Second order differential equations 4

Transcription

1 A course in differential equations Chapter 3. Second order differential equations A second order differential equation is of the form y 00 = F (t; y; y 0 ) where y = y(t). We shall often think of t as parametrizing time, y position. In this case the differential equation asserts that at a given moment the acceleration is a function of time, position, and velocity. Such equations arise often in physical situations, as a rendering of Newton s Law F = ma. Second order equations are complicated enough that this chapter is divided into three parts. We shall begin by looking at a physical example, using this as motivation for the second section of a more formal nature. In the third section we shall again take up a large class of questions motivated by physics. Before we begin this scheme, however, we point out one simple basic fact. For first order equations, a solution is completely determined by its value y(t 0 ) at any given moment. For second order equations we have this analogous assertion: In all normal circumstances, given t 0, y 0, v 0 any second order differential equation possesses a unique solution y(t) with y(t 0 )=y 0 y 0 (t 0 )=v 0 : In other words, we must specify two initial conditions to solve an equation completely. In physics, this is the familiar principle that in order to describe how a system evolves we must specify (i) a set of state variables which describe the system completely; (ii) a physical law telling how the state variables evolve in time; and (iii) values of those variables at some moment. For a simple system the state variables are position and velocity. The way they change is governed by the differential equation since in a very short time interval of length dt the position changes from y to y + v dt and velocity from v to v + a dt where acceleration a is determined by the formula a = F (t; y; v) : Part I. Introduction Let s begin with an example to give us some intuition about what the theory can do for us. 1. The pendulum Here is an example we shall consider often, sometimes with a few variations. A pendulum of length ` is swinging on a fixed pivot. The mass of its bob is m. The bob is pulled away from the dead vertical position and released. Describe the subsequent motion.

2 Second order differential equations mg sin ` m mg cos mg The first step is to write down what Newton s Second Law asserts in these circumstances. Gravity pulls downward on the bob, but if we assume that the shaft is rigid the component exerted along the shaft will have no effect. The only effect of gravity will therefore be to accelerate the bob in the direction along the periphery of the swing (tangential to the arc of motion). This component has magnitude jmg sin j if is the angular displacement of the pendulum from equilibrium. If we let y be the linear distance along the arc of the pendulum then y=` =. Since the force acts to decrease the angular displacement, the tangential force is precisely mg sin There will be another force exerted on the bob, namely that of friction, but we will ignore this for the moment. Newton s Law now tells us that my 00 ỳ = mg sin or y 00 = ỳ : ỳ g sin : We therefore obtain a second order differential equation as a direct translation of Newton s Law into mathematics, no more and no less. The differential equation, as I have mentioned before, tells us how the position and velocity change instantaneously. It is not a particularly complicated equation, but nonetheless neither Newton s Law nor the differential equation tell us directly how to describe the motion explicitly. That can be, as it is here, a difficult problem. The problem also tells us something else that is necessary to describe the motion. Let 0 be the initial angle of displacement. The bob is released from rest at that displacement, so the initial velocity is 0. In other words we have the initial conditions at time t =0 y(0) = ` 0 ; y 0 (0) = 0 : Here, as for all second order differential equations, motion subsequent to any moment is completely determined by position and velocity at that moment.. A simple method of numerical solution The physics of the example suggests a simple way to figure out how to estimate the motion. Suppose at a certain time t the position is y and the velocity v. The acceleration at this moment will then be a = g sin(y=`). After a short time interval t the position will be approximately y + v t, the new velocity v + a t. And so on for further intervals. If we start at time t 0 with position y 0 and velocity v 0 we therefore carry out the following procedure to estimate positions at subsequent times.

3 Second order differential equations 3 First we choose a small interval of time t. We will estimate the position and velocity at discrete times t 1 = t 0 +t, t =t 0 +t, etc. Suppose that we have estimates y n, v n for position and velocity at time t n = t 0 + n t. The acceleration at this moment will be a n = g sin(y n =`) : Then at time t n+1 = t n +twe obtain estimates y n+1 = y n + v n t v n+1 = v n + a n t: This is the analogue of Euler s method for second order equations. For a more general second order differential equation y 00 = F (t; y; v) the only thing that changes is the formula for evaluating acceleration: a n = F (t n ;y n ;v n ): All of the earlier methods we saw for approximating the solutions of first order differential equations can in fact be applied to second order as well. Again here, this technique is simple to carry out in a spread sheet. Here is a table of how things go with an initial angular displacement of = and a time interval t =0:1. t y v a We get more accurate estimates if we make the step size t smaller. Here are the graphs of solutions I get in this way with various initial displacements and very small step size.

