DIFFERENTIAL EQUATIONS Laplace Transforms. Paul Dawkins

Transcription

1 DIFFERENTIAL EQUATIONS Laplace Tranform Paul Dawkin

2 Table of Content Preface... Laplace Tranform... Introduction... The Definition... 5 Laplace Tranform... 9 Invere Laplace Tranform... Step Function... 4 Solving IVP with Laplace Tranform... 7 Noncontant Coefficient IVP IVP With Step Function Dirac Delta Function Convolution Integral Paul Dawkin i

3 Preface Here are my online note for my differential equation coure that I teach here at Lamar Univerity. Depite the fact that thee are my cla note, they hould be acceible to anyone wanting to learn how to olve differential equation or needing a refreher on differential equation. I ve tried to make thee note a elf contained a poible and o all the information needed to read through them i either from a Calculu or Algebra cla or contained in other ection of the note. A couple of warning to my tudent who may be here to get a copy of what happened on a day that you mied.. Becaue I wanted to make thi a fairly complete et of note for anyone wanting to learn differential equation I have included ome material that I do not uually have time to cover in cla and becaue thi change from emeter to emeter it i not noted here. You will need to find one of your fellow cla mate to ee if there i omething in thee note that wan t covered in cla.. In general I try to work problem in cla that are different from my note. However, with Differential Equation many of the problem are difficult to make up on the pur of the moment and o in thi cla my cla work will follow thee note fairly cloe a far a worked problem go. With that being aid I will, on occaion, work problem off the top of my head when I can to provide more example than jut thoe in my note. Alo, I often don t have time in cla to work all of the problem in the note and o you will find that ome ection contain problem that weren t worked in cla due to time retriction.. Sometime quetion in cla will lead down path that are not covered here. I try to anticipate a many of the quetion a poible in writing thee up, but the reality i that I can t anticipate all the quetion. Sometime a very good quetion get aked in cla that lead to inight that I ve not included here. You hould alway talk to omeone who wa in cla on the day you mied and compare thee note to their note and ee what the difference are. 4. Thi i omewhat related to the previou three item, but i important enough to merit it own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Uing thee note a a ubtitute for cla i liable to get you in trouble. A already noted not everything in thee note i covered in cla and often material or inight not in thee note i covered in cla. 7 Paul Dawkin ii

4 Laplace Tranform Introduction In thi chapter we will be looking at how to ue Laplace tranform to olve differential equation. There are many kind of tranform out there in the world. Laplace tranform and Fourier tranform are probably the main two kind of tranform that are ued. A we will ee in later ection we can ue Laplace tranform to reduce a differential equation to an algebra problem. The algebra can be mey on occaion, but it will be impler than actually olving the differential equation directly in many cae. Laplace tranform can alo be ued to olve IVP that we can t ue any previou method on. For imple differential equation uch a thoe in the firt few ection of the lat chapter Laplace tranform will be more complicated than we need. In fact, for mot homogeneou differential equation uch a thoe in the lat chapter Laplace tranform i ignificantly longer and not o ueful. Alo, many of the imple nonhomogeneou differential equation that we aw in the Undetermined Coefficient and Variation of Parameter are till impler (or at the leat no more difficult than Laplace tranform) to do a we did them there. However, at thi point, the amount of work required for Laplace tranform i tarting to equal the amount of work we did in thoe ection. Laplace tranform come into it own when the forcing function in the differential equation tart getting more complicated. In the previou chapter we looked only at nonhomogeneou differential equation in which g(t) wa a fairly imple continuou function. In thi chapter we will tart looking at g(t) that are not continuou. It i thee problem where the reaon for uing Laplace tranform tart to become clear. We will alo ee that, for ome of the more complicated nonhomogeneou differential equation from the lat chapter, Laplace tranform are actually eaier on thoe problem a well. Here i a brief rundown of the ection in thi chapter. The Definition The definition of the Laplace tranform. We will alo compute a couple Laplace tranform uing the definition. Laplace Tranform A the previou ection will demontrate, computing Laplace tranform directly from the definition can be a fairly painful proce. In thi ection we introduce the way we uually compute Laplace tranform. Invere Laplace Tranform In thi ection we ak the oppoite quetion. Here a Laplace tranform, what function did we originally have? Step Function Thi i one of the more important function in the ue of Laplace tranform. With the introduction of thi function the reaon for doing Laplace tranform tart to become apparent. 7 Paul Dawkin iii

5 Solving IVP with Laplace Tranform Here how we ued Laplace tranform to olve IVP. Noncontant Coefficient IVP We will ee how Laplace tranform can be ued to olve ome noncontant coefficient IVP IVP with Step Function Solving IVP that contain tep function. Thi i the ection where the reaon for uing Laplace tranform really become apparent. Dirac Delta Function One lat function that often how up in Laplace tranform problem. Convolution Integral A brief introduction to the convolution integral and an application for Laplace tranform. Table of Laplace Tranform Thi i a mall table of Laplace Tranform that we ll be uing here. 7 Paul Dawkin 4

6 The Definition You know, it alway a little cary when we devote a whole ection jut to the definition of omething. Laplace tranform (or jut tranform) can eem cary when we firt tart looking at them. However, a we will ee, they aren t a bad a they may appear at firt. Before we tart with the definition of the Laplace tranform we need to get another definition out of the way. A function i called piecewie continuou on an interval if the interval can be broken into a finite number of ubinterval on which the function i continuou on each open ubinterval (i.e. the ubinterval without it endpoint) and ha a finite limit at the endpoint of each ubinterval. Below i a ketch of a piecewie continuou function. In other word, a piecewie continuou function i a function that ha a finite number of break in it and doen t blow up to infinity anywhere. Now, let take a look at the definition of the Laplace tranform. Definition Suppoe that f(t) i a piecewie continuou function. The Laplace tranform of f(t) i L f t and defined a { } denoted t { f () t } - Ú e () L f t dt () There i an alternate notation for Laplace tranform. For the ake of convenience we will often denote Laplace tranform a, L f t F 7 Paul Dawkin 5 { } With thi alternate notation, note that the tranform i really a function of a new variable,, and that all the t will drop out in the integration proce. Now, the integral in the definition of the tranform i called an improper integral and it would probably be bet to recall how thee kind of integral work before we actually jump into computing ome tranform.

