Ordinary Differential Equations

Transcription

1 II 12/01/2015

2 II Second order linear equations with constant coefficients are important in two physical processes, namely, Mechanical and Electrical oscillations. Actually from the Math point of view, both problems are the same. However, from the Physics point of view they are quite different. For example, the motion of a mass on a vibrating spring, the angular motion of a simple pendulum, the flow of electric current in a simple series circuit and the electrical charge in an electric circuit, are just examples of that difference.

3 II Let s get the situation setup. We are going to start with a spring of length l, called the natural length, and we re going to hook an object with mass m up to it. When the object is attached to the spring the spring will stretch a length of L. Below is sketch of the spring with and without the object attached to it.

4 II Convention As denoted in the sketch we are going to assume that all forces, velocities, and displacements in the downward direction will be positive. All forces, velocities, and displacements in the upward direction will be negative. Also, as shown in the sketch above, we will measure all displacement of the mass from its equilibrium position. Therefore, the u = 0 position will correspond to the center of gravity for the mass as it hangs on the spring and is at rest.

5 II Now, we need to develop a differential equation that will give the displacement of the object at any time t. First, recall Newton s Second Law of Motion. F = ma In this case we will use the second derivative of the displacement, u, for the acceleration and so Newton s Second Law becomes, F (t, u, u ) = mu Here is a list of the forces that will act upon the object.

6 II Gravity, F g The force due to gravity will always act upon the object of course. This force is Spring, F s F g = mg We are going to assume that Hookes Law will govern the force that the spring exerts on the object. This force will always be present as well and is F s = k(l + u) Hookes Law tells us that the force exerted by a spring will be the spring constant, k > 0, times the displacement of the spring from its natural length Ḋr. Marco A Roque Sol

7 II Damping, F d The next force that we need to consider is damping. This force may or may not be present for any given problem. This force works to counteract any movement. This damping force is F d = γu where, γ > 0 is the damping coefficient. External Forces, F (t) If there are any other forces acting on our object we collect them in this term. We typically call F (t) the forcing function.

8 II Putting all of these together gives us the following for Newtons Second Law. Or, upon rewriting, we get, mu = mg k(l + u) γu + F (t) mu + γu + ku = mg kl + F (t) Now, when the object is at rest in its equilibrium position, mg kl = 0 Using this in Newtons Second Law gives us the final version of the differential equation that well work with. mu + γu + ku = F (t)

9 II Along with this differential equation we will have the following initial conditions. u(0) = u 0 Initial displacement OBS u (0) = u 0 Initial velocity If we have a mass m attached to a spring with constant k in a surface with friction c ( or γ) and subject to an external force F (t),

10 II then it satisfy the differential equation Free, Undamped Vibrations mu + cu + ku = F (t) This is the simplest case that we can consider. Free or unforced vibrations means that F (t) = 0 and undamped vibrations means that γ = 0. In this case the differential equation becomes, mu + ku = 0 The characteristic equation has the roots, k r = ± m = ±ω 0i; ω 0 = k m

11 II Where ω 0 is called the natural frequency. Recall as well that m > 0 and k > 0 and so we can guarantee that this quantity will not be complex. The solution in this case is then u(t) = c 1 cos(ω 0 t) + c 2 sin(ω 0 t) We can write the above equation in the following form u(t) = Rcos(ω 0 t δ) (If c 1 = Rcos(δ) c 2 = Rsin(δ) = u(t) = Rcos(δ)cos(ω 0 t)+ Rsin(δ)sin(ω 0 t) = u(t) = Rcos(ω 0 t δ); R 2 = c1 2 + c2 2 ; tan(δ) = c 2 /c 1 where R is the amplitude of the displacement and δ is the phase shift or phase angle of the displacement. T = 2π ω 0 = 2π m k is called the natural period.

12 II Example 55 A 16 lb object stretches a spring 8/9 ft by itself. There is no damping and no external forces acting on the system. The spring is initially displaced 6 inches upwards from its equilibrium position and given an initial velocity of 1 ft/sec downward. Find the displacement at any time t, u(t). Solution We first need to set up the IVP for the problem. We need to find m and k. This is the British system so we ll need to compute the mass. m = W g = = 1 2

13 II Now, lets find k. We can use the fact that mg = kl to find k. We ll use feet for the unit of measurement for this problem. We can now set up the IVP. k = mg L = 16 8/9 = u + 18u = 0; u(0) = 1 2 (6inches), u (0) = 1 Now, the natural frequency, is 18 ω 0 = 1/2 = 6

14 II The general solution, along with its derivative, is then, u(t) = c 1 cos(6t) + c 2 sin(6t) u (t) = 6c 1 sin(6t) + 6c 2 cos(6t) Applying the initial conditions gives 1 2 = u(0) = c 1 = c 1 = = u (0) = 6c 2 = c 2 = 1 6 The displacement at any time t is then u(t) = 1 2 cos(6t) sin(6t)

15 II Now, lets convert this to a single cosine function. First let s get the amplitude, R. R = ( 1 ) Now let s get the phase shift. ( ) 1/6 δ = tan 1 1/2 ( ) = = = From the above equations, we have two angles δ 1 = ; δ 2 = δ 1 + π =

16 II We need to decide which of these phase shifts is correct, because only one will be correct. To do this recall that c 1 = Rcos(δ) = 1/2 < 0; c 2 = Rsin(δ) = 1/6 > 0 This means that the phase shift must be in Quadrant II and so the second angle is the one that we need. Thus, the displacement at any time t is. ( ) 10 1 u(t) = cos(6t δ); δ = tan 1 + π 6 3

17 II Here is a sketch of the displacement for the first 5 seconds.

18 II Free, Damped Vibrations We are still going to assume that there will be no external forces acting on the system, with the exception of damping of course. In this case the differential equation will be mu + γu + ku = 0 where m, γ, and k are all positive constants. Upon solving for the roots of the characteristic equation we get the following. r 1,2 = γ ± γ 2 4mk 2m

19 II We will have three cases here. 1. γ 2 4mk = 0 In this case we will get a double root out of the characteristic equation and the displacement at any time t will be. u(t) = c 1 e γt 2m + c2 te γt 2m Notice that as t the displacement will approach zero. This case is called critical damping and will happen when the damping coefficient is, γ 2 4mk = 0 = γ = 4mk = γ CR

20 II The above value of the damping coefficient is called the critical damping coefficient and denoted by γ CR. 2. γ 2 4mk > 0. In this case let s rewrite the roots r 1,2 = γ ± γ 2 4mk 2m ( ) = γ 1 ± 1 4mk 2m γ 2 Also notice that from our initial assumption that we have, 4mk γ 2 < 1 = 1 4mk γ 2 < 1 = 1 4mk γ 2 < 1

21 II This means that the quantity in the parenthesis is guaranteed to be positive and so the two roots in this case are guaranteed to be negative. Therefore the displacement at any time t is ( ) ( ) u(t) = c 1 e γ mk 2m γ 2 t + c2 e γ 1 1 4mk 2m γ 2 t and will approach zero as t. This case will occur when γ 2 4mk > 0 = γ > 4mk = γ CR and is called over damping.

22 II 3. γ 2 4mk < 0. In this case we will get complex roots out of the characteristic equation. r 1,2 = γ γ 2m ± 2 4mk = λ ± i µ 2m where the real part is guaranteed to be negative and so the displacement is ) u(t) = e γ 4mk γ 2m (c t 2 4mk γ 2 1 cos( t) + c 2 sin( t) 2m 2m u(t) = = Rcos(µt δ) Since λ < 0 the displacement will approach zero as t.

