Section 7.2 Solving initial value problems with the Laplace transform In this section, we will see how to use the Laplace transform to solve initial value problems involving linear equations with constant coefficients, i.e. problems of the form (n) 0 (n−1) any + ::: + a1y + a0y = f(t); y(0) = a0; : : : ; y (0) = an−1: However, before we do this, we must think about one more interesting property of Laplace transforms{the way that they interact with derivatives. To understand Laplace transforms of derivatives, we should begin by recalling the types of functions that have Laplace transforms: Theorem 2, Section 7.1. Suppose that f(t) is defined for t ≥ 0, is piecewise continuous, and that there are constants M, T , and c so that f(t) −M ≤ ≤ M for all t ≥ T: (1) ect Then Lff(t)g exists for all s > c. The theorem says that if 1. f(t) is piecewise continuous, and 2. f(t) does not grow too fast with respect to the exponential function ect for some constant c (f(t) is of exponential order), then we are guaranteed that f(t) has a Laplace transform F (s). There is one interesting side note here: if f(t) is of exponential order, we know that the quantity f(t) ect does not get too large as t ! 1; alternatively, we interpret this comment as saying that f(t) and ect grow at roughly the same rate. Because of this, if we increase c at all, then the denominator of the fraction will automatically grow faster than f(t). As a result, f(t) lim = 0; s > c: t!1 est Derivatives and the Laplace Transform If f(t) is a function that is continuous and of exponential order, then we know that its Laplace transform F (s) exists, and we also know that f(t) lim = 0; s > c t!1 est 1 Section 7.2 for some number c. We would like to understand the Laplace transform Lff 0(t)g of f 0(t) (assuming that the transform exists). We should go back to the definition of the transform: Z 1 Lff 0(t)g = e−stf 0(t) dt: 0 Let's make the calculation, using integration by parts with u = e−st dv = f 0(t) dt du = −se−st dt v = f(t). Z 1 Lff 0(t)g = e−stf 0(t) dt 0 ( Z ) b = lim e−stf 0(t) dt b!1 ( 0 Z ) b = lim e−stf(t) b + s e−stf(t) dt b!1 0 ( ) 0 ( Z ) b = lim e−stf(t) b + lim s e−stf(t) dt !1 0 !1 b ( ) b 0 = lim e−stf(t) b + sF (s) b!1 0 = lim (e−sbf(b) − f(0)) + sF (s) b!1 f(b) − = lim ( − f(0)) + sF (s) b!1 e sb = −f(0) + sF (s) = sF (s) − f(0) since f(b) lim − = 0 for s > c: b!1 e sb In other words, if we know F (s) = Lff(t)g and f(0); then we can easily find the Laplace transform Lff 0(t)g of f 0(t) using the formula Lff 0(t)g = sF (s) − f(0): This idea is stated more carefully in the following theorem: Theorem 1. Suppose that f(t) is continuous for t ≥ 0, that f 0(t) exists at all but a finite number of points on every bounded subinterval of [0; 1), and that f(t) is of exponential order so that there are constants M, T , and c so that f(t) −M ≤ ≤ M for all t ≥ T: (2) ect 2 Section 7.2 Then Lff 0(t)g exists for all s > c, and is given by Lff 0(t)g = sF (s) − f(0): 1 Example. Given that the Laplace transform of f(t) = sin t is F (s) = s2+1 , verify the formula for the Laplace transform of cos t. Since f 0(t) = cos t, the theorem says that the transform of cos t is given by Lf g − cos t = sF((s) f)(0) 1 = s − 0 s2 + 1 s = ; s2 + 1 s > 0. Let's use the Laplace transform of f 0(t) to find the Laplace transform of f 00(t). Using the formula from the theorem, we know that Lff 00(t)g = sLff 0(t)g − f 0(0) = s(sF (s) − f(0)) − f 0(0) = s2F (s) − sf(0) − f 0(0): Thus we begin to see the development of a remarkable property of Laplace transforms: trans- forms rewrite all of the derivatives of f(t) as algebraic quantities involving the transform F (s) of f(t). The corollary below makes this idea precise: Corollary. If each of the functions f, f 0,··· , and f (n−1) are continuous on [0; 1), each of their derivatives exists at all but a finite number of points on every bounded subinterval of [0; 1), and each is of exponential order for the same values of M and c, then Lff (n)(t)g exists for s > c and is given in terms of the Laplace transform F (s) of f(t) by Lff (n)(t)g = snF (s) − sn−1f(0) − · · · − sf (n−2)(0) − f (n−1)(0): (3) Notice that the corollary does not apply to functions that are piecewise continuous, but not actually continuous; we will need to handle such functions separately. 