JUST THE MATHS SLIDES NUMBER 16.1 LAPLACE TRANSFORMS 1 - PDF document

“JUST THE MATHS” SLIDES NUMBER 16.1 LAPLACE TRANSFORMS 1 (Definitions and rules) by A.J.Hobson 16.1.1 Introduction 16.1.2 Laplace Transforms of simple functions 16.1.3 Elementary Laplace Transform rules 16.1.4 Further Laplace Transform rules

UNIT 16.1 - LAPLACE TRANSFORMS 1 DEFINITIONS AND RULES 16.1.1 INTRODUCTION The theory of “Laplace Transforms” is used to solve certain kinds of “differential equation” . ILLUSTRATIONS (a) A “first order linear differential equation with constant coefficients” , a d x d t + bx = f ( t ) , together with the value of x (0). We obtain a formula for x in terms of t which does not include any derivatives. (b) A “second order linear differential equation with constant coefficients” , a d 2 x d t 2 + b d x d t + cx = f ( t ) , together with the values of x (0) and x ′ (0). We obtain a formula for x in terms of t which does not include any derivatives 1

The method of Laplace Transforms converts a calculus problem (the differential equation) into an algebra prob- lem (frequently an exercise on partial fractions and/or completing the square). The solution of the algebra problem is then fed backwards through what is called an “Inverse Laplace Trans- form” and the solution of the differential equation is obtained. ✬ ✩ ✬ ✩ CALCULUS ALGEBRA LAPLACE ✛ ✲ ✛ ✲ ✫ ✪ ✫ ✪ DEFINITION The Laplace Transform of a given function f ( t ), defined for t > 0, is defined by the definite integral � ∞ e − st f ( t ) d t, 0 where s is an arbitrary positive number . Notes (i) The Laplace Transform is usually denoted by L [ f ( t )] or F ( s ). (ii) Although s is an arbitrary positive number, it is oc- casionally necessary to assume that it is large enough to avoid difficulties in the calculations. 2

16.1.2 LAPLACE TRANSFORMS OF SIMPLE FUNCTIONS 1. f ( t ) ≡ t n . � ∞ e − st t n d t = I n say . F ( s ) = 0 Hence, ∞  t n e − st   + n e − st t n − 1 d t = n � ∞ I n = s.I n − 1 .     0 − s s  0 Note: A “decaying exponential” will always have the dominating effect. We conclude that s. ( n − 1) . ( n − 2) .... 2 s. 1 s.I 0 = n ! I n = n s n .I 0 . s s But, ∞  e − st = 1   � ∞ e − st d t = I 0 = s.     0 − s  0 Thus, L [ t n ] = n ! s n +1 . Note: This result also shows that L [1] = 1 s, since 1 = t 0 . 3

2. f ( t ) ≡ e − at . � ∞ � ∞ e − st e − at d t = e − ( s + a ) t d t F ( s ) = 0 0 ∞  e − ( s + a ) t  1 =   = s + a.     − ( s + a )   0 Hence, 1 L [ e − at ] = s + a. Note: 1 L [ e bt ] = s − b, assuming that s > b. 3. f ( t ) ≡ cos at . � ∞ e − st cos at d t F ( s ) = 0  e − st sin at ∞   + s � ∞ e − st sin at d t =     0 a a  0  − e − st cos at ∞   F ( s ) = 0+ s   − s � ∞ e − st cos at d t        ,     0 a a a      0  which gives a 2 − s 2 F ( s ) = s a 2 .F ( s ) . 4

That is, s F ( s ) = s 2 + a 2 . In other words, s L [cos at ] = s 2 + a 2 . 4. f ( t ) ≡ sin at . The method is similar to that for cos at , and we obtain a L [sin at ] = s 2 + a 2 . 16.1.3 ELEMENTARY LAPLACE TRANSFORM RULES 1. LINEARITY If A and B are constants, then L [ Af ( t ) + Bg ( t )] = AL [ f ( t )] + BL [ g ( t )] . Proof: This follows easily from the linearity of an integral. EXAMPLE Determine the Laplace Transform of the function, 2 t 5 − 7 cos 4 t − 1 . 5