23 II We will get this case will occur when γ 2 4mk < 0 = γ < 4mk = γ CR and is called under damping. Example 56 Take the spring and mass system from the example 55 and consider there is a damping force to it that will exert a force of 17 lbs when the velocity is 2ft/s. Find the displacement at any time t, u(t). Solution So, the only difference between this example and the previous example is damping force. So let s find the damping coefficient 17 = γ(2) = γ = 2/17 = 8.5 > γ CR = 4km = 6

24 II So it looks like we have got over damping this time around so we should expect to get two real distinct roots from the characteristic equation and they should both be negative. The IVP for this example is 1 2 u u + 18u = 0; u(0) = 1 2, u (0) = 1 The roots of the characteristic equation are r 1,2 = 17 ± The general solution for this example is u(t) = c 1 e t + c 2 e t 2

25 II and after applying initial conditions, the particular solution is u(t) = 0.52e ( ) t e ( Here is a sketch of the displacement for the first 5 seconds. ) t

26 II Example 57 Take the spring and mass system from the example 55 and consider there is a damping force to it that will exert a force of 12lbs when the velocity is 2ft/s. Find the displacement at any time t, u(t). Solution The damping coefficient is given by 12 = γ(2) = γ = 12/2 = 6 = γ CR So it looks like we have got critical damping this time. The IVP for this problem is 1 2 u + 6u + 18u = 0; u(0) = 1 2, u (0) = 1

27 II The roots of the characteristic equation are 6 r 1,2 = (2)(1/2) = 6 The general solution for this example is u(t) = c 1 e 6t + c 2 te 6t and after applying initial conditions, the particular solution is u(t) = 1 2 e 6t 2te 6t

28 II Here is a sketch of the displacement for this example

29 II Example 58 Take the spring and mass system from the example 55 and consider there is a damping force to it that will exert a force of 5lbs when the velocity is 2ft/s. Find the displacement at any time t, u(t). Solution The damping coefficient is given by 5 = γ(2) = γ = 5/2 = 2.5 < γ CR So it looks like we have got under damping this time. The IVP for this problem is 1 2 u u + 18u = 0; u(0) = 1 2, u (0) = 1

30 II The roots of the characteristic equation are r 1,2 = 5 ± 119 i 2 The general solution for this example is ( ) ( )) u(t) = e (c t 1 cos t + c 2 sin t 2 2 and after applying initial conditions, the particular solution is u(t) = e 5t 2 ( ( ) ( )) cos t 0.046sin t 2 2

31 II Lets convert this to a single cosine as we did in the undamped case. R = ( 0.5) 2 + (.046) 2 = δ = tan 1 = or δ = π = 3.23 ( ) = This means δ must be in the Quadrant II ( why?) and so the second angle is the one that we want. The displacement is then ( ) u(t) = 0.502e 5t cos t

32 II Here is a sketch of the displacement for this example

33 II Undamped, Forced Vibrations We will first take a look at the undamped case. The differential equation in this case is mu + ku = F (t) This is just a nonhomogeneous differential equation and we know how to solve these. The general solution will be u(t) = u c + U P There is a particular type of forcing function that we should take a look at since it leads to some interesting results. Lets suppose that the forcing function is a simple periodic function of the form F (t) = F 0 cos(ωt) or F (t) = F 0 sin(ωt)

34 II For the purposes of this discussion we will use the first one. Using this, the ODE becomes, mu + ku = F 0 cos(ωt) The solution of the associate homogeneus, as pointed out above, is just u c (t) = c 1 cos(ω 0 t) + c 2 sin(ω 0 t) where ω 0 is the natural frequency. We will need to be careful in finding a particular solution. The reason for this will be clear if we use undetermined coefficients. With undetermined coefficients our guess for the form of the particular solution would be,

35 II U P (t) = Acos(ωt) + Bsin(ωt) Now, this guess will have problems if ω = ω 0. So, we will need to look at this in two cases. 1. ω ω 0 In this case our initial guess is okay since it wont be the complementary solution. Upon differentiating the guess and plugging it into the differential equation and simplifying we get, mu + ku = F 0 cos(ωt) m (Acos(ωt) + Bsin(ωt)) + k (Acos(ωt) + Bsin(ωt)) = F 0 cos(ωt)

36 II m ( ω 2 Acos(ωt) ω 2 Bsin(ωt) ) + k (Acos(ωt) + Bsin(ωt)) = F 0 cos(ωt) ( mω 2 A + ka ) cos(ωt) + ( mω 2 B + kb ) sin(ωt) = F 0 cos(ωt) Setting coefficients equal gives us, cos(ωt) ( mω 2 + k ) A = F 0 = A = F 0 k mω 2 sin(ωt) ( mω 2 + k ) B = 0 = B = 0

37 II The particular solution is then F 0 k mω 2 cos(ωt) = F 0 m (k/m ω 2 ) cos(ωt) = F 0 m ( ω 2 0 ω2)cos(ωt) Note that we rearranged things a little. Depending on the form that you d like the displacement to be in we can have either of the following. u(t) = c 1 cos(ω 0 t) + c 2 sin(ω 0 t) + u(t) = Rcos(ω 0 t δ) + F 0 m ( ω 2 0 ω2)cos(ωt) F 0 m ( ω 2 0 ω2)cos(ωt) If we used the sine form of the forcing function we could get a similar formula.

38 II 2. ω = ω 0 In this case we will need to add in a t to the guess for the particular solution. U P (t) = Atcos(ωt) + Btsin(ωt) Differentiating our guess, plugging it into the differential equation and simplifying gives us the following. ( mω k ) Atcos(ωt) + ( mω k ) Btsin(ω)t +... but mω 0 Bcos(ωt) 2mω 0 Asin(ωt) = F 0 cos(ωt) ( mω k ) = m ( ω0 2 + k/m ) = m ( ω0 2 + ω0 2 ) = 0

39 II So, the first two terms actually drop out and this gives us cos(ωt) 2mω 0 B = F 0 = B = F 0 2mω 0 In this case the particular will be, sin(ωt) 2mω 0 A = 0 = A = 0 F 0 2mω 0 tsin(ω 0 t) The displacement for this case is then u(t) = c 1 cos(ω 0 t) + c 2 sin(ω 0 t) + F 0 2mω 0 tsin(ω 0 t)

40 II depending on the form that you prefer for the displacement. u(t) = Rcos(ω 0 t δ) + F 0 2mω 0 tsin(ω 0 t) So, what was the point of the two cases here? Well in the first case, our displacement function consists of two cosines and is nice and well behaved for all time. In contrast, the second case, will have some serious issues at t increases. The addition of the t in the particular solution will mean that we are going to see an oscillation that grows in amplitude as t increases. This case is called resonance and we would generally like to avoid this at all costs.

41 II

42 II In this case resonance arose by assuming that the forcing function was, F (t) = F 0 cos(ω 0 t) We would also have the possibility of resonance if we assumed a forcing function of the form. F (t) = F 0 sin(ω 0 t) We should also take care to not assume that a forcing function will be in one of these two forms. Forcing functions can come in a wide variety of forms. If we do run into a forcing function different from the one that used here you will have to go through undetermined coefficients or variation of parameters to determine the particular solution.

43 II Example 59 A 3 kg object is attached to spring and will stretch the spring 392 mm by itself. There is no damping in the system and a forcing function of the form F (t) = 10cos(ωt) is attached to the object and the system will experience resonance. If the object is initially displaced 20 cm downward from its equilibrium position and given a velocity of 10 cm/sec upward find the displacement at any time t. Solution Since we are in the metric system we wont need to find mass as its been given to us. Also, for all calculations we will be converting all lengths over to meters. The first thing we need to do is find k.

44 II k = mg L = (3)(9.8) = 75 Now, we are told that the system experiences resonance so let s go ahead and get the natural frequency. k 75 ω 0 = m = 3 = 5 The IVP for this is then 3u + 75u = 10cos(5t); u(0) = 0.2, u (0) = 0.1 The complementary solution is the free undamped solution which is easy to get and for the particular solution we can just use the formula that we derived above. The general solution is then, u(t) = c 1 cos(5t) + c 2 sin(5t) tsin(5t)

45 II Applying the initial conditions gives u(t) = 1 5 cos(5t) 1 50 sin(5t) tsin(5t) The last thing that we ll do is combine the first two terms into a single cosine. (1 ) 2 R = + 5 ( 1 50 ) 2 ( ) 1/50 = δ 1 = tan 1 = /5 δ 2 = δ 1 + π = In this case c 1 > 0 is positive and c 2 < 0. This means that the phase shift needs to be in Quadrant IV and so the first one is the correct phase shift this time.

