Engineering Mathematics 1 Fall 2017 Lecture 1, August 21, 2017 What is a differential equation? A differential equation is an equation relating a function (known sometimes as the unknown) to some of its derivatives. Solving it means finding the unknown function. Differential equations come in two flavors, ordinary and partial. In an ordinary differential equation the unknown is a function of a singe variable. In this course we will mostly deal with ordinary differential equations (ODE s). In most cases (not always) our textbook uses the symbol y to denote the unknown function and x the independent variable. However, in a lot of situations the independent variable represents time, then it is better to denote it by t and one can use x (or y, or z, or u) for the dependent unknown. Here are some examples of differential equations: 1. y = 0. This is the absolute simplest possible differential equation. It has all the necessary ingredients: A derivative of an unknown function and an equal sign. The solution is y = φ(x) where φ(x) = C, a constant. Or, simply, y(x) = C. 2. Another simple example is y = x 2. The solution here is y = 1 3 x3 + C, where C is an arbitrary constant. 3. A trifle harder is the equation y = x. It requires two integrations; the first one gives y = 1 2 x2 +C, the second gives the solution as y = 1 6 x3 +Cx+D, where C, D are arbitrary constants. These could be called trivial differential equations because the unknown function itself appears only once; it can be solved by simple integration. But it already illustrates one peculiarity of solutions. If the equation is of order 1 (the highest derivative is the first derivative), the solution will be known up to an arbitrary constant; if the order is 2, two arbitrary constants appear in the solution. In other words, there are multiple solutions; every choice of values for the constants gives a different solution. Here is a more interesting example, showing also how some equations come up. A population of mice left alone increases at a rate that is proportional to the amount of mice present. So if the population at time t is denoted by p(t), that means that (roughly) as time changes from t to t + t the population increases by rp(t) t, where r > 0 is the rate. Unfortunately for the mice there are owls about and in a period of length T the owls consume M field mice. Assuming T = k t this means that in time t the owls consume M/k = MT t mice we get the following difference equation: p(t + t) p(t) = r t p(t) MT t. The idea is that this equation gets better and better the smaller the time period t is so, forgetting that there is no such thing as an infinitesimal mouse, we divide by t and let t 0 to get the differential equation p = rp MT.
2 In the text it is assumed that the time unit is a month, r = 0.5 and MT = 450. So the equation is p = 0.5p 450. We will see soon enough how to solve equations such as this one, so right now I will only comment that the solution is p(t) = 900 + ce t/2. The constant c is arbitrary; any c gives a solution. It can depend, for example, on the initial mouse population. For example, if at time 0 we have 1000 mice, then 1000 = p(0) = 900 + c so c = 100 and the solution is p(t) = 900 + 100e t/2. As t we have p(t) ; in other words, a mice population obeying this model will keep on growing without bounds and eventually take over the whole earth. If, on the other hand, p(0) = 800, we get 800 = 900 + c so c = 100, the solution is p(t) = 900 100e t/2 as t ;. Since mice, once there are none available, do not keep on becoming negative in numbers, the model breaks down once p(t) = 0. Solving for t in 900 100e t/2 = 0 gives t = 2 ln 9 4.39; the population of mice becomes extinct in somewhat over 4 months. The owls should then feel sorry for having eaten so many mice each day; now they might face extinction. An interesting case is if the initial population is p(0) = 900. Then we get 900 = p(0) = 900 + c so that c = 0 and p(t) 900; is constant. 900 is an equilibrium solution, a solution that does not change. As many mice are added as are being subtracted. It is an unstable equilibrium because if we start with one more mouse, p(0) = 901, the population grows without bounds; one fewer mouse, at p(0) = 899, we have extinction. Here are a few other examples of differential equations. In this course, following the textbook, assume that the unknown function is denoted by the symbol that appears differentiated. If no independent variable is mentioned explicitly, and the unknown is y, assume it is x. But if the unknown is x, assume the independent variable is t. 1. y + y 5 = x 3. 2. y = yy y. 3. y + 4 sin y = 0. 4. u + (cos t)u + e t u = u 2. How do differential equations originate? As I mention in the syllabus, when we observe nature we observe rates of change; the laws of nature, the laws of mechanics (classical and quantum), of electricity, even of biology and finance, are laws relating rates; i.e. derivatives. Here is an additional example: Newton s second law states that force equals mass times acceleration. Let us try to apply this law to describe the motion of a rocket shot straight up from the earth. We must make some simplifications; we assume that air resistance is negligible and that it is only affected (once launched) by the earth s gravity. Let y(t) be its height at time t, with y(0) = 0.
