18.03 Differential Equations, Lecture Note 11

18.03 Class 11 Second order linear equations: Physical model, solutions in homogeneous case. Characteristic polynomial, distinct real roots. [1] [2] [3] [4] [5] Springs and masses Dashpots Second order linear equations Solutions in homogeneous case: Superposition I Exponential solutions: characteristic polynomial [1] Second order equations are the basis of analysis of mechanical and electrical systems. We'll build this up slowly. A spring is attached to a wall and a cart: spring mass | || |-------> || | || ___|___ || | | ||---VVVVVVV---| | || |_______| || O | O || | |-------> | x F_ext Set up the coordinate system so that at x = 0 which means that it is exerting no force. the spring is relaxed, In addition to the spring, suppose that there is another force acting on the cart -- an "external force," maybe wind blowing on a sail attached to it, maybe gravity, or some other force. Then mx" = F_spr + F_ext The spring force is characterized by the fact that it depends only on position. In fact: If If If x > 0 x = 0 x < 0 , , , F_spr(x) < 0 F_spr(x) = 0 F_spr(x) > 0 I sketched a graph of F_spr(x) as a function of x . The simplest way to model this behavior (and one which is valid in general for small x , by the tangent line approximation) is F_spr(x) = -kx k > 0 the "spring constant." "Hooke's Law" This is another example of the linear approximation that Linn was discussing on Monday. So we get mx" + kx = F_ext . I displayed a weight on a rubber band. This is not a spring, as you usually think of one, but it behaves like one, at least in a range. Lay a rubber band laid out on a table. Fix the right end of it and set where the left end is in a relaxed state, then the graph of the force exerted by the rubber band looks something like this ... x = 0 | | || .------- fastened here || | || linear spring | || | | || | | .---- end of unstretched band | \ | | | | \ /__/ | | | \ \ V V | \ | \ *=========* | \ ______________| \__________.__________ ________________ \ | \ | \ | linear spring -----> \ | \ | \ | \ | || || || || || | | | | ^ | | the rubber band breaks! ----- [2] Any real mechanical system has friction. Friction takes many forms. It is characterized by the fact that it depends on the motion of the mass. We will suppose that it depends only on the velocity of the mass and not on its position. Often the damping is controlled by a "dashpot." This is a cylinder filled with oil, that a piston moves through. Door dampers and car shock absorbers often actually work this way. Write F_dash(x'). It acts against the velocity: If If If x' > 0 , x' = 0 , x' < 0 , F_Dash(x') < 0 F_Dash(x') = 0 F_Dash(x') > 0 The simplest way to model this behavior (and one which is valid in general for small x' , by the tangent line approximation) is F_fric(x) = -bx b > 0 This is "linear damping." mx" + bx' + kx = the "damping constant." Altogether the equation is F_ext Diagrammatically: spring mass dashpot | || |-------> F_ext || || | || || ___|___ _____________ || || | | _____ | || ||---VVVVVVV---| |------|_____| |-------|| || |_______| _____________| || || O | O || || | || |-------> | x [3] A linear differential equation is one of the following form: a_n x^(n) + a_{n-1} x^(n-1) + ... + a_1 x' + a_0 = q(t) (no constant term on the left hand side!) The a_k's are the "coefficients." They may depend upon t (but not on x ). If a_n is not zero, this is "of order n." Our spring system is an example of a *second order* linear equation. (Two springs in series will give a fourth order equation.) The left hand side represents the SYSTEM, the spring/mass/dashpot system. The right hand side encodes the INPUT SIGNAL, an external force at work. The coefficients represent parameters of the system -- for example, mass, damping constant, spring constant. In general they may depend upon time: maybe the force is actually a rocket, and the fuel burns so m decreases. Maybe the spring gets softer as it ages. Maybe the honey in the dashpot gets stiffer with time. Most of the time we will assume that the coefficients are CONSTANT: the timescale of their variation is much longer than the timescale of the