46 II The displacement then becomes, u(t) = cos ( 5t + tan 1 ( 1 10 )) tsin(5t) Here is a sketch of the displacement for this example

47 II Example 60 Solve the initial value problem and plot the solution. Solution u + u = 0.5cos(0.8t), u(0) = 0, u (0) = 0 The general solution of is u = c 1 cos(ω 0 t) + c 2 sin(ω 0 t) + Applying initial conditions, we obtain F 0 c 1 = m(ω0 2 ω2 ) ; c 2 = 0 F 0 m(ω 2 0 ω2 ) cos(ωt)

48 II and the particular solution of the IVP is u = F 0 m(ω 2 0 ω2 ) (cos(ωt) cos(ω 0t)) This is the sum of two periodic functions of different periods but the same amplitude. Making use of the trigonometric identities for cos(a ± B) with A = (ω 0 + ω)t/2 and B = (ω 0 ω)t/2, we can write the above equation in the form [ ( )] ( ) F 0 u = m(ω0 2 ω2 ) sin (ω0 ω)t (ω0 + ω)t sin 2 2

49 II If ω 0 ω is small, then ω 0 + ω is much greater than it. Consequently, sin(ω 0 + ω)t/2 is a rapidly oscillating function compared to sin(ω 0 ω)t/2. Thus the motion is a rapid oscillation with frequency (ω 0 + ω)/2 but with a slowly varying sinusoidal amplitude F 0 m ω 2 0 ω2 ( ) sin (ω0 ωt) 2 This type of motion, possessing a periodic variation of amplitude, exhibits what is called a beat. In this case ω 0 = 1, = 0.8, and F 0 = 0.5, so the solution of the given problem is u(t) = [2.77sin(0.1t)] sin(0.9t)

50 II Here is a sketch of the displacement for this example.

51 II Forced, Damped Vibrations This is the full blown case where we consider every last possible force that can act upon the system. The differential equation for this case is, mu + γu + ku = F (t) The displacement function this time will be, u(t) = u c + U P where the complementary solution will be the solution to the ( free, damped) homogeneous case and the particular solution will be found using undetermined coefficients or variation of parameters.

52 II There are a couple of things to note here about this case. First, from our work back in the free, damped case we know that the complementary solution will approach zero as t. Because of this, the complementary solution is often called the transient solution in this case. Also, because of this behavior the displacement will start to look more and more like the particular solution as t increases and so the particular solution is often called the steady state solution or forced response.

53 II Example 61 Take the system from the example 59 and add in a damper that will exert a force of 45 Newtons when then velocity is 50 cm/sec. Solution So, all we need to do is compute the damping coefficient for this problem then pull everything else down from the previous problem. The damping coefficient is F d = γu = 45 = γ(0.5) = γ = 90

54 II The IVP for this problem is. 3u + 90u + 75u = 10cos(5t); u(0) = 0.2, u (0) = 0.1 The complementary solution for this example is u(t) = c 1 e ( )t + c 2 e ( )t For the particular solution we the form will be, U P = Acos(5t) + Bsin(5t) Plugging this into the differential equation and simplifying gives us, 405Bcos(5t) 450Asin(5t) = 10cos(5t)

55 II Setting coefficient equal gives, The general solution is then U P = 1 45 sin(5t) u(t) = c 1 e ( )t + c 2 e ( )t sin(5t) Applying the initial condition gives u(t) = e ( )t e ( )t sin(5t)

56 II Here is a sketch of the displacement for this example.

57 II Electric Circuits A second example of the occurrence of second order linear differential equations with constant coefficients is their use as a model of the flow of electric current in the simple series circuit shown below

58 II The current I, measured in amperes (A), is a function of time t. The resistance R in ohms (Ω), the capacitance C in farads (F ), and the inductance L in henrys (H) are all positive and are assumed to be known constants. The impressed voltage E in volts (V ) is a given function of time. Another physical quantity that enters the discussion is the total charge Q in coulombs (C) on the capacitor at time t. The relation between charge Q and current I is I = dq dt

59 II The flow of current in the circuit is governed by Kirchhoff s second law: (Gustav Kirchhoff ( ) was a German physicist and professor at Breslau, Heidelberg, and Berlin. britannica.com/biography/gustav-robert-kirchhoff ) vspace3mm In a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the circuit. According to the elementary laws of electricity, we know that

60 II The voltage drop across the resistor is IR. The voltage drop across the capacitor is Q/C. The voltage drop across the inductor is LdI /dt. Hence, by Kirchhoffs law, L di dt RI 1 Q + E(t) = 0 = LdI C dt + RI + 1 C Q = E(t)

61 II The units for voltage, resistance, current, charge, capacitance, inductance, and time are all related: 1volt = 1ohm 1ampere = 1coulomb/1farad = 1henry 1ampere/1seco Substituting dq dt for I, we obtain the differential equation L d 2 Q dt + R dq dt + 1 C Q = E(t) for the charge Q. The initial conditions are Q(t 0 ) = Q 0, Q (t 0 ) = I (t 0 ) = I 0

62 II Thus we must know the charge on the capacitor and the current in the circuit at some initial time t 0. Alternatively, we can obtain a differential equation for the current I by differentiating the above equation with respect to t, and then substituting dq/dt for I. The result is L d 2 I dt 2 + R di dt + 1 C I = E (t) with the initial conditions I (t 0 ) = I 0, I (t 0 ) = I 0 it follows from the equation for Q(t) that I 0 = E(t 0) RI 0 (1/C)Q 0 L

63 II The most important conclusion from this discussion is that the flow of current in the circuit is described by an initial value problem of precisely the same form as the one that describes the motion of a springmass system.

64 II Second Order Lineat Differential Equations. Reduction of Order In the final part of this section, we will consider the general second order linear differential equation y + p(t)y + q(t)y = g(t) and introducing new variables, the idea will be to reduce the above equation of second order into something of first order. In that way we propose x 1 = y x 2 = y

65 II these two varaibles satisfy x 1 = y = x 2 x 2 = y = p(t)y q(t)y + g(t) = p(t)x 2 q(t)x 1 + g(t) therefore, we obtain x 1 = x 2 x 2 = p(t)x 2 q(t)x 1 + g(t) In this way, we have made a reduction of order, but now instead of a single equation, we have a system linear first oder differential equations!!!!!

66 II Thus, we have showed that a second order linear differential equations can always be transformed into a system of two linear first order differential equations. OBS 1. The above proposition can be generalized for the n dimensional case.... An n th order linear differential equation is equivalent to a system of n linear first order differential equations...

67 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters General Theory of n th Order Linear Equations Example 4.2 Determine whether the functions f 1 (t) = 1, f 2 (t) = 2 + t, f 3 (t) = 3 t 2, and f 4 (t) = 4t + t 2 are linearly independent or dependent on the interval I : < t <. Solution Form the linear combination k 1 f 1 (t) + k 2 f 2 (t) k 3 f 3 (t) + k 4 f 4 (t) = k 1 (1) + k 2 (2 + t) k 3 (3 t 2 ) + k 4 (4t t 2 ) = 0 And differentiating the above formula three times we have k 2 2tk 3 + k 4 (4 2t) = 0 2k 3 2k 4 = 0 0 = 0

68 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters General Theory of n th Order Linear Equations Thus, can show that the system has infinitely solutions and therefore k 1, k 2, k 3, and k 4 are not necesarily all of them equal to zero. Hence, the set of functions is linearly dependent. OBS f 4 (t) = 4t t 2 = (3 t 2 ) 4(2 + t) + 5(1) = f 3 (t) 4f 2 (t) + 5f 1 (t) therefore the set f 1 (t) = 1, f 2 (t) = 2 + t, f 3 (t) = 3 t 2, f 4 (t) = 4t + t 2 cannot be linearly indepent.

69 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters General Theory of n th Order Linear Equations Theorem 4.3 If y 1 (t),..., y n (t) is a fundamental set of solutions of the equation L[y](t) = d n y dt n + p 1(t) d n 1 y dt n p dy n 1 dt + p n(t)y = 0 on an interval I, then y 1 (t),..., y n (t) are linearly independent on I. Conversely, if y 1 (t),..., y n (t) are linearly independent solutions of the above equation on I, then they form a fundamental set of solutions on I.