3 It s acceleration, due to the attraction of the earth is y (t) (minus because it tends to pull the rocket down). By Newton s law of universal attraction, the force that causes this acceleration is directly proportional to the product of the mass of the rocket (m) and the mass of the earth (M), which we ll consider constant (the rocket is merely launched, has no motor), and inversely proportional to the distance between the center of mass of the rocket and the earth. We assume that the rocket s size is negligible compared to the size of the earth so that this distance is y(t) + R, where R is the radius of the earth. Newton s second law now tells us that my (y) = mm (y(t) + R) 2 or simply y = M (y + R) 2. Let us get a bit more formal. Definition 1 A differential equation is an equation of the form F (x, y, y,..., y (n) ) = 0, or an equation that can be written in that form. Here F is a function of n + 1 variables. The order of the equation is the order of the highest derivative appearing in the equation. But we will always assume that we can solve for the highest order derivative and that we can write our differential equation y (n) = f(x, y,..., y (n 1) ). Can does not mean must. For example, the equation y + y + y 2 = x 3 can be written y = y y 2 + x 3 but we don t explicitly have to write
4 in such a way. On the other hand the equation (y ) 2 y = 0 will be considered as two equations; namely y = y and y = y. The following definition could be important. Recall that an open interval is a subset I of the real line of the form I = (α, β), where α < β ; that is I = (α, β) = {x : α < x < β}. In this course we will always assume α < β so as not to be dealing with the silly case of an interval with no points. Definition 2 A function y = φ(x) is a solution of the equation of order n y (n) = f(x, y,..., y (n 1) ). in the open interval I if (and only if) φ(x) is (at least) n times differentiable in I and φ (n) (x) = f(x, φ (x),..., φ (n 1) (x)) holds at all points of I. An important consideration here is that solutions are ALWAYS defined in some open interval. Consider the equation y y = 0. Let y(x) = x 4. Then y (x) = 12x 2 and at 0 we get y (0) y(0) = 0. Can we say y(x) = x 4 solves the equation at 0? NO!!. Solutions must be defined in open intervals, they come with an open interval in which they solve the equation. Now consider the function y(x) = e x. In this case y (x) = e x and y (x) y(x) = e x e x = 0 for all real values of x.we can say that y(x) = e x is a solution of this equation in the interval (, ). Another example. Consider the equation y = xy 2. Verify that 2 y(x) = solves the equation and determine the intervals in which it is a 1 x2 solution. Solving differential equations is hard; and it can be very hard. But verifying that something is a solution only requires that one know how to differentiate. In this case, writing y = 2(1 x 2 ) 1 we see that while y = 2(1 x 2 ) 2 ( 2x) = 4x(1 x 2 ) 2 xy 2 = x[2(1 x 2 ) 1 ] 2 = 4x(1 x 2 ) 2. Clearly y = xy 2 for all x for which these computations made sense. They don t make sense where y is undefined, namely for x = ±1. It follows that the function in question is a solution in the intervals (, 1), ( 1, 1) and (1, ). A partial differential equation (PDE) is an equation relating a function of several variables to its partial derivatives. Here are some examples; in each case the unknown function is u and the variables are denoted by x, y or x, y, z. 1. 2. 3. u x + u y = x2 + y 2. 2 u x 2 + 2 u y 2 + 2 u z 2 = 1. u x 2 u x y 3 u z 3 = 0.
5 In this course, in all likelihood, no further PDE a will be encountered (except by accident). An important mathematical concept is linearity. The world is sadly nonlinear but, locally, looking at it in the small, it is sort of linear. An example is the earth, we know that it is round, but we can t see that except from far away. To us, it seems flat; a plane, a linear object. Non linear phenomena are hard to study and a a favorite method is to have linear approximations. The best definition of what constitutes a linear equation is: The equation F (x, y, y,..., y (n) ) = 0 is linear if F is a linear function in the variables y, y,..., y (n). In detail: A differential equation of order 1 is linear if and only if it can be written a(x)y + b(x)y = g(x), where a(x), b(x), g(x) are given functions of x, possibly constant. A differential equation of order 2 is linear if and only if it can be written a(x)y + b(x)y + c(x)y = g(x), where a(x), b(x), c(x), g(x) are given functions of x, possibly constant. A differential equation of order 3 is linear if and only if it can be written a(x)y + b(x)y + c(x)y + d(x)y = g(x), where a(x), b(x), c(x), d(x), g(x) are given functions of x, possibly constant. In general, a differential equation of order n is linear if and only if it can be written a n (x)y (n) + a n 1 (x)y (n 1) + + a 1 (x)y + a 0 (x)y = g(x), where a k (x) for k = 0,..., n, g(x) are given functions of x, possibly constant. If in any of these equations the function g(x) equals 0 identically, the equation is said to be linear homogeneous. Which equations can be so written? The answer is that the equation must be of the form expression = expression where each of the two expression is a sum of terms, each term being either a function of x (assuming x is the independent variable), a function of x times y, or a function of x times a derivative of y. If you have a term with a factor of y 2, or yy or sin(y) or 1/y, there is no linearity. The only terms allowed are of the form h(x), h(x)y, h(x)y (k), for functions h(x) of x.