7 Example If c, evaluate the following integral. Ú e ct dt Solution Remember that you need to convert improper integral to limit a follow, Ú ct Now, do the integral, then evaluate the limit. Ú Ú e dt lim e næ ct e dt lim Ú næ n n e ct ct dt Ê ct ˆ lim Á e næ Ëc dt Ê cn ˆ lim Á e - næ Ëc c Now, at thi point, we ve got to be careful. The value of c will affect our anwer. We ve already aumed that c wa non-zero, now we need to worry about the ign of c. If c i poitive the exponential will go to infinity. On the other hand, if c i negative the exponential will go to zero. So, the integral i only convergent (i.e. the limit exit and i finite) provided c<. In thi cae we get, Ú e ct dt - provided c< c () Now that we remember how to do thee, let compute ome Laplace tranform. We ll tart off with probably the implet Laplace tranform to compute. Example Compute L{}. Solution There not really a whole lot do here other than plug the function f(t) into () L {} Ú e - t dt Now, at thi point notice that thi i nothing more than the integral in the previou example with c-. Therefore, all we need to do i reue () with the appropriate ubtitution. Doing thi give, L {} - Ú e t dt - provided - < - Or, with ome implification we have, L {} provided > n 7 Paul Dawkin 6

8 Notice that we had to put a retriction on in order to actually compute the tranform. All Laplace tranform will have retriction on. At thi tage of the game, thi retriction i omething that we tend to ignore, but we really houldn t ever forget that it there. Let do another example. at Example Compute L { e } Solution Plug the function into the definition of the tranform and do a little implification. at t at - a-t L e e e dt e dt { } Ú Ú Once again, notice that we can ue () provided c a-. So let do thi. L at ( - ) { e } Ú e a t dt - a- provided a- < - a provided > a Let do one more example that doen t come down to an application of (). Example 4 Compute L{in(at)}. Solution Note that we re going to leave it to you to check mot of the integration here. Plug the function into the definition. Thi time let alo ue the alternate notation. L in at F { } Ú lim - t e næ Ú n in e -t at dt in at dt Now, if we integrate by part we will arrive at, n Ê Ê n -t ˆ ˆ -t F( ) lim -Á e co( at) - co( at) dt næ Á a aú e Ë Ë Now, evaluate the firt term to implify it a little and integrate by part again on the integral. Doing thi arrive at, n Ê n n Ê - Ê -t ˆ ˆˆ -t F( ) lim ( -e co( an) )- Á e in( at) + in ( at) dt næ Áa aá a aú e Ë Ë Ë Now, evaluate the econd term, take the limit and implify. 7 Paul Dawkin 7

9 Ê n -n Ê -n -t ˆˆ F( ) lim ( co( an) ) in( an) in ( at) dt næ Á -e - Á e + a a a aú e Ë Ë Ê - t ˆ - Á in ( at) dt a a aú e Ë - a a Ú - t e in at dt Now, notice that in the limit we had to aume that > in order to do the following two limit. -n lime co an næ ( an) -n lime in næ Without thi aumption, we get a divergent integral again. Alo, note that when we got back to the integral we jut converted the upper limit back to infinity. The reaon for thi i that, if you think about it, thi integral i nothing more than the integral that we tarted with. Therefore, we now get, F( ) - F ( ) a a Now, imply olve for F() to get, L { in( at) } F( ) a provided + a > A thi example how, computing Laplace tranform i often mey. Before moving on to the next ection, we need to do a little ide note. On occaion you will ee the following a the definition of the Laplace tranform. L t { f () t } - Ú e () - f t dt Note the change in the lower limit from zero to negative infinity. In thee cae there i almot alway the aumption that the function f(t) i in fact defined a follow, Ï if t < f () t Ì Ó f () t if t In other word, it i aumed that the function i zero if t<. In thi cae the firt half of the integral will drop out ince the function i zero and we will get back to the definition given in (). A Heaviide function i uually ued to make the function zero for t<. We will be looking at thee in a later ection. 7 Paul Dawkin 8

10 Laplace Tranform A we aw in the lat ection computing Laplace tranform directly can be fairly complicated. Uually we jut ue a table of tranform when actually computing Laplace tranform. The table that i provided here i not an incluive table, but doe include mot of the commonly ued Laplace tranform and mot of the commonly needed formula pertaining to Laplace tranform. Before doing a couple of example to illutrate the ue of the table let get a quick fact out of the way. Fact Given f(t) and g(t) then, for any contant a and b. L { af ( t) + bg( t) } af( ) + bg( ) In other word, we don t worry about contant and we don t worry about um or difference of function in taking Laplace tranform. All that we need to do i take the tranform of the individual function, then put any contant back in and add or ubtract the reult back up. So, let do a couple of quick example. Example Find the Laplace tranform of the given function. -5t t (a) f ( t) 6e + e + 5t -9 [Solution] (b) g( t) 4co( 4t) 9in( 4t) co( t) (c) ht inh( t) + in( t) [Solution] t t (d) g( t) + co( 6t) - co( 6t) - + [Solution] e e [Solution] Solution Okay, there not really a whole lot to do here other than go to the table, tranform the individual function up, put any contant back in and then add or ubtract the reult. We ll do thee example in a little more detail than i typically ued ince thi i the firt time we re uing the table. f t 6e + e + 5t -9-5t t (a) F! ( 5) [Return to Problem] 7 Paul Dawkin 9

