Chapter1. Ordinary Differential Equations

Transcription

1 Chapter1. Ordinary Differential Equations In the sciences and engineering, mathematical models are developed to aid in the understanding of physical phenomena. These models often yield an equation that contains some derivatives of an unknown function. Such an equation is called a differential equation. Two examples of models developed in calculus are the free fall of a body and the decay of a radioactive substance. In the case of free fall, an object is released from a certain height above the ground and falls under the force of gravity. Newton s second law, which states that an object s mass times its acceleration equals the total force acting on it, can be applied to the falling object. This leads to the equation m d2 x dt 2 = mg where m is the mass of the object, x is the height above the ground, d2 x dt 2 is its acceleration, g is the (constant) gravitational acceleration, and -mg is the force due to gravity. This is a differential equation containing the second derivative of the unknown height x as a function of time. Fortunately, the above equation is easy to solve for x. All we have to do is divide by m and integrate twice with respect to t. That is, d 2 x dt 2 = g so and dx dt = gt + c 1 x = gt2 2 + c 1t + c 2 In the case of radioactive decay, we begin from the premise that the rate of decay is proportional to the amount of radioactive substance present. This leads to the equation dq dt = kq, k > 0 where is the unknown amount of radioactive substance present at time t and k is the proportionality constant. To solve this differential equation, we rewrite it in the form And integrate to obtain Solving for A yields, 1 dq kdt Q ln Q c1 kt c2 Q Q() t e e e Ce lnq kt C2 C1 kt The number of examples about DE can be increased. Mathematically, a differential equation is an equation for a function that relates the values of the function to the values of its derivatives. Many real life problems can be formulated as differential equations. For example 1

2 d 2 y dy + 2x + y = sin(x) dx2 dx is a differential equation. Sometimes one uses shortened notations to write the same equation as or using the differential operator D = dy y + 2xy + y = sin(x) dx D 2 y + 2xDy + y = sin(x) There are two classes of differential equations: O.D.E. (ordinary differential equations): linear and non-linear; An ordinary differential equation (ode) is a differential equation for a function of a single variable, e.g., x(t), P.D.E. (partial differential equations): A PDE equation is a differential equation for a function of several variables, e.g., (x, y, z, t). An ode contains ordinary derivatives and a pde contains partial derivatives. Typically, pde s are much harder to solve than ode s. For example, are ordinary differential equations. y + 4y + y = 0 y = x 2 y + sin(x) u xx + u yy = 0 u 2 x + u 2 y = ln(u) are partial differential equations. (Partial Differential Equations are usually much more difficult). 1. Origin and Application of Differential Equations Having classified differential equations in various ways, let us now consider briefly where, and how, such equations actually originate. In this way we shall obtain some indication of the great variety of subjects to which the theory and methods of differential equations may be applied. Differential equations occur in connection with numerous problems that are encountered in the various branches of science and engineering. We indicate a few such problems in the following list, which could easily be extended to fill many pages. The problem of determining the motion of a projectile, rocket, satellite, or planet. The problem of determining the charge or current in an electric circuit. The problem of the conduction of heat in a rod or in a slab. The problem of determining the vibrations of a wire or a membrane. The study of the rate of decomposition of a radioactive substance or the rate of growth of a population. The study of the reactions of chemicals. The problem of the determination of curves that have certain geometrical properties. The mathematical formulation of such problems give rise to differential equations. But just how does this occur? In the situations under consideration in each of the above problems the objects involved obey certain scientific laws. These laws involve various rates of change of one or more quantities with respect to other quantities. Let us recall that such rates of change are expressed mathematically by derivatives. In the 2

