1 Mathematical Models

Transcription

1 Intro to Math Modeling Math ODE for Engineers Lecture 1-1/9/18 Abstract In today s lecture we discuss mathematical modeling in the context of natural sciences. The goal is to derive a class of mathematical models which will lead naturally into the study of Differential Equations. 1 Mathematical Models An important step in the scientific and engineering process is the construction of a Mathematical Model. A mathematical model is fundamentally an attempt to formalize the patterns and regularity observed in the context of a scientific and engineering problem. In many applications a math model is formulated using either continuous mathematics or discrete mathematics. Morevoer the model may reflect either a stochastic process or deterministic process. Immediately prior to constructing a mathematical model, scientists and engineers formulate many concepts which qualitatively describe the phenomenon they are studying. For example many scientific and engineering systems demonstrate properties akin to equilibrium, diffusion, reaction, flow/transport, vibrations/oscillations, control and dispersion. Of course this list is not exhasutive, but these particular properties of a system have been modeled with much success using a class of mathematical models collectively known as Differential Equations. The class of Differential Equations we aim to study arise in the modeling of deterministic continuouss processes. Once formulated, a differential equation can be studied mathematically (since it is a mathematical object) and conclusions may be drawn which give insight and knowledge about the scientific and engineering problem. The final step is to verify any predictions with physical or computational experiment. A rough flowchart below gives the essential steps in the formulation, analysis and application of mathematical models. Figure 1: Outline of Mathematical Modeling Formulation of the Scientific/Engineering Concept Construction of Mathematical Model Apply Math Methods: Exact Methods Perturbative Methods Numerical Methods Geometric Methods Use mathematical results to gain insight into the Scientific and Engineering Problem Testing Scientific and Engineering Prediction with Data To begin our discussion we wil first attempt to understand how a scientist or engineer may go about trying to construct a mathematical model. 1

2 Intro to Math Modeling Math ODE for Engineers Lecture 1-1/9/18 Remark 1.1. Experiments have to be designed and data has to be collected to gain initial insight into any system. The strength of the experimental apparatus and the efficiency of data collection are topics we leave to be discussed in your Scientific and Engineering classes. Let us just remark that in many modern experiments, data analysis algorithms play a crucial role to help scientists and engineers formulate reliable mathematical models. The interesting point for us may be that many of the most important data analysis tools and techniques are themselves mathematical models. Many of these models for data analysis and visualization are discrete models, allowing for their implementation on computational devices, but it is interesting to point out that many of the modern applications of data analysis implement known scientific models (such as the biological theory of learning) to help design more intelligent data analysis algorithms. In this way their is a mutually beneficial relationship between data analysis and mathematical modeling. In any case either it be a scientist/engineer or a computational algorithm, one has to sift through the data to find regulariteis and patterns which may be formalized and serve as a basis for the construction of interesting mathematical models, which for us will be differential equations. To derive a differential equation it is often useful to start from a general consevation law and some specific constitutive equations. For physical systems many conservation laws are known, and for a wide variety of scientific and engineering systems many constitutive equations have been determiend by experiment. Let us focus our attention on two distinct models and derive some important differential equations. The first example is a classical problem in mechanics, which is to understand the dynamics of an oscillating pendulum. Example 1.2. Consider an arbitrary point mass attached to one end of a rigid and weightless rod of length L and the other end fixed at the origin. Consider the system initially displaced by angle θ from the vertical direction. We would like to study its motion and dynamics. The first step will be to derive the system s equation of motion. Figure 2: Swinging Pendulum Consider the general conservation law for energy, which states that the total energy of a physical system must always be conserved. For a mechanical system modeled by point particles and no dissipation of energy, the conservation of energy is a statement about the conservation of the sum total of kinetic and potential energy. Let v(t) denote the velocity of the pendulum then, Kinetic Energy = 1 2 mv(t)2 (1.1) 2

3 Intro to Math Modeling Math ODE for Engineers Lecture 1-1/9/18 To understand Potential Energy, we introduce the constitutive equation modeling the force of gravity on the surface of the Earth, with g 9.8 meters per second squared, F g = mg (1.2) Let h(t) be the height above the equilibrium position (when the Pendulum is at rest), then the potental energy is defined to be the gravitational potential energy, The total energy is defined to be, Potential Energy = mgh(t) (1.3) Total Energy = E(t) = 1 2 mv(t)2 + mgh(t) (1.4) The mathematical statement of conservation of energy is that for all times the energy of the system is a fixed constant value. Hence this implies, d E(t) = 0 (1.5) dt Since we have chosen to model our system by the single parameter, θ, we need to find mathematical expressions for kinetic and potential energy in terms of θ. Since the length of the rod is fixed, the Pendulum traces out a circular arc of length s(t), s(t) = Lθ(t) The velocity of the pendulum is measured by how fast the arclength is changing. We find, Hence the kinetic energy is, d dt s(t) = L d dt θ(t) ( ) 2 1 d 2 ml2 dt θ(t) Moreover when the pendulum makes an angle θ with the vertical, using properties of right triangles we find that the height above equilibrium is, h(t) = L L cos θ So the total energy of the system can be written as, E(t) = 1 2 ml2 ( d dt θ(t) ) 2 + mg (L L cos θ) (1.6) Differentiating this expression with respect to time and setting it equal to zero, we find the systems equation of motion, d 2 dt 2 θ(t) + g sin θ = 0 (1.7) L The second example models the growth of bacteria in a controlled environment called a Chemostat. Example 1.3. Please refer to the uploaded pdf document on mycourses explaining the mathematical model of bacteria growth in a chemostat. In this case the conclusion is that we arrive at a system of differential equations. 3

