follow L(f (t = e t f (t dt = f(te t ( e t f(t dt = L(f(t f( or L(f (t = L(f(t f( 7 Given a function f(t find L(f (t can be eaily computed by uing the previou formula L(f (t = L(f (t f ( = ( L(f(t f( f ( = 2 L(f(t f( f ( So we have L(f (t = 2 L(f(t f( f ( 8 To compute the Laplace tranform of f(t = in(bt we will ue two of the previou formula L(in(bt = b L(co(bt = [L(co(bt co(] b = [ ( ] b ( 2 + b 2 = [ ] 2 b ( 2 + b 2 = [ ] 2 ( 2 + b 2 b ( 2 + b 2 b = ( 2 + b 2 Therefore L(in(bt = b ( 2 + b 2 Let u conider a few example of finding Laplace tranform Example 42 L(2t 4 = 2 4! 5 = 48 5 3
2 L(t 2 + 6t 3 = L(t 2 + 6L(t 3L( = 2 3 + 6 2 3 3 L(2 co(3t + 3 in(2t 3e 7t = 2L(co(3t + 3L(in(2t 6L(e 7t = 2 2 + 9 + 6 2 + 4 6 ( + 7 4 L(2e t + 6e 3t = 2 ( + + 6 ( 3 Example 43 Find the Laplace tranform of f(t = ( + e 2t 2 To do thi we firt note that f(t = + 2e 2t + e 4t o we have L(f(t = L( + 2e 2t + e 4t = + 2 ( 2 + ( 4 Example 44 Find the Laplace tranform of f(t = (co(t + in(t 2 To do thi we firt note that o we have f(t = + 2 in(t co(t = + in(2t L(f(t = L( + in(2t = + 2 ( 2 + 4 In order to do the next example we need one of the addition formula from trig in(α ± β = in(α co(β ± in(β co(α co(α ± β = co(α co(β in(α in(β Example 45 Find the Laplace tranform of f(t = in(t + π/2 f(t = in(t co(π/2 + in(π/2 co(t = co(t Therefore L(f(t = L(co(t = ( 2 + 42 The Invere Laplace Tranform Given a function f(t the operation of taking the Laplace tranform i denoted by L(f(t = F ( and the invere proce i denoted by L (F ( = f(t The proce of computing 4
the Laplace tranform of a function turn out to be more challenging than mot tudent like It involve lot of algebra and uing a table of Laplace tranform backward For example, if we were aked to find L (3/ 3 we would write L (3/ 3 = 3 2 L (2/ 3 = 3 2 t2 ince we know that L(t 2 = 2/ 3 and we can adjut the contant to work out Mot generally thi proce will require the ue of the method of partial fraction Partial Fraction Thee note are concerned with decompoing rational function P ( Q( = a M M + a M M + + a + a N + b N N + + b + b Note: We can (without lo of generality aume that the coefficient of N in the denominator i Alo in our intended application we will alway have M < N By the fundamental theorem of algebra we know that the denominator factor into a product of power of linear and quadratic term where the quadratic term correpond to complex root Namely, it can be written in the form ( r m ( r k m k ( 2 2α + α 2 + β 2 p ( 2 2α l + αl 2 + βl 2 p l, k l where m j + 2 p j = n j= j= The proce referred to a Partial Fraction i a method to reduce a complex rational function into a um of much impler term of the form c ( r j or c + d ( 2 2α + α 2 + β 2 j The mot important point i to learn how to deal with certain type of term that can appear But firt there i a pecial cae that arie and i worth a pecial attention Thi i the cae of non-repeated linear term 5
I Nonrepeated Linear Factor If Q( = ( r ( r 2 ( r n and r i r j for i j P ( Q( = A ( r + A 2 ( r 2 + + A n ( r n II Repeated Linear Factor If Q( contain a factor of the form ( r m then you mut have the following term A ( r + A 2 ( r 2 + + A m ( r m III A Nonrepeated Quadratic Factor If Q( contain a factor of the form ( 2 2α + α 2 + β 2 = ( α 2 + β 2 then you mut have the following term A + B ( 2 2α + α 2 + β 2 IV Repeated Quadratic Factor If Q( contain a factor of the form ( 2 2α + α 2 + β 2 m then you mut have the following term A + B ( 2 2α + α 2 + β 2 + A 2 + B 2 ( 2 2α + α 2 + β 2 2 + + A m + B m ( 2 2α + α 2 + β 2 m Let conider ome example of computing invere Laplace tranform: ( Example 46 Find L = ( 2! 3 2! L = 2 t2 ( ( 2 Find L = L 2 = t 2 ( ( (3 + 7 3 Find L = 3L 2 + 6 2 + 4 + 7 4 = 3 co(4t + 7 2 4 2 + 4 2 4 in(4t 6 3
