Lecture Notes for Laplace Transform

Transcription

1 Lecture Note for Laplace Tranform Wen Shen April 9 NB! Thee note are ued by myelf. They are provided to tudent a a upplement to the textbook. They can not ubtitute the textbook. Laplace Tranform i ued to handle piecewie continuou or impulive force. 6.: Definition of the Laplace tranform () Topic: Definition of Laplace tranform, Compute Laplace tranform by definition, including piecewie continuou function. Definition: Given a function f(t), t, it Laplace tranform F () = L{f(t)} i defined a F () = L{f(t)}. = e t f(t) dt. = lim A A e t f(t) dt We ay the tranform converge if the limit exit, and diverge if not. Next we will give example on computing the Laplace tranform of given function by definition. Example. f(t) = for t. A F () = L{f(t)} = lim e t dt = lim A A e t Example. f(t) = e t. F () = L{f(t)} = lim A = lim A a A e t e at dt = lim A A ( e ( a)a ) =, ( > a) a A = lim [ e A ] =, ( > ) A e ( a)t dt = lim A a e ( a)t A

2 Example 3. f(t) = t n, for n integer. F () = lim A A = + n lim A So we get a recurive relation e t t n dt = A lim A { t n e t A e t t n dt = n L{tn }. A } nt n e t dt L{t n } = n L{tn }, n, which mean L{t n } = n L{t n }, L{t n } = n L{t n 3 }, By induction, we get L{t n } = n L{tn } = n (n ) L{t n } = n = = n (n ) (n ) n! L{} = n Example 4. Find the Laplace tranform of in at and co at. (n ) (n ) L{t n 3 } = n!, ( > ) n+ Method. Compute by definition, with integration-by-part, twice. (lot of work...) Method. Ue the Euler formula By Example we have e iat = co at + i in at, L{e iat } = L{co at} + il{in at}. L{e iat } = ia = ( + ia) ( ia)( + ia) = + ia + a = Comparing the real and imaginary part, we get L{co at} = + a + i a + a. + a, L{in at} = a, ( > ). + a Remark: Now we will ue A intead of lim A, without cauing confuion. For piecewie continuou function, Laplace tranform can be computed by integrating each integral and add up at the end. Example 5. Find the Laplace tranform of {, t <, f(t) = t, t.

3 We do thi by definition: F () = = e t e t f(t) dt = t= e t dt + + (t ) e t = (e ) + ( ) + t= e t A (t )e t dt t= e t dt = (e ) + e 3

4 6.: Solution of initial value problem (4) Topic: Propertie of Laplace tranform, with proof and example Invere Laplace tranform, with example, review of partial fraction, Solution of initial value problem, with example covering variou cae. Propertie of Laplace tranform:. Linearity: L{c f(t) + c g(t)} = c L{f(t)} + c L{g(t)}.. Firt derivative: L{f (t)} = L{f(t)} f(). 3. Second derivative: L{f (t)} = L{f(t)} f() f (). 4. Higher order derivative: L{f (n) (t)} = n L{f(t)} n f() n f () f (n ) () f (n ) (). 5. L{ tf(t)} = F () where F () = L{f(t)}. Thi alo implie L{tf(t)} = F (). 6. L{e at f(t)} = F ( a) where F () = L{f(t)}. Thi implie e at f(t) = L {F ( a)}. Remark: Note property and 3 are ueful in differential equation. It how that each derivative in t caued a multiplication of in the Laplace tranform. Property 5 i the counter part for Property. It how that each derivative in caue a multiplication of t in the invere Laplace tranform. Property 6 i alo known a the Shift Theorem. A counter part of it will come later in chapter 6.3. Proof:. Thi follow by definition.. By definition L{f (t)} = e t f (t)dt = e t f(t) ( )e t f(t)dt = f() + L{f(t)}. 4

5 3. Thi one follow from Property. Set f to be f we get L{f (t)} = L{f (t)} f () = (L{f(t)} f()) f () = L{f(t)} f() f (). 4. Thi follow by induction, uing property. 5. The proof follow from the definition: F () = d d e t f(t)dt = 6. Thi proof alo follow from definition: L{e at f(t)} (e t )f(t)dt = e t e at f(t)dt = ( t)e t f(t)dt = L{ tf(t)}. e ( a)t f(t)dt = F ( a). By uing thee propertie, we could find more eaily Laplace tranform of many other function. Example. From L{t n } = n! n+, we get L{eat t n } = n! ( a) n+. Example. From L{in bt} = b + b, we get L{eat in bt} = b ( a) + b. Example 3. From L{co bt} = + b, we get L{eat co bt} = a ( a) + b. Example 4. L{t 3 + 5t } = L{t 3 } + 5L{t} L{} = 3! Example 5. L{e t (t 3 + 5t )} = 3! ( ) ( ). 5

