Laplace Transforms e st f ( t ) dt . Definition 1 (Laplace - PDF document

Laplace Transforms � ∞ e − st f ( t ) dt . Definition 1 (Laplace Transform) . L [ f ( t )] = 0 We’ll often write Y ( s ) = L [ y ( t )]. Because e − st = 1 e st is very small is s > 0 and t is large, the Laplace Transform will be defined, for large enough s , for almost all functions we run across. Algebraic Properties of Laplace Transforms The algebraic properties of Laplace Transforms are key to their use in solving differential equations. Laplace Transforms are linear: L [ f ± g ] = L [ f ] ± L [ g ] L [ cf ] = c L [ f ] Transforms of Derivatives The real key is the Laplace Transform of a derivative can be ex- pressed in terms of the Laplace Transform of the original function: � dy � L = s L [ y ] − y (0) dt This may be proven from the definition of the Laplace Transform if we integrate by parts: Transforms of Derivatives � ∞ � ∞ � dy � e − st dy e − st dy L = dt dt = dt 0 t =0 � ∞ u � y ( t )( − s ) e − st dt � = lim u →∞ e − st y ( t ) − � � 0 0 � ∞ = lim u →∞ y ( u ) e − su − y (0) e 0 + s y ( t ) e − st dt 0 = 0 − y (0) + s L [ y ] = s L [ y ] − y (0) This can be used to obtain a formula for the Laplace Transform of any order derivative in terms of the Laplace Transform of the original function. Transforms of Higher Order Derivatives Suppose we apply the formula L [ y ′ ] = s L [ y ] − y (0) to find L [ y ′′ ]: L [ y ′′ ] = s L [ y ′ ] − y ′ (0) = s ( s L [ y ] − y (0)) − y ′ (0) = s 2 L [ y ] − sy (0) − y ′ (0) We can apply the formula above to find the Laplace Transform for the third derivative: 1

2 L [ y ′′′ ] = s L [ y ′′ ] − y ′′ (0) = s ( s 2 L [ y ] − sy (0) − y ′ (0)) − y ′′ (0) = s 3 L [ y ] − s 2 y (0) − sy ′ (0) − y ′′ (0) This can be parlayed into a more general formula: L [ y ( n ) ] − s n L [ y ] − s n − 1 y (0) − s n − 2 y ′ (0) − · · · − sy ( n − 2) (0) − y ( n − 1) (0) Solving Differential Equations We can use the properties of Laplace Transforms along with tables for transforms of standard functions to solve a variety of linear differential equations with constant coefficients, including differential equations containing initial conditions. n a i y ( n − i ) = g ( t ), we write � Given a differential equation of the form i =0 � n � � a i y ( n − i ) = L [ g ( t )]. L i =0 Using the properties of Laplace Transforms, this becomes a linear equation in L [ y ]. We can solve it for L [ y ] and then find a function having that as its Laplace Transform. That function will be a solution of the differential equations satisfying the given initial conditions. Example Consider a mass-spring system with a mass of 2, spring constant � 1 when 0 ≤ t ≤ 1 4, and forcing function g ( t ) = Suppose also we 0 otherwise . � have initial conditions x (0) = 1, dx � = 0. � dt � t =0 We may write g ( t ) = u 0 ( t ) − u 1 ( t ), so we may write the differential equation as 2 x ′′ + 4 x = u 0 ( t ) − u 1 ( t ). Applying the Laplace Transform, we get: L [2 x ′′ + 4 x ] = L [ u 0 ( t ) − u 1 ( t )] 2 L [ x ′′ ] + 4 L [ x ] = L [ u 0 ( t )] − L [ u 1 ( t )] Writing X = L [ x ], we get 2[ s 2 X − sx (0) − x ′ (0)] + 4 X = 1 s − e − s s s − e − s 2[ s 2 X − s · 1 − 0] + 4 X = 1 s 2 s 2 X − 2 s + 4 X = 1 s − e − s s s − e − s 2 s 2 X + 4 X = 1 s + 2 s s − e − s 2( s 2 + 2) X = 1 s + 2 s

3 e − s X = 1 s ( s 2 + 2) − 1 1 s s ( s 2 + 2) + s 2 + 2. 2 2 We thus have solution � � � e − s � � � x ( t ) = 1 1 − 1 s 2 L − 1 2 L − 1 + L − 1 . s ( s 2 + 2) s ( s 2 + 2) s 2 + 2 s ( s 2 + 2) = 1 1 � 1 � s Using Partial Fractions, we calculate , so s − s 2 + 2 2 � e − s s − se − s x ( t ) = 1 � 1 � 1 s �� − 1 � 1 �� � s � 2 L − 1 2 L − 1 + L − 1 s − s 2 + 2 s 2 + 2 s 2 + 2 2 2 � e − s s x ( t ) = 1 � 1 � +3 � � − 1 � e − s � +1 � s 4 L − 1 4 L − 1 4 L − 1 4 L − 1 s 2 + 2 s 2 + 2 s s Using the table of Laplace Transforms, � 1 � L − 1 = 1 s √ � � s L − 1 = cos ( 2 t ) s 2 + 2 � e − s � L − 1 = u 1 ( t ) s � e − s s � e − s s √ � � L − 1 2 t ), so L − 1 = u 1 ( t ) f ( t − 1), where f ( t ) = cos( = s 2 + 2 s 2 + 2 √ u 1 ( t ) cos( 2( t − 1)). √ √ We thus get: x ( t ) = 1 4 · 1 + 3 2 t ) − 1 4 u 1 ( t ) + 1 4 cos( 4 u 1 ( t ) cos( 2( t − √ √ 1)) = 1 4 + 3 2 t ) − 1 4 u 1 ( t ) + 1 4 cos( 4 u 1 ( t ) cos( 2( t − 1)).