3 Section 7.2 Solving initial value problems using Laplace transforms With the tools that we have developed in this and in the previous section, we now have the ability to solve initial value problems involving linear equations with constant coefficients. To solve the initial value problem (n) 0 (n−1) any + ::: + a1y + a0y = f(t); y(0) = a0; ··· ; y (0) = an−1 using Laplace transforms, follow the steps below: 1. Apply the Laplace transform to the entire equation. 2. The resulting equation will involve only functions of s and the Laplace transform Y (s) of y(t). Use algebraic methods to solve for Y (s). 3. Since the inverse Laplace transform of Y (s) is unique{it must be y(t){we can apply the inverse Laplace transform to the entire equation to return to an equation for y(t). Example. Use Laplace transforms to solve the initial value problem y00 + y = sin 2t; y(0) = 2; y0(0) = 1: Let's start by applying the Laplace transform to both sides of the equation: y00 + y = sin 2t 2 (s2Y (s) − sy(0) − y0(0)) + Y (s) = s2 + 4 2 (s2Y (s) − 2s − 1) + Y (s) = s2 + 4 2 s2Y (s) − 2s − 1 + Y (s) = s2 + 4 The transformed version of our equation is thus 2 s2Y (s) − 2s − 1 + Y (s) = : s2 + 4 Notice that the equation is algebraic{there are no derivatives left{and involves only the variable s and the unknown function Y (s). Let's solve the equation for Y (s): 2 s2Y (s) − 2s − 1 + Y (s) = s2 + 4 2 Y (s)(s2 + 1) − 2s − 1 = s2 + 4 2 Y (s)(s2 + 1) = + 2s + 1 s2 + 4 2 2s 1 Y (s) = + + : (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 4 Section 7.2 We now have a formula for Y (s), the Laplace transform of the unknown function y(t). As we saw in the previous section, Y (s) is the Laplace transform of precisely one function y(t). So if we can find the inverse Laplace transform of the equation 2 2s 1 Y (s) = + + ; (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 we will have a formula for y(t)! We know that y(t) = L−1fY (s)g 2 2s 1 = L−1f + + g (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 2 2s 1 = L−1f g + L−1f g + L−1f g: (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 Notice that we can find the inverse transforms of the second two fractions using a table of transforms: 2s s L−1f g = 2L−1f g = 2 cos t s2 + 1 s2 + 1 and 1 L−1f g = sin t: s2 + 1 However, computing L−1f 2 g (s2 + 1)(s2 + 4) is a bit trickier{this fraction doesn't show up on our table of transforms. There are many possible techniques for handling such a situation; the most natural one is to try to decompose the fraction using partial fractions. If the resulting fractions have forms that appear on our table of transforms, then we'll be able to compute the answer. The denominator of the fraction 2 (s2 + 1)(s2 + 4) is factored as far as possible, into irreducible quadratics. Thus the form of the decomposed fraction must be 2 B s + C B s + C = 1 1 + 2 2 : (s2 + 1)(s2 + 4) s2 + 1 s2 + 4 Let's add the fractions: B s + C B s + C (B s + C )(s2 + 4) + (B s + C )(s2 + 1) 1 1 + 2 2 = 1 1 2 2 s2 + 1 s2 + 4 (s2 + 1)(s2 + 4) B s3 + C s2 + 4B s + 4C + B s3 + C s2 + B s + C = 1 1 1 1 2 2 2 2 (s2 + 1)(s2 + 4) Since 2 B s + C B s + C B s3 + C s2 + 4B s + 4C + B s3 + C s2 + B s + C = 1 1 + 2 2 = 1 1 1 1 2 2 2 2 ; (s2 + 1)(s2 + 4) s2 + 1 s2 + 4 (s2 + 1)(s2 + 4) 5 Section 7.2 we know that 3 2 3 2 B1s + C1s + 4B1s + 4C1 + B2s + C2s + B2s + C2 = 2: We can determine the constants by equating factors: 3 s B1 + B2 = 0 2 s C1 + C2 = 0 1 s 4B1 + B2 = 0 0 s 4C1 + C2 = 2 Using the second and fourth equations, we have 4C1 + C2 − C1 − C2 = 2 − 0 3C1 = 2 2 C = ; 1 3 − − 2 so that C2 = C1 = 3 .