Differential Equations Brannan Copyright © 2010 by John Wiley & Sons, Inc. All rights reserved. Chapter 05: The Laplace Transform

Chapter 5 The Laplace Transform

Chapter 5 - The Laplace Transform 5.1 Definition of the Laplace Transform 5.2 Properties of the Laplace Transform 5.3 The Inverse Laplace Transform 5.4 Solving Differential Equations with Laplace Transforms 5.5 Discontinuous Functions and Periodic Functions 5.6 Differential Equations with Discontinuous Forcing Functions 5.7 Impulse Functions 5.8 Convolution Integrals and Their Applications 5.9 Linear Systems and Feedback Control

5.1 Definitions and Examples DEFINITION 5.1.1 The Laplace Transform Let f be a function on [0, ∞). The Laplace transform of f is the function F defined by the integral (1), The domain of F(s) is the set of all values of s for which the integral in Eq. (1) converges. The Laplace transform of f is denoted by both F and L{ f }.

Examples 1. Let f (t) = 1, t ≥ 0. Then, 2. Let f (t) = eat, t ≥ 0 and a real. Then,

Examples 3. Let f (t) = e(a+bi)t , t ≥ 0. Then

THEOREM 5.1.2 Linearity of the Laplace Transform. Suppose that f1 and f2 are two functions whose Laplace transforms exist for s > a1 and s > a2, respectively. In addition, let c1 and c2 be real or complex numbers. Then, for s greater than the maximum of a1 and a2, L{c1f1(t)+c2f2(t)} = c1L{f1(t)} + c2L{f2(t)}.

Examples 1. Find the Laplace transform of f (t) = sin at, t ≥ 0. f(t)=2 + 5e−2t − 3 sin 4t, t ≥ 0.

Example 3. Find the Laplace transform of f (t) = t cos at, t ≥ 0.

DEFINITION 5.1.3 Piecewise Continuous Functions A function f is said to be piecewise continuous on an interval α ≤ t ≤ β if the interval can be partitioned by a finite number of points α = t0 < t1 < ・・・ < tn = β so that: 1. f is continuous on each open subinterval ti−1 < t < ti , and 2. f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.

Example 1. Find the Laplace transform of f(t)={ e2t ,0 ≤t<1

Functions of Exponential Order. THEOREM 5.1.4 If f is piecewise continuous for t ≥ a, if |f(t)| ≤ g(t) when t ≥ M for some positive constant M, and if converges, then also converges. On the other hand, if f (t) ≥ g(t) ≥ 0 for t ≥ M, and if diverges, then also diverges. DEFINITION 5.1.5 A function f (t) is of exponential order (as t → +∞) if there exist real constants M ≥ 0, K > 0, and a such that |f(t)| ≤ Keat when t ≥ M.

Example Determine which of the following functions are of exponential order: f (t) = cos at, f (t) = t2, and f (t) = exp(t2). Answer: (a) Yes (b) Yes (c) No.

Existence of the Laplace Transform THEOREM 5.1.6 Suppose i. f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A, and ii. f is of exponential order, that is, there exist real constants M ≥ 0, K > 0, and a such that | f (t)| ≤ Keat when t ≥ M. Then the Laplace transform L{ f (t)} = F(s), defined by 5.1.1, exists for s > a. COROLLARY 5.1.7 If f (t) satisfies the hypotheses of Theorem 5.1.6, then |F(s)| ≤ L/s for some constant L as s→∞. Thus Lim F(s) = 0. s→∞

5.2 Properties of the Laplace Transform THEOREM 5.2.1 - Laplace Transform of ectf(t) If F(s) = L{ f (t)} exists for s > a, and if c is a constant, then L{ectf(t)}=F(s−c), s > a + c. Example: Find the Laplace transform of g(t) = e−2tsin4t. Answer: Using Theorem 5.2.1 with c=−2, the Laplace transform of g is G(s) = L{e−2tf(t)} = F(s + 2) = 4/((s + 2)2 + 16) for s > −2.

THEOREM 5.2.2 Laplace Transform of Derivatives of f (t). Suppose that f is continuous and f' is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that f and f‘ are of exponential order with a as specified in Theorem 5.1.6. Then L{f'(t)} exists for s > a, and moreover L{f'(t)} = sL{ f (t)} − f (0). Example: Verify that Theorem 5.2.2 holds for g(t) = e−2t sin 4t and its derivative g'(t) = −2 e−2t sin 4t + 4 e−2t cos 4t.