11 (b) g( t) 4co( 4t) - 9in( 4t) + co( t) G (c) ht inh( t) + in( t) ( 4) ( 4) ( ) H t t (d) g( t) e + co( 6t) -e co( 6t) G [Return to Problem] [Return to Problem] [Return to Problem] Make ure that you pay attention to the difference between a normal trig function and hyperbolic function. The only difference between them i the + a for the normal trig function become a - a in the hyperbolic function! It very eay to get in a hurry and not pay attention and grab the wrong formula. If you don t recall the definition of the hyperbolic function ee the note for the table. Let do one final et of example. Example Find the tranform of each of the following function. f t tcoh t [Solution] (a) (b) ht t in( t) [Solution] (c) g() t t [Solution] f t (d) () (e) f ( t) tg ( t) Solution f t tcoh t (a) t [Solution] [Solution] Thi function i not in the table of Laplace tranform. However we can ue # in the table to compute it tranform. Thi will correpond to # if we take n. 7 Paul Dawkin

12 L { }, where coh( ) F tg t - G g t t So, we then have, Uing # we then have, (b) ht t in( t) + 9 G( ) G ( ) F + 9 ( -9) ( ) [Return to Problem] Thi part will alo ue # in the table. In fact we could ue # in one of two way. We could ue it with n. H L tf t - F, where f t tin t { } Or we could ue it with n. H L t f t F, where f t in t { } Since it le work to do one derivative, let do it the firt way. So uing #9 we have, 4-6 F( ) F ( ) The tranform i then, (c) g() t t H -6 ( + 4) [Return to Problem] Thi part can be done uing either #6 (with n ) or # (along with #5). We will ue # o we can ee an example of thi. In order to ue # we ll need to notice that t t Ú vdv t fi t vdv Ú Now, uing #5, we get the following. () f t t F Ê p ˆÊˆ p Á Á Ë Ë 4 5 G p Thi i what we would have gotten had we ued #6. [Return to Problem] 7 Paul Dawkin

13 f t t (d) () For thi part we will ue #4 along with the anwer from the previou part. To ee thi note that if then () g t t ( ) f t g t Therefore, the tranform i. (e) f ( t) tg ( t) F Ê ˆ GÁ Ë Ê Á p Á Á Ê ˆ Á 4 Á Ë Ë p ˆ [Return to Problem] Thi final part will again ue # from the table a well a #5. d L{ tg () t } - L{ g } d d - { G ( )- g ( ) } d - G + G - ( ) G( ) G ( ) - - Remember that g() i jut a contant o when we differentiate it we will get zero! [Return to Problem] A thi et of example ha hown u we can t forget to ue ome of the general formula in the table to derive new Laplace tranform for function that aren t explicitly lited in the table! 7 Paul Dawkin

14 Invere Laplace Tranform Finding the Laplace tranform of a function i not terribly difficult if we ve got a table of tranform in front of u to ue a we aw in the lat ection. What we would like to do now i go the other way. We are going to be given a tranform, F(), and ak what function (or function) did we have originally. A you will ee thi can be a more complicated and lengthy proce than taking tranform. In thee cae we ay that we are finding the Invere Laplace Tranform of F() and ue the following notation. - f t L F { } A with Laplace tranform, we ve got the following fact to help u take the invere tranform. Fact Given the two Laplace tranform F() and G() then L af + bg al F + bl G for any contant a and b. { } { } { } So, we take the invere tranform of the individual tranform, put any contant back in and then add or ubtract the reult back up. Let take a look at a couple of fairly imple invere tranform. Example Find the invere tranform of each of the following. 6 4 (a) F( ) [Solution] 9 7 (b) H( ) [Solution] 6 (c) F( ) [Solution] 8 (d) G( ) [Solution] Solution I ve alway felt that the key to doing invere tranform i to look at the denominator and try to identify what you ve got baed on that. If there i only one entry in the table that ha that particular denominator, the next tep i to make ure the numerator i correctly et up for the invere tranform proce. If it in t, correct it (thi i alway eay to do) and then take the invere tranform. If there i more than one entry in the table ha a particular denominator, then the numerator of each will be different, o go up to the numerator and ee which one you ve got. If you need to correct the numerator to get it into the correct form and then take the invere tranform. So, with thi advice in mind let ee if we can take ome invere tranform. 7 Paul Dawkin

15 (a) F( ) From the denominator of the firt term it look like the firt term i jut a contant. The correct numerator for thi term i a o we ll jut factor the 6 out before taking the invere tranform. The econd term appear to be an exponential with a 8 and the numerator i exactly what it need to be. The third term alo appear to be an exponential, only thi time a and we ll need to factor the 4 out before taking the invere tranform. So, with a little more detail than we ll uually put into thee, F( ) H (b) 5 () () 8 t 6 4( t - e + e ) f t 6- e + 4e 8t t [Return to Problem] The firt term in thi cae look like an exponential with a - and we ll need to factor out the 9. Be careful with negative ign in thee problem, it very eay to loe track of them. The econd term almot look like an exponential, except that it got a intead of jut an in the denominator. It i an exponential, but in thi cae we ll need to factor a out of the denominator before taking the invere tranform. The denominator of the third term appear to be # in the table with n 4. The numerator however, i not correct for thi. There i currently a 7 in the numerator and we need a 4! 4 in the numerator. Thi i very eay to fix. Whenever a numerator i off by a multiplicative contant, a in thi cae, all we need to do i put the contant that we need in the numerator. We will jut need to remember to take it back out by dividing by the ame contant. So, let firt rewrite the tranform. 9 7 H( ) ! 4! ! ! So, what did we do here? We factored the 9 out of the firt term. We factored the out of the denominator of the econd term ince it can t be there for the invere tranform and in the third term we factored everything out of the numerator except the 4! ince that i the portion that we need in the numerator for the invere tranform proce. 4+ Let now take the invere tranform. () ht 7 9e - e + t 4 5t -t 4 [Return to Problem] 7 Paul Dawkin 4