4 Intro to Math Modeling Math ODE for Engineers Lecture 1-1/9/18 Having now derived a couple of differential equations from conservation and constitutive equations, we can now begin to undertake a systematic understanding of various types of differential equations. The next order of business will be to mathematically define a differential equation and then to understand in a more mathematical way what it means to solve a differential equation. What is clear from the outset is that problems in science and engineering lead to many different types of differential equations, and it will be our job to first identify some important classes of equations and then to develop systematic tools that will help us solve these equations. 2 Suggested Problems Question 1. Section a, 22, 23, 24a, 25a ; Section , 31 Question 2. Consider the ideal Spring-Mass system with spring constant k Figure 3: Spring-Mass System Suppose we compress/stretch the point mass a distance x from equilibrium and then track its oscillations. Assuming the constitutive relationship (Hooke s Law) F s = kx (2.1) use the conservation of energy to derive the equations of motion for this system. (Hint: You will need to derive the potential energy for the system from the physical definition of Work) 4

5 Intro to Differential Equations Math ODE for Engineers Lecture 2-1/11/18 Abstract In today s lecture we continue our disucssion of differential equations and understand what it means to find a solution. We offer a general classification for ordinary differential equations and highlight some of the major examples we will consider in this course. 1 Differential Equations Last time we discussed mathematical modeling and then derived differential equations using conservation laws and constitutive equations. Let me point out that an important check that a model is consistent is to check the dimensions (units) of each term appearing in the model. This is known as dimensional analysis. Note the care given in the Bacteria example to checking the dimesions of each term in the differential equation. As a simpler example one should check if the pendulum equation we derived in the last lecture is also dimensionally consistent. Question 1. Let mass (M) be measured in kilograms, length (L) measured in meters, and time (T) measured in seconds. We write this as, [M] = kilograms [L] = meters [T] = seconds Check that the equation of motion for the oscillating pendulum is dimensionanlly consistent. Question 2. Consider the ideal spring-mass system introduced in the last lecture and the units introduced in the last question. Use the equation of motion and dimensional analysis to deduce the units of measurement for the spring constant k (this is known as the stiffness of the spring).. Turning our focus to differential equations, it is important to first define precisely what we mean by a differential equation and then to define precisely what it means to solve a differential equation. Most of the course will be concerned with solving differential equations, and for this we will develop various types of mathematical methods to understand solutions. But before exploring various mathematical methods, it is important to pause and think about what we are actually looking for. Definition 1.1. A differential equation is a relationship between an unknown function and its derivatives. This is actually a rather vauge but hopefully intuitive defintion. Mathematically this means that the relationship could be either algebriac or transcendental. Moreover the definition makes no mention of the number of independent or dependent variables you need to define a differential equation. If in the relationship, the number of independent variables is one, then the differential equation is called an ordinary differential equation. If the number of independent variables in the equation is greater than 1, then we may have a relationship involving partial derivatives of the unknown function, hence a partial differential equation. Remark 1.2. For ordinary differential equations the following notations are all equivalent to each other: and so on for higher order derivatives. d dt y(t) = y (t) = ẏ(t) d 2 dt 2 y(t) = y (t) = ÿ(t) 1

6 Intro to Differential Equations Math ODE for Engineers Lecture 2-1/11/18 Also if in the relationship there is one dependent variable (namely one unknown function) then the differential equation is called a scalar differential equation. If there are more than one dependent variables then we have a system of differential equations. Finally we define the order of a differential equation to be the order of the highest derivative appearing in the equation. Example 1.3. The oscillating pendulum and the ideal spring-mass system are both second order ordinary differential scalar equations. Example 1.4. The bacteria harvesting system is a first order ordinary differential system of equations Example 1.5. Consider the unknown function u(x, y) which is a function of the two independent variables x and y. The following are examples of second order partial differential equations. In the first example x and y represent spatial coordinates, and in the last two examples the y variable represents time: u xx + u yy = 0 Laplace Equation (1.1) u y = u xx Heat Equation (1.2) u yy = u xx Wave Equation (1.3) Let us point out that the focus of this course is on ordinary differential equations. We will discuss both scalar ordinary differential equations and systems of ordinary differential equations. Lets now give a general defintion for an ordinary differential equation of arbitrary order n. Definition 1.6. An ordinary differential equation of the nth order is an equation, Example 1.7. For the pendulum equation, F (t, y, y, y,..., y n ) = 0 (1.4) F (t, θ, θ, θ ) = θ + g L sin θ = 0 Remark 1.8. A second order partial differential equation is an equation, F (x, y, u, u x, u y, u xy, u xx, u yy ) = 0 Compare that for a second order ordinary differential equation F is a function of four variables, while for a second order partial differential equation, F is a function of eight variables Using this general defintion it becomes a lot easier to check if a given differential equation satisfies important properties. The defining property in this course will be the concept of linearity. We will see this concept appear in a variety of ways when we discuss differential equations. Of course you have already seen the concept of linearity appear in previous courses in Calculus and Linear Algebra. For us the defining property of linearity will be the following, 2