4 Second order differential equations 4 = = t! Figure.1. Graphs of pendulum displacement. Conservation of energy requires that the motion be periodic. The period lengthens as the initial angular displacement approaches, and for small initial displacements the period tends to a fixed size. We shall see later what the limiting period is. What we have seen here is the extension of Euler s method to second order equations. The improved Euler s and Runge-Kutta methods can also be extended to second order equations, but for technical reasons we shall do this in the Chapter on systems. Part II. Linear equations If x is small then sin x is approximately equal to x. Therefore if the pendulum displacement y is small sin(y=`) : = y=` and the differential equation for the pendulum can be written approximately as y 00 = gy=` : This equation is linear in y, and is called a linear differential equation. equation (of second order) is one of the form More generally, a linear differential y 00 = a(t)y 0 + b(t)y + f (t) : Linear differential equations play an important role in the general theory of differential equations because, as we have just seen for the pendulum Differential equations can be approximated near equilibrium by linear ones.

5 Second order differential equations 5 We saw this also in the case of the first order differential equation describing how height changed in a filling tank, where solutions to a linear equation gave a good approximation near the equilibrium height. For higher order equations this sort of thing becomes even more important, because they are far more difficult to solve explicitly, and we need all the help we can get. 3. Linearity We saw that for arbitrary first order linear equations there exists an integral formula y 0 = a(t)y + f (t) y = Ce A(x) + e A(x) Z x e A(t) f(t)dt for the general solution. For second order equations we shall see a somewhat similar formula, although it will not turn out to be quite so explicit. The overall structure of the formula will be the same. Recall that the first term in the first order formula is the general solution of the homogeneous equation where f (t) vanishes, and the constant C is there because if y(t) is a solution to a first order linear equation then so is Cy(t). Something similar happens for arbitrary homogeneous linear differential equations. The expression y 00 + a(x)y 0 + b(x)y may be thought of as a kind of operator: it transforms a function y(x) into another function of x. For example, the operator y 7! y 0 transforms y into its first derivative. The point here is that this is a linear operator. I will explain this formally in just a moment, but first let me give linearity some intuitive content by mentioning that the concept makes sense whenever we can talk about quantitative input and output. Here the input is y and the output is what we get when we transform y by the operator. Linearity in this context encapsulates two features: Scalability. Output is proportional to input. Thus if input is doubled, so is output. Superposition. If we add together two inputs, we get out the sum of their respective outputs. This means that distinct inputs have an essentially independent effect. The reason linearity is an important concept is that Many physical systems have essentially linear response near equilibrium. We shall see why this is true in the next section. Linearity in a physical structure, amounts to a generalization of Hooke s Law, which asserts that stress (output) is proportional to strain (input). This is scalability. Superposition is just an extension of this to complicated systems where stress and strain have several components. In our context, a linear operator is an operator y 7! Ly that satisfies these two conditions: If c is a constant then L(cy) =cl(y). If y 1 and y are two functions then L(y 1 + y )=L(y 1 )+L(y ). These two principles can be combined into one: If c 1 and c are constants, y 1 and y functions then L(c 1 y 1 + c y )=c 1 L(y 1 )+c L(y ). In this course there are several linear operators: Linear differential operators which take the function y to a linear combination of its derivatives whose coefficients are functions of x: y 7! c 0 (x)y + c 1 (x)y 0 + +c n (x)y (n) :