( ( + Example 47 Find L 3 We need to firt expand the numerator to get So we have ( + 3 4 4 = 3 + 3 2 + 3 + 4 = + 3 2 + 3 3 + 4 ( ( ( + L 3 = L + 3 + 3 + = + 3t + 3 4 2 3 4 2 t2 + 6 t3 ( Example 48 Find L We have (4 + ( L = 4 ( (4 + L = ( + /4 4 e t/4 ( 2 6 Example 49 Find L For thi example we have ( 2 + 9 ( ( 2 6 L = 2L ( 2 + 9 ( 2 + 9 2L ( 3 ( 2 + 9 = 2 co(3t 2 in(3t ( Example 4 Find L 4 For thi example we firt note that ( 2 +2 3 = ( 2 + 2 3 ( + 3( o we have ( ( L 4 = L ( 2 + 2 3 4 ( + 3( In order to carry out thi invere Laplace tranform we mut ue partial fraction Since the denominator conit of non-repeated term we can ue the firt box (in the formula for partial fraction to write 4 ( 2 + 2 3 = A ( + 3 + B ( To find A and B we can ue the cover-up method to find that A = 3 and B = o that ( L 4 ( 2 + 2 3 ( ( 3 = L + L = 3e 3t + e t ( + 3 ( 7
( Example 4 Find L 2 for thi we need to do partial fraction ( 2( 3( 6 Namely we have 2 ( 2( 3( 6 = A ( 2 + B ( 3 + C ( 6 = ( 2 + 2 ( 3 + ( 6 So f(t = e 2t 2e 3t + e 6t ( 5 Example 42 Find L for thi we need to do partial fraction Namely we 3 + 5 have 5 3 + 5 = A + B + C 2 + 5 = + 2 + 5 So f(t = co( 5t ( Example 43 Find L (2 4 for thi we need to do partial fraction Namely ( + ( 2 + we have (2 4 ( + ( 2 + = A + B ( + + C + D ( 2 + = A( + (2 + + B( 2 + + (C + D( + ( + ( 2 + = (A + B + C3 + (A + C + D 2 + (A + B + D + (A ( + ( 2 + A + B + C = A + C + D = A + B + D = 2 A = 4 Uing A = 4 thee equation become B + C = 4 C + D = 4 B + D = 6 8
Taking time the firt equation added to the third give C + D = 4 C + D = 2 and adding thee together give 2D = 6 o that D = 3 But then we have C = 4 D = So we have A = 4, C = and D = 3 o we can finally find B from the firt equation a B = A C = 4 = 3 Thu we have o f(t = 4 + 3e t + co(t + 3 in(t 4 + 3 ( + + + 3 ( 2 + Now let olve an initial value problem uing the formula for the Laplace tranform of a derivative In thee example we will ue the horthand notation Y = L(y Example 44 Find the olution of y y = with y( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y L(y = L( (Y Y = Y = ( So to find y we need to take the invere Laplace tranform which require we do partial fraction ( y = L (Y = L ( ( ( = L + L = + e t ( Example 45 Find the olution of y y = with y( =, y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get 2 Y Y = (2 Y = + = + Y = + ( 2 = ( So to find y we need to take the invere Laplace tranform which require we do partial 9
fraction ( y = L (Y = L ( ( ( = L + L = + e t ( Example 46 Find the olution of y + 5y + 4y = with y( = 3 and y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y + 5L(y + 4L(y = ( 2 Y 3 + 5(Y 3 + 4Y = Solving for Y we get Y = 3( + 5 ( 2 + 5 + 4 Y = 3( + 5 ( + 4( + So to find y we need to take the invere Laplace tranform which require we do partial fraction ( 3( + 5 y = L (Y = L ( + 4( ( ( 4 = L + L ( + 4 ( + = e 4t + 4e t Example 47 Find the olution of y + 5y + 6y = 6 with y( = and y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y + 5L(y + 6L(y = L(6 ( 2 Y + 5(Y + 6Y = 6 Solving for Y we get Y = 2 + 6 + 6 ( 2 + 5 + 6 Y = 2 + 6 + 6 ( + 2( + 3 So to find y we need to take the invere Laplace tranform which require we do partial fraction Y = 2 + 6 + 6 ( + 2( + 3 = + ( + 2 ( + 3
y = L (Y = + e 2t e 3t Example 48 Find the olution of y + y = 2 in( 2t with y( = and y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y + L(y = L( 2 in( 2t ( 2 Y + Y = 2 2 + 2 ( 2 + Y = 2 2 + 2 + = 3 + 2 + 2 ( 2 + 2 Y = 3 + 2 + 2 ( 2 + 2( 2 + So to find y we need to take the invere Laplace tranform which require we do partial fraction (thi time I give the anwer and let you do the work Y = 3 + 2 + 2 ( 2 + 2( 2 + = + 2 ( 2 + 2 ( 2 + 2 y = co(t + 2 in(t 2 in( 2t 43 The Shift Theorem Two of the mot important and ueful reult we need to dicu are the firt and econd hift (or Tranlation theorem Theorem 4 (Firt Shift Theorem If L(f(t = F ( then L(e at f(t = F ( a for any real number a Proof L(e at f(t = e t e at f(t dt = e ( at f(t dt = F ( a Let conider ome example Example 49 Example 42 L(t 2 e 3t = L(t 2 3 = L(e t in(2t = L(in(2t + = 2 ( 3 3 2 ( + 2 + 4