7 Intro to Differential Equations Math ODE for Engineers Lecture 2-1/11/18 Definition 1.9. An nth order ordinary differential equation is linear homogenous if for two functions y 1 (t) and y 2 (t) and any two constants a and b, F (t, (ay 1 + by 2 ), (ay 1 + by 2 ),..., (ay 1 + by 2 ) n ) = af (t, y 1, y 1, y 2,..., y n 1 ) + bf (t, y 2, y 2, y 2,..., y n 2 ) (1.5) This is the most general definition and the only way to appreciate it is to try it out on some examples. Example Consider the differential equation with k a constant, We rewrite this in the form, y (t) = ky(t) F (t, y, y ) = y (t) ky(t) = 0 Consider two functions y 1 (t) and y 2 (t) and two arbitrary constants a and b. Define Plugging this into the differential equation we find, z(t) = ay 1 (t) + by 2 (t) F (t, z, z) = F (t, ay 1 (t) + by 2 (t), (ay 1 (t) + by 2 (t)) ) = (ay 1 (t) + by 2 (t)) k(ay 1 (t) + by 2 (t)) Hence this differential equation is linear homogenous. = ay 1(t) kay 1 (t) + (by 2(t) kby 2 (t)) = a(y 1(t) ky 1 (t)) + b(y 2(t) ky 2 (t)) = af (t, y 1, y 1) + bf (t, y 2, y 2) The simplest example of an ordinary differential equation that is not linear homogenous is the following, Example Consider the differential equation for an arbitrary function f(t), Doing as before we easily find that y (t) = y(t) + f(t) F (t, ay 1 (t) + by 2 (t), (ay 1 (t) + by 2 (t)) ) af (t, y 1, y 1) + bf (t, y 2, y 2) Examples like the previous one shall be called linear inhomogenous. The distinction between the two examples above is the same as the difference between the two algebraic equations for m and b constants, y = mx y = mx + b We still want to say that both examples are linear but their is a finer distinction to be made because of the additional b term or (f(t) term). Finally, if we are neither linear homogenous nor linear inhomogenous, then we call the differential equation nonlinear. The simplest example of a nonlinear ordinary differential equation is the following, 3

8 Intro to Differential Equations Math ODE for Engineers Lecture 2-1/11/18 Example Consider the differential equation, Doing as before we find, y (t) = y(t) 2 F (t, z, z) = F (t, ay 1 (t) + by 2 (t), (ay 1 (t) + by 2 (t)) ) = (ay 1 (t) + by 2 (t)) (ay 1 (t) + by 2 (t)) 2 Notice that the first term will split as before, but the second term has a cross term of the form, So it follows that, 2aby 1 (t)y 2 (t) F (t, ay 1 (t) + by 2 (t), (ay 1 (t) + by 2 (t)) ) af (t, y 1, y 1) + bf (t, y 2, y 2) Later when we develop some more differential equations models from science and engineering systems we will explore a more conceptual understanding of the concept of linearity. At the moment it will be important to practice with the algebraic definition and keep in mind what linearity means for algebraic equations. Finally the last topic to introduce for our general discussion of differential equations is the concept of a solution. Let us precisely define a solution to a differential equation. Definition A solution of the nth order ordinary differential equation on an interval α < t < β, is a function, ϕ(t) with all derivatives up to the nth order defined on the interval α < t < β, and satisfying on the same interval, F (t, ϕ, ϕ, ϕ,..., ϕ (n) ) = 0 (1.6) Of course intuitively we know that a solution to a differential equation is a function which when substituted into the differential equation gives us an identity on both sides of the equation. The only technical point is that functions come with domains and so we can only say a function is a solution to a differential equation on its domain of definition. Example Consider the function, Verify that y(x) solves the differential equation, on the interval 5 < x < 5 y(x) = 25 x 2 y (x) = x y This can be checked in a straightforward manner by plugging in the function into the differential equation. The above is an example of an explicit solution because we have a function ϕ(t) defined on a concrete interval which we can plug directly into the differential equation and check to see if it satisfies the equation. On the other hand, sometimes you will have to check if an implicit relatonship satisfies a differential equation. What is meant by this is that sometimes you will have a relationship between the variables x and y in which 4

9 Intro to Differential Equations Math ODE for Engineers Lecture 2-1/11/18 one of the variables is in fact a function of the other variable. Sometimes you can use algebraic manipulations to solve for one function in terms of the other, but most times you cannot. All you have to remember is that theoretically there is an interval on which one function can be solved for the other. This is the content of the implicit function theorem which you will see in your vector calculus course. In this case we say that the implicit relationship is an implicit solution of the differential equation. Let us look at an example to see how to work with implict relationships. Example Consider the implicit relationship x 2 + y 2 = 25. Show that this is an implicit solution of y (x) = x y. Of course we know that we can solve for y in terms of x, but if we could not, then we could check that it still satisfies the differential equation on some interval. To do this you can implicitly differentiate to find, 2x + 2yy = 0 This implies after some algebra y = x y Hence the relationship x 2 + y 2 = 25 is an implicit solution to the differential equation y = x y. So to sum up, if you have an implicit relationship in which you cannot solve for one variable in terms of the other, just know that theoretically it can be done on some interval, and on that particular interval we will check if that relationship solves a differential equation via implicit differentiation. Having now discussed the general features of ordinary differential equations we conclude by listing the four class of equations we will study in this course. We will focus on in chronological order, 1) First order differential equations in normal form 2) Linear nth order homogenous and inhomogenous equations 3) System of n first order linear equations 4) System of two first order nonlinear equations y (t) = f(t, y) (1.7) a 0 (t)y (n) (t) + a 1 (t)y (n 1) (t) + + a n (t)y(t) = f(t) (1.8) Next time we begin our study with first order differential equations. 2 Suggested Problems Question 3. Section

