33B Mathematics This is intended to be an outline for the entire course and a guide for studying for the final exam. I. First Order Equations (Sections 2.1,2.2, 2.4-2.9) We covered quite a bit here, but it really all comes under the following topics: i) Solving y + a(t)y = g(t) for the unknown function y(t) (Section 2.4). ii) Exact equations in general, and y = f (t)h(y) in particular (Section 2.2 and 2.6). iii) Asymptotic stability and instability of solutions to autonomous equations (Section 2.9). iv) Learning to use these differential equations in simple applications (Section 2.5, 3.1 and 3.3). v) The dreaded existence and uniqueness theory (Section 2.1 and 2.7). With regard to (2.4) you should definitely know how to explain that y + a(t)y = g(t) is equivalent to (eA(t) y) = eA(t) g(t), where A(t) = a(t)dt is any anti-derivative of a(t), as I asked some of you to do on the first hour exam. For y = f (t)h(y) be on the lookout for values of y where h(y) = 0. Remember, if h(y0 ) = 0, then the constant function y(x) = y0 will be a solution that you will NOT find by the standard method of reducing the equation to (h(y))-1 dy = f (t)dt. There are some basic facts from Section 2.6 which you need to know: a) The statement that y(x) is a solution to y = f (x, y) is equivalent to the statement that the differential form = (x, y)dy-(x, y)f (x, y)dx is zero on y = y(x). When = (F/x)dx + (F/x)dy, then the level curves F (x, y(x)) = C are solutions to y = f (x, y). b) The differential form = P dx + Qdy equals (F/x)dx + (F/x)dy for some function F on a rectangle if and only if P/y = Q/x. When this happens, is called "exact". c) When is exact, you can find F in the form F = P dx + h(y) or F = Qdy + g(x),
where you have to choose h so that F/y = Q, or you have to choose g so that F/x = P . On the first hour exam I asked some of you to explain why that works. I might ask again.
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d) If is not exact, you can always multiply it by an integrating factor (x, y) to make it exact. Unfortunately, finding is usually very difficult. It is not difficult to look for integrating factors which just depend on one variable, and I might ask you to do that. For all first order equations remember that two solutions of the same equation "cannot cross" each other, meaning if two solutions agree at a point, y(t1 ) = z(t1 ), then they must agree on their whole interval of existence, that means y(t) z(t). This follows from the uniqueness part of the existence and uniqueness theorem, and it can fail when the hypotheses of that theorem are not satisfied. That happens in examples like y = y 1/3 with y(t0 ) = 0. The proof of the existence and uniqueness theorem is well beyond the scope of this course (even though the proof of the uniqueness half is posted on the home-page), but you should know what it says. II. Second Order Equations i) General existence and uniqueness theory for second order equations the Wronskian and all that (Section 4.1). ii) Constant Coefficient Homogeneous Equations y + py + qy = 0 (Section 4.3). iii) Constant Coefficient Inhomogeneous Equations of the form y + py + qy = f (t), where f is a sum of terms of the form P (t)ert with P (t) a polynomial. (Section 4.5). iv) General Linear Inhomogeneous Second Order Equations y +p(t)y +q(t)y = g(t) (Section 4.6). v) Applications of Second Order Equations to Vibration Problems (Sections 4.4 and 4.7). Topic ii) is very straightforward: the roots r1 and r2 of the characteristic polynomial 2 + p + q tell you everything. Remember, since we always assume that p and q are real numbers, there are three cases: real distinct roots, roots 1 = a + ib and 2 = a - ib, and the double real root case 1 = 2 . To see if you really understand this apply it to solve y = 0. Topic i) is not so straightforward. Let me try to organize it a little bit. There are really just three main points here: a) The existence and uniqueness theorem for second order linear equations (Theorem 1.1.7, p.140). Know what that says.