Example 42 Example 422 L(e 3t (t + 2 = L(t 2 + 2t + 3 = 2 ( 3 + 2 3 ( 3 + 2 ( 3 L(t(e t + e 2t 2 = L(t(e 2t + 2e 3t + e 4t = L(te 2t + 2L(te 3t + L(te 4t = ( 2 2 + 2 ( 3 2 + ( 4 2 Next we conider ome example of invere Laplace tranform uing the hift theorem ( Example 423 Find L We note that the denominator doe not factor 2 + 2 + 5 ince the dicriminant i (2 2 4((5 < o we know the reult will involve ine and coine We proceed by completing the quare in the denominator ( L = 2 ( 2 2 + 2 + 5 L ( + 2 + 2 2 = ( 2 2 e t L = 2 + 2 2 2 e t in(2t ( 2 + 5 Example 424 Find L Once again we note that the denominator doe 2 + 6 + 34 not factor ince the dicriminant i (6 2 4((34 < o we know the reult will involve ine and coine We proceed by completing the quare in the denominator ( ( 2 + 5 L = L 2 + 6 + 34 2 + 5 ( + 3 2 + 5 2 So we ee that the denominator ha a hifted value but the numerator doe not To fix thi we write So we have 2 + 5 = 2[( + 3 3] + 5 = 2( + 3 ( ( 2 + 5 2( + 3 L = L 2 + 6 + 34 ( + 3 2 + 5 2 2
( ( + 3 = 2L 5 ( 5 ( + 3 2 + 5 2 L ( + 3 2 + 5 ( 2 = 2e 3t L ( 5 2 + 5 2 5 e 3t L 2 + 5 2 2e 3t co(5t 5 e 3t in(5t The Second Shift (Tranlation Theorem Conider finding the following Laplace tranform e 2t for t < f(t = for t The only way to do thi up to thi point i to apply the definition L(f = = e t f(t dt = e t e 2t dt e ( (+2t dt = e( (+2t ( + 2 = e (+2 ( + 2 We will ow introduce one of the mot important tool in thi block of material that give a much impler way to do the above problem Namely we will introduce the econd hift theorem In order to preent the econd hift theorem we firt need to dicu the unit tep function (or Heaviide function t < u(t = t or the hifted unit tep function t < a u(t a = t a 3
With thi we can tate the econd hift theorem Theorem 42 (Second Shift Theorem If L(f(t = F ( then L(u(t af(t a = e a F ( for any a > Proof L(u(t af(t a = ( et τ = t a = e t u(t af(t a dt = a e t f(t a dt e (τ+a f(τ dτ = e a e τ f(τ dτ =e a e t f(τ dt = e a F ( A very ueful modification of the econd hift theorem how how to take the Laplace tranform of a unit tep function time a function that i not hifted Theorem 43 (Modified Second Shift Theorem L(u(t ag(t = e a L(g(t + a for any a > Proof Let f(t = g(t + a then we have f(t a = g(t and we can ue the econd hift theorem to write L(u(t ag(t = L(u(t af(t a = e a L(f(t = e a L(g(t + a One of the mot important thing you need to come to grip with i the ue of the unit tep function to expre a function defined in piece Let me give a very generic example Example 425 Suppoe you want to take the Laplace tranform of a general function 4
given by f (t, t < t f(t = f 2 (t, t t < t 2 f 3 (t, t t 2 The firt tep i to expre thi function uing the unit tep function a follow f(t = f (t + u(t t (f 2 (t f (t + u(t t 2 (f 3 (t f 2 (t Then you could proceed to compute the Laplace tranform by applying Theorem 43 Here i an example 2, t < 3 f(t = t, 3 t < 4 e t, t 4 The firt tep i to expre thi function uing the unit tep function a follow f(t = 2 + u(t 3(t 2 + u(t 4(e t t So we need to find L(f(t = L(2 + L(u(t 3(t 2 + L(u(t 4(e t t Let u do thee tranform one at a time L(2 = 2 2 For the econd term we ue Theorem 43 with g(t = t 2 and a = 3 which give g(t + 3 = t + o we have L(u(t 3(t 2 = e 3 L(t + = e 3 ( 2 + 3 For the third term we could ue Theorem 43 but it i eaier to break it up into two 5
