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# Note for MATH 3321 at UH

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Date Created: 02/06/15
43 Inverse Laplace Transforms and InitialValue Problems In Section 42 we saw that the Laplace transform of the solution y ofthe initial value problem 24 W by 1 240 a MO B is given by F s as B my aim Yltsgt where Fs fz is the Laplace transform of 1 Now that we know yz the obvious question is What is The general problem of nding a function with a given Laplace transform is called the inversion problem The inversion problem and its application to solving initial value problems is the topic of this section If f is continuous on 0 00 and if the Laplace transform fm Fs exists for s gt A then the function F is uniquely determined by 1 that is the operator L is itself a function Our rst result states that E is a oneto one function A proof of this result is beyond the scope of this introductory treatment THEOREM 1 If f and g are continuous functions on 0 00 and if fm gz then f E 9 that is x gz for all z E 0 00 The following de nition gives the terminology and notation used in treating the inversion problem DEFINITION If Fs is a given transform and if the function 1 continuous on 0 00 has the property that fz Fs then f is called the inverse Laplace transform of Fs and is denoted by 1 00 71F8 The operator E 1 is called the inverse operator of E There is a general formula for the inverse operator E l corresponding to 1 Section 41 but use of the formula requires a knowledge of complex valued functions a topic which is treated in more advanced courses The relationship between E and E 1 is given by the following equations r1 Ei z M r E 1Fs Fs for all functions 1 continuous on 0 00 such that fz For convenience here we reproduce the table of Laplace transforms given at the end of Section 41 129 Table of Laplace Transforms 1W F8 Elfl 1 1 7 s gt 0 s 1 50m 7 s gt Oz 37 a s cos z 82327 sgt0 t 5 sin Bx W7 3 gt 0 M s 704 e cos z 870027 327 sgta 5 sin x 3 7 s gt 04 s7 a 32 l mn7 n1727 8717 sgt0 nl mnem7 n1727 W7 sgtr 82 732 gt 0 z cos Bx 82 32 s t 233 0 z s1n Bx 82 32 s gt A simple way to interpret Theorem 1 is that the table can be read either from left to right or from right to left That is7 the table is simultaneously a table of Laplace transforms and of inverse Laplace transforms Example 1 a If rum Fs 877 then f e4w b If mm Fs then me cos 3x 3 2 s 2 8 72 c If rum Fs 32738713 7873279 787712772797 then fx 5 21 cos 3m The properties of the Laplace transform operator E can be used to derive corresponding properties of its inverse operator 71 For our purposes7 the most important property is that of linearity THEOREM 2 The operator E 1 is linear that is 1Fs 13 E 1Fs 1Gslt and 130 E 1CF3 cE 1Fs c any constant The proof is left as an exercise Example 2 Find E 1Fs if E 1Fs rill 7 7L 1 2 7K1 73E 1 33 94 Now reading the table from right to left we see that 1 71 7 731 71 7 A E 8BJ75 and E 824J7sin2m Therefore E 1 7 7 6 7e 3h3sin 2m l 33 324 39 The translation property of E Theorem 5 Section 42 is also useful in nding inverse transforms The analog of Theorem 5 for inverse transforms is THEOREM 3 If f is continuous on 0 00 and if Efm Fs exists for s gt A then for any real number r E71Fs 7 7 em x The following examples illustrate the kinds of manipulations that typically occur in calculating inverse Laplace transforms The basic strategy is to try to re Write a given expression Fs as sum of terms which appear in the table Example 3 Find E 1Fs if SOLUTION From the table To put in a form in the table we complete the square in the denominator 32 7 2s 10 and adjust the numerator 1 1 1 1 3 5272s105272s1937129 3712939 From the table 1 1 3 1 7 1 l w L 8272810 3E J as sin 3m Putting the two results together we have 71 4 43555 em sin 3x I Example 4 Find 1Fs if 2s1 FltSgtm39 SOLUTION By factoring the denominator we can write 2s1 2s1 F 8 8272878 s2si4 Now by partial fraction decomposition L s2874 574 82 Therefore Example 5 Find L 1Fs if 2s4 FSsi23274s839 SOL UTION The quadratic factor in the denominator cannot be factored into linear factors By partial fraction decomposition 284 2 7286 F S 3723274s8 372327438 Next we complete the square in the denominator of the second term 2 72s6 7 2 72s6 7 2 72s6 372 3274387372 3274s44isi2 3722439 132 Finally we adjust the numerator of the second term so that we can use the Table 2 72s6 2 7237227 2 72372 2 87237224isi2 37224isi2 372243722439 K1 1 372 2 J 72 372 37224 37224 2 e2m 7 2 e2m cos 2x e2m sin 2x I Solution of InitialValue Problems Here we complete the