10 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Abstract In today s lecture we start our discussion of first order ordinary differential equations. We first focus on geometric methods to understand solutions and then solve analytically for solutions of separable first order equations. 1 First Order Equations We start by studying first order equations in normal form, y (t) = f(y, t) (1.1) Of course if f(y, t) = f(t), then the problem we are solving is to find the antiderivative of y (t). As long as f(t) is a continuous function we know that the solution to the problem can be found by integration, y(t) = f(t) dt. This we know from the fundamental theorem of calculus. Example 1.1. Consider the differential equation, Integrating both sides we find, y (t) = sin t y(t) = cos t + C Integration introduces the arbitary constant C. We call cos t + C the general solution of the ordinary differential equation. Graphically we have an infinite number of curves, so we conclude that this differential equaton has an infintite number of solutions, namely one for each different value of C. Each solution is known as an integral curve for the differential equation. Notice that when you graph the infinite number of solutions, no two integral curves intersect each other. In fact if you fixed one value of y(t) then it would only go through one integral curve. In this case you would have a unique solution to the differential equaton. Example 1.2. Suppose in the previous example we knew that y(0) = 1. condition. Then we find that the constant C has a fixed value, This is known as an initial 1 = y(0) = cos(0) + C = C = 0 A differential equation which comes with an initial condition, is known as an initial value problem. Let us now look at the next simplest example of a first order differential equaiton, which is of the form, y (t) = f(y) (1.2) These first order ordinary differential equations are known as autonomous equations. 1

11 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Example 1.3. The simplest and most important example of an autonomous equation is, y (t) = y(t) Luckily this is an easy differential equation to solve. We know that the solution is an exponential function, multiplied by an arbitrary constant A, y(t) = Ae t Drawing the integral curves we have an infinite number of exponential functions with no two curves intersecting. What will be extremely helpful is to have a way to plot points and see geometrically how the solution behaves. Looking again at the first order differential equation in normal form, we find that the equation is telling us that for a fixed point (t, y(t)) we know the derivative of the solution at that point. Recall that the derivative is the best linear approximation to the function at nearby points, hence if we draw small lines at each point in the (t, y(t)) plane with the correct direction and magnitude we should be able to visualize how solutions to the differential equation behave. Such a plot is known as a direction field. Practically it is helpful to find large collection of points that have the same slope. For autonomous equations we do this by looking at the slope of the function at different values of y. For example in the previous example it is helpful to look at the derivative values when y = 1, y = 2, etc., A line in the plane where all points have the same slope are called isoclines. As you can see in the first plot (See Figure 1), the features of the cosine function are rather apparent and one can trace out the graph of the cosine function from the direction field. In the second example we can see that the exponential behavior is apparent and similarly trace out the integral curves using isoclines (See Figure 2). Moreover notice how the function y(t) = 0 satsfies the differential equation. We call solutions to the differential equation for which y (t) = 0 for every value of t, equilibrium solutions. These solutions are the zeros of the function f(y) appearing on the right hand side of the autonomous equation. The zeros of f(y) are also called critical points. Furthermore notice how all the integral curves are diverging away from the equilibrium solution y(t) = 0 as t goes to infinity. In this case we call the equilibrium solution an unstable equilibrium solution. If on the other hand, all the integral curves are converging to the equilibrium solution, then we call such solutions, asymptotically stable solutions. Question 1. Consider the differential equation, y (t) = y(t) Show by plotting the direction fields, that y(t) = 0 is an asymptotically stable solution. Let us now look at an example where we do not yet know how to explicitly solve for the solution of the equation. Example 1.4. Consider the first order autonomous differential equation, y (t) = y 2 (1.3) We can plot its direction field and some integral curves to understand how the solution behaves (See Figure 3). Here we do not at the moment have an explicit solution but we can deduce some qualitative behavior for the solution. As before, y(t) = 0 is an equilibrium solution. In this case, integral curves starting at y values above the equilibrium solution diverge from 0 while integral curves starting from y values below 0 converge. In this case we say that the equilibrium solution is a saddle point or semistable solution. 2

