First Order Differential Equations

Transcription

1 2 First Order Differential Equations We have seen ho to solve simple first order differential equations using Simulin. In particular e have solved initial value problems for the equations dy dt dy dt dx dt = 2 y, y() =, (2.) t = 2 t y + t2, y() =, (2.2) = 2 sin 3t 4x, x(0) = 0. (2.3) The Simulin models ere provided in Figures.6,.20, and.4, respectively. In this chapter e solve a fe more first order equations in the form of applications. These ill include groth and decay, Neton s La of Cooling, pursuit curves, free fall and terminal velocity, the logistic equation, and the logistic equation ith delay. 2. Exponential Groth and Decay The simplest differential equations are those governing groth and decay. As an example, e ill discuss population models. Let P(t) be the population at time t. We see an expression for the rate of change of the population, dp. Assuming that there is no migration dt of population, the only ay the population can change is by adding or subtracting individuals in the population. The equation ould tae the form dp dt = Rate In Rate Out. The Rate In could be due to the number of births per unit time and the Rate Out by the number of deaths per unit time. The simplest forms for these rates ould be given by terms proportional to the population: Rate In = bp and Rate Out = mp. Here e have denoted the birth rate as b and the mortality rate as m. This gives the total rate of change of population as dp dt = bp mp P, (2.4)

2 26 solving differential equations using simulin here = b m. Equation (2.4) is easily modeled in Simulin. All of the needed blocs are under the Commonly Used Blocs group. We need an Integrator, Constant, Gain, and a Scope bloc. The output from the Integrator can be feed into a Gain control, hich represents, and the output from the Gain, P, can then be used as an input to the Integrator. We add the Scope in order to plot the solution. The model is shon in Figure 2.. Note that a Constant bloc as added to provide an external input of the initial condition Gain P' xo s P Figure 2.: Simulin model for exponential groth and decay. The initial value, P(0) = 0, is set in the Constant bloc and = 0.5 is set in the Gain. 0 Integrator Scope Constant The solution for exponential decay ith P(0) = 0 and = 0.5 is shon in Figure 2.2. The simulation time as set at 0s. Figure 2.2: Solution for the exponential decay ith P(0) = 0 and = 0.5. The simulation time as set at 0. The exact solution is easily found noting that Equation (2.4) is a separable equation. Rearranging the equation, its differential form is Integrating, e have dp P dp P = dt. = dt

3 first order differential equations 27 Next, e solve for P(t) through exponentiation, ln P = t + C. (2.5) P(t) = e t+c P(t) = ±e t+c = Ae t. (2.6) Here e define the arbitrary constant, A = ±e C. If the population at t = 0 is P 0, i.e., P(0) = P 0, then the solution gives P(0) = Ae 0 = A = P 0. So, the solution of the initial value problem is P(t) = P 0 e t. In the Simulin model, the initial value as given as P(0) = 0 and = 0.5. Therefore, the solution in Figure 2.2 is of the function P(t) = 0e 0.5t. Equation (2.4) is the familiar exponential model of population groth: Malthusian population groth. dp dt = P. We obtained solutions exhibiting exponential groth ( > 0) or decay ( < 0). This Malthusian groth model has been named after Thomas Robert Malthus ( ), a clergyman ho used this model to arn of the impending doom of the human race if its reproductive practices continued. Later e modify this model to account for competition for resources, leading to the logistic differential equation. 2.2 Neton s La of Cooling If you tae your hot cup of tea, and let it sit in a cold room, the tea ill cool off and reach room temperature after a period of time. The la of cooling is attributed to Isaac Neton ( ) ho as probably the first to state results on ho bodies cool. The main idea is that a body at temperature T(t) is initially at temperature T(0) = T 0. It is placed in an environment at an ambient temperature of T a. The goal is to find the temperature at a later time, T(t). We ill assume that the rate of change of the temperature of the body is proportional to the temperature difference beteen the body and its surroundings. Thus, e have Neton s 70 La of Cooling is an approximation to ho bodies cool for small temperature differences (T T a T) and does not tae into account all of the cooling processes. One account is given by C. T. O Sullivan, Am. J. Phys (990) p dt dt T T a. The proportionality is removed by introducing a cooling constant, here > 0. dt dt = (T T a), (2.7)