6 Example 6. L{(t + 4)e t e t co t} = ( ) ( + ) +, becaue L{t + 4} = 3 + 4, L{(t + 4)e t } = ( ) Next are a few example for Property 5. Example 7. Given L{e at } = ( ) a, we get L{teat } = = a ( a) Example 8. ( ) b L{t in bt} = = b + b ( + b ). Example 9. ( ) L{t co bt} = = = b + b ( + b ). 6

7 Invere Laplace tranform. Definition: Technique: find the way back. Some imple example: Example. { } 3 L + 4 L {F ()} = f(t), if F () = L{f(t)}. = L { 3 } = 3 { } + L = 3 in t. + Example. { } { } L = L ( + 5) = { } 3! ( + 5) 4 3 L = { } 3! ( + 5) 4 3 e 5t L = 4 3 e 5t t 3. Example. { } + L + 4 { } = L + { } + 4 L = co t in t. + 4 Example 3. L { + 4 } = L { } + ( )( + ) Here we ued partial fraction to find out: { 3/4 = L + /4 } = et + 4 e t. + ( )( + ) = A + B, A = 3/4, B = /4. + 7

8 Solution of initial value problem. We will go through one example firt. Example 4. (Two ditinct real root.) Solve the initial value problem by Laplace tranform, y 3y y =, y() =, y () =. Step. Take Laplace tranform on both ide: Let L{y(t)} = Y (), and then L{y (t)} = Y () y() = Y, L{y (t)} = Y () y() y () = Y. Note the initial condition are the firt thing to go in! L{y (t)} 3L{y (t)} L{y(t)} = L{}, Y 3(Y ) Y =. Now we get an algebraic equation for Y (). Step : Solve it for Y (): ( 3 )Y () = = +, Y () = + ( 5)( + ). Step 3: Take invere Laplace tranform to get y(t) = L {Y ()}. The main technique here i partial fraction. Y () = + ( 5)( + ) = A + Compare the numerator: B 5 + C + A( 5)( + ) + B( + ) + C( 5) =. ( 5)( + ) + = A( 5)( + ) + B( + ) + C( 5). The previou equation hold for all value of. Set = : we get A =, o A =. 5 Set = 5: we get 35B =, o B =. 35 Set = : we get 4C = 8, o C = 4. 7 Now, Y () i written into um of term which we can find the invere tranform: y(t) = AL { } + BL { 5 } + CL { + } = e5t e t. 8

9 Structure of olution: Take Laplace tranform on both ide. You will get an algebraic equation for Y. Solve thi equation to get Y (). Take invere tranform to get y(t) = L {y}. Example 5. (Ditinct real root, but one matche the ource term.) Solve the initial value problem by Laplace tranform, y y y = e t, y() =, y () =. Take Laplace tranform on both ide of the equation, we get L{y } L{y } L{y} = L{e t }, Y () Y () Y () =. Solve it for Y : ( )Y () = + =, Y () = ( )( ) = ( ) ( + ). Ue partial fraction: Compare the numerator: ( ) ( + ) = A + + B + C ( ). = A( ) + B( + )( ) + C( + ) Set =, we get A =. Set =, we get C =. Set = (any convenient value of 9 3 can be ued in thi tep), we get B =. So 9 and Y () = ( ) y(t) = L {Y } = 9 e t + 9 et + 3 tet. Compare thi to the method of undetermined coefficient: general olution of the equation hould be y = y H + Y, where y H i the general olution to the homogeneou equation and Y i a particular olution. The characteritic equation i r r = (r + )(r ) =, o r =, r =, and y H = c e t + c e t. Since i a root, o the form of the particular olution i Y = Ate t. Thi dicuion conclude that the olution hould be of the form y = c e t + c e t + Ate t for ome contant c, c, A. Thi fit well with our reult. 9