16 F (c) In thi part we ve got the ame denominator in both term and our table tell u that we ve either got #7 or #8. The numerator will tell u which we ve actually got. The firt one ha an in the numerator and o thi mean that the firt term mut be #8 and we ll need to factor the 6 out of the numerator in thi cae. The econd term ha only a contant in the numerator and o thi term mut be #7, however, in order for thi to be exactly #7 we ll need to multiply/divide a 5 in the numerator to get it correct for the table. The tranform become, F ( 5) ( 5) ( 5) 5 ( 5) Taking the invere tranform give, f t t 5 t () 6co( 5 ) + in( 5 ) [Return to Problem] G (d) In thi cae the firt term will be a ine once we factor a out of the denominator, while the econd term appear to be a hyperbolic ine (#7). Again, be careful with the difference between thee two. Both of the term will alo need to have their numerator fixed up. Here i the tranform once we re done rewriting it. 8 G( ) ( 4) Notice that in the firt term we took advantage of the fact that we could get the in the numerator that we needed by factoring the 8. The invere tranform i then, 4 g() t in( t) + inh( 7t) 7 [Return to Problem] So, probably the bet way to identify the tranform i by looking at the denominator. If there i more than one poibility ue the numerator to identify the correct one. Fix up the numerator if needed to get it into the form needed for the invere tranform proce. Finally, take the invere tranform. 7 Paul Dawkin 5

17 Let do ome lightly harder problem. Thee are a little more involved than the firt et. Example Find the invere tranform of each of the following. 6-5 (a) F( ) + 7 [Solution] - (b) F( ) + 8+ [Solution] - (c) G( ) -6- [Solution] + 7 (d) H( ) -- [Solution] Solution F (a) From the denominator of thi one it appear that it i either a ine or a coine. However, the numerator doen t match up to either of thee in the table. A coine want jut an in the numerator with at mot a multiplicative contant, while a ine want only a contant and no in the numerator. We ve got both in the numerator. Thi i eay to fix however. We will jut plit up the tranform into two term and then do invere tranform. 6 5 F 5 () 6co( 7 )- in( 7 ) 7 Paul Dawkin f t t t 7 Do not get too ued to alway getting the perfect quare in ine and coine that we aw in the firt et of example. More often than not (at leat in my cla) they won t be perfect quare! [Return to Problem] F (b) In thi cae there are no denominator in our table that look like thi. We can however make the denominator look like one of the denominator in the table by completing the quare on the denominator. So, let do that firt ( ) Recall that in completing the quare you take half the coefficient of the, quare thi, and then add and ubtract the reult to the polynomial. After doing thi the firt three term hould factor a a perfect quare.

18 So, the tranform can be written a the following. - F( ) Okay, with thi rewrite it look like we ve got #9 and/or # from our table of tranform. However, note that in order for it to be a #9 we want jut a contant in the numerator and in order to be a # we need an a in the numerator. We ve got neither of thee o we ll have to correct the numerator to get it into proper form. In correcting the numerator alway get the a firt. Thi i the important part. We will alo need to be careful of the that it in front of the. One way to take care of thi i to break the term into two piece, factor the out of the econd and then fix up the numerator of thi term. Thi will work, however it will put three term into our anwer and there are really only two term. So, we will leave the tranform a a ingle term and correct it a follow, - ( + 4-4) F( ) ( ) ( ) ( ) ( + ) We needed an + 4 in the numerator, o we put that in. We jut needed to make ure and take the 4 back out by ubtracting it back out. Alo, becaue of the multiplying the we needed to do all thi inide a et of parenthei. Then we partially multiplied the through the econd term and combined the contant. With the tranform in thi form, we can break it up into two tranform each of which are in the table and o we can do invere tranform on them, G (c) F ( ) ( ) -4t -4t () - e co( 5 ) + e in( 5 ) f t t t 5 [Return to Problem] Thi one i imilar to the lat one. We jut need to be careful with the completing the quare however. The firt thing that we hould do i factor a out of the denominator, then complete the quare. Remember that when completing the quare a coefficient of on the term i needed! So, here the work for thi tranform. 7 Paul Dawkin 7

19 G - ( - - ) ( ) 4 So, it look like we ve got # and # with a corrected numerator. Here the work for that and the invere tranform. ( - + ) - G( ) - - ( ) ( ) ( - ) ( ) ( ) Ê + Á Ë Ê t Ê ˆ 5 t Ê ˆˆ g() t e coh t + inh t Á e Á Á Ë Ë Ë In correcting the numerator of the econd term, notice that I only put in the quare root ince we already had the over part of the fraction that we needed in the numerator. [Return to Problem] H (d) Thi one appear to be imilar to the previou two, but it actually in t. The denominator in the previou two couldn t be eaily factored. In thi cae the denominator doe factor and o we need to deal with it differently. Here i the tranform with the factored denominator. + 7 H( ) + -5 The denominator of thi tranform eem to ugget that we ve got a couple of exponential, however in order to be exponential there can only be a ingle term in the denominator and no in the numerator. To fix thi we will need to do partial fraction on thi tranform. In thi cae the partial fraction decompoition will be A B H( ) Don t remember how to do partial fraction? In thi example we ll how you one way of getting ˆ 7 Paul Dawkin 8

