Chapter 4. The Laplace Transform Method

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1 Chapter 4. The Laplace Tranform Method The Laplace Tranform i a tranformation, meaning that it change a function into a new function. Actually, it i a linear tranformation, becaue it convert a linear combination of function into a linear combination of the tranformed function. Even more intereting, the Laplace Tranform convert derivative into multiplication. Thee two propertie make the Laplace Tranform very ueful to olve linear differential equation with contant coefficient. The Laplace Tranform convert uch differential equation for an unknown function into an algebraic equation for the tranformed function. Uually it i eay to olve the algebraic equation for the tranformed function. Then one convert the tranformed function back into the original function. Thi function i the olution of the differential equation. Solving a differential equation uing a Laplace Tranform i radically different from all the method we have ued o far. Thi method, a we will ue it here, i relatively new. The Laplace Tranform we define here wa firt ued in 9, but it ue grew rapidly after 92, pecially to olve differential equation. Tranformation like the Laplace Tranform were known much earlier. Pierre Simon de Laplace ued a imilar tranformation in hi tudie of probability theory, publihed in 82, but analogou tranformation were ued even earlier by Euler around 737. δ n 3 δ 3 (t) 2 δ 2 (t) δ (t) 3 2 t

2 2 4.. Introduction to the Laplace Tranform The Laplace tranform i a tranformation it change a function into another function. Thi tranformation i an integral tranformation the original function i multiplied by an exponential and integrated on an appropriate region. Such an integral tranformation i the anwer to very intereting quetion: I it poible to tranform a differential equation into an algebraic equation? I it poible to tranform a derivative of a function into a multiplication? The anwer to both quetion i ye, for example with a Laplace tranform. Thi i how it work. You tart with a derivative of a function, y (t), then you multiply it by any function, we chooe an exponential e t, and then you integrate on t, o we get y (t) e t y (t) dt, which i a tranformation, an integral tranformation. And now, becaue we have an integration above, we can integrate by part thi i the big idea, y (t) e t y (t) dt = e t y(t) + e t y(t) dt. So we have tranformed the derivative we tarted with into a multiplication by thi contant from the exponential. The idea in thi calculation actually work to olve differential equation and motivate u to define the integral tranformation y(t) Ỹ () a follow, y(t) Ỹ () = e t y(t) dt. The Laplace tranform i a tranformation imilar to the one above, where we chooe ome appropriate integration limit which are very convenient to olve initial value problem. We dedicate thi ection to introduce the precie definition of the Laplace tranform and how i ued to olve differential equation. In the following ection we will ee that thi method can be ued to olve linear contant coefficient differential equation with very general ource, including Dirac delta generalized function Oveview of the Method. The Laplace tranform change a function into another function. For example, we will how later on that the Laplace tranform change a f(x) = in(ax) into F (x) = x 2 + a 2. We will follow the notation ued in the literature and we ue t for the variable of the original function f, while we ue for the variable of the tranformed function F. Uing thi notation, the Laplace tranform change a f(t) = in(at) into F () = 2 + a 2. We will how that the Laplace tranform i a linear tranformation and it tranform derivative into multiplication. Becaue of thee propertie we will ue the Laplace tranform to olve linear differential equation. We Laplace tranform the original differential equation. Becaue the the propertie above, the reult will be an algebraic equation for the tranformed function. Algebraic equation are imple to olve, o we olve the algebraic equation. Then we Laplace tranform back the olution. We ummarize thee tep a follow, [ ] differential L eq. for y. () Algebraic eq. for L[y]. (2) Solve the algebraic eq. for L[y]. (3) Tranform back to obtain y. (Ue the table.)