10 Example 6. (Complex root.) y y + y = e t, y() =, y () =. Before we olve it, let ue the method of undetermined coefficient to find out which term will be in the olution. o the olution hould have the form: The Laplace tranform would be r r + =, (r ) + =, r, = ± i, y H = c e t co t + c e t in t, Y = Ae t, y = y H + Y = c e t co t + c e t in t + Ae t. Y () = c ( ) + + c ( ) + + A + = c ( ) + c ( ) + + A +. Thi give u idea on which term to look for in partial fraction. Now let ue the Laplace tranform: Y () = L{y}, L{y } = Y y() = Y, L{y } = Y y() y() = Y. Y Y + Y = +, ( + )Y () = + + = + + Y () = Compare the numerator: + ( + )( + ) = + ( + )(( ) + ) = A B( ) + C + + ( ) + + = A(( ) + ) + (B( ) + C)( + ). Set = : 5A =, A = 5. Compare coefficient of -term: A + B =, B = A = 5. Set any value of, ay = : = A B + C, C = A + B = 9 5. We ee thi fit our prediction. Y () = ( ) ( ) + y(t) = 5 e t 5 et co t et in t.

11 Example 7. (Pure imaginary root.) y + y = co t, y() =, y () =. Again, let firt predict the term in the olution: o r + =, r, = ±i, y H = c co t + c in t, Y = A co t and the Laplace tranform would be y = y H + Y = c co t + c in t + A co t, Y () = c + + c + + A + 4. Now, let take Laplace tranform on both ide: Comparing numerator, we get Y + Y = L{co t} = + 4 ( + )Y () = = Y () = ( + 4)( + ) = A + B + + C + D = (A + B)( + 4) + (C + D)( + ). One may expand the right-hand ide and compare term to find A, B, C, D, but that take more work. Let try by etting into complex number. Set = i, and remember the fact i = and i 3 = i, we have i + 9i + 4 = (Ai + B)( + 4), 3 + 7i = 3B + 3Ai, B =, A = 7 3. Set now = i: 6i 4 + 8i + 4 = (Ci + D)( 3), + i = 3D 6Ci, D =, C = 3. So and Y () = y(t) = 7 3 co t + in t co t. 3

12 A very brief review on partial fraction, targeted toward invere Laplace tranform. Goal: rewrite a fractional form P n() P m() impler term. We aume n < m. (where P n i a polynomial of degree n) into um of The type of term appeared in the partial fraction i olely determined by the denominator P m (). Firt, fact out P m (), write it into product of term of (i) a, (ii) + a, (iii) ( a ) + b. The following table give the term in the partial fraction and their correponding invere Laplace tranform. term in P M () from where? term in partial fraction invere L.T. real root, or a g(t) = e at A a Ae at double root, ( a) or r = a and g(t) = e at A a + B Ae at + Bte at ( a) double root, ( a) 3 and g(t) = e at A a + B ( a) + C Ae at + Bte at + C ( a) 3 t e at imaginary root or + µ g(t) = co µt or in µt complex root, or ( λ) + µ g(t) = e λt co µt(or in µt) A + B + µ A co µt + B in µt A( λ) + B ( λ) + µ e λt (A co µt + B in µt) In ummary, thi table can be written = P n () ( a)( b) ( c) 3 (( λ) + µ ) A a + B b + B ( b) + C c + C ( c) + C 3 ( c) + D ( λ) + D 3 ( λ) + µ.

13 6.3: Step function () Topic: Definition and baic application of unit tep (Heaviide) function, Laplace tranform of tep function and function involving tep function (piecewie continuou function), Invere tranform involving tep function. We ue tep function to form piecewie continuou function. Unit tep function(heaviide function): for c. A plot of u c (t) i below: u c u c t = {, t < c,, c t. c t For a given function f(t), if it i multiplied with u c (t), then {, < t < c, u c tf(t) = f(t), c t. We ay u c pick up the interval [c, ). Example. Conider A plot of thi i given below u c (t) = {, t < c,, c t. u c c t 3

14 We ee that thi function pick up the interval [, c). Example. Rectangular pule. The plot of the function look like u a u b a b t for a < b <. We ee it can be expreed a and it pick up the interval [a, b). Example 3. For the function g(t) = u a (t) u b (t) { f(t), a t < b, otherwie We can rewrite it in term of the unit tep function a ( ) g(t) = f(t) u a (t) u b (t). Example 4. For the function in t, t <, ft = e t, t < 5, t 5 t, we can rewrite it in term of the unit tep function a we did in Example 3, treat each interval eparately ( ) ( ) f(t) = in t u (t) u (t) + e t u (t) u 5 (t) + t u 5 (t). 4