12 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Figure 1: Example 1.1 3

13 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Figure 2: Example 1.2 4

14 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Figure 3: Example 1.3 Remark 1.5. The most important takeaway from using a direction field plot, is that we have an intuitive feel for how solutions behave in the long run. We are able to deduce certain global properties about solutions to many differential equations. From the perspective of dynamics, the differential equation is a statement about derivatives of a solution. Remember that to define the derivative of a function you only need to know how a function behaves nearby to that point. When we are modeling science and engineering systems it is important to know how that system evolves in time. If we waited only an infinitesimal amount of time, nothing interesting would really happen. It takes a bit of time for interesting dynamics to appear. The direction field plot is our first tool that predicts interesting behavior after a sufficiently long time. 2 Population Dynamics One of the first models for how population evovles was proposd by Thomas Malthus. He proposed that the rate at which population changes is proportional to the current population. That is for r a constant (called the rate of growth), y (t) = ry(t) (2.1) 5

15 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 The rate of growth r should be thought of as the difference between the birth rate and death rate of a population. This model works well in the short-run but in the long-run resources are limiited and competition within species results. A basic model that takes into consideration the limitation of resources is the logistic equation. Consider for r and a constants, Example 2.1. Consider the logistic equation, y (t) = (r ay)y (2.2) y (t) = (1 y)y Lets first solve for the critical points. Here we find that f(y) = y(1 y). Setting it equal to 0, yields the equilibrium solutions, y(t) = 0 y(t) = 1 Note also that f(y) is a downward facing parabola with maximum value at y = 1 2 (You can check this by differentiating f(y) and setting it equal to zero). Now conisder in the interval 0 < y < 1, y (t) is positive, hence y(t) is increasing, similarlty if y < 0 or y > 1, than y (t) < 0, hence y(t) is decreasing on those intervals. To determine the shape of the integral curves (concavity) we can look at y (t). For autonomous equations by the chain rule, y (t) = f (y)y (t) Hence we know that y (t) is positive on the intervals y(t) > 1 and 0 < y(t) < 1 2 (because the product f (y)y (t) is positive). Moreover, y (t) is negative on the intervals 1 2 < y(t) < 1 and y(t) < 0. Finally we have an inflection point at y(t) = 1 2 (because f ( 1 2 ) = 0). The direction fields and integral curves for the logistic equation are plotted on the next page. Note that y = 0 is an unstable equilibrium and y = 1 is an asymptotically stable equilibrium. In the application to population dynamics, y = 1 is known as the saturation level or carrying capacity. Finally on the next page we introduce a phase line which tells us the behavior of integral curves around equilibrium solutions. The arrows indicate if integral curves are converging or diverging from the equilibrium solutions. The following are some problems to try that modify the logistic model. Question 2. Sketch the graph of f(y) versus y, the phase line, determine and classify equilibrium points, and sketch some integral curves for, y (t) = (1 y)y This equation models population dynamics with a critical threshold. It is a model for a biological population that has to sustain a population of at least some value or it will go extinct. Question 3. Sketch the graphy of f(y) versus y, the phase line, determine and classify equilibrium points, and sketch some integral curves for, ( y (t) = (1 y) 1 y ) y 2 This equation models Logistic Growth with a critical threshold. In the biological population model with critical threshold, if populations are above the critical threshold then they experience exponential growth. We would still like to say that their is a carrying capacity in the environment, hence if above the critical threshold populations should still approach this carrying capacity. Remark 2.2. After trying to do the last two exercises, have a look at section 2.5 in you textbook to understand the interpretation of your solutions in the context of population dynamics. 6

16 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Figure 4: Example 2.1 Figure 5: Example 2.1 Phase Line 7

17 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 3 Separation of Variables In this section we introduce our first analytic method to solve explicitly for solutions to first order differential equations in normal form i.e., y (t) = f(t, y) In order to apply this method, we assume that f(t, y) can be written as the product of two functions of a single variable. Namely we assume that there exists two functions g(t) and h(y) such that, f(t, y) = g(t)h(y) (3.1) Note that first order autonomous equations are a special case where g(t) = 1. rearrange your differential equation, y (t) h(y) = g(t) Now we integrate both sides with resepect to the independent variable t, y (t) h(y) dt = g(t) dt To solve for y(t) simply 1 The left hand side is simply the chain rule. We are looking for an antiderivative of h(y). Call it H(y). Then by the chain rule we know that, H (y) = 1 h(y y (t). So the left hand side evaluates to, y (t) dt = H(y) + C h(y) Similarly we have to find an antiderivative for the right hand side, call it G(t). Then we find after integration and collecting the arbitrary constants to one side, H(y) = G(t) + C To see this in action lets do the simplest example to get an idea. Example 3.1. Use separation of variables to solve the differential equation, y (t) = y Separating variables we find, y (t) y = 1 Integrate both sides with respect to t to find, y (t) dt = y 1 dt To evaluate the left hand side we are looking for an antiderivative for 1 y. This we know is ln y. So the left hand side evaluates to ln y + C 8