6 Second order differential equations 6 Any integral expression y 7! Z x x0 c(t)y(t) dt : We shall need these simple properties of linear operators: If L(y 1 )=0and L(y )=0and c 1 and c are constants then L(c 1 y 1 + c y )=0. If L(y 1 )=f 1 and L (y )=f and then L(y 1 + y )=f 1 +f. These are straightforward and formal to verify. A linear differential equation y 00 + a(x)y 0 + b(x)y = c(x) is said to be homogeneous when the right hand side c(x) is 0. Linearity implies: Linear combinations of solutions to a linear homogeneous differential equation are again solutions of the same equation. If y 1 and y are solutions of the differential equation the y 1 y 00 + a(x)y 0 + b(x)y(x) =c(x) y is a solution of the associated homogeneous equation. If we have found one solution y of the differential equation y 00 + a(x)y 0 + b(x)y(x) =c(x) then every solution of the same equation is of the form y + z where z is a solution of the associated homogeneous equation. The second assertion is true because The last is just a rephrasing of the second. L(y 1 y )=L(y 1 ) L(y )=c(x) c(x)=0: An example of the way things work is the basic formula for solutions of a linear first order differential equation. In that formula Ce A(x) is the general solution of the associated homogeneous equation, and the integral is a particular solution of the inhomogeneous one. There is a practical effect of these considerations on the structure of solutions to a linear equation. We have seen that solutions of an equation of second order depend on two parameters, the position and velocity at any given moment. This and linearity have the consequence that the solutions of a second order linear homogeneous equation are in some sense a two dimensional vector space. In the next section we shall do this more precisely in terms of a possible coordinate system on the set of solutions. 4. Why linearity is ubiquitous Linearity is ubiquitous for a very simple purely mathematical reason. Suppose we are considering a system with one degree of freedom in an equilibrium state, and suppose for the moment that friction is negligible in the system. Suppose also that the physics of the system is not changing in time. Let y be a variable measuring displacement from equilibrium. The equilibrium position is determined by the fact that it must be where potential energy P is a minimum. By elementary calculus, we know that as a function of the variable y the first derivative of P with respect to y vanishes. The Taylor expansion of P near equilibrium then asserts that P (y) =P 0 +y P + P 0 + P y

7 Second order differential equations 7 where the terms left out are of order higher than in y. As in any potential field, the force restoring the system to equilibrium is given by the formula P y: Newton s second law then gives us the differential equation my which for small displacements y is approximately the linear equation my 00 = P y: If there is friction, its exact nature will be rather complicated, but we can approximate it in a plausible way according to the following reasoning: Friction depends only on rate of change in the system, and its magnitude does not depend on the direction of change. It acts in opposition to change. From these two facts we can deduce that the force due to friction is of the form '(v) where v = y 0 is rate of change, '( v) = '(v), and ' 0 for v 0. The function ' must be odd, in other words. If we expand it in a Taylor series around v =0we will get only odd terms, and for small values of v we will get a good linear approximation '(v) = cv with c 0. The differential equation for the system is precisely my 00 = '(y and if y and y 0 are small it is approximated by the linear equation my 00 = cy 0 P y: The total energy E stored in the linear system is the sum of kinetic and potential energies E = mv + P y + PE 0 (v = y 0 ) and if we differentiate this expression we get from the differential equation de dt = mvv0 + E yy 0 = v(my 00 + E y)= cv : This says that the energy of the system decreases only through friction losses. 5. Fundamental solutions of a homogeneous linear equation Suppose we are given a second order differential equation which is linear and homogeneous. Fix a point t 0.If y is any solution of the equation then we can assign to y as coordinates the numbers y(t 0 ) and y 0 (t 0 ), which we know to distinguish y among all solutions. This reinforces the idea that the set of solutions is two dimensional. Some solutions have simple coordinates. We can find a solution y 0 (t) of the differential equation satisfying the simple initial conditions y 0 (t 0 )=1 y 0 0 (t 0)=0

8 Second order differential equations 8 and another y 1 such that y 1 (t 0 )=0 y 0 1 (t 0)=1: If y is any other solution of the differential equation, let c 0 = y(t 0 ), c 1 = y 0 (t 0 ). Then because of linearity c 0 y 0 + c 1 y 1 will be a solution of the differential equation, and furthermore by the definition of the y i it will satisfy the same conditions at t 0 that y does. Therefore this linear combination of y 0 and y 1 must be the same as y, which means that y = c 0 y 0 + c 1 y 1 ; c 0 = y(t 0 ); c 1 = y 0 (t 0 ) : The coordinates of a solution will vary with the choice of t 0. No problem, but keep it in mind. We shall see the solutions y 0 and y 1 again. They are called the fundamental solutions of the equation determined by the point t 0. In order to help keep track of the fact that they depend on the particular value of t 0 at hand, I shall write them y 0;t 0 (t); y 1;t0 (t) : The index t 0 here is a parameter of the solutions, meaning that for each choice of t 0 we get two solutions of the equation with certain properties related to t 0. We shall see some examples in just a moment. 6. Linear equations with constant coefficients I have said above that we can write down a formula for the general solution of any linear second differential equation, but that it would not be so explicit as the formula for first order equations. In fact, it is a formula that is almost useless unless we make some special assumption about the equation. The reason for this is that the formula involves the general solution to the homogeneous equation. For this there is bad news and good news: There is no general formula for solving second order homogeneous linear differential equations. There is, however, a way to find all solutions of a linear homogeneous equation with constant coefficients. What the formula does is solve the inhomogeneous case in terms of the homogeneous case, but we are usually stopped right there. There is one large category of equations which can be solved, however those linear homogeneous equations where the coefficients a(t) and b(t) are constants. Even in this case there is much to be said, and we shall spend a fair amount of time in this course on this topic. Let s now consider such an equation, which I write in a form slightly different from earlier: y 00 + ay 0 + by =0: To get some feel for what will happen, let s look at an explicit example This can be written in the old form y 00 y =0: y 00 = y so that solutions of this differential equation are those functions y(t) whose double derivatives are equal to themselves. There is certainly one function y(t) =e t which comes to mind quickly. Another and similar function is y(t) =e t, since for this function y 0 = y, y 00 = ( y) =y. Now the solutions to any linear homogeneous equation satisfy the linearity principle, so we can obtain lots more solutions by taking linear combinations ae t + be t of the ones we already have. The converse turns out also to be true. Any solution of the differential equation y 00 y =0