15 Laplace tranform of u c (t): by definition L{u c (t)} = e t u c (t) dt = c e t dt = e t Shift of a function: Given f(t), t >, then { f(t c), c t, g(t) =, t < c, i the hift of f by c unit. See figure below. t=c = e c = e t, ( > ). f g t c t Let F () = L{f(t)} be the Laplace tranform of f(t). Then, the Laplace tranform of g(t) i L{g(t)} = L{u c (t) f(t c)} = e t u c (t)f(t c) dt = c Let y = t c, o t = y + c, and dt = dy, and we continue L{g(t)} = e t f(t c) dt. e (y+c) f(y) dy = e c e y f(y) dy = e c F (). So we conclude L{u c (t)f(t c)} = e c L{f(t)} = e c F (), which i equivalent to L {e c F ()} = u c (t)f(t c). Note now we are only conidering the domain t. So u (t) = for all t. 5

16 In following example we will compute Laplace tranform of piecewie continuou function with the help of the unit tep function. Example 5. Given { in t, t < π f(t) = in t + co(t π), π t. 4 4 It can be rewritten in term of the unit tep function a (Or, if we write out each interval f(t) = in t + u π 4 (t) co(t π 4 ). 4, f(t) = in t( u π 4 (t)) + (in t + co(t π 4 ))u π 4 (t) = in t + u π 4 (t) co(t π 4 ). which give the ame anwer.) And the Laplace tranform of f i F () = L{in t} + L{u π 4 (t) co(t π 4 )} = + + π e 4 +. Example 6. Given f(t) = { t, t <,, t. It can be rewritten in term of the unit tep function a The Laplace tranform i f(t) = t( u (t)) + u (t) = t (t )u (t). L{f(t)} = L{t} L{(t )u (t)} = e. Example 7. Given f(t) = {, t <, t + 3, t. We can rewrite it in term of the unit tep function a The Laplace tranform i f(t) = (t + 3)u (t) = (t + 5)u (t) = (t )u (t) + 5u (t). L{f(t)} = L{(t )u (t)} + 5L{u (t)} = e + 5e. 6

17 Example 8. Given g(t) = {, t <, t, t. We can rewrite it in term of the unit tep function a Oberve that we have g(t) = ( u (t)) + t u (t) = + (t )u (t). t = (t + ) = (t ) + 4(t ) + 4 = (t ) + 4(t ) + 3, The Laplace tranform i g(t) = + ( (t ) + 4(t ) + 3 ) u (t). L{g(t)} = ( + e ). Example 9. Given, t < 3, f(t) = e t, 3 t < 4,, 4 t. We can rewrite it in term of the unit tep function a The Laplace tranform i f(t) = e t (u 3 (t) u 4 (t)) = u 3 (t)e t 3 e 3 u 4 (t)e t 4 e 4. L{g(t)} = e 3 e 3 e4 e 4 = [ e 3( ) e 4( )]. 7

18 Invere tranform: We ue two propertie: L{u c (t)} = e c, and L{u c(t)f(t c)} = e c L{f(t)}. In the following example we want to find f(t) = L {F ()}. Example. We know that L { 3 } = t, o we have F () = e 3 = 3 e 3. f(t) = L {F ()} = t u (t) (t ) = { t, t <, t (t ), t.. Example. Given F () = e 3 + = e 3 ( + 4)( + 3) = e 3 By partial fraction, we find A = 7 and B = 7. So ( A B ). 3 f(t) = L {F ()} = u 3 (t) [ Ae 4(t 3) + Be 3(t 3)] = 7 u 3(t) [ e 4(t 3) + e 3(t 3)] which can be written a a p/w continuou function {, t < 3, f(t) = 7 e 4(t 3) + 7 e3(t 3), 3 t. Example. Given So F () = e = + e ( + ) + = [ + ( + ) ]. ( + ) + f(t) = L {F ()} = u (t) [ e (t ) co(t ) e (t ) in(t ) ] which can be written a a p/w continuou function {, t <, f(t) = e (t ) [co(t ) in(t )], t. 8

19 6.4: Differential equation with dicontinuou forcing function () Topic: Solve initial value problem with dicontinuou force, example of variou cae, Decribe behavior of olution, and make phyical ene of them. Next we tudy initial value problem with dicontinuou force. We will tart with an example. Example. (Damped ytem with force, complex root) Solve the following initial value problem y + y + y = g(t), g(t) = {, t <,, t,, y() =, y () =. Let L{y(t)} = Y (), o L{y } = Y and L{y } = Y. Alo we have L{g(t)} = L{u (t)} = e. Then Y + Y + Y = e, which give e Y () = ( + + ) Now we need to find the invere Laplace tranform for Y (). We have to do partial fraction firt. We have ( + + ) = A + B + C + +. Compare the numerator on both ide: Set =, we get A =. = A( + + ) + (B + C) Compare -term: = A + B, o B = A =. Compare -term: = A + C, o C = A =. So We work out ome detail ( Y () = e + ) = + ( + ) + ( = 3 ) 9 ( + ) ( + ) + (, 3 )