problem We have L(u(t 4(e t t = L(u(t 4e t L(u(t 4t For the econd term we ue Theorem 43 to get L(u(t 4t = e 4 L(t + 4 = e 4 ( 2 + 4 where we have ued g(t = t and g(t + 4 = t + 4 For the remaining term we apply the firt hift theorem (Theorem 4 a follow L(u(t 4e t = L(u(t 4 = e 4( ( Finally then we have L(f = 2 ( + e 3 + ( e 4 2 + 4 + e 4( 2 ( Example 426 Find the Laplace tranform of t, t < f(t =, t Firt we write f(t = t u(t t and then compute L(f(t = L(t u(t t = L(t L(tu(t = 2 e L((t + = e [ L(t + L( ] = [ 2 2 e + ] 2 Example 427 Find the Laplace tranform of, t < π f(t = in(t, t π 6
Firt we write f(t = u(t π in(t and then compute where we have ued the addition formula L(f(t = L(u(t π in(t = e π L(in(t + π = e π L( in(t = e π ( 2 + in(α + β = in(α co(β + in(β co(α to get in(t + π = in(t co(π + in(π co(t = in(t Example 428 Find the Laplace tranform of in(2t, f(t = in(t, t < π t π Firt we write f(t = in(2t + u(t π(in(t in(2t and then for the firt term we compute L(in(2t = 2 2 + 4, and for the econd term we apply Theorem 43 Namely we et g(t = (in(t in(2t and we need g(t + π = (in(t + π in(2(t + π Applying the addition formula in(α + β = in(α co(β + in(β co(α we get and in(t + π = in(t co(π + in(π co(t = in(t in(2(t + π = in(2t + 2π = in(2t co(2π + in(2π co(2t = in(2t 7
So we have ( L(u(t π(in(t in(2t = e π L(in(t + in(2t = e π 2 + + 2 2 + 4 Finally then the anwer i f(t = 2 ( 2 + 4 e π 2 + + 2 2 + 4 Example 429 Find the Laplace tranform of 2, t < 3 f(t = 2, t 3 Firt we write f(t = 2 4u(t 3 and then compute L(f(t = 2 4L(u(t 3 = 2 4e 3 Example 43 Find the Laplace tranform of f(t = u(t e 2t t Ue the firt hift theorem to get rid of the e 2t and then apply the econd hift theorem L(f(t = L(u(t e 2t t = L(u(t t ( 2 = [ e L((t + ] [ ] ( 2 = e ( 2 ( 2 + 2 ( 2 ( e Example 43 Find L Since thi function of contain an exponential e a 2 + we know the invere Laplace tranform mut involve the econd hift theorem Namely we have ( ( e L = L e 2 + 2 + So by the econd hift theorem backward we have ( ( e L = u(t L 2 + 2 + (t 8
In order to continue we need to ue partial fraction to implify matter Combining thee reult we have ( ( L = L 2 + = e t ( + ( e L = u(t [ e (t ] 2 + ( e Example 432 Find L 5 Since thi function of contain an exponential e a 2 + 4 we know the invere Laplace tranform mut involve the econd hift theorem Namely we have ( ( e L 5 = L e 2 + 4 Now let olve an initial value problem = u(t 5 co(2(t 5 2 + 4 Example 433 Solve y y = 2u(t and y( = 4, y ( = 2 Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y L(y = 2 e ( 2 Y 4 4 Y = 2 e Solving for Y we get Y = 2e (4 + 2 + ( 2 ( 2 So to find y we need to take the invere Laplace tranform which require we do partial fraction But ince partial fraction only applie to rational function we need to do two eparate partial fraction Namely 2 ( ( + = A + B ( + C ( + We eaily find A = 2, B = and C = We alo need to do (4 + 2 ( ( + = D ( + E ( + Once again thi i very eay and we get D = 3 and E = So we have 9
( y = L (Y = L 2 e (4 + 2 + ( ( + ( 2 ( ( = L 2 e + L e ( + 3 ( + ( + = u(t [ e (t + e (t 2 ] + 3e t + e t + L ( e ( + Example 434 Solve y + 4y = 4u(t and y( = 4, y ( = 2 Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y + 4L(y = 4 e ( 2 Y 4 2 + 4Y = 4 e Solving for Y we get Y = 4 e ( 2 + 4 + 4 + 2 2 + 4 So to find y we need to take the invere Laplace tranform which require we do partial fraction on the term multiplying e Namely 4 ( 2 + 4 = A + B + C ( 2 + 4 We eaily find A =, B = and C = o we have ( 4 e y = L (Y = L ( 2 + 4 + +4 + 2 2 + 4 ] = L ( e [ + ( 2 + 4 + 4 + 2 2 + 4 = u(t [ + co(2(t ] + 4 co(2t + in(2t Example 435 Solve y 6y +9y = 27t and y( =, y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y 6L(y + 9L(y = 27 2 ( 2 Y 6(Y + 9Y = 27 2 2