9 Second order differential equations 9 can be expressed as a linear combination ae t + be t : This is another way of saying that we can assign coordinates, here the constants a and b, to every solution of the equation. The proof involves a calculation we shall repeat often. The point is to construct first the fundamental solutions associated to t = 0from the solutions e t, e t. The fundamental solution y 0 for example satisfies y 0 (0) = 1 y 0 0 (0) = 0 : We speculate that y 0 is a linear combination of e t and e t. Set y 0 = ae t + be t ; y 0 0 = aet be t : Then y 0 (0) = a + b; y 0 0 (0) = a b: We must see if we can choose a and b so that a + b =1 a b=0: This gives us two equations in the two unknowns a and b, and has the solution a =1=; b=1=: Therefore Similarly we can calculate that y 0 = y 0;0 = et + e t : y 1 = y 1;0 = et But now if we are given any solution of the differential equation we can write it as a linear combination of y 0 and y 1, according to reasoning I presented earlier, and then we can write it in turn as a linear combination of e t and e t. Much of what I have said here applies to more general linear differential equations with constant coefficients. The first step is to realize that solving such equations always involves exponential functions of some kind. To see why this should be so, and to understand why exponential functions arise in this context, is important and interesting in its own right. Any homogeneous linear differential equation with constant coefficients has at least one solution of the form for a suitable choice of. y = e t In solving the equation, we will therefore simply have to find what has to be. It turns out that this reduces to a simple problem in algebra. But before I discuss the algebra involved, I will discuss in the next section why this very general result is true. This discussion will illustrate an important way of thinking about linear differential equations with constant coefficients, and introduces for them an important principle which distinguishes them from linear differential equations with variable coefficients. e t :

10 Second order differential equations Time invariance and the translation principle Systems which give rise to linear second order equations with constant coefficients are time invariant. I have said that a differential equation is often the mathematical formulation of Newton s Law. The coefficients which occur are physical quantities associated to the system, and they will be constant precisely when the physical situation that the differential equation encodes is one which does not vary with time. The system is time invariant, but this is not the same as saying that the solutions of the system are time-invariant. What it does mean is that if we shift the initial conditions in time, then the solutions will likewise be shifted in time. Mathematically, a shift by time interval h means replacing a function f (t) by f (t h), and the principle of time invariance for constant coefficient equations is this: If y(t) is a solution of a linear homogeneous differential equation with constant coefficients, then for any constant h so is y(t h). Keep in mind: it is not the particular motions of a system which are time invariant, but the underlying physical law which controls its dynamics. This statement about time shifts is the way to formulate this precisely. A very closely related fact is this: If y(t) is a solution of a linear homogeneous differential equation with constant coefficients, then so is its derivative y 0 (t). The way in which these two principles are related is that y 0 y(t + h) = lim h!0 h y(t) so that the derivative of a function is the limit of a linear combination of y and its shifts. Both of these are straightforward to prove. The second, for example: Suppose that y is a solution of the linear equation y 00 + a(t)y 0 + b(t)y =0 where for the moment I don t assume a and b to be constants. Take the derivative of both sides: y a 0 (t)y 0 + a(t)y 00 + b 0 (t)y + b(t)y 0 =0: But now if we assume a(t) and b(t) are constant, then their derivatives vanish, and this becomes which means that y 0 is also a solution. (y 0 ) 00 + a(t)(y 0 ) 0 + b(t)(y 0 )=0 So we are now in the following circumstances: we have an operation y 7! y 0 on the set of all solutions of the differential equation, and the set of solutions is a two dimensional vector space. The operator is linear. This means that if we express it in terms of coordinates, we get a matrix. But now one thing we know about matrices is that they always have at least one eigenvalue. In our circumstances, this means that there exists at least one solution y of the differential equation which is taken into a multiple of itself by the differentiation operator. y 0 = y for some constant, the eigenvalue. This is a first order differential equation, and we know that it has as solution the exponential function e t. Linearity together with time invariance imply that we can find exponential functions which are solutions. 8. Finding explicit solutions