3 mathematical formulation of each of the above situations, the various rates of change are thus expressed by various derivatives and the scientific laws themselves become mathematical equations involving derivatives, that is, differential equations. A natural question now is the following: How does one obtain useful information from a differential equation? The answer is essentially that if it is possible to do so, one solves the differential equation to obtain a solution; if this is not possible, one uses the theory of differential equations to obtain information about the solution. To understand the meaning of this answer, we must discuss what is meant by a solution of a differential equation; this is done in the next section. 2. Some concepts related to differential equations: Order: The order of a differential equation is the order of the highest derivative that occurs in the equation. A first order differential equation contains y 0, y and x so it is of the form F(x; y; y 0 ) = 0 or y 0 = f(x; y). For example, the following differential equations are first order: While these are second order: y = x 2 y + e x xy = (1 + y 2 ) y 2 = 4xy + cos(x) y x 2 y + y = 1 + sin(x)e x y + 6yy = (1 + x 3 ) Thus, a general form for an nth-order equation with x independent, y dependent, can be expressed as where F is a function that depends on x, y, and the derivatives of y up to order n; that is, on x,y,...,.d n y/dx n. General and Particular Solutions: A general solution of a differential equation involves arbitrary constants. In a particular solution, these constants are determined using initial values. As an example, consider the differential equation y = 2x. y = x 2 + c is a general solution, y = x is a particular solution, Example: Find the general solution of the differential equation y = 0. Then, find the particular solution that satisfies y(0) = 5; y (0) = 3. y = 0 y = c y = cx + d. This is the general solution. y 0 = 3 c = 3, y 0 = 5 d = 5 thus y = 3x + 5 Therefore y = 3x + 5. This is the particular solution. Explicit and Implicit Solutions: y = f(x) is an explicit solution, F(x; y) = 0 is an implicit solution. We have to solve equations to obtain y for a given x in implicit solutions, whereas it is straight forward for explicit solutions. For example, y = e 4x is an explicit solution of the equation y = 4y. 3

4 x 3 + y 3 = 1 is an implicit solution of the equation y 2 y + x 2 = 0. It is much easier to verify that a function (given either implicitly or explicitly) is a solution of a given differential equation. What is a solution? Solution is a function that satisfied the equation and the derivatives exist. Example: Verify that y(t) = e at is a solution of the IVP (initial value problem) y = ay, y(0) = 1. Here y(0) = 1 is called the initial condition. Let s check if y(t) satisfies the equation and the initial condition: They are both OK. So it is a solution. y = ae at = ay, y(0) = e 0 = 1. Example: Verify that y(t) = 10 c e t with c a constant, is a solution to y + y = 10. y = ce t = ce t, y + y = ce t + 10 ce t = 10, thus it is verified. Example: Verify that y = c 1 sin 2x + c 2 cos(2x) is a solution of y + 4y = 0. Since the second derivative is involved, we differentiate y twice to get and substituting these in the given equation, Example: Verify that the function x 2 = 2 y 2 ln(y) solves the equation y = We differentiate both sides of the given solution to get xy (x 2 +y 2 ). 4

5 x 2x Exercise: The function () x c e c e is a solution to 1 2 d 2 y dx 2 dy dx 2y = 0, with y 0 = 2, y 0 = 3. Determine the constants c 1 and c 2. Linear or non-linear equations: Let y(t) be the unknown. Then, a 0 t y n + a 1 t y n a n t y = g t (1) is a linear equations. If the equation can not be written as Eq. 1, then it s a non-linear. Two things you must know: identify the linearity and order of an equation. Example: The following ordinary differential equations are both linear. In each case y is the dependent variable. Observe that y and its various derivatives occur to the first degree only and that no products of y and/or any of its derivatives are present. The following ordinary differential equations are all nonlinear: The first equation is nonlinear because the dependent variable y appears to the second degree in the term 6y 2. The second equation owes its nonlinearity to the presence of the term 5(dy/dx) 3, which involves the third power of the first derivative. Finally, the last equation is nonlinear because of the term 5y(dy/dx), which involves the product of the dependent variable and its first derivative. Linear ordinary differential equations are further classified according to the nature of the coefficients of the dependent variables and their derivatives. For example, an equation is said to be linear with constant coefficients, while an equation is linear with variable coefficients. 5