Solution L [2 t 5 + 7 cos 4 t − 1] = 2 . 5! s 2 + 4 2 − 1 s s 6 + 7 . s = 240 s 2 + 16 − 1 7 s s 6 + s. 2. THE LAPLACE TRANSFORM OF A DERIVATIVE (a) L [ f ′ ( t )] = sL [ f ( t )] − f (0) . Proof: � ∞ � ∞ � ∞ e − st f ′ ( t ) d t = e − st f ( t ) e − st f ( t ) d t � L [ f ′ ( t )] = 0 + s 0 0 using Integration by Parts. Thus, L [ f ′ ( t )] = − f (0) + sL [ f ( t )] . as required. (b) L [ f ′′ ( t )] = s 2 L [ f ( t )] − sf (0) − f ′ (0) . Proof: Treating f ′′ ( t ) as the first derivative of f ′ ( t ), we have L [ f ′′ ( t )] = sL [ f ′ ( t )] − f ′ (0) . This gives the required result on substituting the ex- pression for L [ f ′ ( t )]. 6

Alternative Forms (Using L [ x ( t )] = X ( s )) : (i)  d x    = sX ( s ) − x (0) . L   d t (ii)  d 2 x    = s 2 X ( s ) − sx (0) − x ′ (0) or s [ sX ( s ) − x (0)] − x ′ (0) . L     d t 2 3. THE (First) SHIFTING THEOREM L [ e − at f ( t )] = F ( s + a ) . Proof: � ∞ � ∞ e − ( s + a ) t f ( t ) d t. L [ e − at f ( t )] = e − st e − at f ( t ) d t = 0 0 Note: L [ e bt f ( t )] = F ( s − b ) . EXAMPLE Determine the Laplace Transform of the function, e − 2 t sin 3 t . Solution First, we note that 3 3 L [sin 3 t ] = s 2 + 3 2 = s 2 + 9 . Replacing s by ( s + 2), the First Shifting Theorem gives 7

3 L [ e − 2 t sin 3 t ] = ( s + 2) 2 + 9 . 4. MULTIPLICATION BY t L [ tf ( t )] = − d d s [ F ( s )] . Proof: It may be shown that d ∂ � ∞ ∂s [ e − st f ( t )]d t d s [ F ( s )] = 0 � ∞ − te − st f ( t ) d t = − L [ tf ( t )] . = 0 EXAMPLE Determine the Laplace Transform of the function, t cos 7 t. Solution L [ t cos 7 t ] = − d s     s 2 + 7 2   d s = − ( s 2 + 7 2 ) . 1 − s. 2 s s 2 − 49 = ( s 2 + 49) 2 . ( s 2 + 7 2 ) 2 8

A TABLE OF LAPLACE TRANSFORMS f ( t ) L [ f ( t )] = F ( s ) K K (a constant) s 1 e − at s + a n ! t n s n +1 a sin at s 2 + a 2 a sinh at s 2 − a 2 s cos at s 2 + a 2 s cosh at s 2 − a 2 1 te − at ( s + a ) 2 2 as t sin at ( s 2 + a 2 ) 2 ( s 2 − a 2 ) t cos at ( s 2 + a 2 ) 2 2 a 3 sin at − at cos at ( s 2 + a 2 ) 2 9

16.1.4 FURTHER LAPLACE TRANSFORM RULES 1.  d x    = sX ( s ) − x (0) . L   d t 2.  d 2 x    = s 2 X ( s ) − sx (0) − x ′ (0) L     d t 2 or  d 2 x    = s [ sX ( s ) − x (0)] − x ′ (0) . L     d t 2 3. The Initial Value Theorem lim t → 0 f ( t ) = lim s →∞ sF ( s ) , provided that the indicated limits exist. 4. The Final Value Theorem t →∞ f ( t ) = lim lim s → 0 sF ( s ) , provided that the indicated limits exist. 5. The Convolution Theorem �� t � 0 f ( T ) g ( t − T ) d T = F ( s ) G ( s ) . L 10