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b) The following simple, but sometimes overlooked, fact: if y1 +p(t)y1 +q(t)y1 = g1 (t) and y2 + p(t)y2 + q(t)y2 = g2 (t), then for any constants c1 and c2 , the function y = c1 y1 + c2 y2 will satisfy the equation y + p(t)y + q(t)y = c1 g1 (t) + c2 g2 (t). This has turned up in lots of places, sometimes with g1 (t) = g2 (t) = g(t) and many times with g1 (t) = g2 (t) = 0. It is the essence of linearity. Learn the simple computation which is the proof of this fact.
c) The form of the general solution. The basic result is this: given two solutions y1 (t) and y2 (t) to y + p(t)y + q(t)y = 0, you can solve the initial value problem
y + p(t)y + q(t)y = 0, y(t0 ) = y0 , y (t0 ) = y0 with y(t) = c1 y1 (t)+c2 y2 (t) for every choice of y0 and y0 if and only if the Wronskian
W (t0 ) = det
y1 (t0 ) y2 (t0 ) y1 (t0 ) y2 (t0 )
= 0.
This is just a fact about simultaneous linear equations that really has nothing to do with differential equations. However, there is one surprise: the Wronskian W (t) is either never zero or always zero on any interval, I = {a < t < b}, where p(t) is continuous. That comes from the formula, W (t) = C exp(- p(t)dt) (see Prop. 1.26, pp.142-143). When the Wronskian is not zero we say that {y1 , y2 } is a fundamental set of solutions. Moreover, when the Wronskian is zero, there is either a constant C such that y1 (t) = Cy2 (t) for all t I or y2 (t) 0 on I. So this gives you an easy way to check whether W (t) 0 without even differentiating y1 and y2 . One point that should be clarified here: in linear algebra one usually says that two functions, y1 (t) and y2 (t), are linearly dependent on I if there are constants c1 and c2 not both zero such that c1 y1 (t) + c2 y2 (t) 0 on I. I usually stated that as either y1 (t) Cy2 (t) or y2 (t) 0 on I. It is worth thinking about that enough to see that these two definitions amount to the same thing. Topic iii) is the method of undetermined coefficients. The underlying idea here has not received the attention it deserves because it has been easier just to give you rules. What is really going on is this: you have to begin with a set of linear combinations of functions such that derivative any of a function in the set will be in the set, too. There are not a lot of sets like that. Some of the simplest ones are: {A + Bt + Ct2 }, {A cos bt + B sin bt}, and {Aert }. More generally, for each choice of a, b and n you have the set {P (t)eat cos bt + Q(t)eat sin bt, P (t) and Q(t) general polynomials of degree n}. In most cases all you have to do is pick the smallest set of this form that includes the inhomogeneous term in the equation. Unfortunately, when r is a simple root (or a double root) of the characteristic polynomial, you have to pick P (t) and Q(t) to have degrees one (or two) higher than the polynomials in the inhomogeneous term. The best way to get good at this is to practice.
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Topic iv) is the method of variation of parameters. The idea here is a little odd. You look for a solution to the inhomogeneous equation in the form y(t) = c1 (t)y1 (t) + c2 (t)y2 (t), where y1 and y2 are a fundamental set of solutions for the homogeneous equation, but then you need to add the condition c (t)y1 (t) + c (t)y2 (t) = 0 1 2 The point of that condition is that it simplifies the equations that c1 and c2 will have to satisfy if y is a solution. The next step is simply to substitute y = c1 y1 + c2 y2 in the equation y + p(t)y + q(t)y = g(t) (1). Using y1 and y2 are solutions of the homogeneous equation and the condition c (t)y1 (t) + c (t)y2 (t) = 0, 2 1 you find that (1) will hold if
c (t)y1 (t) + c (t)y2 (t) = g(t). 1 2
So you have to solve the system of equations c (t)y1 (t) + c (t)y2 (t) = 0 1 2
c (t)y1 (t) + c (t)y2 (t) = g(t), 1 2
to find c and c . You will always be able to solve it because the Wronskian of 1 2 y1 and y2 is not zero. The method of variation of parameters is more useful as a way of finding integral formulas for solutions, than as a way of solving equations explicitly. Topic v) will not be on the final exam ... explicitly. Just as on the second hour exam I will not give you questions where you need to convert units or set up differential equations for the motion of vibrating systems. However, the equations themselves might turn up, and you might be asked to solve them using undetermined coefficients.