Solving for Y we get omething that require a partial fraction with repeated term Y = 2 + 27 2 ( 3 = A 2 + B + C 2 ( 3 + D ( 3 2 After a bit of tediou work (which I ugget you do for practice we get 2 + 3 + 2 2 ( 3 + 4 ( 3 2 y = L (Y = 2 + 3t 2e 3t + 4te 3t Example 436 Solve y y = 2e t co(t and y( =, y ( = Firt we take the Laplace tranform of both ide (uing Y = L(y to get L(y L(y = 2( ( 2 + (2 Y = 2( ( 2 2 + 2 Solving for Y we get omething that require a partial fraction Namely 2( Y = ( ( 2 2 + 2 = 2 ( 2 2 + 2 = A + B + C 2 2 + 2 = A(2 2 + 2 + (B + C ( 2 2 + 2 = (A + B2 + ( 2A + C + 2A ( 2 2 + 2 From thi we ee that 2A = 2 o that A = Then 2A + C = o C = 2 And finally A + B = o B = and we have Y = 2 2 2 + 2 Next we need to find the invere Laplace tranform The firt term i very eay but for the econd we mut ue the Firt Shift Theorem ( Theorem 4 Firt we ue completing the quare to write the problem a ( ( 2 ( L = L 2 2 + 2 ( 2 + 2
Notice that and ( ( L = e t co(t ( 2 + ( L = e t in(t ( 2 + So after applying Theorem 4 we obtain y = e t co(t e t in(t Example 437 Next conider an example of computing an invere Laplace tranform uing Theorem 4 Find ( 2 + 5 L 2 + 6 + 34 To do thi we will apply the Firt Shift Theorem ( Theorem 4 after completing the quare in the denominator and uing the adding zero trick in the numerator Namely we have ( ( 2 + 5 L = L 2 + 5 2 + 6 + 34 ( + 3 2 + 5 ( 2 2[( + 3 3] + 5 = L ( + 3 2 + 5 ( 2 2( + 3 = L ( + 3 2 + 5 ( 2 2( + 3 = L 5 ( 5 ( + 3 2 + 5 2 L ( + 3 2 + 5 2 = 2e 3t co(5t 5 e 3t in(5t 44 Additional Operational Propertie In thi ection we cover everal very ueful theorem for Laplace tranform Theorem 44 If L(f(t = F ( then L(t n f(t = ( n dn d n F ( Proof L(t n f(t = e t t n f(t dt = te t [t (n f(t] dt = d d L(t(n f(t 22
Repeating thi calculation n time we arrive at L(t n f(t = d d L(t(n f(t = ( 2 d2 d 2 L(t(n 2 f(t = ( n dn d n L(t( f(t = ( n dn d n F ( Example 438 Find the Laplace tranform of f(t = t in(2t Applying Theorem 44 we have L(t in(2t = d ( 2 d ( 2 + 4 ( 4 = = ( 2 + 4 2 4 ( 2 + 4 2 Example 439 Find the Laplace tranform of f(t = te 2t co(t In thi cae we firt apply the firt hift theorem to write L(f(t = L(te 2t co(t = L(t co(t ( 2 So we next need to compute L(t co(t and for thi we apply Theorem 44 to obtain L(t co(t = d ( ( ( 2 + (2 = d ( 2 + ( 2 + 2 Finally then we arrive at L(f(t = ( 22 + (( 2 2 + 2 Example 44 Find the invere Laplace tranform f(t of F ( = ln Theorem 44 a follow: o we have df d = d (ln( + 2 ln( 5 = d + 2 5 = 2 + ( 2 + 2 ( + 2 We ue 5 L(tf(t = df ( d = + 2 5 23 tf(t = e 5t e 2t
or f(t = e5t e 2t t Example 44 Find the invere Laplace tranform f(t of F ( = ln Theorem 44 a follow: o we have df d = d (ln( + 2 ln( 5 = d + 2 5 ( + 2 We ue 5 L(tf(t = df ( d = + 2 5 tf(t = e 5t e 2t or f(t = e5t e 2t t Example 442 Find the invere Laplace tranform f(t of F ( = tan ( above theorem a follow: df d = d ( ( tan = d 2 + We ue the o we have L(tf(t = df ( d = 2 + tf(t = in(t or f(t = in(t t ( Theorem 45 If L(f(t = F ( then L f(τ dτ = F ( Proof Let g(t = f(τ dτ o that g (t = f(t L(f(t = L(g (t = L(g(t g( = L(g(t Therefore, dividing by we have ( L f(τ dτ = L(g(t = L(f(t = F ( 24
Example 443 Find the invere Laplace tranform f(t of F ( = previou reult and the fact that e 5t = L (/( 5 we have ( ( L = L ( 5 ( 5 ( = L L(e5t = = 5 [ e 5τ ] t = 5 [ e 5t ] e 5τ dτ Uing the ( 5 Convolution and their Application Definition 4 (Convolution Given two function f(t and g(t we define the convolution product by (f g(t = f(τg(t τ dτ (3 Theorem 46 (Two Important Propertie (f g(t = (g f(t, ie, Convolution multiplication i commutative 2 If F ( = L(f(t and G( = L(g(t, then L(f g( = F (G( Proof To how that (f g(t = (g f(t we begin with the definition (f g(t = f(τg(t τ dτ (Let w = t τ dw = dτ, τ = t w = = t f(t wg(w ( dw f(t wg(w dw = (g f(t 2 If F ( = L(f(t and G( = L(g(t, then we want to how that L(f g( = F (G( 25
