Math 1270 Honors ODE I Fall, 2008 Class notes # 1

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1 Math 1270 Honors ODE I Fall, 2008 Class notes # 1 Tentative Schedule for the next three weeks Week 1 1. Dene \initial value problems", and give examples where we introduce an important method for solving some ode's, called \separation of variables". 2. Discuss dierent types of ode's, and which kinds can be solved and which cannot. 3. Give a second method for solving some simple ode's 4. Introduce a graphical method for studying ode's, called the \Direction eld", with examples. 5. If there is still time left, move on to a third solution method, for some more dicult equations. Week 2. What do we do when no solution method works (which is most of the time)? We begin our use of mathematical analysis to prove rigorously that many initial value problems have a unique solution, even if we cannot nd a formula for the solution. Completing this task will probably take about three class days. Week 3. We return to the subject of \exact solutions", continuing with an important class of equations where there are many physical applications. We will discuss several of these as time permits, and also try to show how, by restricting ourselves to equations which can be explicitly solved, we are making important physical simplications. 1

2 1 Initial value problems An initial value problem consists of two parts: (i) An ordinary dierential equation, involving one or more derivatives of an unknown function, which for the moment we will denote by y (t). 1 Example: Often we would write this as either y 0 (t) = e y(t) : (1) y 0 = e y ; or dy dt = ey : (2) (ii) An initial condition, which xes the value of y (t) at some particular value of t: Example: y (0) = 3: (3) The goal is to nd a specic function y (t) satisfying both equations. We want y 0 (t) = e y(t) for every value of t (or at least for an interval of values of t that includes the starting point t = 0), and we want the initial condition (3) to be satised. We are lucky in this case, because it is possible to nd a specic formula for such a function. This can be done for this example by the method of \separation of variables.", which is the most important method for solving rst order ode's (those involving only a rst derivative of the unknown function). Separation of variables makes use of the chain rule. Since e y(t) cannot equal zero, we can divide the equation (1) by e y(t) : y 0 (t) e y(t) = 1: (4) Here is where the chain rule comes in. We have the simple integration formula Z e y dy = e y : 1 As remarked in the notes on uniform convergence, it is more correct to refer to the function as \y", because \y (t)" denotes the value of the function y at the number t. But we sometimes refer to the \function y (t)" to emphasize that in this section, the independent variable will usually be denoted by \t ". 2

3 Therefore, using the chain rule, d dt e y(t) = e y(t) y 0 (t) : This is exactly the left side of the equation 4. Hence, nding the indenite integral of both sides of (4) ; and remembering to include a constant of integration, we get e y(t) = t + c: We only need "c" on the right. Why? We can evaluate c from the initial condition, by setting t = 0 and y (t) = 3. get e 3 = c: Hence We e y(t) = t e 3 : (5) There remains to solve for y if possible. (If it is not possible, we say that (5) gives an \implicit solution". In many cases, this is an acceptable answer.) We can solve for y in this case: y = log e 3 t y (t) = log e 3 t 1 = log : (6) e 3 t It is always good to check our answer. Dierentiating (6), we get, y 0 (t) = while also from (6) we see that 1 e 3 t ( 1) = 1 e 3 t (7) e y(t) = e log(e 3 t) = 1 e 3 t : (8) Getting the same result in (7) and (8) is what we mean when we say that y (t) is a solution of (1). We should also use (6) to check the initial condition: so (3) is also satised. y (0) = log 1 e 3 = 3; 3

4 Now we examine our solution more closely. We see, for example, that the solution in (6) doesn't work for all values of t. The equation (6) for y only makes sense if t < 1 : This includes all negative values of t. We say that the interval of existence e 3 of the solution is 1; 1 e : It is important that this interval include the starting 3 value t = 0: Here is an easier example: y 0 = y 2 y (0) = 1: This time I will use dierent notation, treating y 0 = dy dt Now substitute the initial condition: so c = 1; and the nal solution is y (t) = dy dt = y2 dy y = dt Z 2 Z dy y = dt 2 1 y = t + c y = 1 t + c : y (0) = 1 = 1 c ; 1 t 1 = 1 1 t : as a fraction: To be sure about the validity of this, we could repeat it using \prime" notation, as in our rst example. Also, we should check the nal answer. y (t) = 1 1 t = (1 t) 1 ; y 0 (t) = (1 t) 2 ( 1) = y (0) = 1: 1 (1 t) 2 = y (t)2 4