4 28 solving differential equations using simulin This differential equation can be solved by first reriting the equations as d dt (T T a) = (T T a ). This no taes the form of exponential decay of the function T(t) T a. The solution is easily found as T(t) T a = (T 0 T a )e t, or T(t) = T a + (T 0 T a )e t. Example 2.. A cup of tea at 90 o C cools to 85 o C in ten minutes. If the room temperature is 22 o C, hat is its temperature after 30 minutes? Using the general solution ith T 0 = 90 o C, T(t) = 22 + (90 22)e = e t, e then find using the given information, T(0) = 85 o C. We have 85 = T(0) = e 0 63 = 68e 0 e 0 = = ln = ln min. This gives the solution for this model as T(t) = e t. No e can anser the question. What is T(30)? T(30) = e (30) = 76 o C. 0. Product - Gain 60 T' s xo T Scope Figure 2.3: Simulation model for Neton s La of Cooling, T = (T T a ), T(0) = T0. Here e set = 0. s, T a = 20 o C, and T 0 = 60 o C. T0 Integrator 20 Ta Neton's La of Cooling T' = - (T-T0) Next e model Equation (2.7) in Simulin. The input for the integrator is simply (T T a ). We need to define the constants and T a. We ill

5 first order differential equations 29 externally input the initial condition, T(0) = T 0. The simple model is shon in Figure 2.3. In this case e set = 0. s, T a = 20 o C, and T 0 = 60 o C. Running the simulation for 00 s, e obtain the solution shon in Figure 2.4. Figure 2.4: Solution of Neton s La of Cooling example. Ho good is the solution? We can solve the problem by hand for this set of parameters. Hoever, e ill tae this opportunity to introduce the idea of a subsystem and set up a model in hich e can interactively modify the constants and get Simulin to automatically provide the exact solution for comparison. Product - T' Gain 2 T s xo T(t) T(0) Figure 2.5: Creating a subsystem for the Neton s La of Cooling model. Integrator 3 T ambient Neton's La of Cooling T' = - (T-T0) We begin by replacing the scope ith an output bloc. The Out bloc can be found in the Sin group. The input to the subsystem ill be the three parameters,, T 0, and T a. Each of these constant blocs in Figure 2.3 ill be replaced by an In bloc, found in the Sources group. In Figure 2.5 the three inputs and one output are no oval blocs. Double-clic each of the three input blocs, one at a time, and set the Port Number of, T 0, and T a, to, 2, and 3, respectively. Finally, rename each of these controls using the labels that mae sense, such as for. In Figure 2.5 e sho the subsystem that e have created. No highlight the entire subsystem using CTRL-A. In the menu sys- Creating a subsystem.

6 30 solving differential equations using simulin 2 T(0) 3 T ambient T(0) T(t) T ambient Subsystem T(t) Figure 2.6: Subsystem for Neton s La of Cooling, T = (T T a ), T(0) = T0. tem, loo for Create Subsystem from Selection. In the 205 version, this is under the menu item Diagram and subitem Subsystem & Model Reference. Rearranging the resulting subsystem, one has something lie the subsystem bloc in Figure 2.6. This is the equivalent of a blac box ith three inputs and one output. Next, e can mae use of the subsystem just created. Replace the three input ports ith constant blocs. Rename the Constant blocs ith the parameter name and fill each bloc ith a value. The output port can be replaced ith a Scope bloc, or any other form of output desired. This can be seen in Figure 2.7. Before finishing ith this model, e ill build in the exact solution. Recall that the general solution can be ritten in terms of the parameters as T(t) = T a + (T 0 T a )e t. So, e can feed the values of the parameters in the model into a Fcn bloc and output the exact solution for comparison. We ill also need a time value. So, e ill need the Cloc bloc as ell. 0. Neton's La of Cooling T' = - (T-T0) Figure 2.7: Using a user-created subsystem for Neton s La of Cooling. 60 T(0) T(t) IC Scope 20 Ta T ambient Cooling System The entire model is shon in Figure 2.8. The subsystem is labeled Cooling System The top portion is a repetition of the Neton s La of Cooling model implemented previously. We have added a Fcn bloc from the User-Defined Functions group. The input ill be a vector containing all of the variables in the exact solution. This is accomplished by adding a Mux (or Multiplex) bloc. Doubleclic the Mux bloc and set the number of inputs to 4. No, double-clic the Fcn bloc and enter the exact solution in the form u()+u(2)*exp(-u(3)*u(4)) Here e have assumed that the variables are fed into the Mux bloc in the order T a, T 0 T a,, and t. In Figure 2.8 one can see ho the values are