We have ( F (G( = = ( e τ f(τ dτ e w g(w dw ( e (τ+w f(τg(w dτ dw = f(τ ( e (τ+w g(w dw dτ (Let t = τ + w dt = dw and w = t τ ( = f(τ e t g(t τ dt dτ τ ( Change the order of integration ( = e t f(τg(t τ dτ dt = L((f g(t Remark 4 The econd part of the theorem give a very imple proof of an earlier reult ( L f(τ dτ = L(( f(t = L(L(f(t = F ( Thi reult can be ueful in finding invere Laplace tranform ince ( F ( L = Example 444 Find the invere Laplace tranform of For the firt problem we have f(τ dτ ( ( L = L ( 2 + L(in(t = ( 2 + and 2 ( 2 + in(τ dτ = co(τ t = co(t And, uing the above we can olve the econd problem a follow ( ( L = L 2 ( 2 + L( co(τ 26
= ( co(τ dτ = (τ in(τ t = t in(t Example 445 Show that ( 2k L 3 = in(kt kt co(kt (4 ( 2 + k 2 2 For thi problem will ue one of the addition formula from trig: co(a ± b = co(a co(b in(a in(b which implie that in(a in(b = [co(a b co(a + b] 2 By our main property of the convolution we have ( ( ( 2k L 3 = 2k L k k ( 2 + k 2 2 ( 2 + k 2 ( 2 + k 2 = 2k (in(kt (in(kt = in(kτ in(k(t τ dτ = k [co(k(2τ t co(kt] dτ [ ] = k in(kt t co(t = in(kt kt co(kt k Example 446 Let u conider an example of finding a convolution by two different method The firt method will be to ue the definition of convolution Let f(t = t and g(t = e t and find h(t = (f g(t Recall the f g = g f o we have h(t = (f g(t = (t ve v dv = ve t v dv = e t ve v dv [ = e t v ( e v ] dv 27
[ t = e t ve v = e t [ te t + ( e v dv ] e v dv = e t [ te t ( e t ] = t + e t ] Next we compute the ame reult uing Laplace tranform We have h(t = (f g(t o that We ue partial fraction to find H( = F (G( = L(tL(e t = ( h(t = L 2 ( 2 ( ( A = L + B + C 2 ( We eaily find A =, B =, C = ( = L + + 2 ( = t + e t Example 447 Conider another example of finding a convolution Let f(t = in(t and g(t = co(t and find h(t = (f g(t One way to compute the reulting integral i to ue a trig identity in(a ± b = in(a co(b ± in(b co(a which, upon adding the two equation (one for plu and one for minu together, implie in(a co(b = 2[ in(a + b + in(a b ] So for thi example we et a = v and b = t v and in(v co(t v = 2[ in(t + in(2v t ] 28
So we have h(t = (f g(t = in(v co(t v dv [ ] in(t + in(2v t dv 2 = 2 t in(t + 2 in(2v t dv Let u = 2v t du = 2dv = 2 t in(t + 4 = 2 t in(t t in(u du ince the econd integral i If we try to compute the ame reult uing Laplace tranform We have h(t = (f g(t o that H( = F (G( = L(in(tL(co(t = ( 2 + 2 Thi i already in partial fraction form o partial fraction will not help The only way out here i to ue a formula in the book (Example, page 22, Cullen and Zill, Advanced Engineering math, 4th Ed L(t in(kt = In the preent cae we then have (with k = 2k ( 2 + k 2 2 ( h(t = L = 2 ( 2 ( 2 + 2 L = ( 2 + 2 2 t in(t Example 448 Let u conider an example olving an initial value problem y + y = 2 co(t, y( =, y ( = Applying Laplace tranform we get ( 2 + Y = 2 2 + 29
or Y = 2 ( 2 + 2 To olve the initial value problem we need to find the invere Laplace tranform ( y = L 2 ( 2 + 2 Thi problem i related to the above Example 447 Indeed we have ( y = L 2 = 2(in(t co(t ( 2 + 2 = 2 = = t in(t in(t τ co(τ dτ [in(t + in(t 2τ] dτ where on the third we have ued the addition formula in(a co(b = 2 [ in(a + b + in(a b ] with a = t τ and b = τ to write 2 in(t τ co(τ = in(t + in(t 2τ and on the lat tep we have a above in(t 2τ dτ = Volterra Integral Equation Laplace tranform method are particularly ueful in olving Volterra Integral equation An example of a Volterra equation i an equation in the form y(t = f(t + k(t τ y(τ dτ 3
