Special Mathematics Laplace Transform

Transcription

1 Special Mathematics Laplace Transform March 28

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3 Nature laughs at the difficulties of integration. Pierre-Simon Laplace 4 Laplace Transform Motivation

4 Properties of the Laplace transform the Laplace transform of a function f(t) is defined by: L[f(t)](p) = f(s)e sp ds and it transforms a function f(t) into another depending on p usually denoted with L[f(t)](p) or F (p). sometimes the function f(t) is called the source function. it is a linear transform: L[af(t) + bg(t)](p) = al[f(t)](p) + bl[g(t)](p) the Laplace transform is one to one: L[f] = L[g] = f = g. change of scale property: L[f(at)](p) = ( p ) a L[f(t)] a exponential scaling: time delay: L[e at f(t)](p) = L[f(t)](p a) L[f(t a)](p) = e ap L[f(t)](p) transform of the integral: [ t ] L f(s)ds (p) = p L[f(t)](p) Convolution of two functions: is another function defined as the integral: (f * g)(t) = and it provides the nice property: sometimes called Borel s theorem. transform the derivative: t f(t s)g(s)ds L[(f * g)(t)](p) = L[f(t)](p) L[g(t)](p) L[f (n) (t)](p) = p n L[f(t)](p) p n f() p n 2 f ()... f (n ) () derivate the transform: (L[f(t)](p)) (n) = ( ) n L[t n f(t)](p) the Laplace transform can be used to compute improper integrals according to the formula: [ ] f(t) F (p)dp = L (p) t if we recognize the integrand as being a Laplace transform of some source function f(t). 2

5 A formula for the inverse L : f(t) = L [F (p)](t) = Res(F (p)e pt ) all poles of F (p) Example: Find the inverse transform of F (p) = (p + 3) 2 (p ) In this case the function F (p) has a pole of order 2 in p = 3 and one of order in p =. One has the formulae: Res(F (p)e pt e pt, ) = lim(p ) p (p + 3) 2 (p ) = e t ( + 3) 2 (in the above residue t behaves like a parameter) Res(F (p)e pt, 3) = lim ( (p + 3) 2 e pt ) p 3 (p + 3) 2 = te 3t e 3t (p ) thus: [ f(t) = L (p + 3) 2 (p ) (derivation with respect to p) ] (t) = et 6 te 3t 4 e 3t, t. 6 Initial value theorem: lim t t> f(t) = lim p L[f(t)](p) p Final value theorem: lim f(t) = lim p L[f(t)](p) t p 3

6 Function Time t-domain f(t) Laplace (frequency) domain F (p) unit impulse δ(t) delayed impulse δ(t a) e ap unit step u(t) p ramp function t u(t) p 2 n-th power t n u(t) n! p n+ α-th power t α u(t) Γ(α+) p α+ exponential e ωt u(t) p ω general exponential a ωt u(t) p ln ω sine sin(ωt) u(t) ω p 2 +ω 2 cosine cos(ωt) u(t) p p 2 +ω 2 hyperbolic sine sinh(ωt) u(t) ω p 2 ω 2 hyperbolic cosine cosh(ωt) u(t) p p 2 ω 2 exp. decaying sine wave e at sin(ωt) u(t) ω (p+a) 2 +ω 2 exp. decaying cosine wave e at cos(ωt) u(t) p+a (p+a) 2 +ω 2 exp. decaying sinh wave e at sinh(ωt) u(t) ω (p+a) 2 ω 2 exp. decaying cosh wave e at cosh(ωt) u(t) p+a (p+a) 2 ω 2 4

7 Solved problems Problem. Find the source functions for the following Laplace transforms: a) F (p) = p 2 3p + 2 b) G (p) = p 2 (p 2 + ). Solution: a) We decompose the Laplace transform in: F (p) = making use of the formula: p 2 3p + 2 = (p ) (p 2) = p 2 p L [ e at] (p) = p a obtained from the table, one gets the source function: f (t) = e 2t e t. b) We decompose again the function making of the theory of rational functions: Ap + B Cp + D p 2 (p 2 = + ) p 2 + p 2 +, The identity yields, after identification, the coefficients : which implies Using now the formulae: L [t n ] (p) = one gets the source function A =, B =, C =, D =, G (p) = p 2 p 2 +. n! p n+, L [sin ωt] (p) = ω p 2 + ω 2, g (t) = t sin t. Problem 2. Solve the Cauchy problem: x + 2x + 5x = x () = x () =, using the Laplace transform. 5

8 Solution: Using the Laplace transform we have: which implies: thus: L [x] (p) = F (p), L [x ] (p) = pf (p) x () = pf (p), L [x ] (p) = p 2 F (p) p x () x () = p 2 F (p) p, F (p) = p 2 F (p) p + 2 [pf (p) ] + 5F (p) =, = and finally one gets: p + 2 p 2 + 2p + 5 = p + (p + ) (p + ) p + (p + ) (p + ) x (t) = e t cos 2t + 2 e t sin 2t ( = e t cos 2t + ) 2 sin 2t. Problem 3. Solve the initial value problem x (t) + x (t) 2x(t) = t where x () = x () = and x () =. Solution: The above equation is a linear nonhomogeneous equation. apply the Laplace transform to both terms: We L [x] (p) = F (p), L [x ] (p) = p 2 F (p) p x () x () = p 2 F (p) L [x ] (p) = p 3 F (p) p 2 x () p x () x () = p 3 F (p) +, L [t] (p) = p 2, Using the linearity of the Laplace transform the above identities provide an algebraic equation with the unknown F (p): Now: p 3 F (p) + +p 2 F (p) 2F (p) = p 2 F (p) = p 2 p 2 (p 3 + p 2 2) = ( p) ( + p) p 2 (p ) (p 2 + 2p + 2) = p + p 2 (p 2 + 2p + 2) = 2 p (p + ) 2 + 6