20 the value of the contant and after thi example we ll review how to get the correct form of the partial fraction decompoition. Okay, o let get the contant. There i a method for finding the contant that will alway work, however it can lead to more work than i ometime required. Eventually, we will need that method, however in thi cae there i an eaier way to find the contant. Regardle of the method ued, the firt tep i to actually add the two term back up. Thi give the following. + 7 A( - 5) + B( + ) Now, thi need to be true for any that we hould chooe to put in. So, ince the denominator are the ame we jut need to get the numerator equal. Therefore, et the numerator equal. + 7 A B + Again, thi mut be true for ANY value of that we want to put in. So, let take advantage of that. If it mut be true for any value of then it mut be true for -, to pick a value at random. In thi cae we get, 5 5 A( - 7) + B( ) fi A- 7 We found A by appropriately picking. We can B in the ame way if we choe 5. A( ) + B( 7) fi B 7 Thi will not alway work, but when it doe it will uually implify the work coniderably. So, with thee contant the tranform become, H( ) We can now eaily do the invere tranform to get, 5 - t ht - e + e 7 7 5t () [Return to Problem] The lat part of thi example needed partial fraction to get the invere tranform. When we finally get back to differential equation and we tart uing Laplace tranform to olve them, you will quickly come to undertand that partial fraction are a fact of life in thee problem. Almot every problem will require partial fraction to one degree or another. Note that we could have done the lat part of thi example a we had done the previou two part. If we had we would have gotten hyperbolic function. However, recalling the definition of the hyperbolic function we could have written the reult in the form we got from the way we worked our problem. However, mot tudent have a better feel for exponential than they do for hyperbolic function and o it uually bet to jut ue partial fraction and get the anwer in 7 Paul Dawkin 9

21 term of exponential. It may be a little more work, but it will give a nicer (and eaier to work with) form of the anwer. Be warned that in my cla I ve got a rule that if the denominator can be factored with integer coefficient then it mut be. So, let remind you how to get the correct partial fraction decompoition. The firt tep i to factor the denominator a much a poible. Then for each term in the denominator we will ue the following table to get a term or term for our partial fraction decompoition. Factor in denominator ax+ b ( ax+ b) k Term in partial fraction decompoition A ax+ b A A Ak + + L + ax+ b ax+ b ax+ b ax + bx+ c Ax+ B ax + bx+ c k ax + bx+ c Ax + B Ax + B Ax k + Bk + + L + ax + bx+ c ax + bx+ c ax + bx+ c Notice that the firt and third cae are really pecial cae of the econd and fourth cae repectively. So, let do a couple more example to remind you how to do partial fraction. Example Find the invere tranform of each of the following (a) G( ) [Solution] Solution (a) G( ) (b) F( ) (c) G( ) - 5 ( - 6)( + ) 5 ( ) ( + )( -4)( 5-) [Solution] [Solution] Here the partial fraction decompoition for thi part. A B C G( ) Now, thi time we won t go into quite the detail a we did in the lat example. We are after the 7 Paul Dawkin k k

22 numerator of the partial fraction decompoition and thi i uually eay enough to do in our head. Therefore, we will go traight to etting numerator equal A B C A with the lat example, we can eaily get the contant by correctly picking value of A -7-6 fi A- 4 Ê6ˆÊ 9 ˆ - CÁ Á- fi C Ë 5 Ë B 7 9 fi B So, the partial fraction decompoition for thi tranform i, 5 G( ) Now, in order to actually take the invere tranform we will need to factor a 5 out of the denominator of the lat term. The corrected tranform a well a it invere tranform i. G( ) (b) F( ) - 5 ( - 6)( + ) t -t 4t 5 () - e + e + e g t 5 [Return to Problem] So, for the firt time we ve got a quadratic in the denominator. Here the decompoition for thi part. A B+ C F( ) Setting numerator equal give, - 5 A + + B+ C - 6 Okay, in thi cae we could ue 6 to quickly find A, but that all it would give. In thi cae we will need to go the long way around to getting the contant. Note that thi way will alway work, but i ometime more work than i required. The long way i to completely multiply out the right ide and collect like term. - 5 A + + B+ C A A B 6B C 6C A+ B + - 6B+ C + A-6C In order for thee two to be equal the coefficient of the, and the contant mut all be equal. 7 Paul Dawkin

23 So, etting coefficient equal give the following ytem of equation that can be olved. : A+ B Ô : - 6B+ C -5 fi A-, B, C :A- 6C Ô Notice that I ued to denote the contant. Thi i habit on my part and in t really required, it jut what I m ued to doing. Alo, the coefficient are fairly mey fraction in thi cae. Get ued to that. They will often be like thi when we get back into olving differential equation. There i a way to make our life a little eaier a well with thi. Since all of the fraction have a denominator of 47 we ll factor that out a we plug them back into the decompoition. Thi will make dealing with them much eaier. The partial fraction decompoition i then, Ê ˆ F( ) Á- + 47Ë Ê Ë Á ˆ The invere tranform i then t f t Ê 8co t in t 47 Á- e + - Ë () ˆ [Return to Problem] (c) G( ) 5 ( ) With thi lat part do not get excited about the. We can think of thi term a ( ) - and it become a linear term to a power. So, the partial fraction decompoition i A B C D+ E G( ) Setting numerator equal and multiplying out give. 5 A B C D+ E 4 A+ D + 4A+ B+ E + 5A+ 4B+ C + 5B+ 4C + 5C Setting coefficient equal give the following ytem. 7 Paul Dawkin