11 Second order differential equations 11 The basic idea in solving a linear homogeneous differential equation with constant coefficients y 00 + ay 0 + by =0 is to set y = e t and substitute it into the equation. We have y = e t y 0 = e t y 00 = e t and y 00 + ay 0 + by =( +a + b)e t : To obtain a solution of the equation we must set equal to a root of the characteristic equation + a + b =0: Example. Here the characteristic equation is y 00 y =0 =1 and we get the same solutions we got before. e t ; e t Example. The characteristic equation is y 00 3y 0 +y =0 3 +=0 with roots =, 1and we get solutions e t ; e t The simplest case is when the characteristic equation has two distinct roots 1 solution to the equation is c 1 e 1t + c e t : 6=. Then the general Example. Characteristic equation y 00 + y =0 +1=0

12 Second order differential equations 1 Here the roots are = i, and we get solutions What good are these? We recall Euler s formula and write e it ; e it : e it = cos t + i sin t e it = cos t + i sin t e it = cos t i sin t: We now recall that linear combinations of solutions are solutions. Adding and then subtracting the expressions above we get solutions e it + e it = cos t e it e it =isin t Dividing by constants is also allowed, so we also get solutions e it + e it e it e it i = cos t = sin t and these are the solutions which are of interest to us. When we have complex roots of the characteristic equation we also have as solutions their real and imaginary parts, which we calculate according to Euler s formula. More explicitly, if = a + ib is a root of the characteristic equation then the general solution of the equation is c 1 e at cos bt + c e at sin bt : Example. For the roots are and the solutions are p e t= cos( 3t=); y 00 + y 0 + y =0 p p e t= sin( 3t=) Example. Here the characteristic equation is y 00 =0 =0 so we get a single root =0, and a solution y = e 0 t =1. How do we get another solution? It is trivial to see here that it is y = t.

13 Second order differential equations 13 When the characteristic equation has just a single root then the general solution to the equation is c 1 e t + c te t : When there is only a single root, the characteristic polynomial must be of the form ( c) and the differential equation looks like y 00 cy 0 +c =0: In this case you can verify explicitly that te ct does satisfy the equation. 9. Initial conditions The procedure described above gives us two different solutions of the differential equation. All other solutions will be linear combinations of these two. Solving for initial conditions amounts to solving two equations for the two unknown coefficients. Example. To find the solution of such that y(0)=0, y 0 (0) = 1 we set Then y 0 = Ae t +Be t and we get y 00 3y 0 +y =0 y = Ae t + Be t : A + B =0 A+B=1 which leads to A = 1, B =1. The solution we want is e t e t. 10. More about fundamental solutions If y 00 + a(t)y 0 + b(t)y =0 is any second order linear homogeneous differential equation, then the fundamental solutions with respect to the point t 0 are the two solutions y 0 (t), y 1 (t) such that y 0 (t 0 )=1 y 0 0 (t 0)=0 y 1 (t 0 )=0 y 0 1 (t 0)=1: I recall that since these solutions depend on the parameter t 0, we write them as y 0;t 0, y 1;t0. One reason they are useful is that if you want to solve the equation above with initial conditions y(t 0 )=y ; y 0 (t 0 )=v then you just set y = y y 0;t 0 + v y 1;t 0 : But usually it is more work to find y and y 0;t 0 1;t0 than to solve for the initial conditions directly.