70 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters General Theory of n th Order Linear Equations The Nonhomogeneous Equation Now consider the nonhomogeneous eq L[y](t) = d n y dt n + p 1(t) d n 1 y dt n p dy n 1 dt + p n(t)y = g(t) If Y 1 and Y 2 are any two solutions of the above equation, then it follows immediately from the linearity of the operator L that L[Y 1 Y 2 ](t) = L[Y 1 ](t) L[Y 2 ](t) = g(t) g(t) = 0

71 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters General Theory of n th Order Linear Equations Hence the difference of any two solutions of the nonhomogeneous equation is a solution of the homogeneous equation. Since any solution of the homogeneous equation can be expressed as a linear combination of a fundamental set of solutions y 1,..., y n, it follows that any solution of nonhomogeneus equation can be written as y(t) = c 1 y 1 (t) + c 2 y 2 (t) c n y n (t) + Y (t) where Y is some particular solution of the nonhomogeneous equation. The above linear combination is called the general solution of the nonhomogeneous equation.

72 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients Let s take the n th order linear homogeneous differential equation d n y L[y](t) = a 0 dt n + a d n 1 y 1 dt n a dy n 1 dt + a ny = 0 where a 0, a 1,..., a n are real constants and a 0 0. Again we proposed that y = e rt is a solution of the above equation. As a matter of fact, L[e rt ] = e rt ( a 0 r n + a 1 r n a n 1 r + a n ) = e rt Z(r) for all r, where Z(r) = a 0 r n + a 1 r n a n 1 r + a n. For those values of r for which Z(r) = 0, it follows that L[e rt ] = 0 and y = e rt is a solution of homogeneous equation. The polynomial Z(r) is called the characteristic polynomial, and the equation Z(r) = 0 is the characteristic equation.

73 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients Since a 0 0, we know that Z(r) is a polynomial of degree n and therefore as n zeros, say, r 1, r 2,..., r n, some of which may be equal. Hence we can write the characteristic polynomial in the form (Fundamental Theorem of Algebra: Every non-constant single-variable polynomial with complex coefficients has at least one complex root. ) Z(r) = a 0 (r r 1 )(r r 2 )... (r r n ). In general there are three cases, namely 1) Real and Different Roots. If the roots of the characteristic equation are different, then we have n distinct solutions e r 1t, e r 2t,..., e rnt. These functions are linearly independent, then the general solution is

74 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients y(t) = c 1 e r 1t + c 2 e r 2t c n e rnt 2) Complex Roots. If the characteristic equation has complex roots, they must occur in conjugate pairs, λ ± i µ, since the coefficients a 0, a 1, a 2,..., a n are real numbers. Provided that none of the roots is repeated. Now, just as for the second order equation, we can replace the complex-valued solutions z 1 = e (λ+i µ)t and z 2 = e (λ i µ)t by the real-valued solutions e λt cos(µt); e λt sin(µt) obtained as the linear cobinations 1 2 (z 1 + z 2 ) and 1 2i (z 1 z 2 )

75 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients 3) Repeated Roots. Let s assume that some of the roots are repeated. For an equation of order n, if a root of Z(r) = 0, say r = r 1, has multiplicity s ( that is, it is repeated s times) (where s n ), then e r 1t, te r 1t t n 1 e r 1t are corresponding solutions of the differential equation. If a complex root λ + i µ is repeated s times, the complex conjugate λ i µ is also repeated s times. Corresponding to these 2s complex-valued solutions, we can find 2s real-valued linearly independent solutions:

76 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients e λt cos(µt), e λt sin(µt); te λt cos(µt), te λt sin(µt);......; t n 1 e λt cos(µt), t n 1 e λt sin(µt) Hence the general solution of the homogeneous equation can always be expressed as a linear combination of n real-valued solutions. Example 4.3 Find the general solution of the IVP y (4) + y 7y + 6y = 0; y(0) = 1, y (0) = 0, y (0) = 2, y (0) = 1

77 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients Solution The characteristic equation is r 4 + r 3 7r + 6 = 0 The roots of this equation are r 1 = 1, r 2 = 1, r 3 = 2, and r 4 = 3. Therefore, the general solution is y = c 1 e t + c 2 e t + c 3 e 2t + c 4 e 3t and aplying initial conditions we have c 1 + c 2 + c 3 + c 4 = 1 c 1 c 2 + 2c 3 3c 4 = 0 c 1 + c 2 + 4c 3 + 9c 4 = 2 c 1 c 2 + 8c 3 27c 4 = 1

78 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients By solving this system of four linear algebraic equations, we find that c 1 = 11 8, c 2 = 5 12, c 3 = 2 3, c 4 = 1 8 Thus the solution of the initial value problem is y = 11 8 et e t 2 3 e2t 1 8 e 3t

79 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients Example 4.4 Find the general solution of Solution The characteristic equation is y (4) + y = 0 r = 0 = r 4 = 1 In this way, we need to find the four roots of -1. Now 1, thought of as a complex number, is 1 + 0i. It has magnitude 1 and polar angle π (r = Re θi ). Thus 1 = cos(π) + i sin(π) = e πi

80 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients Moreover, the angle is determined only up to a multiple of 2. Thus 1 = cos(π + 2mπ) + i sin(π + 2mπ) = e (π+2mπ)i where m is an integer. Thus ( 1) 1/4 = e (π/4+2mπ/4)i = cos ( π 4 + 2mπ ) ( π + i sin mπ ) 4 The four fourth roots of 1 are obtained by setting m = 0, 1, 2, and 3;

81 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients m = 4 = cos m = 0 = cos m = 1 = cos m = 2 = cos m = 3 = cos ( π 4 + 4π 2 ( π ) ( π ) + i sin 4 4 ( π 4 + π ) ( π + i sin π ) 2 ( π ) ( π ) 4 + π + i sin 4 + π ( π 4 + 3π 2 ) ( π + i sin 4 + 4π 2 ) ( π + i sin 4 + 3π 2 ) = cos ) ( π ) ( π ) + i sin!!! 4 4

82 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Homogeneous Equations with Constant Coefficients and the solutiones are 1 + i, 1 + i 2 2, 1 i, 1 i 2 2 The general solution is ( y(t) = e t 2 c 1 cos ( ) ( )) t t + c 2 sin ( ( ) ( )) e t t t 2 c 3 cos + c 4 sin 2 2

83 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Undetermined Coefficients We can find a particular solution Y P of the nonhomogeneous n th order linear equation with constant coefficients d n y L[y](t) = a 0 dt n + a d n 1 y 1 dt n a dy n 1 dt + a ny = g(t) using The Method of Undetermined Coefficients, provided that g(t) is of an appropriate form. We have to be careful when the roots of the characteristic polynomial equation have multiplicity, because now this could be greater than 2. Example 4.5 Find the general solution of y 3y + 3y y = 4e t

84 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Undetermined Coefficients Solution The characteristic polynomial is r 3 3r 2 + 3r 1 = (r 1) 3 so the general solution of the homogeneous equation is y c (t) = c 1 e t + c 2 te t + c 3 t 2 e t To find a particular solution Y P (t) of the nonhomogeneous equation we start by assuming that Y P (t) = Ae t.