COROLLARY 5.2.3 Suppose that i. the functions f, f' , . . . , f (n−1) are continuous and that f (n) is piecewise continuous on any interval 0 ≤ t ≤ A, and ii. f, f', . . . , f (n−1), f (n) are of exponential order with a as specified in Theorem 5.1.6. Then L{ f(n)(t)} exists for s > a and is given by L{f(n)(t)} = snL{ f (t)} − sn−1f(0) − ・・・ − sf (n−2)(0) −f (n−1)(0). EXAMPLE Assume that the solution of the following initial value problem satisfies the hypotheses of Corollary 5.2.3. Find its Laplace transform. y'' + 2y' + 5y = e−t , y(0) = 1, y'(0) = −3.

Laplace Transform of tn f(t) THEOREM 5.2.4 Suppose that f is (i) piecewise continuous on any interval 0≤t≤A, and (ii) of exponential order with a as specified in Theorem 5.1.6. Then for any positive integer n L{tn f(t)} = (−1)nF(n)(s), s > a. COROLLARY 5.2.5 For any integer n ≥ 0

5.3 The Inverse Laplace Transform Corollary 5.2.3 provide the tools to convert linear differential equations with constant coefficients into algebraic equations in the s-domain. For example, applying these tools to the initial value problem, y'' + 2y' + 5y = e−t , y(0) = 1, y'(0) = −3. led to the algebraic equation (s2 + 2s + 5)Y (s) − s + 1 = 1/(s + 1) The solution of this is Y (s) = s2/(s + 1)(s2 + 2s + 5), is presumably the Laplace transform of the solution of the initial value problem. Determining the function y = φ(t) corresponding to the transform Y(s) is the the main difficulty in solving initial value problems by the transform technique.

Existence of the Inverse Transform THEOREM 5.3.1 If f (t) and g(t) are piecewise continuous and of exponential order on [0,∞) and F = G where F = L{ f } and G = L{g}, then f (t) = g(t) at all points where both f and g are continuous. In particular, if f and g are continuous on [0, ∞), then f (t) = g(t) for all t ∈ [0,∞).

Inverse Laplace transform DEFINITION 5.3.2 If f (t) is piecewise continuous and of exponential order on [0,∞) and L{ f (t)} = F(s), then we call f the inverse Laplace transform of F, and denote it by f = L−1{F}. Examples: Determine L−1{F}, where (a) F(s) = 4/(s2 + 16) (b) F(s) = 6/(s + 2)4 , (c) F(s) = (s + 1)/(s2 + 2s + 5).

Table 5.3.1. - Elementary Laplace Transforms

Table 5.3.1. - Elementary Laplace Transforms

L−1{c1F1 + c2F2} = c1L−1{F1} + c2L−1{F2} Linearity of L−1. THEOREM 5.3.3 Assume that f1 = L−1{F1} and f2 = L−1{F2} are piecewise continuous and of exponential order on [0,∞). Then for any constants c1 and c2, L−1{c1F1 + c2F2} = c1L−1{F1} + c2L−1{F2} =c1f1+ c2f2. Example: Determine L−1{2/(s + 2)4+ 3/(s2 + 16)+ 5(s + 1)/(s2 + 2s + 5)} Answer: 1/3 e−2t t3 + ¾ sin 4t + 5e−t cos 2t.

Partial Fractions Most of the Laplace transforms that arise in the study of differential equations are rational functions of s, that is, functions of the form F(s) = P(s)/Q(s) where Q(s) is a polynomial of degree n and P(s) is a polynomial of degree less than n. The method of partial fractions involves the decomposition of rational functions into an equivalent sum of simpler rational functions that have numerators which are of degree one or zero and denominators of degree one or two, possibly raised to an integer power. The inverse Laplace transforms of these simpler rational functions can then often be found in a Laplace transform table.

