CHAPTER 9. Inverse Transform and. Solution to the Initial Value Problem

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1 A SERIES OF CLASS NOTES FOR TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES A COLLECTION OF HANDOUTS ON SCALAR LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODE") CHAPTER 9 Invere Tranform and Solution to the Initial Value Problem. The Invere Laplace Tranform. Technique for Uing the Laplace Tranform to Solve IVP 3. Table of Laplace Tranform that Need Not be Memorized Ch. 9 Pg.

2 Handout # THE INVERSE TRANSFORM Profeor Moeley We wih to etablih a ubpace of T where the Laplace Tranform i a one-to-one mapping. Thi mean that different function f(t) get mapped (or tranformed) into different function F(). Specifically, if f g, then {f} {g}. The contrapoitive of thi tatement, which i equivalent, i that if {f} = {g}, then f = g. Thi i the tandard definition of one-toone. For a linear mapping between vector pace, there i an eay tet to determine if a mapping (like the Laplace Tranform) i or i not one-to-one (-). THEOREM. If T i a linear operator from the vector pace V to the vector pace W and it null pace N T i { 0 }, then T i a one-to-one mapping; that i, if x, y ε V and T( x ) = T( y ), then x = y. PROOF: We tart by proving the following: Lemma. If N T i { 0 } and T( x ) = 0, then x = 0. Proof of lemma: We begin by recalling the definition of the null pace N T. DEFINITION (Null Space). The null pace i the et of all vector x that atify the linear homogeneou equation T( x ) = 0 ; that i N T = { x εv: T( x ) = 0 }. Thu if T( x ) = 0, then by the definition of N T we have that x ε N T. However, we alo aume that N T = { 0 }. Hence ince x εn T, we have that x = 0. QED for lemma. Ret of proof of theorem (For variety, thi time we write the proof in paragraph form): Now aume that x, y ε V and T( x ) = T( y ). Hence we have that T( x ) T( y ) = 0. Hence ince T i a linear operator we have that T( x y ) = 0. Since we alo are auming that N T = { 0 }, we have by the lemma that x - y = 0. Hence x = y. Hence T i a one-to-one mapping. QED for Theorem. Recall that C[0,) denote the et of continuou function on [0.). LERCH'S THEOREM. If f,g C[0,), and { f(t) }() = { g(t) }() > a, then f(t) = g(t) t 0. Lerch theorem ay that the Laplace tranform i a one-to-one mapping on the vector pace C[0,) and hence ha only the zero function in it null pace. Different function in C[0,)T are mapped to different function in the pace of tranform F. Thu our table give the unique tranform for each of the continuou function in the table. Although it i not true that i one- Ch. 9 Pg.

3 to-one on all function in T, if we throw out the null function, it i one-to-one. For example, i one-to-one on T pcexp = PC a [0,) E xp, the et of all piecewie continuou function of exponential order where we have taken the average value at the point of dicontinuity. Since i a one-to-one mapping on T pcexp, it invere mapping exit. (Similar to limiting the domain of in(x) o that we can define Sin (x), but the et we are mapping from and to are et of function, not et of number.) We denote the invere Laplace Tranform of a function F() by f(t) = - { F() }. THEOREM. The invere Laplace Tranform i linear; that i, - { c F() + c G() } = c - { F() } + c - { G() }. PROOF. Let F() = { f(t) } and G() = { g(t) } o that, ince i one-to-one we have f(t) = - { F() } and g(t) = - { G() }. Then by the linearity of we have { c f(t) + c g(t) } = c { f(t) } + c { g(t) }. Since i one-to-one, we have - { c { f(t) } + c { g(t) } } = c f(t) + c g(t) Rewriting thi equation we have - { c F() + c G() } = c - { F() } + c - { G() } a required. Q.E.D. EXAMPLE #3. If F() =, find f(t) = - { F() }. 3 Solution. 3 = = 3- (-) From the table we recall that a { e at in(ωt) } =, { e at co(ωt) } = (- a) (- a) -+ - So = = = (-) (-) - = + (-) (-) (-) Ch. 9 Pg. 3

4 Hence - { - } = - { + } 3 (-) (-) - = - { } + - { } (-) (-) = e t co( t ) + e t in( t ) EXERCISES on The Invere Tranform Find f(t) = - {F()} (t) if ) F() = ) F() = 3) F() = 4) F() = 5) F() = 6) F() = 7) F() = 8) F() = 9) F() = 0) F() = ) F() = ) F() = 3) F() = 4) F() = 5) F() = 6) F() = 7) F() = 8) F() = 9) F() = Ch. 9 Pg. 4

5 Handout # TECHNIQUE FOR USING LAPLACE Profeor Moeley TRANSFORMS TO SOLVE IVP'S EXAMPLE. Uing Laplace tranform, olve the IVP y" - y' - y = 0, y(0) =, y'(0) = 0. Solution. We tranform the problem into the (complex) frequency domain: For convenience let {y} = Y. { y" - y' - y } = { 0 } { y" } - { y'} - { y } = 0 { y" } - { y'} - { y } = 0 ( Y - y(0) - y'(0) ) - ( Y - y(0) ) - Y = 0 ( Y - () - (0) ) - ( Y - () ) - Y = 0 Y - - Y + - Y = 0 Y - Y - Y = - ( - - ) Y = - A Y = = = + (-)( +) - B + where we may ue partial fraction to obtain A and B. Multiplying by ()(+) we obtain - = A( + ) + B( S - ) = A + A + B - B Equating coefficient ) = A + B 0 ) - = A - B = 3 B B = /3 A = - B = /3 / 3 Y = = + (-)( +) - / 3 + y(t) = /3 e t + /3 e -t Ch. 9 Pg. 5

6 Handout #3 TABLE OF LAPLACE TRANSFORMS Profeor Moeley THAT NEED NOT BE MEMORIZED f(t) = - {F()} F() = {f(t)} Domain F() ))))))))))) )))))))))) ))))))))) t n where n i a poitive integer n! n > 0 inh (at) a a > *a * coh (at) a > *a * e at in (bt) (- a) > a e at co(bt) a (- a) > a t n e at n = poitive integer n! ( a) n > a u(t) > 0 u(t - c) e -c > 0 e ct f(t) F( - c) f(ct) c > 0 c F( c ) δ(t) δ(t - c) e -c Ch. 9 Pg. 6