dynamical variable x . But the external force F_ext(t) can certainly vary (maybe sinusoidally). We can see physically that there are many solutions with fixed initial value of x , x(t_0) = x_0 . So many different solution curves of a second order equation pass through a given point. But, also on physical grounds, the initial position together with the initial vecocity determine the solution: "initial position" data for a second order equation consist of both position and velocity at a give instant. This is a version of the existence/uniqueness theorem. In an n-th order equation, you need to know n numbers: x(t_0), x'(t_0), ... x^(n-1)(t_0) . [4] We'll begin with F_ext = 0 : the system is allowed to evolve on its own, without outside interference. mx" + bx' + kx = 0 , m nonzero (*) Think of a saloon door swinging. This is a *homogeneous* equation. Special case (from recitation): the harmonic oscillator: b = 0, k > 0 : x" + (k/m) x = 0 Solutions come from simple facts about sines and cosines: cos(omega t) ---> - omega sin(omega t) ---> - omega^2 cos(omega t) sin(omega t) ---> omega cos(omega t) ---> - omega^2 sin(omega t) So if omega^2 = k/m , so that our equation is then x = cos(omega t) and x" + omega^2 x = 0 , x = sin(omega t) are solutions: ANOTHER FUNDAMENTAL FACT TO MEMORIZE! In fact, you showed that any sinusoid of circular frequency omega, x = a cos(omega t) + b sin(omega t) = A cos(omega t - phi) is also a solution. In fact these are the only solutions, because x(0) = a x'(0) = omega b and so you can solve (uniquely) for a and initial condition: you don't need more. Q11.1. b to give any desired What is the period of a nonzero solution of 1. Depends upon the solution 2. 2 3. pi 4. 4 5. 2pi 6. pi/2 Blank. Don't know. x" + 4x = 0 ? (*) Ans: : think of what t has to do to take Or use P = 2pi/omega, with omega = 2 . The fact that cos(omega t) and (*) is also a solution, via Superposition I: If x_1 and linear equation, then so is x = (2t) from 0 Ans: 3: pi. sin(omega t) x_2 to 2pi. are solutions implies that are two solutions of the homogeneous c_1 x_1 + c_2 x_2 Check (in second order case mx" + bx' + kx = 0 ): m(c_1x_1 + c_2x_2)" + b(c_1x_1 + c_2x_2) + k(c_1x_1 + c_2x_2) = m(c_1x_1" + c_2x_2") + b(c_1x_1' + c_2x_2') + k(c_1x_1 + c_2x_2) = c_1(mx_1" + bx_1' + kx_1) + c_2(mx_2" + bx_2' + kx_2) = c_1 ( 0 ) + c_2 ( 0 ) = 0 . [5] The equation mx" + bx' + kx = 0 , for m, b, k constant, is a lot like x' + kx = 0 , which has as solution x = e^{-kt} (and more generally multiples of this). It makes sense to try for exponential solutions of (*): e^{rt} for some as yet undetermined constant and r . To see which r might work, plug x = e^{rt} into (*). Organize the calculation: the k] , b] , m] are flags indicating that I should multiply the corresponding line by this number. k ] b ] m ] 0 = x = c e^{rt} x' = c r e^{rt} x" = c r^2 e^{rt} ___________________ mx" + bx' + kx = c ( mr^2 + br + k ) e^{rt} An exponential is never zero, so we can cancel to see that c e^{rt} solution to (*) for any c exactly when r is a root of the "characteristic polynomial" p(s) Example. = is a ms^2 + bs + k x" + 5x' + 4x = 0 The characteristic polynomial s^2 + 5s + 4 I wanted to write out the polynomial was to roots by factoring it. This one factors as so the roots are r = -1 and r = -4 . The solutions are e^{-t} and e^{-4t} . By superposition, the "linear combination" . We want the roots. One reason remember that you can find (s + 1)(s + 4) corresponding exponential x(t) = c_1 e^{-t} + c_2 e^{-4t} is a solution as well. This is the general solution. Suppose we know also that x(0) = 2 condition we'll need to know and x'(0) = - 5 . To use the second x(t) = - c_1 e^{-t} - 4 c_2 e^{-4t} Then 2 = c_1 + c_2 - 5 = - c_1 - 4 c_2 -------------------3 = - 3 c_2 so c_2 = 1 and then c_1 = 1: x(t) = e^{-t} + e^{-4t} General picture: A linear equation of degree n a_n x^(n) + ... + a_1 x' + a_0 x = q(t) with constant coefficients has a characteristic polynomial, p(s) = a_n s^n + ... + a_1 s + a_0 e^{rt} is a solution if and only if r is a root of p(s) : p(r) = 0 . By superposition, any linear combination of these exponentials is also a solution.