85 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Undetermined Coefficients However, since e t, te t, and t 2 e t are all solutions of the homogeneous equation, we must multiply this initial choice by t 3. Thus our final assumption is Y P (t) = At 3 e t = 6Ae t = 4e t = A = 2 3 Thus, the general solution is y c (t) = c 1 e t + c 2 te t + c 3 t 2 e t t3 e t

86 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Undetermined Coefficients Example 4.6 Find a particular solution of Solution The characteristic equation is y 4y = t + 3cos(t) + e 2t r 3 4r = 0 the roots are r = 0, ±2 and the general solution of the homogeneous equation is. y(t) = c 1 + c 2 e 2t + c 3 e 2t

87 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Undetermined Coefficients To find a particular solution Y P (t) of the nonhomogeneous equation we propose that Y P (t) = (A 0 t + A 1 )t + A 3 cost(t) + A 4 sin(t) + A 5 e 2t t The constants are determined by substituting into the differential equation. They are A 0 = 1/8, A 1 = 0, A 3 = 0, A 4 = 3/5, and A 5 = 1/8. Hence a particular solution is Y P (t) = 1 8 t3 3 5 sin(t) te 2t

88 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters The method of variation of parameters for determining a particular solution of the nonhomogeneous n th order linear differential equation L[y](t) = d n y dt n + p 1(t) d n 1 y dt n p dy n 1 dt + p n(t)y = g(t) Suppose then that we know a fundamental set of solutions y 1, y 2,..., y n of the homogeneous equation. Then the general solution of the homogeneous equation is y(t) = c 1 y 1 (t) + c 2 y 2 (t) c n y n (t)

89 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters The Method of Variation of Parameters for determining a particular solution of the nonhomogeneous equation, Y P (t), rests on the possibility of determining n functions u 1, u 2,..., u n such that Y P (t) is of the form Y P (t) = u 1 (t)y 1 (t) + u 2 (t)y 2 (t) u n (t)y n (t) Since we have n functions to determine, we will have to specify n conditions. One of these is clearly that Y P the ODE. The other n 1 conditions are chosen arbitrarily. as to make the calculations as simple as possible. Following the same idea used in the second order case we have that the first derivative of Y P is given by

90 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters Y P (t) = ( u 1 (t)y 1(t) + u 2 (t)y 2(t) u n (t)y n(t) ) + ( u 1 (t)y 1 (t) + u 2(t)y 2 (t) u n(t)y n (t) ) Thus the first condition that we impose is that u 1(t)y 1 (t) + u 2(t)y 2 (t) u n(t)y n (t) = 0 We continue this process by calculating the successive derivatives Y,..., Y (n 1). After each differentiation we set equal to zero the sum of terms involving derivatives of u 1,..., u n. In this way we obtain n 2 further conditions similar to the above and put them together we have

91 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters u 1(t)y (m) 1 (t) + u 2(t)y (m) 2 (t) u n(t)y (m) n (t) = 0; m = 1, 2,..., n 1 As a result of these conditions, it follows that the expressions for Y P, Y P (n 1),..., Y P reduce to Y (m) P = u 1 (t)y (m) 1 (t) + u 2 (t)y (m) 2 (t) u n (t)y (m) n (t) = 0; m = 1, 2, 3,..., n 1 Finally, we need to impose the condition that Y P must be a solution of the nonhomogeneous equation. By differentiating Y (n1) P from the above equation, we obtain

92 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters Y (n) P ( = (Y (n 1) P ) = u 1 (t)y (n) 1 (t) + u 2 (t)y (n) ( u 1(t)y (n 1) 1 (t) + u 2(t)y (n 1) 2 (t) u n(t)y n (n 1) 2 (t) u n (t)y n (n) ) (t) = g(t) plug into the equation and considering that L[y i ] = 0; i = 1, 2,..., n, then the ramaining terms yield the relation u 1(t)y (n 1) 1 (t) + u 2(t)y (n 1) 2 (t) u n(t)y n (n 1) (t) = g(t) Thus, we have a system of n simultaneos linear nonhomogeneus algebraic equations for u 1, u 2,..., u n : ) (t) +

93 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters u 1(t)y 1 (t) + u 2(t)y 2 (t) u n(t)y n (t) = 0 u 1(t)y 1(t) + u 2(t)y 2(t) u n(t)y n(t) = 0 u 1(t)y 1 (t) + u 2(t)y 2 (t) u n(t)y n (t) = 0. u 1(t)y (n 1) 1 (t) + u 2(t)y (n 1) 2 (t) u n(t)y n (n 1) (t) = g(t)

94 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters y 1 y 2... y n y 1 y 2... y n y 1 y 2... y n. y (n 1) 1 y (n 1) 2... y n (n 1) u 1 u 2 u 3 =. u n g(t) The above system, is a linear algebraic system for the unknown quantities u 1, u 2,..., u n. The determinant of coefficients is precisely W (y 1, y 2,..., y n ), and it is nowhere zero since y 1,..., y n is a fundamental set of solutions of the homogeneous equation. Hence it is possible to determine u 1, u 2,..., u n using Cramer s Rule :

95 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters u m(t) = g(t)w m(t) ; m = 1, 2,..., n W (t) where W (t) = W (y 1, y 2,..., y n )(t) ( The Wronskian ) and W m (t) is the determinant obtained from W by replacing the m th column by the column (0, 0,..., 0, 1). W = y 1 y 2... y n y 1 y 2... y n y 1 y 2... y n y (n 1) 1 y (n 1) y n (n 1)

96 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters m th y y n y y n W m = y y n. y (n 1) y n (n 1) And integrating the above equations we have that the particular solution is given by n t Y P (t) = y m m=1 t 0 g(s)w m W (s) ds

97 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters where t 0 is an arbitrary point. As we can see, things get more complicated than in the second order case. In some cases the calculations may be simplified to some extent by using Abel s identity [ W (t) = W (y 1, y 2,..., y n )(t) = cexp ] p 1 (t)dt The constant c can be determined by evaluating W at some convenient point.

98 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters Example 4.7 Given that y 1 (t) = e t, y 2 (t) = te t, and y 3 (t) = e t are solutions of the homogeneous equation corresponding to y y y + y = g(t) determine a particular in terms of an integral. Solution Let s determine first the Wronskian W = W (e t, te t, e t e t te t e t )(t) = e t (t + 1)e t e t e t (t + 2)e t e t

99 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters Factoring e t from each of the first two columns pause and e t from the third column, we obtain W = W (e t, te t, e t )(t) = e t 1 t 1 1 (t + 1) 1 1 (t + 2) 1 Then, by subtracting the first row from the second and third rows, we have W = W (e t, te t, e t )(t) = e t 1 t

100 II General Theory of n th Order Linear Equations Homogeneous Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters The Method of Variation of Parameters Finally, evaluating this determinant by minors using the first column, we find that W (t) = 4e t ; W 1 (t) = 2t 1; W 2 (t) = 2; W 3 (t) = e 2t Thus, the particular solution is given by 3 t g(s)w m Y P (t) = y m W (s) Y P (t) = e t t Y P (t) = 1 4 m=1 g(s)( 1 2s) t 0 4e s t t 0 + te t t 0 t t 0 g(s)(2) 4e s + e t t [ e t s ( 1 + 2(t s)) + e (t s)] g(s)ds t 0 g(s)e 2s 4e s +

101 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Among the tools that are very useful for solving linear differential equations are integral transforms. An integral transform is a relation of the form F (s) = β α K(s, t)f (t)dt where K(s, t) is a given function, called the kernel of the transformation, and the limits of integration α and β are also given. It is possible that α = or β = or both. The relation, introduced above, transforms the function f into another function F, which is called the transform of f.

102 II Definition of The Solution of Initial Value Problems Step Functions Definition of The There are several integral transforms that are useful in applied mathematics, but we consider only the ( ) (... Napoleon asked Laplace where God fit into his mathematical work Traite de mecanique celeste, and Laplace famously replied Sir, I have no need of that hypothesis... ). Let f (t) be given for t 0. Then the Laplace transform of f, which we will denote by L {f (t)} = F (s), is defined by the equation L {f (t)} = F (s) = whenever this improper integral converges. 0 e st f (t)dt

103 II Definition of The Solution of Initial Value Problems Step Functions Definition of The The Laplace transform makes use of the kernel K(s, t) = e st. In particular for linear second order differential equations with constant coeficients is particular useful, since the solutions are based on the exponential function. The general idea in using the Laplace transform to solve a differential equation is as follows: 1. Use the relation L {f (t)} = F (s) to transform an initial value problem for an unknown function f in the t-domain (time domain) into an algebraic problem for F in the s-domain (frequency domain).