Notice that the the integral term i a convolution, o if we apply the Laplace tranform to thi equation with Y ( = L(y, F ( = L(f and K( = L(k we have Y ( = F ( + K(Y ( which implie that Y ( = F ( K( So to find y(t we only need to take the invere Laplace tranform of the right hand ide of the above equation Example 449 Conider the equation y(t = t + Applying the Laplace tranform we have y(t τ in(τ dτ Y ( = ( + Y ( 2 2 + which implie Y ( = (2 + 4 = 2 4 Applying the invere Laplace tranform to both ide we get ( ( y(t = L L = t 2 4 6 t3 Example 45 Conider a mixed integro-differential equation y (t = Applying the Laplace tranform we have y(t τe 2τ dτ, y( = Y ( = ( Y ( + 2 3
which implie or ( + Y ( = + 2 + ( 2 + 2 + Y = + + 2 Solving for Y and noting that 2 + 2 + = ( + 2 we arrive at Y = ( + ( + 2 ( + 2 = ( + 2 ( + = 2 + So we have Y ( = 2 + Applying the invere Laplace tranform to both ide we get y(t = 2 e t Example 45 Conider the equation y(t = 2t Applying the Laplace tranform we have y(t τe τ dτ Y ( = 2 ( Y ( 2 which implie or ( + Y = 2 2 ( Y = 2 2 or Y ( = 2( 3 = 2 2 2 3 32
Applying the invere Laplace tranform to both ide we get y(t = 2t t 2 Example 452 Conider the equation y(t = 2t Applying the Laplace tranform we have y(t τe τ dτ Y ( = 2 ( Y ( 2 which implie or ( + Y = 2 2 ( Y = 2 2 or Y ( = 2( 3 = 2 2 2 3 Applying the invere Laplace tranform to both ide we get y(t = 2t t 2 Example 453 Conider the equation y(t + Applying the Laplace tranform we have (t τy(τ dτ = t Y ( + 2 Y ( = 2 33
which implie or or ( + 2 Y ( = 2 ( 2 + 2 Y ( = 2 Y ( = 2 + Applying the invere Laplace tranform to both ide we get y(t = in(t Example 454 Conider the equation y(t Applying the Laplace tranform we have y(τ dτ = e 2t Y ( + Y ( = 2 which implie or or Y ( = ( + Y ( = 2 ( + Y ( = 2 ( + ( 2 = + 2 2 Applying the invere Laplace tranform to both ide we get y(t = 2e 2t e t 34
Example 455 Conider the equation y (t = Applying the Laplace tranform we have y(τ dτ, y( = Y ( = Y ( which implie or or ( + Y ( = + ( 2 + Y ( = + 2 + = Y ( = + 2 + + 2 + Applying the invere Laplace tranform to both ide we get y(t = co(t + in(t Periodic Function Theorem 47 If f(t i a periodic function with period T >, ie, f(t + T = f(t for all t > then L(f(t = T e t f(t dt e T Proof Aume that f(t i a periodic function with period T >, ie, f(t + T = f(t for all t > Then we have L(f(t = = T e t f(t dt e t f(t dt + T e t f(t dt 35
( Set t = τ + T dt = dτ = = = T T T e t f(t dt + e (τ+t f(τ + T dτ e t f(t dt + e T e τ f(τ dτ e t f(t dt + e T L(f(t From thi we obtain L(f(t = T e t f(t dt e T, for < t < Example 456 Conider the function f(t = for t < 2 all t Then f(t i periodic with period T = 2 and we have and f(t + 2 = f(t for L(f(t = 2 e t t dt e 2 In order to evaluate the integral in the numerator we need to ue integration by part 2 e t f(t dt = = e t e t dt = ( e So we have L(f(t = e ( e 2 = ( + e, for < t < Example 457 Conider the function f(t = for t < 2 all t Then f(t i periodic with period T = 2 and we have and f(t + 2 = f(t for L(f(t = 2 e t t dt e 2 36
In order to evaluate the integral in the numerator we need to ue integration by part 2 e t f(t dt = e t dt = e t e t = ( e 2 (( e 2 2 2 e t dt = ( + (e e 2 So we have L(f(t = ( e 2 ( e 2 = ( e ( + e Example 458 Conider the function f(t = t for < t < 2 and f(t + 2 = f(t for all t Then f(t i periodic with period T = 2 and we have L(f(t = 2 e t t dt e 2 In order to evaluate the integral in the numerator we need to ue integration by part 2 2 ( e e t t t dt = t dt ( e t 2 = t + 2 e t dt = 2e 2 + ( e t 2 = 2e 2 e 2 2 + 2 = 2e 2 e 2 2 So we have L(f(t = 2e 2 e 2 2 ( e 2 in(t, for < t < π Example 459 Conider the function f(t = for π t < 2π and f(t+2π = f(t 37