20 o { } ( + L = e 3 t co + + t + ) 3 in 3 t. We conclude ( [ y(t) = u (t) [ e 3 3 (t ) co (t ) in (t ) )]+e t co Remark: There are other way to work out the partial fraction. Extra quetion: What happen when t? 3 t + ] 3 in 3 t. Anwer: We ee all the term with the exponential function will go to zero, o y in the limit. We can view thi ytem a the pring-ma ytem with damping. Since g(t) become contant for large t, and the particular olution (which i alo the teady tate) with on the right hand ide i, which provide the limit for y. Further obervation: We ee that the olution to the homogeneou equation i ] e 3 3 [c t co t + c in t, and thee term do appear in the olution. Actually the olution conit of two part: the forced repone and the homogeneou olution. Furthermore, the g ha a dicontinuity at t =, and we ee a jump in the olution alo for t =, a in the term u (t).

21 Example. (Undamped ytem with force, pure imaginary root) Solve the following initial value problem, t < π, y + 4y = g(t) =, π t < π, y() =, y () =., π t, Rewrite So g(t) = u π (t) u π (t), L{g} = e π e π. Y + 4Y = ( e π e π). Solve it for Y : Work out partial fraction So Y () = e π e π ( + 4) = e π ( + 4) e π ( + 4) ( + 4) = A + B + C + 4, A = 4, B = 4, C =. Now we take invere Laplace tranform of Y L { ( + 4) } = 4 co t. 4 y(t) = u π (t)( 4 4 co (t π)) u π(t)( 4 co (t π)) + co t 4 = (u π (t) u π ) ( co t) + co t { 4 ( co t), π t < π, = co t + 4, otherwie, = homogeneou olution + forced repone

22 Example 3. In Example 4, let, t < 4, g(t) = e t, 4 5 < π,, 5 t. Find Y (). Rewrite g(t) = e t (u 4 (t) u 5 (t)) = u 4 (t)e t 4 e 4 u 5 (t)e t 5 e 5, o G() = L{g(t)} = e 4 e 4 e5 e 5. Take Laplace tranform of the equation, we get ( + 4)Y () = G() +, Y () = ( e 4 e 4 e 5 e 5) ( )( + 4) Remark: We ee that the firt term will give the forced repone, and the econd term i from the homogeneou equation. The tudent may work out the invere tranform a a practice.

23 Example 4. (Undamped ytem with force, example from the book p. 334), t < 5, y + 4y = g(t), y() =, y () =, g(t) = (t 5)/5, 5 5 <,, t. Let firt work on g(t) and it Laplace tranform g(t) = t 5 5 (u 5(t) u (t)) + u (t) = 5 u 5(t)(t 5) 5 u (t)(t ), Let Y () = L{y}, then G() = L{g} = 5 e 5 5 e ( + 4)Y () = G(), Y () = G() + 4 = 5 e 5 ( + 4) 5 e ( + 4) Work out the partial fraction: H(). = one get A =, B =, C =, D =. So 4 8 h(t) =. { } L ( + 4) Go back to y(t) ( + 4) = A + B C + D = L { 4 8 y(t) = L {Y } = 5 u 5(t)h(t 5) 5 u (t)h(t ) = [ 5 u 5(t) 4 (t 5) ] in (t 5) 8 5 u (t) =, t < 5, (t 5) in (t 5), 5 5 <, 4 (in (t 5) in (t )), t. 4 4 } = + 4 t in t. 8 [ 4 (t ) in (t ) 8 Note that for t, we have y(t) = + R co(t + δ) for ome amplitude R and phae δ. 4 The plot of g and y are given in the book. Phyical meaning and qualitative nature of the olution: The ource g(t) i known a ramp loading. During the interval < t < 5, g = and initial condition are all. So olution remain. For large time t, g =. A particular olution i Y = 4. Adding the homogeneou olution, we hould have y = 4 + c in t + c co t for t large. We ee thi i actually the cae, the olution i an ocillation around the contant 4 for large t. 3 ]