104 II Definition of The Solution of Initial Value Problems Step Functions Definition of The 2. Solve this algebraic problem to find F. 3. Recover the desired function f from its transform F. This last step is known as inverting the transform (which in general involve complex integration) and denoted by L 1 {F (s)} (= lim ω 1 2πi σ+iω σ iω F (s)est ds). OBS The full power of becomes available only when we regard F (s) as a function of a complex variable. However, for our purposes it will be enough to consider only real values for s. The Laplace transform F of a function f exists if f satisfies certain conditions:

105 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Theorem 6.1 Suppose that 1. f is piecewise continuous on the interval 0 t A for any positive A. 2. f (t) Ke at when t M. In this inequality, K, a, and M are real constants, K and M necessarily positive. Then the Laplace transform L {f (t)} = F (s), defined by exists for s > a. L {f (t)} = F (s) = 0 e st f (t)dt

106 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Remember that a function, f (t), is piecewise continuous on the interval α t β if the interval can be partitioned by a finite number of points α = t 0 < t 1 <... < t n = β so that 1. f is continuous on each open subinterval t i 1 < t < t i. 2. f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.

107 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Example 6.1 Find the Laplace transform for f (t) = 1, t 0 Solution L {1} = F (s) = L {f (t)} = F (s) = 0 0 e st dt = lim A e st f (t)dt e st s A = 1 0 s ; s > 0

108 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Example 6.2 Find the Laplace transform for f (t) = e at, t 0 Solution L {e at } = F (s) = L {f (t)} = F (s) = lim A 0 e (s a)t s a 0 e at e st dt = A = 1 0 s a ; e st f (t)dt 0 e (s a)t dt = s > a

109 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Example 6.3 Find the Laplace transform for 1 0 t < 1 f (t) = k t = 1 0 t > 1 where k is a constant. In engineering contexts f (t) often represents a unit pulse, perhaps of force or voltage. Solution L {f (t)} = F (s) = 0 e st s f (t)e st dt = 1 = 1 e s 0 s 1 0 e st dt =

110 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Example 6.4 Find the Laplace transform for f (t) = sin(at), t 0 Solution L {f (t)} = F (s) = 0 e st f (t)dt = [ F (s) = lim e st cos(at) A A a s 0 a F (s) = 1 a s2 a 2 0 A 0 0 sin(at)e st dt = Int by Parts ] e st cos(at)dt = Int by Parts sin(at)e st dt = 1 a s2 a 2 F (s)

111 II Definition of The Solution of Initial Value Problems Step Functions Definition of The Hence, solving for F(s), we have a F (s) = s 2 + a 2 Now, the Laplace transform is a linear operator, that is, suppose that f 1 and f 2 are two functions whose Laplace transforms exist for s > a 1 and s > a 2, respectively. Then, for s > max{a 1, a 2 } L {c 1 f 1 + c 2 f 2 } = 0 e st {c 1 f 1 + c 2 f 2 }dt = c 1 e st f 1 dt + c 2 e st f 2 dt = c 1 L {f 1 } + c 2 L {f 2 } 0 Thus, we have 0 L {c 1 f 1 + c 2 f 2 } = c 1 L {f 1 } + c 2 L {f 2 }

112 II Definition of The Solution of Initial Value Problems Step Functions Definition of The OBS Example 6.5 L 1 {d 1 F 1 + d 2 F 2 } = d 1 L 1 {F 1 } + d 2 L 1 {F 2 } Find the Laplace transform for f (t) = 5e 2t 3sin(4t), t 0 Solution L {5e 2t 3sin(4t)} = 5L {e 2t } 3L {sin(4t)} = L {5e 2t 3sin(4t)} = 5 s s ; s > 0 L 1 { 5 s s } = 5L 1 { 1 s + 2 } 3L 1 4 { s } = 5e 2t 3sin(4t)

113 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions To see how we can apply the method of Transform Laplace to solve linear differential equations with constant coefficients. We establish the following results. Theorem 6.2 Suppose that f is continuous and f is piecewise continuous on any interval 0 t A. Suppose further that there exist constants K, a, and M such that f (t) Ke at for t M. Then proof L {f } = sl {f } f (0) L {f, (t)} = 0 A e st f (t)dt = lim A e st f (t)dt = 0

114 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions lim A [ t1 0 t2 e st f (t)dt + e st f (t)dt + t 1 ] t n 1 t n = Ae st f (t)dt and integrating by parts, we have s = t3 lim A {e st f (t) t e st f (t) t e st f (t) t 1 [ t1 0 t2 e st f (t)dt + e st f (t)dt t 1 t 2 e st f (t)dt tn=a t n 1 tn=a + t n 1 e st f (t)dt ]} =

115 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions lim A [e sa f (A) f (0) + s In his way we obtain A 0 ] e st f (t)dt = s L {f } = sl {f } f (0) As a corollary, we have the following A 0 e st f (t)dt f (0) Corollary Suppose that the functions f, f,..., f (n 1) are continuous and that f (n) is piecewise continuous on any interval 0 t A.

116 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Suppose further that there exist constants K, a, and M such that f (t) Ke at, f (t) Ke at,..., f (n 1) (t) Ke at for t M. Then L {f (n) (t)} exists for s > a and is given by L {f (n) (t)} = s n L {f (t)} s n 1 f (0) s n 2 f (0)... sf (n 2) (0) f (n 1) (0) We use the previous results to solve IVP s using Laplace Transform. It is most useful for problems involving nonhomogeneous differential equations. However, just to illustrate the method we will start with a homogeneus case

117 II Solution of Initial Value Problems Example 6.6 Consider the IVP Definition of The Solution of Initial Value Problems Step Functions Solution y y 2y = 0; y(0) = 1, y (0) = 0 Using the traditional method we find that the general solution is y(x) = c 1 e t + c 2 e 2t and applying initial conditions we get c 1 = 2/3 and c 2 = 1/3. Hence, the particular solution is y(x) = 2 3 e t e2t

118 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Now, let us solve the same problem by using the Laplace transform. We start off with the differential equation Applying the y y 2y = 0 because of linearity L {y y 2y = 0} and using corollary L {y } L {y } 2L {y} = 0 s 2 L {y} sy(0) y (0) [sl {y} y(0)] 2L {y} = 0

119 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Taking Y (s) = L {y} and applying initial coditions (y(0) = 1, y (0) = 0), we obtain s 2 Y (s) sy(0) y (0) [sy (s) y(0)] 2Y (s) = 0 = Y (s) = s 1 s 2 s 2 = s 1 (s 2)(s + 1) The above can be written, using partial fractions, as Y (s) = 1/3 s 2 + 2/3 s + 1 Now, applying the inverse Laplace transform

120 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions { 1/3 y(t) = L 1 {Y (s)} = L 1 s 2 + 2/3 } s + 1 y(t) = L 1 {Y (s)} = 1 { } 1 3 L { } 1 s 2 3 L 1 s + 1 but we know that L {e at } = 1 s a or equivalently L 1 { 1 s a } = eat y(t) = 1 3 e2t e t The same procedure can be applied to the general second order linear equation with constant coefficients ay + by + cy = f (t)

121 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions obtaining Y (s) = (as + b)y(0) + ay (0) as 2 + bs + c + F (s) as 2 + bs + c ; F (s) = L {f (t)} The main difficulty that occurs in solving initial value problems by the transform method lies in the problem of determining the function y = y(t), corresponding to the inverse transform of Y (s). Since we will not deal with the formula for the inverse transform, because it requires complex integration, we will use a table of common for basic functions (http: //

122 II Solution of Initial Value Problems Example 6.7 Consider the IVP Solution Definition of The Solution of Initial Value Problems Step Functions y + y = sin(2t); y(0) = 2, y (0) = 1 Let us solve the problem by using the Laplace transform. We start off with the differential equation Applying the y + y = sin(2t) L {y + y = sin(2t)}

123 II Solution of Initial Value Problems because of linearity and using corollary Definition of The Solution of Initial Value Problems Step Functions L {y } + L {y} = L {sin(2t)} s 2 L {y} sy(0) y (0) + L {y} = 2 s taking Y (s) = L {y} and applying initial coditions (y(0) = 2, y (0) = 1), we obtain s 2 Y (s) 2s 1 + Y (s) = 2 s = Y (s) = 2s3 + s 2 + 8s + 6 (s 2 + 1)(s 2 + 4)