14 Second order differential equations 14 Example. Take y 00 y =0; t=0: All solutions are combinations of e t and e t,soweset y = c 1 e t + c e t and solve first the system to get Next we solve to get y(0) = c 1 + c =1 y 0 (0) = c 1 c =0 y 0;0 =(1=)(e t + e t ) : y(0) = c 1 + c =0 y 0 (0) = c 1 c =1 y 1;0 =(1=)(e t e t ) : Example. For we have y 00 + y =0; t=0 y 0;0 = cos t y 1;0 = sin t: 11. Fundamental solutions and the translation principle The graph of the function y(t) when shifted to the right by t 0 is the graph of the function y(t t 0 ). This new function of x is called the translation of y by t 0. Just below, for example, are the graphs of y = t and y =(t 3). I recall the translation principle for equations with constant coefficients. If y(t) is a solution of the differential equation y 00 (t) +a(t)y 0 (t)+b(t)y(t)=0 then also for any t 0 y 00 (t t 0 )+a(t t 0 )y 0 (t t 0 )+b(t t 0 )y(t t 0 )=0: In particular, if a(t) and b(t) are constants we see that if y(t) is a solution of a differential equation with constant coefficients then so are all translations y(t t 0 ). I also recall that this is related to the principle that derivatives of solutions are solutions, since if y(t) is a solution so is y(t+h), and then according to linearity so is [y(t+h) y(t)]=h. But then y 0 y(t + h) y(t) (t) = lim : h!0 h

15 Second order differential equations 15 If z(t) =y(t coefficients t 0 )then z(t 0 ) =y(0) and z 0 (t 0 ) =y 0 (0). Therefore for differential equations with constant y 0;t 0 = y 0;0(t t 0 ) y 1;t 0 = y 1;0(t t 0 ) In other words, to find the fundamental solutions for any point t 0 we have only to find those for t 0 =0and translate them. 1. Wronskians Suppose we have found a pair of independent solutions of a second order differential equation, say z 1 and z.if we want to find the fundamental solutions for a point t = t 0 then we set y = c 1 z 1 + c z and solve to get y 0;t 0, and set y = d 1z 1 + d y and solve c 1 z 1 (t 0 )+c z (t 0 )=1 c 1 z 0 1 (t 0)+c z 0 (t 0)=0 d 1 z 1 (t 0 )+d z (t 0 )=0 d 1 z 0 1 (t 0)+d z 0 (t 0)=1 to get y. These equations say that 1;t 0 c1 d 1 c d is the matrix inverse to z1 (t W (t 0 ) z (t 0 ) 0 )= z 0 1 (t 0) z 0 (t 0) which is called the Wronskian matrix at t 0. Since this is only a matrix we know that the inverse of W (t) is W 1 (t) = 1 z 0 (t) z (t) z 0 1 (t) z 1(t) where is called the Wronskian determinant at t. In short we have: =z 1 (t)z 0 (t) z (t)z 0 1 (t) For the fundamental solutions with respect to t y 0;t 0 = c 1z 1 + c z y 1;t 0 = d 1z 1 + d z where c 1 = z 0 (t 0)= c = z 0 1 (t 0)= d 1 = z (t 0 )= d = z 1 (t 0 )= where is the Wronskian determinant at t 0. Note that this is valid even when the equation does not have constant coefficients.

16 Second order differential equations A formula for solving second order inhomogeneous equations I will now lay out a recipe for solving the inhomogeneous equation & initial conditions y 00 + a(t)y 0 + b(t)y = c(x); y(t 0 )=y 0 ;y 0 (t 0 )=v 0 : Find the fundamental solutions y 0;t 0, y 1;t0 and a formula for all y 1;s. The formula we want is Z t y(x) =y 0 y 0;t 0 + v 0 y 1;t 0 + y 1;s (t)c(s) ds : t0 Example. If we want to solve y 00 + y = c(t); y(0)=;y 0 (0) = 1 we know that y 0;0 = cos t, y 1;0 = sin t. Then y 0 = v 0 = 1 y 1;s = sin(t s) y = cos t sin t + = cos t sin t + Z t sin(t 0 Z t 0 (sin t cos s Z t s)c(s) ds cos t sin s)c(s) dt = cos t sin t + sin t c(s) cos sds cos t c(s) sin sds Z t 0 0 If you put the formula in this section together with the formulas for y 1;s in the section on Wronskians, you will arrive at a version of this formula usually called the variation of parameters formula. 14. Analogies with the formula for first order linear equations The formula for solving the first order linear equation y 0 + a(t)y = b(t); y(t 0 )=y 0 is where Z t y = y 0 e A(t) + e A(t) e A(s) b(s) ds t0 Z t A(t) = a(s)ds : t0 The function e A(t) is the solution of the homogeneous equation taking the value 1 at t 0. For any other point s, y(t) =e A(s) e A(t) will be the solution of the homogeneous which takes the value 1 at s. In other words y 0;s = e A(s) A(t)