5 Now, the solution y = 1 appears to exist everywhere except at the single point 1 t t = 1; with the following graph: However, this is a misinterpretation of what we mean by a solution. Any solution of a dierential equation must be dierentiable, and hence continuous, in its entire interval of existence. The interval of existence of this solution is therefore ( 1; 1), and no t 1 is allowed. By denition of a solution, its graph has only one piece. We choose the piece to the left, because the time in the initial condition, t = 0; is in the interval ( 1; 1). The graph of the solution is 5

6 Later we will consider initial value problems where there is no way to nd a formula for the solution. An example was given in the syllabus: y 0 = y 3 + sin t y (0) = 0: It's not the initial condition that is the problem (it rarely is). we can't do any algebra which puts the equation into the form The problem is that y 0 (t) h (y (t)) = f (t) ; or, in dierent notation, h (y) dy = f (t) dt as we did with our rst example. No other trick (that either I or Maple know) works either. Hence, we can't integrate to nd even an implicit solution. For an initial value problem which can't actually be solved, we need to ask: Is there a solution at all? I can indeed give an initial value problem where there is no solution: y 0 = 1 y y (0) = 0: This is pretty obvious. But how do we draw the line between those problems which have solutions and those which do not? This is what we will discuss next week. 2 Classication of equations It is important to be able to recognize when we can solve an equation, at least implicitly, and when we can't, and if we can solve it, to know what method will work. For this reason, we put equations into various categories, to help us decide how to approach them. Here are the two main distinctions: 1. First order, second order, n 0 th order. The \order" of the equation refers to the highest derivative which appears. Equation (1) is rst order. The equation y y 0 + y = 0 is second order. The equation y (iv) = y 3 is fourth order, etc. 6

7 2. Linear or nonlinear. The \linearity" of a dierential equation refers to how it depends on y; y 0 ; y 00 ; etc., not to how it depends on t. are both linear, while are all nonlinear. y 00 + t 3 y 0 + y = 0, and y 0 = (cos t) y + sin t y 00 + ty 02 + y = 0 y yy y 02 = 0 y 00 + sin y = 0 y 0 = e y Within the class of linear ode's there are further important distinctions: The equation can be \homogeneous" or \nonhomogeneous": The equation is linear and homogeneous, while y 0 = t 3 y y 0 = 3y + 1 (9) is linear and nonhomogenous. You can tell if a linear ode is homogenous by determining if y (t) = 0 is a solution of the equation. Also, a linear equation can have "constant coecients" or "variable coecients." Here are two second order examples: y 00 3y 0 + 5y = 0 y 00 3ty 0 + (cos t) y = 0: I imagine you can gure out which one has "constant coecients". For nonlinear equations, dierent terminology is usually used: y 0 = e y is called \autonomous", while y 0 = e ty 7

8 is called \non-autonomous". (We are assuming that "t" is the independent variable, so y = y (t).) Now here are some rules of thumb for when we can solve and when we can't: (i) All rst and second order linear equations with constant coecients can be solved. (ii) All rst order linear equations can be solved, at least in terms of an integral. 2 (iii) Second or higher order linear ode's with variable coecients usually cannot be solved. (iv) Third or higher order linear odes with constant coecients can be solved occasionally. The diculty in solving them is doing the algebra. 3 (v) First order nonlinear ode's can sometimes be solved. For example, separation of variables might work. (vi). Second and higher order nonlinear odes can rarely be solved. 2 Example: is A solution of y 0 = p 1 + t 5 y y = e R p 1+t 5 dt Any antiderivative of p p 1 + t 5 gives a solution. Unfortunately, we can't nd any antiderivatives of 1 + t5! Still, this sort of formula can often be useful. If we use a denite integral (thus xing the constant of integration), we get something like y = e R t 0 p 1+s5 ds : (You should check by dierentiating this expression, using the chain rule and the fundamental theorem of calculus.) From this, for example, we can see that lim t!1 y (t) = 1: 3 For example, consider y y 00 + y = 0:This is hard to solve because there is no formula for solving the \algebraic" equation r 5 + 3r = 0 for r. (We will see the connection between these a little later.) 8

