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Unformatted text preview: The Laplace Inansform Here we give a very brieﬂ introduction to the Laplacd ftransforml a iiseful devicd for so VinQ bertairﬂ kinds bfl differential eguationsl
Suppose f(t) is a piecewisg continuous function deﬁned fori ﬂ : WhiChl grows at mosﬁ exponentiallyiasitjgthatis,ﬁwhiclihhereexistconstantsUzandagllsnchthat
IWWEQﬁl (H
W Zaplacd transform M i is ﬁfe function Elsi deﬁned for 3 S a Q ms) = /°° Wm) at. (21 L KM View of Q], the integra cowerges fori s S a Q comparison to K Wk“ The Iaplacd
transform itselfl is the operation that takes f E El Mien it is useful to indicate this operation
explicitly, we denote iii by El and write E = Zllfl 01‘ E = 3W0] Let’d compute a fevxi examp ed to get started: ;=/Edt=1sﬂ=§i (ﬁzg. @
0 _
Zl[cos bt] — /00 6’“ cos bt dt —€iSt(3 cos bt : bsiri bit] m —g— (3 > 0] (4
— L 2—52 W [L a E? '
. _ W _St . : _€iSt . __ II b
;@@_LLsmwui 82+b2dembﬁbcosht)ﬂ J”, b3 (pp). (5 The lutilitﬂ bfl the Laplace transform is dud E some general lfactd about the way it operates.
Elrstl it is obxions that the Laplacg m is a linear operationj mlcifi k2f2l=mlf1l mlbl (6i More interestingly, the Laplacd transforml converts differentiationl intd a Simple a geBraic
operationi lndeedl suppose [fl is differentiable, is continuous, and botlﬂ satisfyl the estimate K1],sothaﬁ£i[ﬂandﬂwarebothmfﬁisgau1henweham If E = Z[fli them Zlf’l = 8F(S) — f(0) (7i
TlTe [proof is a W integration E partsi
rims) = ﬂ e‘5tf’Ktl dt = «reswam — ﬂewSW) d: : —f(0) s K «re“m dt
= —f(0) E SUNS) he_5tfltl vanishes at ti z oo Becausd [ﬂ satisﬁes {1) m .51 S g] ll If ﬂ had were derivativesl can] apply this resnlii repeatedly to gﬂ formulas for their
llaplace transforms in terms of E = L[ f]. For the second] derivativel we have Zllf"l(sl = sﬁlf’KS) — f’(0) = Slsﬁlflls) — f(0)] — f’KOJ
= 82F(8) — 8f(0) — f’(0) [Ehdpattermshmildmmbecleari memamnﬂ] e
4&8) = 8”F(8> — s"‘1f(0l — ls”‘2f’(0) —    — f("‘1)(0) (9) The next genera formula is, in a sense; dual to Just as U converts differentiationl into
multiplication byl s [with an adjustment for thd initiaﬂ value W0», it Eonverts @ultiplication
byl t inte (withl a minus Sign]: ME = Ule WW Eltﬂtﬂ = ﬂ (10) (a W prooﬂ is easX2 if] we take for granted the fact that differentiation with lrespecﬂ to .51
Eommutes with integration with respect to t (notl completely obvious2 but true): Ewa=ﬁ/ anon: wﬂeﬂﬂwa=jflvwwa=4mnaw. LJS— L
Here’s one useful general formulai
TﬂFzﬂmzthenﬂeatfltl =F(s—a). (11) This is bbvious: [l[e“tf(t)] 2 AT e’Steatfltlﬂ = A? 67(8’“)tf(t ﬂ = lF(s — a). This formula also has a dual2 which Ell let you ﬁgure m in a homework problem.
(lomBining K10) and (11) with [307(5), we can enlarge mm dictionary Laplacg trans—
m E examplel aﬁ=ﬁ% (E
unﬂ=em§k%=ﬂﬁﬁa e:
U[e“tcosbt] = (8 32am {g
ewmﬂ : WM (15 0% E and absolutelyl crucial general fﬁ is that a m is completelyi determined
byl its Laplace m That E2 Eff andg are piecewisg WW that @7711) at
m exponentially ﬂ inﬁnity kind Ll = QEL then I = 11 We shall not attempt to prove
HIE m Let’a [heat the Laplace ttahsferm can [be used tQ selxe differential Equatmns. Emmpld ﬂ. Let’sJ Start With a problem that we already know how to solvei the [homoget
neous seconEI—oraer constant—coefﬁcient equation with ﬁnitia Eonaition§ MHMHE=Q Mj=m ﬂ@=ﬁ
Applying HE liaplacé transform turna m into HE algeBmz'c bquationl
bQY—M—Q+ang—wamzo for H = The so utioh is M2y08__y0+ay. (ﬂ that the guadratig polynomiall 53 It is the same Qne that enters intd the method
thehltiQnJ E Section] 18.4 thhe text Iﬂﬁt has tam real mete 5&2, we can kid a partiall
fraction Elecompositioh bﬂ {ﬂ td oBtaiﬂ Cl 92 s—E"S—E’ where g and @ depeng] QIl the yalues ya and Undoing the Laplace m ﬂay
E1 m3 thain V: t
y=quwﬂﬂ If the guadratiq 1m (16) has complex r00t§ 04 :H M (16] can be rewritten in the form] :m__2 c: Cr: oz
W 4E:QX_2 (t %,a @3(d )mM% yzclea’tces t Cgeatsjnl t. Finally, E (13 2 4b SQ that the guadratic tie — r)2 with r): = ﬂ me rewrite (16) as MZQG—ﬂ+&: g 67 (3—7‘)2 3L7“ Kat—7“)?