24 4 : A+ D Ô : 4A+ B+ E Ô 4 :5A+ 4B+ C fi A, B- 4, C 5, D-, E : 5B+ 4C Ô Ô : 5C 5Ô Thi ytem look mey, but it eaier to olve than it might look. Firt we get C for free from the lat equation. We can then ue the fourth equation to find B. The third equation will then give A, etc. When plugging into the decompoition we ll get everything with a denominator of 5, then factor that out a we did in the previou part in order to make thing eaier to deal with. Ê ˆ G( ) Á Ë Note that we alo factored a minu ign out of the lat two term. To complete thi part we ll need to complete the quare on the later term and fix up a couple of numerator. Here that work. Ê ˆ G( ) Á Ë Ê 5 ( + - ) + 4 ˆ Á ( + ) + Ë ( + ) Ê Ë!! 5Á The invere tranform i then. Ê 5 t co t g t Á - t+ t - e - t - e - in t 5Ë () () () ˆ ˆ [Return to Problem] So, one final time. Partial fraction are a fact of life when uing Laplace tranform to olve differential equation. Make ure that you can deal with them. 7 Paul Dawkin

25 Step Function Before proceeding into olving differential equation we hould take a look at one more function. Without Laplace tranform it would be much more difficult to olve differential equation that involve thi function in g(t). The function i the Heaviide function and i defined a, Ï if t < c uc () t Ì Ó if t c Here i a graph of the Heaviide function. Heaviide function are often called tep function. Here i ome alternate notation for Heaviide function. u t u t- c H t- c c We can think of the Heaviide function a a witch that i off until t c at which point it turn on and take a value of. So what if we want a witch that will turn on and take ome other value, ay 4, or -7? Heaviide function can only take value of or, but we can ue them to get other kind of witche. For intance 4u c (t) i a witch that i off until t c and then turn on and take a value of 4. Likewie, -7u c (t) will be a witch that will take a value of -7 when it turn on. Now, uppoe that we want a witch that i on (with a value of ) and then turn off at t c. We can ue Heaviide function to repreent thi a well. The following function will exhibit thi kind of behavior. Ï- if t < c () t - uc Ì Ó - if t c Prior to t c the Heaviide i off and o ha a value of zero. The function a whole then for t < c ha a value of. When we hit t c the Heaviide function will turn on and the function will now take a value of. We can alo modify thi o that it ha value other than when it i one. For intance, 7 Paul Dawkin 4

26 ( t) - u c will be a witch that ha a value of until it turn off at t c. We can alo ue Heaviide function to repreent much more complicated witche. Example Write the following function (or witch) in term of Heaviide function. Ï- 4 if t < 6 Ô5 if 6 t < 8 f () t Ì Ô6 if 8 t < Ô Ó if t Solution There are three udden hift in thi function and o (hopefully) it clear that we re going to need three Heaviide function here, one for each hift in the function. Here the function in term of Heaviide function. f t u t -9u t - 6u t It fairly eay to verify thi. 6 8 In the firt interval, t < 6 all three Heaviide function are off and the function ha the value f t - 4 Notice that when we know that Heaviide function are on or off we tend to not write them at all a we did in thi cae. In the next interval, 6 t < 8 the firt Heaviide function i now on while the remaining two are till off. So, in thi cae the function ha the value. f ( t ) In the third interval, 8 t < the firt two Heaviide function are one while the lat remain off. Here the function ha the value. f ( t ) In the lat interval, t all three Heaviide function are one and the function ha the value. f ( t ) So, the function ha the correct value in all the interval. All of thi i fine, but if we continue the idea of uing Heaviide function to repreent witche, we really need to acknowledge that mot witche will not turn on and take contant value. Mot witche will turn on and vary continually with the value of t. So, let conider the following function. 7 Paul Dawkin 5

27 We would like a witch that i off until t c and then turn on and take the value above. By thi we mean that when t c we want the witch to turn on and take the value of f() and when t c + 4 we want the witch to turn on and take the value of f(4), etc. In other word, we want the witch to look like the following, Notice that in order to take the value that we want the witch to take it need to turn on and take f t- c! We can ue Heaviide function to help u repreent thi witch a well. the value of Uing Heaviide function thi witch can be written a g t u t f t- c () c Okay, we ve talked a lot about Heaviide function to thi point, but we haven t even touched on Laplace tranform yet. So, let tart thinking about that. Let determine the Laplace tranform of (). Thi i actually eay enough to derive o let do that. Plugging () into the definition of the Laplace tranform give, t { c() (- )} e - c() (- ) Ú L u t f t c u t f t c dt Ú - t c e f t-c dt 7 Paul Dawkin 6

28 Notice that we took advantage of the fact that the Heaviide function will be zero if t < c and otherwie. Thi mean that we can drop the Heaviide function and tart the integral at c intead of. Now ue the ubtitution u t c and the integral become, L u ( c) { c() } e 7 Paul Dawkin u - c u t f t c f u du Ú Ú e e f u du The econd exponential ha no u in it and o it can be factored out of the integral. Note a well that in the ubtitution proce the lower limit of integration went back to. -c -u L { uc() t f ( t- c) } e Ú e f ( u) du Now, the integral left i nothing more than the integral that we would need to compute if we were going to find the Laplace tranform of f(t). Therefore, we get the following formula -c L u t f t- c e F () { c } In order to ue () the function f(t) mut be hifted by c, the ame value that i ued in the Heaviide function. Alo note that we only take the tranform of f(t) and not f(t-c)! We can alo turn thi around to get a ueful formula for invere Laplace tranform. L - e -c F u t f t - c () { } We can ue () to get the Laplace tranform of a Heaviide function by itelf. To do thi we will conider the function in () to be f(t). Doing thi give u -c -c -c L{ uc() t } L{ uc() t g} e L {} e e Putting all of thi together lead to the following two formula. -c -c e -Ïe L{ uc() t } L Ì uc() t (4) Ó Let do ome example. Example Find the Laplace tranform of each of the following. (a) () () () ( - t g t u t t 6 u t 7 ) u () t e [Solution] 6 4 (b) f ( t) tu ( t) co( t) u ( t) - + [Solution] 5 4 Ï t if t < 5 Ô (c) ht () Ì 4 Ê t ˆ [Solution] Ôt + iná - if t 5 Ó Ë ÏÔ t if t < 6 (d) f () t Ì 8 ( t 6) [Solution] ÔÓ if t 6 c