Example

Partial Fraction Decomposition 1. Nonrepeated Linear Factors. If the denominator Q(s) admits the factorization Q(s) = (s − s1)(s − s2) ・ ・ ・ (s − sn), where s1, s2, . . . , sn are distinct, then F(s) can be expanded as 2. Repeated Linear Factors. If any root sj of Q(s) is of multiplicity k, that is, the factor (s−sj) appears exactly k times in the factorization of Q(s), then the jth term in the right-hand side of above equation must be changed to

Partial Fraction Decomposition 3. Quadratic Factors. Complex conjugate roots of Q(s), μj + iνj and μj − iνj where νj ≠ 0, give rise to quadratic factors of Q(s) of the form [s − (μj + iνj )][s − (μj − iνj )] = (s − μj)2 + ν2j. If the roots μj + iνj and μj − iνj are of multiplicity k, that is, k is the highest power of (s − μj)2 + ν2j that divides Q(s), then the partial fraction expansion of F(s) must include the terms. Constants in the numerators in all 3 cases must be determined.

Example Answer: Find a, b, c in the partial fraction decomposition and then apply inverse Laplace Transform.

5.4 Solving Differential Equations with Laplace Transforms Method Using the linearity of L, its operational properties (for example, knowing how derivatives transform), and a table of Laplace transforms if necessary, the initial value problem for a linear constant coefficient differential equation is transformed into an algebraic equation in the s-domain. Solving the algebraic equation gives the Laplace transform, say Y(s), of the solution of the initial value problem. This step is illustrated by Examples 3, 4, and 5 in Section 5.2. The solution of the initial value problem, y(t) = L−1{Y (s)}, is found by using partial fraction decompositions, the linearity of L−1, and a table of Laplace transforms. Partial fraction expansions and need for a table can be avoided by using a computer algebra system or other advanced computer software functions to evaluateL−1 {Y (s)}.

Example Find the solution of the initial value problem y'' + 2y' = sin 4t, y(0) = 1. Answer: Apply the Laplace transform to both sides of the differential equation and solve for Y(s). Find the partial fraction expansion of Y(s). Use L−1 to find the solution of the given initial value problem.

Characteristic Polynomials and Laplace Transforms of Differential Equations Consider the nth order linear equations with constant coefficients, any(n)+ an-1y(n-1) + ・・・ + a1y' + a0y = f (t). Since the characteristic polynomial associated with it is Z(s) = ansn + an−1sn-1 + ・・・ + a1s + a0, the Laplace transform Y(s) = L{y(t)} is given by

Example Find the solution of the differential equation y'''' + 2y'' + y = 0 subject to the initial conditions y(0) = 1, y'(0) = −1, y''(0) = 0, y'''(0) = 2. Answer:

Laplace Transforms of Systems of Differential Equations. The Laplace transform method easily extends to systems of equations. Consider the initial value problem y'1= a11y1 + a12y2+f1(t), y1(0) = y10, y'2= a21y1 + a22y2 + f2(t), y2(0) = y20. Taking the Laplace transform of each equation and simplifying (s − a11)Y1 − a12Y2 = y10+ F1(s), −a21Y1 + (s − a22)Y2 = y20 + F2(s),

Laplace Transforms of Systems of Differential Equations. Using matrix notation, the algebraic system becomes We note that Z(s) = |sI − A| = s2−(a11+a22)s +a11a22−a12a21 is the characteristic polynomial of A. Inverting each component of Y in Eq. (27) then yields the solution of system of diff. eqs.

Example Solution

5.5 Discontinuous Functions and Periodic Functions The Unit Step Function. To deal effectively with functions having jump discontinuities, it is very helpful to introduce a function known as the unit step function or Heaviside function. This function is defined by For a real number c, we define

Example

The Laplace Transform of the Unit Step Function. The Laplace transform of uc with c ≥ 0 is L{uc(t)} = e−cs/s, for s > 0. For the indicator function ucd(t)=uc(t)−ud(t), the linearity of L gives L{ucd(t)}=L{uc(t)}−L{ud(t)}=(e−cs−e−ds)/s,

Laplace Transforms of Time-Shifted Functions. THEOREM 5.5.1

Examples

Periodic Functions. DEFINITION 5.5.2 A function f is said to be periodic with period T > 0 if f (t + T ) = f (t) for all t in the domain of f . THEOREM 5.5.3 If f is periodic with period T and is piecewise continuous on [0, T], then

Examples Find the Laplace transform of the periodic function (sawtooth waveform) described by the graph in Figure.

5.6 Differential Equations with Discontinuous Forcing Functions We consider some examples in which the nonhomogeneous term, or forcing function, of a differential equation is modeled by a discontinuous function. Solutions of the constant coefficient equation, ay'' + by' + cy = g(t), are in fact continuous whenever the input g(t) is piecewise continuous.

Example 01

Example 02 That is f(t) is defined below with the given graph.