9 and finally we get the particular solution of this initial value problem using the table with the common values of the Laplace transform: x (t) = L [F (p)](t) = 2 t + 2 e t sin t. Problem 4. Solve the following system of linear equations: x = x + 2y + t y = 2x + y + t, with the initial data x () = 2, y () = 4. Solution: We apply the Laplace transform to this system: L [x] (p) = X (p), L [y] (p) = Y (p) L [x ] (p) = px (p) x () = px (p) 2 L [y ] (p) = py (p) y () = py (p) 4, L [t] (p) = p 2, which leads to the algebraic linear system: px (p) 2 X (p) 2Y (p) = p 2 2X (p) + py (p) 4 Y (p) = p 2 thus: X (p) + Y (p) = ( 6 + 2p ) p 3 2 X (p) Y (p) = 2 p +, implies: X (p) = 3 p 3 + p 2 (p 3) p + = 3 p 3 9 p 3 p p 3 p + = 28 9 p 3 9 p 3 p 2 p + and leads to the source function: x (t) = 28 9 e3t 9 3 t e t. In order to find the source function y (t) we can use the first equation of the given system of differential equations. 7

10 The derivative of x (t) is: and so: x (t) = 28 3 e3t 3 + e t y (t) = x x t = e3t 9 3 t + e t. Eventually the solution of the given linear system is: x (t) = 28 9 e3t 9 t e t 3 y (t) = 28 9 e3t 9 t + e t 3 Problem 5. Solve the equation x + x = cos t, with x () =, x () = 2. Solution: We try to apply the Laplace transform to this equation. We can not find the Laplace tranform of for the moment, thus we pursue using cos t the expression L [ cos t ] (p) in the right side. For the left side: L [x] (p) = X (p), L [x ] (p) = p 2 X (p) p x () x () = p 2 X (p) 2, provides the equation: with the solution: or equivalently: We ll make use of the formula: to get: In conclusion: p 2 X (p) 2 + X (p) = L. [ ] (p), cos t X (p) = 2 p [ ] p 2 + L (p) cos t [ ] X (p) = 2L (sin t) + L (sin t) L (p) cos t L[f(t)](p) L[g(t)](p) = L[(f * g)(t)](p) x (t) = 2 sin t + sin t * t x (t) = 2 sin t + = 2 sin t + t sin (t τ) cos t cos τ dτ sin t cos τ sin τ cos t dτ cos τ = 2 sin t + t sin t + cos t ln (cos t). 8

11 t Problem 6. Solve the equation x (t) = 2 sin 4t + sin 4 (t u) x (u) du. Solution: The given equation can be written in the equivalent form: t x (t) x (u) sin 4 (t u) du = 2 sin 4t. The Laplace transform of the right part is: L [2 sin 4t] = In the left side Borel s theorem implies: L t One gets the equation: with the solution: X (p) = 8 p 2 + 6, x (u) sin 4 (t u) du (p) = L [x (t) * sin 4t] (p) X (p) X (p) = X (p) 4 p = 8 p 2 + 6, 8 p = 8 p 2 + ( 2 3 ) 2 = p p 2 + ( 2 3 ) 2 which provides the solution of the given integral equation: x (t) = 8 ( 2 3 sin 2 ) 3t = 4 3 ( sin 2 ) 3t. 3 Problem 7. Solve the Cauchy problem: x + tx x = x () = x () =, using the Laplace transform. Solution: We apply the Laplace transform: L [x] (p) = X (p), L [tx ] (p) = [px (p)] + x () = = X (p) px (p) L [x ] (p) = p 2 X (p) px () x () = p 2 X (p), 9

12 in order to transform the equation into another differential equation: X (p) + 2 p2 p X (p) = p. This is a linear nonhomogeneous equation, written in the standard form as: X (p) + a (p) X (p) = b (p) with the general solution: X (p) = e a (p) dp k + b (p) e a (p) dp dp. In our case a (p) = 2 p2 p and b (p) = p lead to: X (p) = k p 2 e 2 p 2 + p 2. Making use of the intial data x () =, the transform X (p) has to be, for the initial value theorem states: lim p px (p) = lim x (t) =. t On has to impose the condition k =, which gives X (p) = p 2. Eventually, the solution of the Cauchy problem is x (t) = t.

13 Proposed problems Problem. Find the source function for the Laplace transform F (p) = p 3 5p 2 + 6p. Problem 2. Solve the Cauchy problem x x 6x = x () = x () =, using the Laplace transform technique. Problem 3. Solve the equation: x + 2x + 2x + x =, with the initial data x () = x () = x () =. Problem 4. Solve the linear system of differential equations: x + 4x + 4y =, x () = 3. y + 2x + 6y =, y () = 5 Problem 5. Solve the integral equation: t x (t) = x (u) cos (t u) du, cu x () =. Problem 6. Solve the Cauchy problem tx + x + x = x () = x () =, using the Laplace transform.

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