29 Solution In all of thee problem remember that the function MUST be in the form u t f t- c c before we tart taking tranform. If it in t in that form we will have to put it into that form! (a) () () () ( - t g t u t + t-6 u t - 7-e ) u () t 6 4 So there are three term in thi function. The firt i imply a Heaviide function and o we can ue (4) on thi term. The econd and third term however have function with them and we need to identify the function that are hifted for each of thee. In the econd term it appear that we are uing the following function, () fi (- 6) ( - 6) f t t f t t and thi ha been hifted by the correct amount. The third term ue the following function, t -( t-4) f t 7-e fi f t e 7-e which ha alo been hifted by the correct amount. - -t With thee function identified we can now take the tranform of the function. - e -6! Ê7 ˆ -4 G( ) + e - + Á - e Ë e e Ê7 + Á Ë ˆ e + -4 [Return to Problem] (b) f ( t) - tu ( t) + co( t) u ( t) 5 Thi part i going to caue ome problem. There are two term and neither ha been hifted by the proper amount. The firt term need to be hifted by and the econd need to be hifted by 5. So, ince they haven t been hifted, we will need to force the iue. We will need to add in the hift, and then take them back out of coure. Here they are. () -( - + ) () + co( ) () f t t u t t u t 5 Now we till have ome potential problem here. The firt function i till not really hifted correctly, o we ll need to ue to get thi hifted correctly. a+ b a + ab+ b The econd term can be dealt with in one of two way. The firt would be to ue the formula co a+ b co a co b - in a in b to break it up into coine and ine with argument of t-5 which will be hifted a we expect. There i an eaier way to do thi one however. From our table of Laplace tranform we have #6 and uing that we can ee that if 7 Paul Dawkin 8

30 co( + 5) fi ( - 5) co( ) g t t g t t Thi will make our life a little eaier o we ll do it thi way. Now, breaking up the firt term and leaving the econd term alone give u, ( 6 9) () co( 5 5) 5() () f t - t- + t- + u t + t- + u t Okay, o it look like the two function that have been hifted here are g t t + 6t+ 9 () co( t+ 5) Taking the tranform then give, Ê 6 9ˆ Ê co( 5) - in( 5) ˆ F( ) - Á + + +Á Ë e Ë + e g t - -5 It mey, epecially the econd term, but there it i. Alo, do not get excited about the co( 5 ) and in( 5 ). They are jut number. 4 Ï t if t < 5 Ô (c) ht () Ì 4 Ê t ˆ Ôt + iná - if t 5 Ó Ë [Return to Problem] Thi one in t a bad a it might look on the urface. The firt thing that we need to do i write it in term of Heaviide function. 4 Ê t ˆ ht () t + u5() t in Á - Ë 4 Ê ˆ t + u5() t iná ( t-5) Ë Since the t 4 i in both term there in t anything to do when we add in the Heaviide function. The only thing that get added in i the ine term. Notice a well that the ine ha been hifted by the proper amount. All we need to do now i to take the tranform. 4! H( ) + -5 ( ) e e [Return to Problem] 7 Paul Dawkin 9

31 ÏÔ t if t < 6 (d) f () t Ì ÔÓ ( t - 6) if t 6 Again, the firt thing that we need to do i write the function in term of Heaviide function. 6() () (- 6) f t t t t u t We had to add in a -8 in the econd term ince that appear in the econd part and we alo had to ubtract a t in the econd term ince the t in the firt portion i no longer there. Thi ubtraction of the t add a problem becaue the econd function i no longer correctly hifted. Thi i eaier to fix than the previou example however. Here i the corrected function. 6() ( 8 ( 6) 6 ( 6) ) 6() ( 4 ( 6) ( 6) ) 6() () + -8-(- 6+ 6) + (-6) f t t t t u t t+ - - t- - + t- u t t+ - - t- + t- u t So, in the econd term it look like we are hifting g t t -t- 4 The tranform i then, Ê 4 ˆ Á Ë e [Return to Problem] F Without the Heaviide function taking Laplace tranform i not a terribly difficult proce provided we have our truty table of tranform. However, with the advent of Heaviide function, taking tranform can become a fairly mey proce on occaion. So, let do ome invere Laplace tranform to ee how they are done. Example Find the invere Laplace tranform of each of the following. -4 e (a) H( ) [Solution] + - (b) G( ) (c) F( ) (d) G( ) -6-5e -e ( + )( + 9) 4 + e - ( - )( + ) ( + ) [Solution] [Solution] e - e + 6e [Solution] 7 Paul Dawkin

32 Solution All of thee will ue () omewhere in the proce. Notice that in order to ue thi formula the exponential doen t really enter into the mix until the very end. The vat majority of the proce i finding the invere tranform of the tuff without the exponential. In thee problem we are not going to go into detail on many of the invere tranform. If you need a refreher on ome of the baic of invere tranform go back and take a look at the previou ection. (a) H( ) e -4 ( + )( -) In light of the comment above let firt rewrite the tranform in the following way H( ) e e F( ) + - Now, thi problem really come down to needing f(t). So, let do that. We ll need to partial fraction F() up. Here the partial fraction decompoition. A B F( ) Setting numerator equal give, ( ) A- + B + We ll find the contant here by electing value of. Doing thi give, 8B fi B A fi A 4 So, the partial fraction decompoition become, F ( ) Notice that we factored a out of the denominator in order to actually do the invere tranform. The invere tranform of thi i then, t - t f () t e + e 4 Now, let go back and do the actual problem. The original tranform wa, -4 H e F Note that we didn t bother to plug in F(). There really in t a reaon to plug it back in. Let jut ue () to write down the invere tranform in term of ymbol. The invere tranform i, 7 Paul Dawkin