for all t Then f(t i periodic with period T = 2π and we have L(f(t = 2 e t t dt e 2 In order to evaluate the integral in the numerator we need to ue integration by part 2π e t f(t dt = π e t in(t dt To do thi integral we need to ue integration by part: π π ( e e t t in(t dt = in(t dt ( e t π = in(t + π e t co(t dt = π ( e t co(t dt = [ ( e t π π ( ] co(t e t ( in(t dt = ( + 2 e π π e t in(t dt 2 So So we have π e t in(t dt = ( + e π 2 L(f(t = ( + e π 2 ( e 2π = 2 (( e π 45 The Dirac Delta Function In thi ection we cover application of Laplace tranform involving the Dirac Delta Function The firt thing you hould know i that the Dirac Delta Function i not a function Thi will become clear from it defining property What i it? It i a limit of function in a 38
pecial ene which we now decribe Let a > be given and define, < t < a δ n (t a = n, a t < a + /n, t a + /n Then we define the Dirac Delta Function upported at t = a, δ(t a, applied to a continuou function f(t by the formula δ(t a f(t dt = lim n δ n (t a f(t dt Theorem 48 For any continuou function f(t we have δ(t a f(t dt = f(a Proof We need to how that lim n δ n (t a f(t dt = f(a To thi end we firt define F (t = f(τ dτ Then, from the fundamental theorem of calculu we have F (t = f(t We alo have lim n δ n (t a f(t dt = lim n n a+/n a f(t dt = lim n F (a + /n F (a /n = F (a = f(a Example 46 Conider a few integral involving the delta function which are eaily com- 39
puted uing the definition and property that δ(t c i zero everywhere except at c b a f(c, if c [a, b] f(tδ(t c dt =, if c [a, b] 2 3 4 3 (t 2 δ(t dt = in(3tδ(t π/2 dt = in(3π/2 e 2t δ(t dt = e 2 (t 2 + 2δ(t 2 dt = becaue t = 2 in not in the interval [3, and δ(t 2 i zero everywhere except at t = 2 Now we conider the Laplace tranform of the delta function Theorem 49 L (δ(t a = e a Proof We only need to apply the reult of Theorem 48 L(δ(t a = e a δ(t a dt = e a Example 46 Conider the differential equation y + y = δ(t π with y( = and y ( = Applying the Laplace tranform we have ( ( 2 Y + Y = e π Y = e π 2 + Applying the invere Laplace tranform and uing the econd hift theorem we have y(t = u(t π in(t π = u(t π in(t 4
Example 462 Conider the differential equation y + 5y + 6y = e t δ(t 2 with y( = 2 and y ( = 5 Applying the Laplace tranform we have ( 2 Y 2 + 5 + 5(Y 2 + 6Y = e 2(+ ( 2 + 5 + 6Y = (2 + 5 + e 2(+ Note that ( 2 Y 2 + 5 = ( + 2( + 3 o we have Y = (2 + 5 ( + 2( + 3 + e 2(+ ( + 2( + 3 Next we apply partial fraction to each term to get (2 + 5 ( + 2( + 3 = ( + 2 + ( + 3 and ( + 2( + 3 = ( + 2 ( + 3 Applying the invere Laplace tranform and uing the econd hift theorem we have y(t = ( e 2t + e 3t + u(t 2 ( e 2(t 2 e 3(t 4/3 Example 463 Conider the differential equation y + y = δ(t π/2 + δ(t 3π/2 with y( = and y ( = Applying the Laplace tranform we have ( 2 Y + Y = e π/2 + e 3π/2 ( 2 + Y = + e π/2 + e 3π/2 So we get Y = ( ( 2 + + e π/2 + e 3π/2 2 + 2 + Applying the invere Laplace tranform and uing the econd hift theorem we have y(t = co(t + u(t π/2 in(t π/2 + u(t 3π/2 in(t 3π/2 Example 464 Conider the differential equation y + 4y + 5y = δ(t 2π with y( = and y ( = 2 Applying the Laplace tranform we have ( 2 Y 2 + 4Y + 5Y = e 2π ( 2 + 4 + 5Y = 2 + e 2π 4
Y = 2 ( 2 + 4 + 5 + e 2π ( 2 + 4 + 5 We need to complete the quare: ( 2 + 4 + 5 = ( + 2 2 + o we have Y = 2 ( + 2 2 + + e 2π ( + 2 2 + Applying the invere Laplace tranform and uing the firt and econd hift theorem we have y(t = 2e 2t in(t + u(t 2πe 2(t 2π in(t 2π 42
Table of Laplace Tranform f(t for t f = L(f = e t f(t dt e at a t n n! (n =,, n+ t a Γ(a + a+ (a > in bt co bt b 2 + b 2 2 + b 2 f (t L(f f( f (t 2 L(f f( f ( f (n (t n L(f (n f( (n 2 f ( f (n ( t n f(t ( n dn F d n ( e at f(t L(f( a t a e a u(t a = t > a (f g(t = u(t af(t a δ(t a f(t τg(τ dτ e a L(f( e a L(f g = L(fL(g 43