# MA1506CHAP4 - CHAPTER 4 THE LAPLACE TRANSFORM Definition...

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Unformatted text preview: CHAPTER 4 THE LAPLACE TRANSFORM Definition. Let f be a function defined for all t ≥ 0. The Laplace transform of f is the function F ( s ) defined by F ( s ) = L ( f ) = Z ∞ e- st f ( t ) dt, (1) provided the improper integral on the right exists. The original function f ( t ) in (1) is called the inverse transform or inverse of F ( s ) and is denoted by L- 1 ( F ); i.e., f ( t ) = L- 1 ( F ( s )) . Notation: Original functions are denoted by lower case letters and their Laplace transforms by the same 1 letters in capitals. Thus F ( s ) = L ( f ( t )), Y ( s ) = L ( y ( t )) etc. Recall that, by definition, for any function h defined on [0 , ∞ ), Z ∞ h ( t ) dt = lim b →∞ Z b h ( t ) dt and the integral is said to converge if this limit exists. Because e- st decreases so rapidly with t, the Laplace transform usually does exist [there are exceptions, however], and then we say that the function f HAS A WELL-DEFINED LAPLACE TRANSFORM. Example 1. Let f ( t ) = e at , when t ≥ 0. Find F ( s ). 2 Solution. F ( s ) = L ( e at ) = Z ∞ e- st e at dt = lim b →∞ Z b e- st e at dt. Now Z b e ( a- s ) t dt = ‰ b if s = a e b ( a- s ) a- s- 1 a- s if s 6 = a . If s < a , a- s > 0 and e ( a- s ) b → ∞ as b → ∞ . Thus when s ≤ a , R ∞ e ( a- s ) t dt diverges. When s > a , a- s < 0, and e ( a- s ) b → 0 as b → ∞ , and then F ( s ) = L ( e at ) = 1 s- a , s > a. (2) Example 2. Let f ( t ) = 1, t ≥ 0. Find F ( s ). 3 Solution. This function is the same as the one in Example 1 with a = 0, thus, L (1) = 1 s , s > . (3) The Laplace transform is a linear operation; i.e., the Laplace transform of a linear combination of func- tions equals the same linear combination of their Laplace transforms. Thus Theorem L ( af ( t ) + bg ( t )) = aL ( f ) + bL ( g ) , (4) where a and b are constants. As a corollary, the inverse Laplace transform also satisfies the linearity property. L- 1 ( aF ( s ) + bG ( s )) = aL- 1 ( F ) + bL- 1 ( g ) . (5) 4 Verification of (4) and (5) is easy. Example 3. Using (4), we obtain L (cosh at ) = L 1 2 ( e at + e- at ) ¶ = 1 2 1 s- a + 1 s + a ¶ = s s 2- a 2 , s > a ≥ . Example 4. If F ( s ) = 3 s + 5 s- 7 , find f ( t ) = L- 1 ( F ). Solution. Using (5), we have L- 1 ( F ) = L- 1 ( 3 s ) + 5 L- 1 ( 1 s- 7 ) = 3 · 1 + 5 · e 7 t = 3 + 5 e 7 t . 5 Example 5. Set a = iw in the formula (2) L ( e iwt ) = L (cos wt + i sin wt ) = L (cos wt ) + iL (sin wt ) = 1 s- iw = s + iw s 2 + w 2 Equating real and imaginary parts, we get L (cos wt ) = s s 2 + w 2 , L (sin wt ) = w s 2 + w 2 . (6,7) Example 6. To show that L ( t n ) = n ! s n +1 , n = 0 , 1 , 2 ,... . Solution. L ( t n ) = Z ∞ e- st t n dt =- 1 s e- st t n | ∞ + n s Z ∞ e- st t n- 1 dt....
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MA1506CHAP4 - CHAPTER 4 THE LAPLACE TRANSFORM Definition...

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