3 The Laplace Tranform. The Laplace tranform i a tranformation, meaning that it convert a function into a new function. We have een tranformation earlier in thee note. In Chapter?? we ued the tranformation L[y(t)] = y (t) + a y (t) + a y(t), o that a econd order linear differential equation with ource f could be written a L[y] = f. There are impler tranformation, for example the differentiation operation itelf, D[f(t)] = f (t). Not all tranformation involve differentiation. There are integral tranformation, for example integration itelf, I[f(t)] = x f(t) dt. Of particular importance in many application are integral tranformation of the form T [f(t)] = b a K(, t) f(t) dt, where K i a fixed function of two variable, called the kernel of the tranformation, and a, b are real number or ±. The Laplace tranform i a tranfomation of thi type, where the kernel i K(, t) = e t, the contant a =, and b =. Definition 4... The Laplace tranform of a function f defined on D f = (, ) i F () = defined for all D F R where the integral converge. e t f(t) dt, (4..) In thee note we ue an alternative notation for the Laplace tranform that emphaize that the Laplace tranform i a tranformation: L[f] = F, that i L[ ] = e t ( So, the Laplace tranform will be denoted a either L[f] or F, depending whether we want to emphaize the tranformation itelf or the reult of the tranformation. We will alo ue the notation L[f(t)], or L[f](), or L[f(t)](), whenever the independent variable t and are relevant in any particular context. The Laplace tranform i an improper integral an integral on an unbounded domain. Improper integral are defined a a limit of definite integral, t g(t) dt = lim t ) dt. g(t) dt. An improper integral converge iff the limit exit, otherwie the integral diverge. Now we are ready to compute our firt Laplace tranform. Example 4..: Compute the Laplace tranform of the function f(t) =, that i, L[]. Solution: Following the definition, L[] = e t dt = lim e t dt. The definite integral above i imple to compute, but it depend on the value of. For = we get lim dt = lim N =. n

4 4 So, the improper integral diverge for =. For we get lim For < we have =, hence e t dt = lim N e t = lim (e N ). lim (e N ) = lim (e N ) =. So, the improper integral diverge for <. In the cae that > we get If we put all thee reult together we get lim (e N ) =. L[] =, >. Example 4..2: Compute L[e at ], where a R. Solution: We tart with the definition of the Laplace tranform, In the cae = a we get L[e at ] = e t (e at ) dt = L[e at ] = dt =, o the improper integral diverge. In the cae a we get L[e at ] = lim = lim = lim Now we have to remaining cae. The firt cae i: e ( a)t dt. e ( a)t dt, a, [ ( ) N] ( a) e ( a)t [ ( ) ] ( a) (e ( a)n ). a < ( a) = a > lim e ( a)n =, o the integral diverge for < a. The other cae i: a > ( a) = a < lim e ( a)n =, o the integral converge only for > a and the Laplace tranform i given by L[e at ] = ( a), > a. Example 4..3: Compute L[te at ], where a R. Solution: In thi cae the calculation i more complicated than above, ince we need to integrate by part. We tart with the definition of the Laplace tranform, L[te at ] = e t te at dt = lim te ( a)t dt.

5 5 Thi improper integral diverge for = a, o L[te at ] i not defined for = a. From now on we conider only the cae a. In thi cae we can integrate by part, [ L[te at ] = lim N ( a) te ( a)t + ] e ( a)t dt, a that i, [ L[te at ] = lim N ( a) te ( a)t In the cae that < a the firt term above diverge, N] ( a) 2 e ( a)t. (4..2) lim ( a) N e ( a)n = lim ( a) N e a N =, therefore L[te at ] i not defined for < a. In the cae > a the firt term on the right hand ide in (4..2) vanihe, ince lim ( a) N e ( a)n =, Regarding the other term, and recalling that > a, lim ( a) 2 e ( a)n =, Therefore, we conclude that L[te at ] = Example 4..4: Compute L[in(at)], where a R. Solution: In thi cae we need to compute L[in(at)] = ( a) t e ( a)t t= =. ( a) 2 e ( a)t t= = ( a) 2. ( a) 2, > a. e t in(at) dt = lim e t in(at) dt The definite integral above can be computed integrating by part twice, e t in(at) dt = [ e t in(at) ] N a [ e t 2 co(at) ] N which implie that ( + a2 2 ) then we get and finally we get e t in(at) dt = e t in(at) dt = a2 2 e t in(at) dt, e t in(at) dt = [ e t in(at) ] N a [ e t 2 co(at) ] N. 2 [ ( 2 + a 2 [ e t in(at) ] N ) a 2 [ e t co(at) ] N 2 [ ( 2 + a 2 [ e N in(an) ] a [ e N ) 2 co(an) ]]. One can check that the limit N on the right hand ide above doe not exit for, o L[in(at)] doe not exit for. In the cae > it i not difficult to ee that ( e t 2 )[ in(at) dt = 2 + a 2 ( ) a ] 2 ( ) ].