6 Example: Let y(t) be the unknown. Identify the order and linearity of the following equations. 3. First-order differential Equations The general first-order differential equation for the function y = (x) is written as dy = f(x, y) (2) dx where f(x, y) can be any function of the independent variable x and the dependent variable y. We first show how to determine a numerical solution of this equation, and then learn techniques for solving analytically some special forms of (2), namely, separable, exact and linear first-order equations. a) Separable equations: A differential equation is said to be separable if it can be rewritten so that terms involving the differential of y is on one side of the equation, and those of x on the other side. One then integrates to get rid of the differentials leading to an equation that implicity or explicitly gives y. Thus, g(y)dy = f(x)dx, it is called a separable equation. We can find the solution by integrating both sides. (Don't forget the integration constant!). g y dy = f x dx + c Example: Solve the initial value problem y = x 3 e y, with y(1) = 0. Exercise: Solve the equation 6

7 Example: Solve y 2 y + 2x = 0. Example: Solve the initial-value problem that consists of the differential equation 2 xsin( y) dx ( x 1)cos( y) dy 0 with y(1) / 2. 7

8 Example: Find the particular solution of the equation y ln y dx xdy = 0 such that when x = 1, y = 2. Rewrite the equation as Exercise: Solve the equation 2 2 ( x 3 y ) dx 2xydy 0 8

9 Homework: Solve the initial-value problem b) Exact equations: A first order equation can often be written in the form M (x; y) dx + N (x; y) dy = 0 This equation is said to be exact if the left hand side is the differential of a function u(x; y) so that u(x, y) x = M x, y, u(x, y) y = N x, y, In other words, du = Mdx + Ndy, so Mdx + Ndy is a total differential. For example, the equation is exact and (4x 3 + 2xy 2 )dx + (4y 3 + 2yx 2 )dy = 0 u x, y = x 4 + x 2 y 2 + y 4 So, the solution of this equation is very simple, if du is zero, u must be a constant, therefore c = x 4 + x 2 y 2 + y 4 Hence, from the equations related with M and N, the condition M y = N x is necessary and sufficient for the equation M(x; y)dx + N(x; y)dy = 0 to be exact. Example: Solve the equation 3y 2 dx + (3y 2 + 6xy)dy = 0 9

10 Example: Solve 2x sin(y) dx + x 2 cos(y) dy = 0. And Integrating Factors Consider the equation P dx + Q dy = 0 that is not exact. If it becomes exact after multiplying by, i.e. if (x)pdx + (x)qdy = 0 is exact, then is called an integrating factor. (Note that P, Q and are functions of x and y). 10

11 Example: Solve (2xe x y 2 )dx + 2ydy = 0. Use integrating factor as e x. Now the equation is exact. We can solve it as we did the previous example and obtain the result How to Find Integrating Factors? x 2 + y 2 e x = c The first-order linear differential equation (linear in y and its derivative) can be written in the form y + p(x)y = g(x), with the initial condition y(x 0 ) = y 0. Linear first-order equations can be integrated using an integrating factor (x). After mathematical calculation, the solution of above equation) satisfying the initial condition (x 0 ) = y 0 is then traditionally written as Example: Solve dy dx + 2y = e x with y(0) = 3/4. Note that this equation is not seperable. With p(x) = 2 and g(x) = e x, we have 11

12 Example: Solve y + ay = b. Exercise: Solve ty + 2y = 4t 2 with initial condition y(1) = 2. c) Bernoulli Equation: The equation y + p x y = g(x)y α is called Bernoulli equation. It is nonlinear. Nonlinear equations are usually much more difficult than linear ones, but Bernoulli equation is an exception. It can be linearized by the substitution u x = [y(x)] 1 α Then, we can solve it as other linear equations. Example: Solve the equation 12