18 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 The right hand side is much simpler, 1 dt = t + C Hence the implicit solution to the differential equation is, In this case we can solve explicitly for y(t) to find, ln y = t + C y(t) = e t+c So we have, after calling e C another arbitrary constant A, Since e t is always a positive function we have either, or y(t) = Ae t y(t) = Ae t, A > 0 y(t) = Ae t, A < 0 Finally lets point out that in order to do separation of variables we had to divide the equation by y. This is only allowed if y 0. As we saw in the direction fields above y(t) = 0 is also a solution to the differential equation (This can be verified by plugging in directly to the differential equation). This solution cannot be found from the separation of variables technique. Instead you have to check directly if it is a solution. To conclude the solutions for the differential equation can be written as, where A is any real number. y(t) = Ae t Let us point out that the solutions found above are all of the integral curves of the differential equation. Recall that if we wanted to choose just one solution for the differential equation, then we need to specify one value for the function y(t). Suppose in addition we impose that y(0) = 1. Then plugging into the general solution we find, 1 = y(0) = Ae 0 = A Hence the solution to the initial value problem, is, y(t) = e t. This is the unique solution. y (t) = y, y(0) = 1 Question 4. Solve by separation of variables the initial value problem, y (t) = y 2, y(0) = y 0 and determine the interval in which the solution is valid. Does the interval depend on your initial point y 0? Let us now solve the Logistic Equation using separation of variables 9

19 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 Example 3.2. We will solve using separation of variables, Separating variables we have, Integrate both sides with respect to t we find, y (t) = (1 y)y, y(0) = y 0 dy dt (1 y)y = 1 dy dt (1 y)y dt = The integral on the right hand side simply evaluates to t + C. The integral on the left hand side can be treated using the method of partial fractions, 1 dt 1 y(1 y) = A y + B 1 y Solving for A and B we find that A = B = 1. Hence the left hand integral decomposes into, The integrals evaluate to, dy dt (1 y) dt + dy dt y dt ln 1 y + ln y + C Combining the arbitrary constants from both sides and combining the logarthmic terms we have, ln y 1 y = t + C Taking the exponential of both sides as before we have the following solutions, y 1 y = Aet, y < 0, A < 0 y 1 y = Aet, 0 < y < 1, A > 0 y 1 y = Aet, y > 1, A < 0 Depending on our initial value y(0) = y 0 we will decide which solution to use. Also note that the separation of variables did not produce for us the equilibrium solutions, y(t) = 0 and y(t) = 1. These must be found as before by direct substitution after determining the critical points of f(y). To solve explicitly for the solution plug in the initial condition to find, y 0 1 y 0 = A Plugging back into the solution and doing some algebra you should find, y(t) = y 0 y 0 + (1 y 0 )e t 10

20 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 You can check the long-time asymptotics of the solution depending on your initial condition. lim y(t) = 1 y 0 > 1 t lim y(t) = 1 0 < y 0 < 1 t Moreover, if y 0 < 0 you can check that y(t) approaches in finite time. In particular, lim y(t) = t y 0 1 y 0 This is an example of a nonlinear affect. Solutions can blow up in finite time. Moreover we should point out that we have always assumed that t 0. If we allow for negative time in our problem then, we can show that for y 0 > 1, lim y(t) = t y 0 1 y 0 Hence sometimes the modeling application gives us a natural domain in which to consider the problem, and other times we have to be a bit more pendatic to identify the domain of our problem. Let us now try to solve a separable differential equation which is nonautonomous. Example 3.3. Use separation of variables to solve, dy dx = x2 1 + y 2 We rearrange to find, (1 + y 2 ) dy dx = x2 Integrate both sides with respect to x, to find, (1 + y 2 ) dy dx dx = x 2 dx Which implies, We find that for an arbitrary constant C, defines our implicit solution. y + y3 3 + C = x3 3 + C y + y3 3 x3 3 + C = 0 11

21 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 4 Suggested Problems Reference: Sections 1.1, 1.2, 2.2, 2.5 Question 5. Consider for constants r, k the first order autonomous equation, y (t) = ry(t) + k a) Use the direction field plot to understand solutions to this equation when: i) r, k > 0 (Applications: Economic and Financial Models regarding Capital and Labor) ii) r > 0 and k < 0 (Applications: Basic Population Growth and Predation Model, Basic RC Circuit) iii) r < 0 and k > 0 (Applications: Newton s Law of Cooling, Letting a ball fall with air resistance, Mixing) iv) r, k < 0 (Applications: Throwing a ball in the air against air resistance) b) Use separation of variables to find the explicit solutions to the differential equation. Question 6. Section 1.1-1, 3, 5, 7, 14, 21b, 24b, 25bcd Question 7. Section 1.2-1, 2, 7-9, 15-17, 19 Question 8. Section 2.2-1, 3, 5, 7, 9, 12, 17, 20, 21, 25, 28, 30 Question 9. Section , 11, 13-15, 20, 22, 23 5 Appendix Here we will do the computation from the logistic example to get the desired solution. Plugging in A = y0 1 y 0 we find, Rearranging terms we get, y 1 y = y 0 1 y 0 e t y 0 y = y 0 e t y e t 1 y 0 1 y 0 Put the term with y on the other side and factor out y(t) to find, ( y 1 + y ) 0 e t = y 0 e t 1 y 0 1 y 0 Divide to find, Combine terms in the denominator to get, y(t) = y 0 1 y 0 e t ( 1 + y0 1 y 0 e t ) y(t) = y 0 1 y 0 e t ( ) 1 y0+y 0e t 1 y 0 12

22 First Order Equations I Math ODE for Engineers Lecture 1-1/16/18 This simplifies to, y(t) = y 0 e t 1 y 0 + y 0 e t Now just factor out a e t term from the denominator to find, y(t) = y 0 e t (1 y 0 )e t + y 0 1 e t This gives us the desired solution, y 0 y(t) = (1 y 0 )e t + y 0 13