6 6 o we obtain the final reult L[in(at)] = a 2 + a 2, >. In Table we preent a hort lit of Laplace tranform. They can be computed in the ame way we computed the the Laplace tranform in the example above. f(t) F () = L[f(t)] D F f(t) = F () = > f(t) = e at F () = ( a) > a f(t) = t n F () = n! (n+) > f(t) = in(at) F () = f(t) = co(at) F () = f(t) = inh(at) F () = f(t) = coh(at) F () = f(t) = t n e at F () = f(t) = e at in(bt) F () = f(t) = e at co(bt) F () = f(t) = e at inh(bt) F () = f(t) = e at coh(bt) F () = a 2 + a 2 > 2 + a 2 > a 2 a 2 2 a 2 > a > a n! ( a) (n+) > a b ( a) 2 + b 2 > a ( a) ( a) 2 + b 2 > a b ( a) 2 b 2 a > b ( a) ( a) 2 b 2 a > b Table. Lit of a few Laplace tranform Main Propertie. Since we are more or le confident on how to compute a Laplace tranform, we can tart aking deeper quetion. For example, what type of function have a Laplace tranform? It turn out that a large cla of function, thoe that are piecewie continuou on [, ) and bounded by an exponential. Thi lat property i particularly important and we give it a name.

7 7 Definition A function f defined on [, ) i of exponential order, where i any real number, iff there exit poitive contant k, T uch that f(t) k e t for all t > T. (4..3) Remark: (a) When the precie value of the contant i not important we will ay that f i of exponential order. (b) An example of a function that i not of exponential order i f(t) = e t2. Thi definition help to decribe a et of function having Laplace tranform. Piecewie continuou function on [, ) of exponential order have Laplace tranform. Theorem 4..3 (Convergence of LT). If a function f defined on [, ) i piecewie continuou and of exponential order, then the L[f] exit for all > and there exit a poitive contant k uch that L[f] k, >. Proof of Theorem 4..3: From the definition of the Laplace tranform we know that L[f] = lim e t f(t) dt. The definite integral on the interval [, N] exit for every N > ince f i piecewie continuou on that interval, no matter how large N i. We only need to check whether the integral converge a N. Thi i the cae for function of exponential order, becaue e t f(t) dt e t f(t) dt Therefore, for > we can take the limit a N, L[f] lim e t ke t dt = k e t f(t) dt k L[e t k ] = ( ). e ( )t dt. Therefore, the comparion tet for improper integral implie that the Laplace tranform L[f] exit at leat for >, and it alo hold that L[f] k, >. Thi etablihe the Theorem. The next reult ay that the Laplace tranform i a linear tranformation. Thi mean that the Laplace tranform of a linear combination of function i the linear combination of their Laplace tranform. Theorem 4..4 (Linearity). If L[f] and L[g] exit, then for all a, b R hold L[af + bg] = a L[f] + b L[g].

8 8 Proof of Theorem 4..4: Since integration i a linear operation, o i the Laplace tranform, a thi calculation how, L[af + bg] = Thi etablihe the Theorem. = a e t[ af(t) + bg(t) ] dt e t f(t) dt + b = a L[f] + b L[g]. Example 4..5: Compute L[3t co(4t)]. e t g(t) dt Solution: From the Theorem above and the Laplace tranform in Table?? we know that Therefore, L[3t co(4t)] = 3 L[t 2 ] + 5 L[co(4t)] ( 2 ) ( ) = , > = L[3t co(4t)] = ( 2, >. + 6) The Laplace tranform can be ued to olve differential equation. The Laplace tranform convert a differential equation into an algebraic equation. Thi i o becaue the Laplace tranform convert derivative into multiplication. Here i the precie reult. Theorem 4..5 (Derivative into Multiplication). If a function f i continuouly differentiable on [, ) and of exponential order, then L[f ] exit for > and L[f ] = L[f] f(), >. (4..4) Proof of Theorem 4..5: The main calculation in thi proof i to compute L[f ] = lim e t f (t) dt. We tart computing the definite integral above. Since f i continuou on [, ), that definite integral exit for all poitive N, and we can integrate by part, [( ) e t f (t) dt = e t N N ] f(t) ( )e t f(t) dt = e N f(n) f() + e t f(t) dt. We now compute the limit of thi expreion above a N. Since f i continuou on [, ) of exponential order, we know that lim e t f(t) dt = L[f], >. Let u ue one more time that f i of exponential order. Thi mean that there exit poitive contant k and T uch that f(t) k e t, for t > T. Therefore, lim e N f(n) lim k e N e N = lim k e ( )N =, >.