9 3 Solving rst order linear equations This is described in section 2.2. The idea is to multiply the equation by an \integrating factor" which allows you to use the product rule for derivatives to simplify the equation. For example, with the linear nonhomogeneous equation (9), we proceed as follows: d dt y 0 = 3y + 1 y 0 3y = 1 (10) e 3t y = 3e 3t y + e 3t y 0 = e 3t (y 0 3y) ; so the integrating factor is (t) = e 3t : Multiply (10) by e 3t and the ode becomes Now integrate: e 3t y 0 = e 3t e 3t y = 3 + c: (Watch your signs, and don't forget the unknown constant, which would be determined if we had an initial condition. ) Hence, the general solution is obtained by dividing by (t): y (t) = ce 3t 1 3 : 4 Direction elds. We now turn to some more subtle questions. In chapter 1 of the text a number of important ode's are discussed. These arise in a wide variety of applied elds, particularly in physics, engineering, biology, and chemistry. In many cases, the independent variable is time, \t", and the goal is to predict future behavior of some process as a function of t: In some others, the independent variable is \x", a measure of distance, and the goal is to predict the change in some quantity as one moves in space. For ode's, one can only move in one direction, giving a single independent variable. If you want to move in (x; y; z) space, you need to study pde's { partial dierential equations. In a mathematical sense, we often do not try to make this physical distinction between time and space. Indeed, our independent variable could be something entirely dierent.. We are looking for functions which satisfy a certain equation 9 e 3t

10 involving derivatives, and we don't care what the independent variable is called. Thus, the general solution to y 0 = 2y (11) could be or y (t) = ce 2t y (x) = ce 2x These are the same function. Often we are concerned with the graphs of solutions. For this example, if c = 1 the graph is We now consider the relation of this graph to the ode. Consider the point (1; y (1)) = (1; e 2 ). The curve has a slope there, given by y 0 (1) = 2e 2. So, we can use the \point-slope" formula 4 to nd the equation of the tangent line: y e 2 = 2e 2 (t 1) ; and graph a piece of it { enough to show the slope { together with the curve: 4 y y 0 = y 0 (t 0 ) (t t 0 ) 10

11 But we could determine this slope without knowing the formula for the solution, because the nature of an ode is to give a derivative in terms of the dependent and independent variables, y and t in this case. So if we know that the solution graph passes through (1; e 2 ) ; then we can use the ode (11) to say that the slope at this point is y 0 = 2y = 2e 2 : How would we know that a solution passed through this point? our \initial condition": y (1) = e 2 : This could be What we call an initial condition need not mean the value of y at t = 0: In this case, it is the value of y at t = 1. Now suppose we graph several solutions, say for c = 1; 2; 3; 4: If we pick any point, on any of these curves, we know the direction of the tangent vector to the curve. For example, at (:75; 3e 1:5 ) ; the tangent has slope y 0 = 2y = 6e 1:5. In the following graph I indicate this slope by again plotting a small piece of the tangent line. Looking at it this way, an ode y 0 = f (t; y) can be considered a set of tangent vectors, at various points in the (t; y) plane. We can nd these vectors without knowing any solutions. Here I plot a lot of them for the example y 0 = 2y. 11

12 You should compare these two graphs, and see how, from the graph of vectors, on the left, you could sketch the graphs of solutions, as on the right. The set of tangent vectors for an ode is called a \direction eld". If you choose any particular point (t 0 ; y 0 ) as an initial condition, the tangent vector gives you the starting direction of a solution curve through that point. Read section 1.1 of the text for other examples. 5 Homework I try to assign about ve problems each week. These will include only a few examples where you try to solve ode's. But the text contains many such exercises, and you should try to do as many as you can, especially when studying for an exam! I always collect homework at the beginning of class, and then immediately go over the more interesting problems in class. For this reason, no late homework is accepted. But in computing the nal homework average, I throw out the lowest grade, such as 0 if you miss an assignment. Homework: Due at the beginning of class on Wednesday, Sept Find the solution of the initial value problem and give its interval of existence. y 0 = 1 + y 2 y (0) = 0 2. pg. 39, #16. (A similar example is in the text.) 12

13 3. Give a general formula in terms of an indenite integral for the integrating factor (t) which can be used to solve an equation of the form y 0 + p (t) y = q (t) : Then, use denite integrals to give the unique solution of y 0 + p (t) y = q (t) y (3) = 2: for general continuous functions p and q. Hint: An earlier footnote might be helpful. But for this problem, is there a better choice for the lower limit of integration? 4. pg. 47, # Follow the directions of problems 1-12, parts (a) and (b), on page 39, for the equation y 0 = y 3 + sin t There is some specialized software available for ode's. One of the most user-friendly packages for direction elds can be found at the link "phase plane applet" on my home page. For this problem you may want to use "DFIELD" on this site. A much more elaborate program, called xppaut, has been written by Professor Ermentrout in our department. If you are considering a class project involving computation, you will probably want to start looking at his program. Information can be found on his home page, at Perhaps (I am not certain), we will eventually reach the point where you can rigorously prove the observations you make to answer parts (a) and (b). But we won't be able to do so using part (c), because as stated before, there is no known formula for solutions of this equation. 13