23 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 Abstract In today s lecture we continued our discussion of separation of variables (end of Lecture 3) and introduced exact differential equations as well as the concept of an integrating factor. 1 Exact Differential Equations Today we continue our discussion of first order differential equations which can be solved by exact methods, i.e., direct integration. So far, using the method of separation of variables, we have been able to solve first order ordinary differential equations of the form, y (t) = f(t) y (t) = g(t)h(y) The next class of equations we will solve by direct integration are called exact differential equations. In this lecture we will study differential equations of the form, M(x, y) + N(x, y) dy dx = 0 (1.1) Remark 1.1. In this lecture we consider the x variable to be the independent variable and the y variable to be the dependent variable. Hence we assume that in some interval y can be written as a function of x. It is important to understand where exact differential equations come from. The motivation for exact differential equations is rooted in the physical concept of a conservative force. We will restrict our discussion to the xy coordinate plane. Recall the spring-mass system from the first lecture. In that example the restoring force had a magnitude of kx(t) and a direction pointing opposite to the particle s displacement. Hence a physical force comes with both magnitude and direction. Thus at a fixed point away from equilibrium we assign a vector to represent the force. Moreover the direction and magnitude of the force varies depending on the displacement from the origin, meaning we assign a positive or negative direction depending on if we compress or stretch the spring-mass system and a magnitude that depends on the distance from the origin. Putting all this together we find that the restoring force can actually be modeled by an infinite collection of vectors, one at each point on the x-axis. Mathematically this collection is known as a vector field. Consider the illustration on the next page for the spring-mass system (See Figure 1). The length of the arrows signifies the magnitude of the force. Since the restoring force only points in the direction of the x-axis we can model the restoring force as, F s = ( kx, 0) where the negative sign is to represent the fact that the force works in the opposite direction to displacement. As long as we are thinking about forces in the plane, it is not hard to imagine that there are probably physical forces which depend on both the x and y coordinates of space, and also point in any direction in the plane (not just in the direction of the x-axis). Such a general force can be represented as, where M(x, y) and N(x, y) are arbitrary functions. F = (M(x, y), N(x, y)) (1.2) Question 1. Consider the force of gravity on the surface of the Earth and restrict it to an xy coordinate system (Gravity on a wall). Draw the vector field and write the mathematical expression for this force. 1

24 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 Lets consider a very simple example of a vector field, F = (x, y) We provide an llustration for this force on the next page (See Figure 2). One of the basic questions in physics to ask is if a given force is conservative. A conservative force is one which conserves total mechanical energy (potential plus kinetic) of a physical system. Examples we have already considered are the oscillating pendulum and the spring-mass system. Mathematically, a force is conservative if there exists a function V (x, y) such that V (x, y) = M(x, y) x and V (x, y) = N(x, y) y This is interesting because all the dynamics of a vector field can be generated from a single function, V (x, y), called the potential function. As will become evident through examples, the question of solving an exact differential equation is equivalent to finding a potential function for a given vector field F = (M(x, y), N(x, y)). Figure 1: Vector Field in Spring-Mass System Let us first define an exact differential equation. Figure 2: Vector Field in Example 1.3 2

25 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 Definition 1.2. Consider a nice domain in the xy plane with no holes in its inerior. Further assume all functions appearing are sufficiently differentiable. A differential equation of the form, M(x, y) + N(x, y) dy dx = 0 is called exact if and only if at each point in the domain, M(x, y) = N(x, y) y x Let us try to understand this definition through an example. Example 1.3. Consider the vector field, F = (M(x, y), N(x, y)) = (x, y) We would like to look for a potential function V (x, y) such that and V (x, y) = x x V (x, y) = y y A theorem from vector calculus tells us that for nice domains in the xy plane with no holes in its inerior the existence of a potential function is equivalent to the condition, M(x, y) = N(x, y) y x Physically this condition is telling you that the vector field F is not rotating. In the language of vector calculus we say that the vector field is curl free. So let us first check if it is even theoretically possible to construct a potential function for this vector field, y M(x, y) = y x = 0 N(x, y) = x x y = 0 Since the condition is verified, we can proceed to try and construct a potential function. To find the potential function we can integrate both equations to find, V (x, y) = V (x, y) dx = x dx = 1 x 2 x2 + g(y) V (x, y) = V (x, y) dy = y dy = 1 x 2 y2 + h(x) 3

26 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 The first integration gave us an arbitrary function of y since the x derivative of this function is 0. Similarly the second integration gave us an arbitrary function of x since the y derivative of this function is 0. Now since we are looking for only one function V (x, y), we see that we can choose V (x, y) = 1 2 x y2 To verify that this is in fact the desired potential function we can differentiate, ( 1 x 2 x2 + 1 ) 2 y2 = x and ( 1 y 2 x2 + 1 ) 2 y2 = y We now try to connect this with the definition of an exact differential equation. Assuming that the second variable y can be written as a function of x we can differentiate V (x, y) with respect to x to find by the chain rule, d V (x, y(x)) = dx x V (x, y) + dy V (x, y) y dx This evaluates in our example to, x + y dy dx How does any of this help us solve differential equations? Suppose we were asked to solve the differential equation, x + y dy dx = 0 Because of our previous work we know that the left hand side of this equation can be written as the total derivative of the potential function V (x, y). Hence the differential equation becomes, d V (x, y(x)) = 0 dx Now integrate with respect to x, d V (x, y(x)) dx = dx This implies that for an arbitrary constant C, 0 dx V (x, y(x)) = C Thus we have found an implicit solution to our differential equation, 1 2 x y2 = C Remark 1.4. We should pause to point out what we have discovered in this example. Suppose the vector field in the previous example models the velocity of fluid flow in a pond. The flow is radially outwards from the origin with increasing magnitude. If you were to place a leaf in this pond, it would flow radially with 4