7 first order differential equations 3 0. Neton's La of Cooling T' = - (T-T0) Figure 2.8: Model of Neton s La of Cooling, T = (T T a ), T(0) = T0, using the subsystem feature. 60 T(0) T(t) IC Scope 20 Ta T ambient Cooling system Exact Solution f(u) Fcn Scope Cloc routed into the Mux bloc. The output can be attached to a second scope, as shon, or can be subtracted from the output of the Cooling System bloc to sho the closeness of the to solutions. One can also send the output to MATLAB using the To Worspace bloc. 2.3 Free Fall ith Drag Consider an object falling to the ground ith air resistance? Free fall is the vertical motion of an object solely under the force of gravity. It has been experimentally determined that an object near the surface of the Earth falls at a constant acceleration in the absence of other forces, such as air resistance. This constant acceleration is denoted by g, here g is called the acceleration due to gravity. The negative sign is an indication that e have chosen a coordinate system in hich up is positive. We are interested in determining the position, y(t), of a falling body as a function of time. The differential equation governing free fall is have ÿ(t) = g. (2.8) Note that e ill occasionally use a dot to indicate time differentiation. We need to model the air resistance. As an object falls faster and faster, the resistive force becomes greater. This drag force is a function of the velocity. The idea is to rite Neton s Second La of Motion F = ma in the form mÿ = mg + f (v), (2.9) here f (v) gives the resistive force and mg is the eight. Note that this applies to free fall near the Earth s surface. Also, for f (v) to be a resistive force, f (v) should oppose the motion. If the body is falling, then f (v)

8 32 solving differential equations using simulin should be positive. If the body is rising, then f (v) ould have to be negative to indicate the opposition to the motion. We ill model the drag as quadratic in the velocity, f (v) = bv 2. Example 2.2. Solve the free fall problem ith f (v) = bv 2. The differential equation that e need to solve is v = v 2 g, (2.0) here = b/m. Note that this is a first order equation for v(t). Formally, e can separate the variables and integrate over time to obtain t + C = v dz z 2 g. (2.) If e can do the integral, then e have a solution for v. One ay to evaluate this integral is to use Partial Fraction Decomposition. When there are to linear factors in the denominator, the integral can be reritten as dx (x a)(x b) = [ b a x a ] dx (2.2) x b The ne integral has to terms hich can be readily integrated. In order to factor the denominator in the current problem, e first have to rerite the constants. We let α 2 = g/ and rite the integrand as z 2 g = z 2 α 2. (2.3) No e use a partial fraction decomposition to obtain z 2 g = [ 2α z α ]. (2.4) z + α The integrand can be easily integrated giving t + C = 2α ln v α v + α. (2.5) Solving for v, e have v(t) = Ae2αt α, (2.6) + Ae2αt here A e C. A can be determined using the initial velocity. There are other forms for the solution in terms of a tanh function, hich the reader can determine as an exercise. One important conclusion is that for large times, the ratio in the solution approaches. Thus, v α = g as t. This means that the falling object ill reach a constant terminal velocity. Equation (2.0) can be modeled in Simulin. The model is shon in Figure 2.9. The solution for = m, hich is found for the above sample computation, is shon in Figure 2.0. We see that terminal velocity is obtained and matches the predicted value, g = 78 m/s.

9 first order differential equations g 0 v' xo s Integrator v Figure 2.9: Model for free fall ith drag as described by v = v 2 g. Constant u 2 Math Function Scope Free Fall ith Drag v' = v^2 - g Figure 2.0: Solution for free fall ith drag ith = starting from rest. 2.4 Pursuit Curves Another application that is interesting is to find the path that a body traces out as it moves toards a fixed point or another moving body. Such curves are no as pursuit curves. These could model aircraft or submarines folloing targets, or predators folloing prey. For example, a ha follos a sparro, a large fish chases a small fish, or a fox chases a rabbit. Example 2.3. A dog at point (x, y) sees a cat traveling at speed v along a straight line. The dog runs toards the cat at constant speed but alays in a direction along line of sight beteen their positions. If the dog starts out at the point (0, 0) at t = 0, hen the cat is at (a, 0), then hat is the path the dog needs to follo? Will the dog catch the cat? We sho the path in Figure 2.2. Let the cat s path be along the