13 Homework: Solve the equation 13

14 4. Mathematical Modeling Differential equations are the natural tools to formulate, solve and understand many engineering and scientific systems. The mathematical models of most of the simple systems are differential equations. The simplest ordinary differential equations can be integrated directly by finding antiderivatives. a) Malthus' law of population Dynamics; We all know that population grows in general, but how fast does the population of a city grow? What will the population of the US be in 2025? Questions of this kind have been of interest to many scientists and the first person to propose a mathematical law was Rev. Thomas Robert Malthus, an English clergy man who laid out his findings in his 1798 writings. Malthusian law says: The rate of change of population is proportional to the actual population at any given time. Notice that this law is very similar in nature to Newton's law. Both laws predict that the rate of change of a quantity is in some sense proportional to the quantity remaining at a given time. The only difference is that Newton's law has to do with the decrease in heat while Malthus law has to do with increase in population. Let the population at a given time t 0 be P 0 and at any time t later be P(t). Then according to Malthus law, dp dt P If k is the constant of proportionality, the above equation becomes dp dt = kp This is a very simple equation whose solution is ln P = k t + c, where c is the constant of integration. Using the initial condition that at t = t 0 ; P = P 0, we get We can then write the solution as that is, ln P 0 = k t 0 + c ln P = k t + ln P 0 k t 0 ln P = k(t t 0 ) + ln P 0 P = P 0 k(t t 0 ) ** The world population in 1965 and in 1970 was billions and billions, respectively. What was the population in 1973? Solution: From equation (2.19), using the data for 1965 and 1970, To find the population in 1973, we have that 14

15 The actual population of the world in 1973 was billions! b) Newton s Law of motion. Consider a mass falling under the influence of constant gravity, such as approximately found on the Earth s surface. Newton s law, F = ma, results in the equation where x is the height of the object above the ground, m is the mass of the object, and g = 9.8 m/s 2 is the constant gravitational acceleration. Find the height as a function of time if the initial height is y 0 and initial speed is v 0. Here, the right-hand-side of the ode is a constant. The first integration, obtained by antidifferentiation, yields where A is the constant of integration, and the second integration gives with B another constant of integration. The two constants of integration A and B can be determined from initial conditions. If we know that the initial height of the mass is x0, and the initial velocity is v0, then the correct initial conditions are Substitution of these initial conditions into the equations for dx/dt and x allows us to solve for A and B to determine the unique solution that satisfies both the ode and the initial conditions: c) Falling Body Problems We shall now consider some examples of a body falling through air toward the earth. In such a circumstance the body encounters air resistance as it falls. The amount of air resistance depends upon the velocity of the body, but no general law exactly expressing this dependence is known. In some instances the law R = kv appears to be quite satisfactory, while in others R = kv 2 appears to be more exact. In any case, the constant of proportionality k in turn depends on several circumstances. In the examples that follow we shall assume certain reasonable resistance laws in each case. Thus we shall actually be dealing with idealized problems in which the true resistance law is approximated and in which certain comparatively negligible factors are disregarded. 15

16 Example: A body weighing 8 lb falls from rest toward the earth from a great height. As it falls, air resistance acts upon it, and we shall assume that this resistance (in pounds) is numerically equal to 2v, where v is the velocity (in feet per second). Find the velocity and distance fallen at time t seconds. 16

17 Homework: A skydiver equipped with parachute and other essential equipment falls from rest toward the earth. The total weight of the man plus the equipment is 160 lb. Before the parachute opens, the air resistance (in pounds) is numerically equal to iv, where v is the velocity (in feet per second). The parachute opens 5 sec after the fall begins; after it opens, the air resistance (in pounds) is numerically equal to V2, where v is the velocity (in feet per second). Find the velocity of the skydiver (A) before the parachute opens, and (B) after the parachute opens. d) Mixture Problems We now consider rate problems involving mixtures. A substance S is allowed to flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further, in one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rat.e; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t. Letting x denote the amount of S present at time t, the derivative dx/dt denotes the rate of change of x with respect to t. If IN denotes the rate at which S enters the mixture and OUT the rate at which it leaves, we have at once the basic equation from which to determine the amount x of S at time t. 17