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III First Order Systems of Equations. Before discussing Chapter 9 I'd like to mention some basic stuff. i) Any single equation y (n) + an-1 (t)y (n-1) + + a1 (t)y + a0 (t)y = g(t) can be written as a first order system of n equations by simply defining x1 = y, x2 = y , . . . , xn = y (n-1) . Then the system looks like x = x2 , x = x3 , . . . , x 1 2 n-1 = xn , x = -an-1 (t)xn - - a1 (t)x2 - a0 (t)x1 - g(t). If the orginal equation was n inhomogeneous (g = 0), the system is inhomogeneous. If the original equation was homogeneous (g = 0), the system will be homogeneous. This is particularly relevant for equations of order two. For the system
y1 = ay1 + by2 y2 = cy1 + dy2 it is sometimes useful to go the other way, substituting y2 = (y1 - ay1 )/b if b = 0 (or y1 = (y2 - dy2 )/c if c = 0), to get a single second order equation for y1 (or y2 ). That's the best thing to do when the characteristic polynomial has a double root and there is only one eigenvector.
ii) The basic existence and uniqueness theory. The existence and uniqueness theorem for first order systems takes the form: if A(t) is an n n matrix with entries that are continuous functions of t on the interval I and g(t) is an n-component vector of continuous functions on I, then, given t0 in I, for every n-component vector x0 , there is a unique vector-valued function x(t) such that x = A(t)x + g(t) and x(t0 ) = x0 . I hope that looks a lot like Theorem 1.1.7 back in Section 4.1 (p. 140). I'll only expect you to know the material in Section 4.1, but this version is important. iii) Constant Coefficient Systems. [Note that, since the coefficients are constant, they are continuous on the whole real line, and the interval I in the existence theorem is just - < t < .] Assume that the n n matrix A has nonzero eigenvectors v1 , . . . , vn , Avj = rj vj , j = 1, . . . , n, and rj = rk when j and k are different. In this case the vj 's are linearly independent (see "Matrices with distinct eigenvalues" on the class homepage ) and xj (t) = erj t vj , j = 1, . . . , n, is a fundamental set of solutions to x = Ax. Until the very last class (last Monday), we only considered n = 2, and there will not be much on the final about higher dimensional systems. Please note that this does not mean that there will be nothing on the final about higher dimensional systems (Section 9.5).
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Chapter 9 The sections 9.1-9.4 on two dimensional systems should be fairly fresh in your minds, since we began this topic February 21. One additional remark (from the very last class, March 16): variation of parameters for first order systems of equations is actually easier than variation of parameters for second order equations. You just look for solutions to the inhomogeneous equation x = A(t)x + g(t) in the form | x(t) = c1 (t)x1 (t) + + xn (t) = x1 (t) | c1 (t) | xn (t) | , | cn (t)
where x1 , . . . , xn is a fundamental set of solutions for the homogeneous equation. Substituting that into the differential equation, you find that it solves the inhomogeneous equation when c (t) | 1 | = x1 (t) c (t) | n -1 | g1 (t) xn (t) | . | gn (t)
As in Section 4.6 this will lead you to integral formulas for solutions of the inhomogeneous equation. When you can do the integrals, the answers will be explicit formulas for the solutions. Notice that the mysterious "extra" equation c y1 + c y2 = 0 1 2 is not longer there when you write second order equations as first order systems! Finally ... Section 10.1 This is the payoff for the detailed analysis of 2 by 2 linear systems in Chapter 9. The interesting systems are not constant coefficient linear, but, if you know how the solve those, you can analyze nonlinear systems near equilibrium points, and often guess what the complete phase plane portrait of a nonlinear system has to be. The only new technique that you need here is linearization of a system at an equilibrium point. Be sure that you know how to do that. THAT'S ALL! BEST OF LUCK ON THE FINAL!