17 Second order differential equations 17 and we can write the first order formula as Z t y = y 0 y 0;t 0 + e A(t)+A(s) b(s) ds t0 Z t = y 0 y 0;t 0 + y 0;s (t)b(s) ds which is analogous to the formula for second order equations. t0 The essential difference between first and second order equations is that for first order equations we can always find solutions of the associated homogeneous equation, whereas for second order this may not be possible. 15. Forgetting the initial conditions If we are only interested in the general solution, we can use the formula for a solution: y(x) =c 1 z 1 +c z +Z t y 1;s (x)c(s) ds We can get by with the indefinite integral because for any fixed t 0 and t 1 the integral Z t 1 t0 y 1;s (x)c(s) ds is a solution to the homogeneous equation. This is because y 1;s (t) =c 1 (s)z 1 (t)+c (s)z (t) for some coefficients c 1 (s) and c (s) and the integral is C 1 z 1 (t)+c z (t)=z 1 (t) Z t 1 c 1 (s)c(s)ds + z (t) Z t 1 t0 t0 c (s)c(s) ds : 16. The structure of the formula The formula as I have stated it gives the solution for the equation y 00 + a(t)y 0 + b(t)y = f (t) with initial conditions y(t 0 )=y 0 It says that The first component of the formula is y 0 (t 0 )=v 0 Z t y = y 0 y 0;t 0 (t) +v 0y 1;t 0 (t)+ y 1;s (t)f (s) ds t0 y 0 y 0;t 0 (t) +v 0y 1;t 0 (t) is a linear combination of the fundamental solutions y 0;t 0 (t) and y 1;t0 (t) of the homogeneous equation, and is consequently itself a solution to the homogeneous equation. Of course its derivative is y 0 y 0 0;t0 (t) +v 0y 0 1;t0 (t)

18 Second order differential equations 18 If you set t = t 0 you will see, by definition of the fundamental solutions, that its value at t 0 is y 0 and the value of its derivative is v 0. The first term in the basic formula can be characterized as the unique solution y of the homogeneous equation with y(t 0 )=y 0 and y 0 (t 0 )=v 0. The formula y = y 0 y 0;t 0 (t)+v 0y 1;t 0 (t) for it is certainly correct, but you should be aware that there may be more efficient ways to find it. The second component of the basic formula is Z t t0 y 1;s (t)f (s) ds : This is the most important and interesting of the two components. It depends linearly on f (t), and in particular vanishes if the equation is homogeneous. The second component of the basic formula is the solution y of the inhomogeneous equation with y(t 0 )=0, y 0 (t 0 )=0. I remind you that, in case the coefficients a(t) and b(t) are constants, in order to find y 1;s we use the translation principle to write y 1;s (t) =y 1;0 (t s) If we want to know the general solution of the inhomogeneous equation, without initial conditions specified, we may replace the first component by the general solution of the homogeneous equation and the second component by an indefinite integral Z t y 1;s (t)f (s) ds : The point here is that We can find the general solution to the inhomogeneous equation by adding the solution of the homogeneous equation to any one solution of the inhomogeneous we can find. Sometimes there is a solution of the inhomogeneous equation much simpler than the rest. 17. When the right hand side is a simple exponential function The basic formula for solving inhomogeneous linear equations in terms of the solutions of the associated homogeneous equation has the virtue of working under just about all circumstances, but it is more complicated to use than necessary in most circumstances. In most circumstances you will meet in practice there is a way to solve an inhomogeneous equation which is simpler than using the basic formula. I will explain this by an example. Suppose we want to find the general solution of the equation where c is an unspecified constant. The general solution to the homogeneous equation is y 00 +3y 0 +y=e ct y = Ae t + Be t

19 Second order differential equations 19 with To find y 1;0 we solve to find y 0 = Ae t Be t A + B =0 A B=1 A =1; B= 1 and y 1;0 (t) =e t e t y 1;s (t) =e (t s) e (t s) = e t e s e t e s Therefore the second component of the basic formula (with indefinite integrals) is Z t y 1;s(t)f (t) ds = Z t e t e s e t e s e cs ds Z t Z t = e t e (c+1)s ds e t e (c+)s ds = e t e (c+1)s (c +1) = ect (c +1) = (c +3c+) This procedure doesn t work if c +1=0or c +=0. These calculations suggest what happens in general. If c is not a root of the characteristic equation e ct e ct (c +) c + ac + b =0 e t e (c+)t (c +) then for the inhomogeneous linear differential equation with constant coefficients the function is a solution. y 00 + ay 0 + y = e ct y = e ct c + ac + b This works in circumstances more varied than you might think. Example. To solve y 00 +3y 0 +y= cos t we recall that cos t is the real part of e it. Therefore we first solve y 00 +3y 0 +y=e it