application of Laplace transforms to the solution of initial value problems For our rst example we finish Example 2 Section 42 Example 6 Find the solution of the initial value problem 1 7 2y 253 y0 72 SOLUTION From the Example if y is the solution then Ely z2yl we gmH le Mliiig selenium Sig s yxiy02 lyl Sig 872 lyl2 Sig seamen 372 Therefore 2 2 Ell95H HS m m Now by partial fraction decomposition 2 2 5 2 5 372s3 lm Therefore 7 25 25 2 7785 25 YltSgtisi2is3isi2isi2is3 and 85 25 1 i yz 872783 7gezmige 3 I 133 Next we nish Example 3 of Section 42 Example 7 Find the solution of the initial value problem 24 4y 36 y0 57 MW 2 SOLUTION From the Example if y is the solution then EllKm 4yz men men S i 2 EllMl 4Llyltmgtl S f 2 82mm 7 8110 7 we 4am S f 2 lt324gt iyltzgtiesselte2gt Sf 32 4 ym S f 2 53 2 Therefore 3 58 7 2 372324 32439 Now by partial fraction decomposition 3 3 3 3 8 85Z 372324 372732439 Therefore W8 3 7 s 5372 1g s in 2 372 324 324 8372 8324 8324 and 3 1 37 s 11 2 E lY pl ifl fifl W l 8 8 372l 8 94 8 94 gezmcos2zisin2z I Example 8 Find the solution of the initial value problem 24 7 52 6y z 71 y0 07 y 0 1 134 SOLUTION If y is the solution then Ely ltzgt75y ltzgt6yltzgti 7 z71i7qzi7wi78127 EllMl 7 sayen mm 1 ew 7 8240 7 Mo 7 5 sag25 7 240 6mm 7 1838 327556 ye71 1823 527556 yz L Therefore yzlYs 1 3 1 1 3 1 323275s6 32753176 82872873 372x373 Now by partial fraction decomposition W lilgtEsi2gtsi3gt and 1 71 1 i 1 372x373 372 373 Therefore 1 1 1 1 3 1 7 1 Y 77 7 7 7 77 7 S 36 9169 4372gt9lts73gt and 1 1 3 7 ymi5milezm egm I In many applications of differential equations it is not required to determine the solutions explicitly lnstead What is needed is information about the solutions Often such information can be obtained by analyzing their Laplace transforms The next example illustrates this type of application Example 9 Consider the differential equation y 7 y 7 6y 25 on 0 00 together with the condition y0 71 From Chapter 3 we know that the general solution of the differential equation has the form 016 2m 025 Ae m Where Cl 02 are arbitrary constants and A is a constant which can be determined 135 The question we want to examine is Can we choose a value for 04 y 0 so that solution of the resulting initial valueproblem y i 2 i 6y 26 240 17 MO a has limit 0 as x a 00 Since 5 2m a 0 and 5 a 0 as x a 00 we want to choose 04 so that the coefficient of the 53m term is 0 If y is the solution of the initial value problem then MW 7 m 7 6yltzgt1 12511 1 1Mltmgt1e 1yltxgt1emuam Si sz lyml 8240 20 S lyMl 240 6 lyl 8 i1 3236 yz3a1 Sil 3236 yx Si11a3 Therefore 2 1043 ElymlY8 mtm 2 1043 313233 323339 Now by partial fraction decomposition 2 A B C 7 g 313233 31 32 33 31 32 33 and Hm D 1 E J T 3233 32 33 32 33 Combining these results we have a1 1 26173 1 1 1 Y 5 52gt 10 lt3BgtT2lt31gt39 Clearly if 204 3 0 that is if 04 32 then the 53w term in is eliminated The resulting solution is WE 6 i This solution has initial values y0 1 y 0 g and lim O I 1400 z 1 i5 136 Exercises 43 Find 1Fs 1 Fs 87 2 Fs 282 3 Fs 8225 4 Fs 7834 5Fs i17 6 Fs 13 7 Fs8 8 1137857 9 Fs im 10 Fs839 Use partial fraction decomposition to nd the inverse Laplace transform 11 113 12 Fs 13 113 14 Fs 821 15 Fs 8832325218 16 Fs 84871 17 Fs 4 323 71si 2 Find the solution of the initial value problem 137 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 y 74y0 y02 y 2yem y01 y 73ye y02 y ysinz y01 y 4y 0 y0 27 MO 2 y 7 y sin 95 y017 20 1 y yem y017 MO 0 y 7 y 7 2y sin 2m y017 yO 1 y 22 y 46 y 32 2y 66m y i 22 5y 352 y017 MO 1 Given the initial value problem gm 7 y 7 6y 2571 See Example 9 What value should be assigned to 04 so that the resulting solution will have limit 0 as maoo What initial conditions should be assigned with the differential equation iz ynye so that lim 0 Where y is the solution 1400 Consider the differential equation y y 7 2y 3sin 2x together with the initial value y0 2 For What values of B y 0 will the resulting solutions be bounded Consider the differential equation y 3g 2y z together with the initial value y 0 72 For What values of Oz y0 will the resulting solutions be bounded The Laplace transform method applies to initial value problems in which the initial values are speci ed at z O Actually7 the method can be applied when the initial 138 x dt dt 7idi71di 71dili dm2 7 dm dm idt dm dzidt dt idtz39 Find the solution of the initial value problem y 7 331 2y m y1 O7 y 1 1 by rst solving the transformed problem Use the Laplace transform method to solve the initial value problem y72y1x y271 139

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