33 where, f(t) i, ( - ) ht u t f t () f t 4 4 e + e 4 t - t Thi i all the farther that we ll go with the anwer. There really in t any reaon to plug in f(t) at thi point. It would make the function longer and definitely meier. We will give almot all of our anwer to thee type of invere tranform in thi form. [Return to Problem] (b) G( ) -6-5e -e ( + )( + 9) Thi problem i not a difficult a it might at firt appear to be. Becaue there are two exponential we will need to deal with them eparately eventually. Now, thi might lead u to conclude that the bet way to deal with thi function i to plit it up a follow, G( ) e -e Notice that we factored out the exponential, a we did in the lat example, ince we would need to do that eventually anyway. Thi i where a fairly common complication arie. Many people will call the firt function F() and the econd function H() and the partial fraction both of them. However, if intead of jut factoring out the exponential we would alo factor out the coefficient we would get, -6 - G( ) 5e -e Upon doing thi we can ee that the two function are in fact the ame function. The only difference i the contant that wa in the numerator. So, the way that we ll do thee problem i to firt notice that both of the exponential have only contant a coefficient. Intead of breaking thing up then, we will imply factor out the whole numerator and get, G( ) ( 5e - e ) ( ( 5e -e ) F( ) ) and now we will jut partial fraction F(). Here i the partial fraction decompoition. A B+ C F Setting numerator equal and combining give u, A B+ C A+ B + B+ C + 9A+ C 7 Paul Dawkin

34 Setting coefficient equal and olving give, : A+ B Ô : B+ C fi A, B-, C :9A+ C Ô Subtituting back into the tranform give and fixing up the numerator a needed give, Ê - + ˆ F( ) Á + Ë+ + 9 Ê ˆ Á Ë A we did in the previou ection we factored out the common denominator to make our work a little impler. Taking the invere tranform then give, Ê -t ˆ f () t Áe - co( t) + in( t) Ë At thi point we can go back and tart thinking about the original problem G 5e -e F 5e F - e F( ) -6 - We ll alo need to ditribute the F() through a well in order to get the correct invere tranform. Recall that in order to ue () to take the invere tranform you mut have a ingle exponential time a ingle tranform. Thi mean that we mut multiply the F() through the parenthei. We can now take the invere tranform, g t 5u t f t-6 -u t f t- where, (c) F( ) 4 + e - ( - )( + ) 6 Ê f t Á t t Ë -t () e - co( ) + in( ) ˆ [Return to Problem] In thi cae, unlike the previou part, we will need to break up the tranform ince one term ha a contant in it and the other ha an. Note a well that we don t conider the exponential in thi, only it coefficient. Breaking up the tranform give, F( ) + e G( ) + e H( ) We will need to partial fraction both of thee term up. We ll tart with G(). A B G( ) Paul Dawkin

35 Setting numerator equal give, 4 A + + B - Now, pick value of to find the contant B fi 8 B 4 A fi 4 A So G() and it invere tranform i, Now, repeat the proce for H(). Setting numerator equal give, 4 8 G t 8 -t g() t e + e H A B B( ) A Now, pick value of to find the contant. - -B fi B- A fi A So H() and it invere tranform i, H t -t ht () e - e Putting all of thi together give the following, - F G + e H where, () () + () (- ) f t g t u t ht 4 8 e + e e - e () and ht () g t t - t t - t [Return to Problem] 7 Paul Dawkin 4

36 (d) G( ) e - e + 6e ( + ) Thi one look meier than it actually i. Let firt rearrange the numerator a little ( - e ) + ( 8e + 6e ) G( ) + In thi form it look like we can break thi up into two piece that will require partial fraction. When we break thee up we hould alway try and break thing up into a few piece a poible for the partial fractioning. Doing thi can ave you a great deal of unneceary work. Breaking up the tranform a uggeted above give, G( ) ( - e ) + ( 8e + 6e ) ( e ) F( ) ( 8e 6e ) H( ) Note that we canceled an in F(). You hould alway implify a much a poible before doing the partial fraction. Let partial fraction up F() firt. Setting numerator equal give, F A B + + A + + B Now, pick value of to find the contant. - -B fi B- A fi A So F() and it invere tranform i, Now partial fraction H(). H F - + -t f () t - e A B C Setting numerator equal give, A + + B + + C 7 Paul Dawkin 5

37 Pick value of to find the contant. - 9C fi C 9 B fi B 4A+ 4B+ C 4A+ fi A- 9 9 So H() and it invere tranform i, 9 9 H t ht () - + t+ e 9 9 Now, let go back to the original problem, remembering to multiply the tranform through the parenthei G F - e F + 8e H + 6e H Taking the invere tranform give, g t f t -u t f t- + 8u t ht- + 6u t ht- 7 7 [Return to Problem] So, a thi example ha hown, thee can be a omewhat mey. However, the me i really only that of notation and amount of work. The actual partial fraction work wa identical to the previou ection work. The main difference in thi ection i we had to do more of it. A far a the invere tranform proce goe. Again, the vat majority of that wa identical to the previou ection a well. So, don t let the apparent meine of thee problem get you to decide that you can t do them. Generally they aren t a bad a they eem initially. 7 Paul Dawkin 6