18 Example: A tank initially contains 50 gal of pure water. Starting at time t = 0 a brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. 1. How much salt is in the tank at any time t > 0? 2. How much salt is present at the end of 25 min? 3. How much salt is present after a long time? Exercise: A large tank initially contains 50 gal of brine in which there is dissolved 10 lb of salt. Brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring, and the stirred mixture simultaneously flows out at the slower rate of 3 gal/min. How much salt is in the tank at any time t > 0? 18

19 e) Exponential growth/decay. Assume the rate of change of Q(t) is proportional to the quantity at time t. We can write where r is the rate of growth/decay and Q(t) is the amount of quantity at time t. If r > 0: exponential growth If r < 0: exponential decay Differential equation: Here r is called the growth rate. By IC, we get Q(0) = C = Q 0. The solution is Two concepts: Doubling time TD (only if r > 0): is the time that Q(T D ) = 2Q 0. Half life (or half time) TH (only for r < 0): is the time that Q(T H ) = (½) Q 0. Note here that T H > 0 since r < 0.. NB! TD, TH do not depend on Q 0. They only depend on r. Example: A radio active material is reduced to 1/3 after 10 years. Find its half life. Model: dq/dt = rq, r is rate which is unknown. We have the solution Q(t) = Q 0 e rt. So 19

20 Q 10 = Q 0 e rt = Q 0, Q 3 0e 10r = Q 0 ln 3, r = 3 10 To find the half life, we only need the rate r T H = ln 2 10 ln 2 = ln 2 = 10 r ln 3 ln 3 Exercise: The rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. Half of the original number of radioactive nuclei have undergone disintegration in a period of 1500 years. a) What percentage of the original radioactive nuclei will remain after 4500 years? b) In how many years will only one-tenth of the original number remain? f) Newton's Law of Cooling The Newton s Law of Cooling states that the rate of change of the temperature of a cooling body is porportional to the difference between the temperature of the body and the constant temperature of the medium surrounding the body. Apply this law to the following problem. Example: A body of temperature 80 F is placed in a room of constant temperature 50 F at time t = 0; and at the end of 5 minutes, the body has cooled to a temperature of 70 F. Determine the temperature of the body as a function of time for t > 0. In particular answer the following questions: 1. What is the temperature of the body at the end of 10 minutes? 2. When will the temperature of the body be 60 F? 3. After how many minutes will the temperature of the body be within 1 F of the constant 50 temperature of the room? 20

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23 Exercises A. 1. Show that the equation is exact, and obtain its general solution. Also, find the particular solution corresponding to the given initial condition as well. Ans: 2. Obtain the general solution, using the methods of this section. Ans: a) x 2-2xy y 2 = C 3. Use separation of variables to find the general solution. Ans: 4. Solve y = (3x 2-1)/(2y), subject to the given initial condition a) y(0) = -3, b) y(0) = -1, c) y(4) = 5, d) y(-1)=0. Ans: 23

24 5. Solve x y' + y= 6x 2 subject to the given initial condition a) y(1) = 0 and b) y(-3) = 18, using any method of this section. Ans: a) y(x) = 2x 2 2/x, b) y(x) = 2x 2 6. Solve the problems in the course books. B. 24

25 References: 1. Fundamentals of Differential Equations, R. K. Nagle, E.B. Saff, A. D. Snider, 8th Edition, (Pearson) (for Lecture 1-5). 2. Ross, S.L. Introduction to Ordinary Differential Equations, 4th edition. Wiley, O'Neil, P.V. Advanced Engineering Mathematics, 5th edition. Thomson,