124 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions The above can be written, using partial fractions, as Y (s) = as + b s cs + d s = (as + b)(s2 + 4) + (cs + d)(s 2 + 1) (s 2 + 1)(s 2 = + 4) (a + c)s 3 + (b + d)s 2 + (4a + c)s + (4b + d) (s 2 + 1)(s 2 + 4) = 2s3 + s 2 + 8s + 6 (s 2 + 1)(s 2 + 4) Then, comparing coefficients of like powers of s, we have a + c = 2; b + d = 1; 4a + c = 0; 4b + d = 6;

125 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Therefore, a = 2, b = 5/3, c = 0, d = 2/3 and Y (s) = Now, taking the inverse, we have 2s s /3 s /3 s y(t) = 2cos(t) sin(t) 1 3 sin(2t)

126 II Solution of Initial Value Problems Example 6.8 Consider the IVP Definition of The Solution of Initial Value Problems Step Functions Solution y (4) y = 0; y(0) = 0, y (0) = 1, y (0) = 0, y (0) = 0 Let us solve the problem by using the Laplace transform. We start off with the differential equation Applying the y (4) y = 0 L {y (4) y = 0}

127 II Solution of Initial Value Problems because of linearity and using corollary L {y (4) } L {y} = 0 Definition of The Solution of Initial Value Problems Step Functions s 4 L {y} s 3 y(0) s 2 y (0) sy (0) y (0) Y (s) = 0 taking Y (s) = L {y} and applying initial coditions (y(0) = 0, y (0) = 1, y (0) = 0, y (0) = 0), we obtain s 4 L {y} s 2 y (0) Y (s) = 0 = Y (s) = s2 s 4 1 = s 2 (s 2 1)(s 2 + 1)

128 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions The above can be written, using partial fractions, as Y (s) = as + b s cs + d s = (as + b)(s2 + 1) + (cs + d)(s 2 1) (s 2 1)(s 2 = + 1) (a + c)s 3 + (b + d)s 2 + (a c)s + (b d) (s 2 1)(s 2 + 1) = s 2 (s 2 1)(s 2 + s) Then, comparing coefficients of like powers of s, we have a + c = 0; b + d = 1; a c = 0; b d = 0;

129 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions therefore, a = 0, b = 1/2, c = 0, and d = 1/2, and and take the inverse, we have Y (s) = 1/2 s /2 s y(t) = sinh(t) + sin(t) 2 = (et e t )/2 + sin(t) 2 = et 4 e t 4 + sin(t) 2

130 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Example 6.9 The Laplace transforms of certain functions can be found conveniently from their Taylor series expansion. Using the Taylor series for sin(t) Let sin(t) = n=0 ( 1) n t 2n+1 (2n + 1)! f (t) = { sin(t) t t 0 1 t = 0

131 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions Find the Taylor series for f about t = 0. Assuming that the Laplace transform of this function can be computed term by term, determine L {f (t)} Solution For t 0 we have that f can written as f (t) = sin(t) t = n=0 ( 1) n t 2n+1 (2n + 1)!t = ( 1) n t 2n (2n + 1)! n=0 n=0 Applying the { } ( 1) n t 2n L {f (t)} = L (2n + 1)!

132 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions because of linearity L {f (t)} = n=0 ( 1) n (2n + 1)! L {t2n } and using the table of s, L {t m } = m! s m+1 L {f (t)} = L {f (t)} = n=0 ( 1) n (2n)! (2n + 1)! s 2n+1 n=0 ( 1) n (2n)! (2n + 1)!s 2n+1 = ( 1) n (2n)! (2n + 1)(2n)!s 2n+1 n=0

133 II Solution of Initial Value Problems Definition of The Solution of Initial Value Problems Step Functions L {f (t)} = but, if we remember Therefore, we have n=0 ( 1) n (2n + 1)s 2n+1 = tan 1 (x) = n=0 n=0 ( 1) n x 2n+1 (2n + 1) L {f (t)} = tan 1 (1/s) ( 1) n (1/s) 2n+1 (2n + 1)

134 II Step Functions Definition of The Solution of Initial Value Problems Step Functions 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 will be denoted by u c and is defined by { 0 t < c u c (t) = 1 t c

135 II Step Functions Definition of The Solution of Initial Value Problems Step Functions The step can also be negative. For instance { 1 t < c u(t) = (1 u c (t)) = 0 t c

136 II Step Functions Definition of The Solution of Initial Value Problems Step Functions In fact, any piecewise-defined function can be written as a linear combination of u c (t) s functions. For instance consider the function 2 0 t < t < 2 f (t) = 2 2 t < t

137 II Step Functions Definition of The Solution of Initial Value Problems Step Functions We start with the function f 1 (t) = 2u 0, which agrees with f (t) on [0, 1). To produce the jump down of one unit at t = 1, we add u 1 (t) to f 1 (t), obtaining f 2 (t) = 2u 0 u 1 (t), which agrees with f (t) on [1, 2). The jump of one unit at t = 2 corresponds to adding u 2 (t), which gives f 3 (t) = 2u 0 u 1 (t) + u 2 (t). Thus we obtain f (t) = f 3 (t) = 2u 0 u 1 (t) + u 2 (t) The Laplace transform of u c for c 0 is easily determined: L {u c } = 0 e st u c dt = c e st dt = e cs, s > 0 s

138 II Step Functions Definition of The Solution of Initial Value Problems Step Functions For a given function f defined for t 0, we will often want to consider the related function g defined by g(t) = u c f (t c) which represents a translation of f a distance c in the positive t direction.

139 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Theorem 6.3 If F (s) = L {f (t)} exists for 0 a < s, and if c is a positive constant, then L {u c f (t c)} = e cs L {f (t)} = e cs F (s), Conversely, if f (t) = L 1 {F (s)}, then L 1 {e cs F (s)} = u c f (t c)

140 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Example 6.10 If the function f is defined { sin(t) 0 t < π/4 f (t) = sin(t) + cos(t π/4) π/4 t find L {f (t)}.

141 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Solution Note that f (t) = sint + g(t), where Thus { 0 0 t < π/4 g(t) = cos(t π/4) π/4 t g(t) = u π/4 cos(t π/4)

142 II Step Functions Definition of The Solution of Initial Value Problems Step Functions and L {f (t)} = L {sin(t)} + L {u π/4 cos(t π/4)} = L {sin(t)} + e πs/4 L {cos(t)} and using the table of s L {f (t)} = 1 s s e πs/4 s = 1 + se πs/4 s 2 + 1

143 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Let s consider the following theorem Theorem 6.4 If F (s) = L {f (t)} exists for 0 a < s, and if c is a constant, then L {e ct f (t)} = F (s c), s > a + c Conversely, if f (t) = L 1 {F (s)}, then L 1 {F (s c)} = e ct f (t)

144 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Example 6.11 Find the inverse transform of Solution H(s) = 1 s 2 4s + 5 First of all the polynomial s 2 4s + 5, has complex roots. By completing the square in the denominator, we can write 1 H(s) = (s 2) 2 = F (s 2) + 1 where F (s) = (s 2 + 1) 1. L 1 {F (s)} = sin(t). It follows from the previous theorem that h(t) = L 1 {H(s)} = L 1 {F (s 2)} = e 2t sin(t)

145 II Step Functions Definition of The Solution of Initial Value Problems Step Functions Shift in the time domain (t domain) If F (s) = L {f (t)} exists for 0 a < s, and if c is a positive constant, then L {u c f (t c)} = e cs F (s) L 1 {e cs F (s)} = u c f (t c) Shift in the frequency domain (s domain) If F (s) = L {f (t)} exists for 0 a < s, and if c is a constant, then L {e ct f (t)} = F (s c), s > a + c L 1 {F (s c)} = e ct f (t)

146 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Considering solving the IVP ay + by + cy = g(t); y(0) = y (0) = 0 Using L {ay + by + cy = g(t)} al {y } + bl {y } + cl {y} = L {g(t)} = a [ s 2 Y (s) sy(0) y (0) ] + b [sy (s) y(0)] + ca [Y (s)] = G(s) Y (s) = G(s) as 2 + bs + c ; where Y (s) = L {y(t)} and g(t) is a piecewise continuous force.