27 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 increasing velocity. On the other hand the integral curves of the potential function are concentric circles. Hence we find that the field lines of the force and the integral curves of the potential function intersect at right angles. This is a general phenomenon for the field lines of conservative forces and the integral curves of its potential function. This is an example of orthognal trajectories. Moreover the integral curves for the potential function represent what trajectory a particle should take if it desries to keep the same potential energy. Question 2. Consider gravity on a wall and solve for the integral field lines for the gravitational potential energy Now that we have motivated the definition of an exact differential equation, lets concentrate solely on solving differential equations. We will check to see if a given differential equation is exact and proceed to solve it using the methods introduced in the previous example. Example 1.5. Determine if the following differential equation is exact, and if so find the solution, (3x 2 2xy + 2) + (6y 2 x 2 + 3) dy dx = 0 To check if the differential equation is exact we first define, M(x, y) = 3x 2 2xy + 2 N(x, y) = 6y 2 x Recall that a differential equation is exact if and only if, M(x, y) = N(x, y) y x Which is true since they each equal 2x. We now proceed to construct the potential function V (x, y). Following the example from before we know that and V (x, y) = M(x, y) x V (x, y) = N(x, y) y Integrating both equations as before we find, V (x, y) = V (x, y) dx = (3x 2 2xy + 2) dx = x 3 x 2 y + 2x + g(y) x V (x, y) = V (x, y) dy = (6y 2 x 2 + 3) dy = 2y 3 yx 2 + 3y + h(x) x Notice that we can choose, This guarantees that, V (x, y) = x 3 + 2x + 2y 3 + 3y x 2 y x V (x, y) = 3x xy 5

28 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 So the differential equation can be written as, Finally we integrate in x to find the implicit solution, y V (x, y) = 6y2 + 3 x 2 d V (x, y(x)) = 0 dx x 3 + 2x + 2y 3 + 3y x 2 y = C 2 Introduction to Integrating Factors Lets consider again the differential equation, Suppose this equation is not exact, namely, M(x, y) + N(x, y) dy dx = 0 M(x, y) N(x, y) y x The question we ask ourselves is if this differential equation can be made exact. The simplest thing we could do is hope to find a function µ(x, y) so that after multiplying each term in the equation, the new differential equation becomes exact. Consider the differential equation upon multiplication by this unknown function µ(x, y), µ(x, y)m(x, y) + µ(x, y)n(x, y) dy dx = 0 Let us pause for a moment and think again about vector fields. What we are doing by introducing an integration factor is changing the underlying vector field by a weight, µ(x, y), so that the strength/magnitude of the vector field is suitably balanced to make it conservative. If you think in terms of a particle moving in this vector field, the factor µ(x, y) is changing the particles velocity through the field (by changing the underlying field strength). We would like to choose the factor so that the particle experiences a conservative vector field and hence we are influencing how fast and how slow the particle should move through the field. Take careful note that the integration factor does not change the shape of the vector fields. The condition for the new differential equation to be exact is, (µ(x, y)m(x, y)) = (µ(x, y)n(x, y)) y x To help make the computation easier to follow we use the following notation, y f(x, y) = f y x f(x, y) = f x 6

29 First Order Equations II Math ODE for Engineers Lecture 4-1/18/18 Remember that µ(x, y) is an unknown function so this condition for exactness will help us determine what function µ(x, y) to use to make the equation exact. We start by expanding both sides of the identity to get by the product rule, µ y M + µm y = µ x N + µn x Hence the function µ that makes the equation exact solves the differential equation, µ y M + µm y µ x N µn x = 0 If you have been following the computation then you know that what we have derived is in fact a partial differential equation. Moreover it can be shown that as long as M and N are reasonable functions then an integration factor µ(x, y) always exists! Hence theoretically all first order differential equations in normal form can be made exact. We will see a more precise statement later. But of course finding one in practice means solving a first order partial differential equation. One way to find a closed form solution for our ordinary differential equation is to assume that µ is a function of only the x variable. Solving for µ we find, d dx µ µ = M y N x N (2.1) Since the left hand side is only a function of x, the right hand side, M y N x N only depends on x and so we can solve the differential equation to find, µ(x) = e u(x) where d dx u = M y N x N A similar statement also holds if the expression, N x M y M only depends on the y variable. Then we can solve the differential equation, to find an integrating factor µ depending only on y, where d dy µ = N x M y M µ µ(y) = e v(y) d dy v = N x M y M Notice the reasoning works in reverse as well. If the above quotients are a function of a single variable then so is the integrating factor. Let us try an example problem. 7