9 9 Thee two reult together imply that L[f ] exit and hold L[f ] = L[f] f(), >. Thi etablihe the Theorem. Example 4..6: Verify the reult in Theorem 4..5 for the function f(t) = co(bt). Solution: We need to compute the left hand ide and the right hand ide of Eq. (4..4) and verify that we get the ame reult. We tart with the left hand ide, L[f ] = L[ b in(bt)] = b L[in(bt)] = b We now compute the right hand ide, o we get L[f] f() = L[co(bt)] = L[f] f() = 2 + b 2. We conclude that L[f ] = L[f] f(). b 2 + b 2 L[f ] = b2 2 + b b 2 = 2 2 b b 2, It i not difficult to generalize Theorem 4..5 to higher order derivative. Theorem 4..6 (Higher Derivative into Multiplication). If a function f i n-time continuouly differentiable on [, ) and of exponential order, then L[f ],, L[f (n) ] exit for > and b2 L[f ] = 2 L[f] f() f () (4..5). L[f (n) ] = n L[f] (n ) f() f (n ) (). (4..6) Proof of Theorem 4..6: We need to ue Eq. (4..4) n time. We tart with the Laplace tranform of a econd derivative, L[f ] = L[(f ) ] = L[f ] f () = ( L[f] f() ) f () = 2 L[f] f() f (). The formula for the Laplace tranform of an nth derivative i computed by induction on n. We aume that the formula i true for n, L[f (n ) ] = (n ) L[f] (n 2) f() f (n 2) (). Since L[f (n) ] = L[(f ) (n ) ], the formula above on f give L[(f ) (n ) ] = (n ) L[f ] (n 2) f () (f ) (n 2) () = (n ) ( L[f] f() ) (n 2) f () f (n ) () = (n) L[f] (n ) f() (n 2) f () f (n ) (). Thi etablihe the Theorem.

10 Example 4..7: Verify Theorem 4..6 for f, where f(t) = co(bt). Solution: We need to compute the left hand ide and the right hand ide in the firt equation in Theorem (4..6), and verify that we get the ame reult. We tart with the left hand ide, L[f ] = L[ b 2 co(bt)] = b 2 L[co(bt)] = b 2 We now compute the right hand ide, o we get 2 L[f] f() f () = 2 L[co(bt)] = 2 2 L[f] f() f () = 2 + b 2. We conclude that L[f ] = 2 L[f] f() f (). 2 + b 2 L[f ] = b2 2 + b b 2 = 3 3 b b 2, Solving Differential Equation. The Laplace tranform can be ued to olve differential equation. We Laplace tranform the whole equation, which convert the differential equation for y into an algebraic equation for L[y]. We olve the Algebraic equation and we tranform back. ] Solve the Tranform back [ differential L eq. for y. () Algebraic eq. for L[y]. (2) b2 algebraic eq. for L[y]. Example 4..8: Ue the Laplace tranform to find y olution of y + 9 y =, y() = y, y () = y. (3) to obtain y. (Ue the table.) Remark: Notice we already know what the olution of thi problem i. Following?? we need to find the root of p(r) = r r +- = ±3 i, and then we get the general olution Then the initial condition will ay that y(t) = c + co(3t) + c - in(3t). y(t) = y co(3t) + y 3 in(3t). We now olve thi problem uing the Laplace tranform method. Solution: We now ue the Laplace tranform method: L[y + 9y] = L[] =. The Laplace tranform i a linear tranformation, L[y ] + 9 L[y] =. But the Laplace tranform convert derivative into multiplication, 2 L[y] y() y () + 9 L[y] =. Thi i an algebraic equation for L[y]. It can be olved by rearranging term and uing the initial condition, ( 2 + 9) L[y] = y + y L[y] = y ( 2 + 9) + y ( 2 + 9).

11 But from the Laplace tranform table we ee that L[co(3t)] = , L[in(3t)] = , therefore, L[y] = y L[co(3t)] + y 3 L[in(3t)]. Once again, the Laplace tranform i a linear tranformation, [ L[y] = L y co(3t) + y ] 3 in(3t). We obtain that y(t) = y co(3t) + y 3 in(3t).

12 Exercie

Practice Problems - Week #7 Laplace - Step Functions, DE Solutions Solutions

Practice Problems - Week #7 Laplace - Step Functions, DE Solutions Solutions For Quetion -6, rewrite the piecewie function uing tep function, ketch their graph, and find F () = Lf(t). 0 0 < t < 2. f(t) = (t 2 4) 2 < t In tep-function form, f(t) = u 2 (t 2 4) The graph i the olid