10 34 solving differential equations using simulin y Figure 2.: A dog at point (x, y) sees a cat at point (a, vt) and alays follos the straight line beteen these points. (a, vt) (x, y) (a, 0) x line x = a. Therefore, the cat is at position (a, vt) at time t. The goal is to find the dog s path, (x(t), y(t)), or y = y(x). First e consider the equation of the line of sight beteen the points (x, y) and (a, vt). Considering that the slope of this line is the same as the slope of the tangent to the path, y = y(x), e have y = vt y a x. The dog is moving at a constant speed, and the distance the dog to travels s given by L = t, here t is the running time from the origin. The distance the dog travels is also given by the arclength of the path beteen (0, 0) and (x, y) : x L = + [y (x)] 2 dx. 0 Eliminating the time using y = vt y a x, e have x + [y (x)] 2 dx = v (y + (a x)y ). 0 Furthermore, e can differentiate this result ith respect to x to get rid of the integral, + [y (x)] 2 = v (a x)y. (2.7) This is the differential equation governing the dog s pursuit. It is modeled later in Simulin as shon in Figure 2.2. Even though Equation (2.7)is a second order differential equation for y(x), it is a first order separable equation in the speed function z(x) = y (x). Namely, v (a x)z = v x + z 2. Separating variables, e find dz v = ln(z + + z 2 ) + z 2 dx a x.

11 first order differential equations 35 sqrt(+u^2) Fcn Figure 2.2: Model for the pursuit curve, (a x)y = v + [y (x)] 2, y(0) = 0, y (0) = 0, for = 2 and v =. v Cloc Product 0 y'(0) xo s Integrator y' 0 y(0) xo s Integrator Scope Constant u(3)*u(2)*u()/(u()^2-u(2)^2) Fcn Display The integrals can be computed using standard methods from calculus. We can easily integrate the right hand side, dx a x = ln a x + c. The left hand side taes a little extra or, or looing the value up in Tables or using a CAS pacage. Recall a trigonometric substitution is in order. We let z = tan θ. Then dz = sec 2 θ dθ. The methods proceeds as follos: dz + z 2 = = sec 2 θ + tan 2 θ dθ sec θ dθ = ln(tan θ + sec θ) + c 2 Putting these together, e have for x > 0, = ln(z + + z 2 ) + c 2. (2.8) ln(z + + z 2 ) = v ln x + C. Using the initial condition z = y = 0 and x = a at t = 0, 0 = v ln a + C, or C = v ln a. Using this value for c, e find ln(z + + z 2 ) = v ln x v ln a ln(z + + z 2 ) = v ln x a

12 36 solving differential equations using simulin ln(z + ( x ) v + z 2 ) = ln a z + ( x ) v + z 2 =. (2.9) a We can solve for z = y, to find y = [ ( x ) v ( x ) v ] 2 a a Integrating, y(x) = a 2 [ ( xa ) + v + v ( xa ) v ] v +. The integration constant,, can be found noing y(a) = 0. This gives 0 = a [ 2 + v ] v + = a [ 2 v ] + v av = 2 v 2. (2.20) The full solution for the path is given by y(x) = a 2 [ ( xa ) + v + v ( xa ) v ] v + av 2 v 2. Can the dog catch the cat? This ould happen if there is a time hen y(0) = vt. Inserting x = 0 into the solution, e have y(0) = av 2 v 2 = vt. This is possible if > v. Figure 2.3: Solution for the pursuit curve.

13 first order differential equations The Logistic Equation In this section e ill explore a nonlinear population model. Typically, e ant to model the groth of a given population, y(t), and the differential equation governing the groth behavior of this population is developed in a manner similar to that done in the section on groth and decay. Recall that a simple population model dy dt = by my, (2.2) here e had defined the birth rate as b and the mortality rate as m. Generally, these rates could depend on the time. In the case that they are both constant rates, e can define = b m and obtain the familiar exponential model of population groth. When populations get large enough, there is competition for resources, such as space and food, hich can lead to a higher mortality rate. Thus, the mortality rate may be a function of the population size, m = m(y). The simplest model ould be a linear dependence, m = m + cy. Then, the previous exponential model taes the form dy dt = y cy2, (2.22) here = b m. This is non as the logistic model of population groth. Typically, c is small and the added nonlinear term does not really ic in until the population gets large enough. Example 2.4. Sho that Equation (2.22) can be ritten in the form The logistic model as first published in 838 by Pierre François Verhulst ( ) in the form dn dt = rn ( N K ), here N is the population at time t, r is the groth rate, and K is hat is called the carrying capacity. Note that in this model r = = Kc. z = z( z) hich has only one parameter. We carry this out be rescaling the population, y(t) = αz(t), here α is to be determined. Inserting this transformation, e have y = y cy 2 αz = αz cα 2 z 2, or z = z ( α c ) z. Thus, e obtain the result, z = z( z), if e pic α = c. The point of this derivation is to sho that there is only one free parameter,, and that many combinations of c and in the original problem lead to essentially the same solution shape. We can model the logistic equation, y = ry( y), ith r = and y(0) = 0. in Simulin. The model is shon in Figure 2.4. Running te model gives the solution in Figure 2.5. It shos the typical sigmoidal curve bounded by the solutions y = 0 and y =.

