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Chapter 6.4: Discontinuous forcing functions We now consider non-homogeneous (forced) equations ay 00 + by 0 + cy = g where g is a discontinuous forcing function. Such equations are everywhere in engineering, where the forcing function might be a voltage which is switched on and then turned off. The only method we know so far which can solve such an equation is the Laplace transform method. 1 Table We often use: L−1(e−csF (s)) = uc(t)f (t − c). Also L[ectf (t)] = F (s − c). 2 An example For the first example, let’s drive an oscillatory system with the force 1, t ≤ 1, g(t) = 1 − u1(t) = 0, t > 1 We thus solve: y 00 + y = g, y(0) = 0 = y 0(0). Taking Laplace transforms, we get: (s2 1 + 1)Y (s) = (1 − e−s). s Then, 1 −s ). (1 − e Y (s) = s(s2 + 1) 3 An example We undo the first factor by partial fractions: 1 s(s2 +1) Bs+C =A + s s2 +1 ⇐⇒ As2 + A + Bs2 + Cs + A = 1 ⇐⇒ A = 1, B = −1, C = 0. Thus, 1 1 s = − . 2 2 s(s + 1) s s +1 So we need: 1 −1 −s L {(1 − e )[ s − 2 ]}. s s +1 Using the translation formula, y = 1 − cos t − u1(t) + u1(t) cos(t − 1). 4 Checking the solution This is hard, because we cannot differentiate u1!! It is not even continuous. It is easy to check that the initial condition holds. Let’s check it formally as if u1 had derivatives. The term cos t is a homogeneous solution so we can ignore it. The factor cos(t − 1) is also a homogeneous solution. All told, we get: y 00+y = 1−u001(t)−u1(t)+u001(t) cos(t−1)+2u01 sin(t−1). The trick is that u0(t), u00(t) = 0 unless t = 1. So we can set t = 1 in cos(t − 1) (getting 1) and in sin(t − 1) (getting 0). We thus have: y 00 + y = 1 − u001(t) − u1(t) + u001(t) = 1 − u1. 5 Second example Solve the initial value problem 00 0 L[y] = 2y + y + 2y = g(t), y(0) = 0 = y 0(0), where 1, 5 ≤ t ≤ 20, g(t) = u5(t) = u20(t) = 0 ≤ t < 5, t ≥ 20. Solution: The equation transforms to: (2s2 e−5s e−20s + s + 2)Y (s) = − . s s 6 Solution Thus, 1 −5s − e−20s]. [e s(2s2 + s + 2) Let us write this as Y (s) = Y (s) = H(s)[e−5s−e−20s], H(s) = 1 . 2 s(2s + s + 2) Let h(t) = L−1[H]. Then the solution is: u5(t)h(t − 5) − u20(t)h(t − 20). To find h we use partial fractions: 1 a bs + c = + 2 . 2 s(2s + s + 2) s 2s + s + 2 We get: bs2 + cs + 2as2 + as + 2a = 1 ⇐⇒ 2a = 1, 2a + b = 0, a + c = 0. Thus, a = 1/2, b = −1, c = −1/2. 7 Solution Thus, s+1 1 −s − 1/2 1 1 2 H(s) = + 2 = − 2 + 15 2s 2s + s + 2 2s 2 (s + 1 ) 4 16 From the table of inverse Laplace transforms, we get: √ √ √ −t/4 1 e 15 15 15 [cos( t) + sin( t)]. h(t) = − 2 2 4 15 4 Recall that the solution is: u5(t)h(t − 5) − u20(t)h(t − 20). 8 Third example This time we use non-trivial initial conditions: Solve the initial value problem: y 00 + 2y 0 + 5y = 10u1(t), y(0) = 3y 0(0) = 0 We have Ly 00 + 2Ly 0 + 5Ly = 10L[u1(t)] =⇒ (s2 + 2s + 5)Y (s) − 3s − 6 = 10e−s/s =⇒ Y (s) = 3s+6 s2+2s+5 + 10e−s . s(s2+2s+5) 9 Solution We use partial fractions on the second term 10 A Bs + C = + s(s2 + 2s + 4) s s2 + 2s + 5 or equivalently A(s2 + 2s + 5) + (Bs + C)s = 10(A + B)s2 + (2A + C)s + 5A = 10. The constant term gives A = 2.Thus B = −2, C = −4 We c complete the square s2 + 2s + 5 = (s + 1)2 + 4 We then have: 2 −2s − 4 3s + 6 −s −s +e +e . Y (s) = (s + 1)2 + 4 s (s + 1)2 + 4 10 Solution Simplifying, we get: s+1+1 −s 2 − 2e−s (s+1)+1 Y (s) = 3 (s+1) 2 +4 + e s (s+1)2 +4 s+1 3 2 −s 2 + + e = 3 (s+1) 2 +4 2 (s+1)2 +4 s s+1 2 −s −2e−s (s+1) − e . 2 +4 (s+1)2 +4 Now use the table of inverse Laplace transforms to get y = 3e−t cos(2t) + 3/2e−t−1 sin(2t) + 2u1 −2u1(t)e−t+1 cos(2t − 2) − u1(t)e−t+1 sin(2t − 2). 11 Fourth example The forcing function is now a ramp function (because of its graph): y 00 + 4y = g(t), with y(0) = y 0(0) = 0, 0, 0 ≤ t < 5, (t − 5)/5, 5 ≤ t < 10 g(t) = 1, t ≥ 10. Thus, g(t) = [u5(t)(t − 5) − u10(t)(t − 10)]/5. 12 Fourth example Taking the Laplace transform, we get: Y (s) = (e−5s − e−10s)H(s)/5, with 1 1 1 1 H(s) = 2 2 = [ 2− 2 ]. s (s + 4) 4 s s +4 Then 1 1 h(t) = t − sin(2t) 4 8 and y = u5(t)h(t − 5) − u10(t)h(t − 10). 13