m M from KB), <01] '2 M): C? : Mt T)y0)z M = he” bte’t [Ehua we recelen the resulta Section 18.42 the nice thingj 1a that the dependence the
henstantsi g and a (III the cenditiene but Qf the computatlon automatically. Exampld 2i Thd idea works foil inhomogeneous WW equatlons
g2 provided that we can hand d HIE Ilap ace transform of Q This is L =
certainly the case Mherﬂ g id ofl one of the typed for whiclﬂ the inethod bflljudicious guessing worksl whiclﬂ are thd same
types whosd Laplace transforms havd beerﬂ Eomputed abovel Td bd Epeciﬁcl let’s solve y l/ 53/ 6 = sid54tl M0] = y’all = 1L Applying HE liaplacd transforml with thesd initial conditions, Kieldg id E W 6 3 83:: m: 25m 5 :(ﬂﬂﬁ (8H3). (3+2 A ratherl gruesome ﬁartiaI—fractiond calculation yields M: 170‘_ 145Lp5s+9a ‘1
9853 n2 ea Ml W121, K14>zand<15lg II] sail l82—25l 7 = 9:6 [170521t — 1456—34 — 259mm — mama]. Examplg ﬂ The Laplace transform id alsd an Efficient tool lfori solving systems of simultal neous differential equationsl [Fori bxamplel herd is a system bfl W Eguationsl in W unknown
m :zﬁ and :02, with initial conditions: 2/1 = —&w — Pym y; = 4% 434%; MW = m y2(0l = 02. Applying the Laplacg transform blields a system bfl W linearl algebraid eguationsl fori Y1 : Zlilyil and W = lely2lJ 8Y1 — cl = P8Yl — 93% OI’ (3 8% 93/; :plz Solving these together yields E—4l01—QL2 (s + 2)2 :I E]: C1 Q lhereforel [by (12) and (13]2 M = 016727 — {Q + 02 be”, 601 + 902
(8 2V ” 8Y2 — 02 = 4Y1 Vii/2; 4Y1 2' C2. M+r+8icﬁ_ p2 401l+6c
A (822 l—SE:Z (sl+2)2' 3g 2 Cge’Ql + (M + 02 teat Exampld Ml The Laplace transform is less eﬁeetive m dealing with trariahleioeﬂieienﬂ
equations, 13W sometimes it can be used] there tool As an example, [let us derive the funda—
mentaﬂ @ﬁ of solutions ofl thd eguation ty” Kt 2M W: (17] that were used irﬂ Problem El LAssignmentj 6 (with t relabeled as Applying the Laplace
transform and setting lﬂ : My]z yo : 34(0)z and ya : t/(O)2 Nve havd d , d
ﬁlmawaitQism—yoi—zism—yonzha
CEhdﬁrsttermiEﬁsE—EHwhndtheseeondorreEYl+Sl/’J e4d_?i[:%
S ’32l— 82—8. (8 — W W (3 MSW = —3y02 or K: Note that ya has tlroppedl bntl The pomﬂ f = O E a singular pomt fort equation (E, m
bannot specifv initial conditions in thd tisuall Way.)  his is a ﬁrst—orden lineaﬂ Equationl for
YJ Since 43—3_s+3(s—1)_ 1 B
ﬁ—s— 8(8—1) 3—1 52 its integrating m is Exp lnl d — 1] 3 [lrﬂ ﬂ 2 £13 — 1). MultiplXing through byt this yieldd — 1)Y]l 2 3y052, 905l::p: .‘ C 2 W .i 1 _W_1_l1l
KEﬂJ—D sl—ll"33t(S_—U a—ll'HLa—ﬂ ti 3i {Some [partial—fractions algebra has beenl Skippedl here. inverting the [Laplace transform nothgived
MZKZJO C)et C(H t £757) il‘aking yo = 12 2 gives the solution a and taking yo = g Q : :2? gives the solution
t3 E g as claimed in the [homework probleml To [harness the fill power E the Laplace m one heedd a way ﬂ inverting it that id
m than merely tompiling a table of functions [ﬂ and their Laplace transforms
U[ f] m M it from W to left, whichl is essentially what we have donel HoweverJ
thd generall formulal for thd inversd Laplacg transform involves thd theoryl of functions ofl
a om le variable tvhich is beyond the W of this course. (It is this general inversion ZDEtherﬂfzpjstatedexamplg ll.) Hi
you’re interested, though2 one ireferenee HE Earthen Analysig and its Applications log Eollangi ...
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This note was uploaded on 10/12/2009 for the course MATH 134 taught by Professor Staff during the Fall '08 term at University of Washington.
 Fall '08
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