147 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Example 6.12 Solve the initial value problem 2y + y + 2y = g(t); y(0) = 0; y (0) = 0 where 0 0 t < 5 g(t) = 1 5 t < t = u 5 u 20

148 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Solution First of all the polynomial s 2 4s + 5, has complex roots. By completing the square in the denominator, we can write H(s) = 1 (s 2 = F (s 2) 2) + 1 where F (s) = (s 2 + 1) 1. L 1 {F (s)} = sin(t), it follows from the previous theorem that h(t) = L 1 {H(s)} = e 2t sin(t)

149 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions The Laplace transform of the above equation is Y (s) = Y (s) = G(s) 2s 2 + s + 2 = (e 5s e 2s )/s 2s 2 + s + 2 e 5s e 2s s (2s 2 + s + 2) = ( e 5s e 2s) F (s) Then, if f (t) = L 1 {F (s)} = L 1 1 { Theorem 6.1 we have s(2s 2 +s+2) y(t) = u 5 f (t 5) u 20 f (t 20) }, and using the Finally, to determine f (t), we use the partial fraction expansion of F (s)

150 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions f (t) = L 1 1 { s (2s 2 + s + 2) } = L 1 { a s } + L 1 bs + c { 2s 2 + s + 2 } and finding the coefficients a, b, and c, we get f (t) = L 1 { 1/2 s } + L 1 ( s) + ( 1/2) { 2s 2 + s + 2 } = L 1 { 1/2 s } ( 1 2 ) L 1 (s + 1/4) + 1/4 { (s + 1/4) /16 }

151 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions but L 1 { 1/2 s } ( ) [ 1 2 L 1 s + 1/4 { (s + 1/4) 2 + ( 15/4) } + 2 ] 1 15/4 L 1 { 15 (s + 1/4) 2 + ( 15/16) } 2 L 1 { 1/2 s } = 1 2

152 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions ( ) [ ] 1 L 1 s + 1/4 { 2 (s + 1/4) 2 + ( 15/4) } = [ ] 1 15/4 2 L 1 { 15 (s + 1/4) 2 + ( 15/4) } = 2 Therefore f (t) = [ e t/4 cos( ] 15t/4) [ ( 15/15)e t/4 sin( ] 15t/4) [ e t/4 cos( 15t/4) + ( 15/15)e t/4 sin( ] 15t/4)

153 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions And the solution is given by y(t) = u 5 f (t 5) u 20 f (t 20) ( 1 y(t) = u [ e (t 5)/4 cos( 15(t 5)/4) 2 + ( 15/15)e (t 5)/4 sin( ]) 15(t 5)/4) ( 1 u [ e (t 20)/4 cos( 15(t 20)/4) 2 + ( 15/15)e (t 20)/4 sin( ]) 15(t 20)/4)

154 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Example 6.13 Solve the initial value problem y + 4y = g(t); y(0) = 0; y (0) = 0 where 0 0 t < 5 g(t) = (t 5)/5 5 t t = 1 5 u 5[t 5] 1 5 u 10[t 10]

155 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Solution G(s) = L { 1 5 u 5[t 5] 1 5 u 10[t 10] = 1 5 e 5s 1 s e 10s 1 s 2 Y (s) = 1 e 5s e 10s 5 s 2 (s 2 + 4) = 1 5 [ e 5s F (s) e 10s F (s) ] y(t) = 1 5 u 5(t)f (t 5) 1 5 u 10(t)f (t 10)

156 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions where F (s) = s 2 (s 2 + 4) = a s + b s 2 + cs + d s = as(s2 + 4) + b(s 2 + 4) + cs 3 s 2 (s 2 + 4) a + c = 0; b + d = 0 4a = 0; 4b = 1 Then, the solution is a = 0, b = 1/4, c = 0, and d = 1/4 F (s) = 1/4 s s 2 + 4

157 II Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Differential Equations with Discontinuous Forcing Functions Finally, we have f (t) h(t) = 1 4 t 1 8 sin(2t) and the solution y(t) is given by y(t) = 1 [ 1 5 u 5 4 (t 5) 1 ] sin2(t 5) 8 1 [ 1 5 u 10 4 (t 10) 1 ] sin2(t 10) 8

158 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Consider the Initial Value Problem ay + by + cy = g(t) where g(t) = { 0 otherwise 1/ɛ t 0 t t 0 + ɛ

159 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral We can solve the problem using technique, but Can we still solve the problem when ɛ 0? Example 6.14 Solve the initial value problem y + y = I 0 g(t); y(0) = 0; y (0) = 0; I 0 = constant where g ɛ (t) = { 1/ɛ 0 < t < ɛ 0 ɛ t = 1 ɛ (1 u ɛ)

160 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Solution G(s) = L { 1 ɛ (1 u ɛ)} = 1 ɛ 1 s 1 1 e ɛs ɛ s = 1 ɛs (1 e ɛs ) Y (s) = G(s) I0 1 s = ɛ s (1 e ɛs ) s 2 = I ɛ (1 e ɛs )F (s) = where I 0 ɛ F (s) I 0 ɛ e ɛs F (s) F (s) = 1 s(s 2 + 1) = 1 s s s 2 + 1

161 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral and Therefore f (t) = 1 cos(t) = f (t ɛ) = 1 cos(t ɛ) y ɛ (t) = L 1 { I 0 ɛ F (s)} L 1 { I 0 ɛ e ɛs F (s)} I 0 ɛ [1 cost(t) u ɛ(t) (1 cos(t ɛ))] y ɛ (t) = I 0ɛ (1 cos(t)) I 0ɛ (cos(t ɛ) cos(t)) 0 t < ɛ ɛ t

162 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Moreover, we have for t ɛ ( if ɛ 0 = t 0) 1 cos(t) lim ɛ 0 ɛ lim 1 cos(ɛ) ɛ 0 ɛ = sin(ɛ) lim ɛ 0 = 0 1 On the other hand, for t ɛ, pause when ɛ 0, we have Thus lim ɛ 0 cos(t ɛ) cos(t) ɛ = lim ɛ 0 sin(t ɛ) 1 = sin(t) y(t) = lim ɛ 0 y ɛ = I 0 sin(t), 0 < t

163 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Unit Impulse The Unit Impulsion Function or Dirac Delta Function ( ) at the point t 0, denoted by δ(t t 0 ) is defined by { t = t0 δ(t t 0 ) = 0 t t 0 such that if a t b, then b a f (t)δ(t t 0 ) = f (t 0 ) In particular, from the above difinition, we have b { 1 a t0 b δ(t t 0 )dt = a 0 otherwise or

164 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral f (t)δ(t t 0 )dt = f (t 0 ) δ(t t 0 )dt = 1 Now, using the definition of δ(t t 0 ), the of the Dirac Delta Function is given by L {δ(t t 0 )} = In particular 0 δ(t t 0 )e st = e st 0 ; t 0 > 0 L {δ(t)} = 1

165 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Example 6.15 Solve the initial value problem y + y = I 0 δ(t); y(0) = 0; y (0) = 0 Solution Y (s) = I 0 s y(t) = I 0L 1 {1 s } = I 0sin(t)

166 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Example 6.16 Solve the initial value problem 2y + y + 2y = δ(t 5); y(0) = 0; y (0) = 0 Solution where Y (s) = G(s) 2s 2 + s + 2 = e 5s 2s 2 + s + 2 = e 5s 2 F (s) F (s) = 1 s s + 1 = 1 (s )

167 II Impulse Functions Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Therefore f (t) = L 1 1 { (s ) } = 16 Hence, the solution is 4 15 e t/4 sin ( 1 15 e t/4 sin 15/16 4 t ( ) 15 4 t y(t) = 1 2 u 5(t)f (t 5) ( ) y(t) = 1 2 u 5(t) 4 15 e (t 5)/4 sin (t 5) = 15 4 { 0 ( 0 t < 5 15 ) 1 2 u 5(t) 4 15 e (t 5)/4 sin 4 (t 5) 5 t ) =

168 II The Convolution Integral Differential Equations with Discontinuous Forcing Functions Impulse Functions The Convolution Integral Convolution can be intuitively described as a function that is the integral or summation of two component functions, and that measures the amount of overlap as one function is shifted over the other