14 38 solving differential equations using simulin Gain Logistic Equation y' s Integrator y Scope Figure 2.4: Logistic equation, y = ry( y). y y' = r y (-y) Product -y Constant Figure 2.5: Solution to the logistic equation, y = ry( y), ith r = and y(0) = The Logistic Equation ith Delay Sometimes the rate of change does not immediately tae place hen the system changes. This can be modeled using differential-delay equations. For example, hen the resources are being depleted, the effects might be delayed. So, a possible model ould be the logistic equation ith delay, y = ry(t)( y(t τ)), here τ is a fixed delay time. The problem ith trying to solve this model at time t is that e need to no something about the solution for earlier times, y(t τ). One ay to tacle the problem is to specify the solution for times [0, τ] and then to solve the equation ith delay using this starting value. So, if y = 2

15 first order differential equations 39 initially, e could let y = 2 for [0, τ]. 2 Alpha >= y' s y Cloc 0 Sitch to enter y=2 for t<= Integrator Scope Solve y'=0 y y(-y(t-)) Logistic Equation ith Delay y' = alpha y(t)(-y(t-)) Product -y(t-) y(t-) Transport Delay Constant The Simulin model is shon if Figure 2.6. A Sitch bloc is used to specify the starting values for times up to τ =. Then, the differential equation solver taes over ith a Delay bloc used to enter the delay term. This model produces the solution 2.7. Figure 2.6: Logistic equation ith delay, y = ry(t)( y(t τ)) Figure 2.7: Logistic equation ith delay, y = ry(t)( y(t τ))

16 40 solving differential equations using simulin 2.7 Exercises. Model the folloing first order differential equations in Simulin and find the solutions for different initial conditions. dy a. dx = ex 2y. dy b. dt = y2 ( + t 2 ). dy y c. dx = 2. x d. xy = y( 2y). e. y (sin x)y = sin x. f. xy 2y = x 2. ds g. dt + 2s = st2. h. x 2x = te 2t. dy i. + y = sin x,. dx dy j. dx 3 x y = x3. 2. Consider the case of free fall ith a damping force proportional to the velocity, f D = ±v ith = 0. g/s. a. Using the correct sign, consider a 50 g mass falling from rest at a height of 00m. Find the velocity as a function of time. Does the mass reach terminal velocity? b. Let the mass be thron upard from the ground ith an initial speed of 50 m/s. Find the velocity as a function of time as it travels upard and then falls to the ground. Ho high does the mass get? What is its speed hen it returns to the ground? 3. A paratrooper, 322 lbs including munitions, jumps from 0,000 ft. Model this free fall ith air resistance f (v) = 5v 2 in Simulin. First, rite don the free fall equation. Use the model to solve for v(t). Is there a terminal velocity? Find the time to land and the impact velocity. 4. Model the folloing problem in Simulin: The temperature inside your house is 70 o F and it is 30 o F outside. At :00 A.M. the furnace breas don. At 3:00 A.M. the temperature in the house has dropped to 50 o F. Assuming the outside temperature is constant and that Neton s La of Cooling applies, determine hen the temperature inside your house reaches 40 o F. 5. Model the folloing problem in Simulin: A body is discovered during a murder investigation at 8:00 P.M. and the temperature of the body is 70 o F. To hours later the body temperature has dropped to 60 o F in a room that is at 50 o F. Assuming that Neton s La of Cooling applies

17 and the body temperature of the person as 98.6 o F at the time of death, determine hen the murder occurred. first order differential equations 4

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