20 Second order differential equations 0 We get e it y = i +3i+ = eit 1+3i = 1 3i e it 1 3i 1+3i (1 3i)(cos t + i sin t) = 10 (cos t + 3 sin t) +i( 3 cos t + sin t) = 10 whose real part, the solution we are looking for, is cos t + 3 sin t : 10 We now know how to solve any equation y 00 + ay 0 + by = e ct when c is not a root of the characteristic polynomial. There are various rules for what happens when c is a root of the characteristic equation, but probably the most straightforward thing to do in this exceptional circumstance is go back to the basic formula. There is one case which is worth while handling explicitly, however. Suppose we are solving y 00 +! 0y = cos!t : The solutions to the associated homogeneous equation are or e i! 0t cos! 0 t; sin! 0 t: If! 6=! 0 we apply the technique explained earlier in this section with right hand side e i!t to get the general solution cos!t c 1 cos! 0 t + c sin! 0 t +! +! 0 : When! =! 0 we have to apply the basic formula. The second fundamental solution to the homogeneous equation is sin! 0 t :! 0 We therefore get Z t sin!(t y = 1! s) cos!s ds Z sin!t t Z = cos cos!t t!s ds sin!s cos!s ds!! Z sin!t t Z 1 + cos s cos!t t sin s = ds ds!! t sin!t sin!t sin!t cos!t cos!t = +! 4! 4! t sin!t cos(t t)! =! 4! t sin!t cos!t = :! 4!

21 Second order differential equations 1 Recall here that cos(x + y) = cos x cos y sin x sin y cos x = cos x 1 sin x = sin x cos x: Also note that the second term cos!t=4! in the expression above is a solution to the homogeneous linear equation, and can be absorbed into the other component. We discover in the end that the solution to the equation y 00 +! y = cos!t (the special case when c = i! is a characteristic root) is equal to c 1 cos!t + c sin!t + t sin!t! 18. The approach to resonance We know that the general solution to is one of two forms: c 1 cos! 0 t + c sin! 0 t + From this it is a simple calculation to get the following: If! 6=! 0 then the solution to y 00 +! 0y = cos!t 8 >< >: cos!t cos! 0 t!! 0 t sin!t! y 00 +! 0 y = cos!t; y(0) = y0 (0) = 0! 6=! 0! =! 0 is cos!t cos! 0 t! +! 0 whereas if! =! 0 it is t sin!t! : The functions cos!t, cos! 0 t are both periodic, although with different periods. The function t sin!t oscillates with increasing amplitude as t!1. It is a little difficult to imagine exactly how the two kinds of solutions relate to each other, and for that reason it is instructive to graph some examples.

22 Second order differential equations Figure The approach to resonance:! 0 =1and! =1:4, 1:1, 1:06, 1. Why do these pictures look like they do? Recall from trigonometry that cos(x + y) = cos x cos y sin x sin y cos(x y) = cos x cos y + sin x sin y: We write!t =!t +! 0t! 0 t =!t +! 0t +!t! 0t!t! 0 t

23 Second order differential equations 3 which lead to cos!t = cos! +!0 cos! 0 t = cos! +!0 cos!t cos! 0 t = sin! +!0 t cos!!0 t cos!!0 t sin!!0 t sin! +!0 t + sin! +!0 t: t sin!!0 t t sin!!0 t If! is near! 0 we can write!! 0 ="where " is small. Then! +!0 =! 0 + " and which is approximately cos!t cos! 0 t! +! 0 sin! 0 t sin "t = (")(! 0 ) sin "t " = sin "t sin(! 0 + ")t (!! 0 )(! +! 0 ) sin! 0 t t sin "t =! 0 "t sin! 0 t since " is small. The first term oscillates rather slowly while the second oscillates very rapidly, so this explains the graphs we see in the first parts of the figure, which portray a rapid oscillation whose amplitude oscillates slowly. Furthermore, as " goes to 0 the second term has t as limit (for any fixed value of t), which explains why the successive figures look more and more like the last one. In real life true resonance can never occur there will always be some friction in the system and we can never match frequencies exactly. But it is a good ideal model for what we do see often in practice.! Weight on a spring One of the simplest physical systems modelled by second order differential equations is a single weight on a spring. k = spring constant m Figure The simplest physical structure. We shall assume that the spring obeys Hooke s Law, which asserts that stress is proportional to strain. The strain here is vertical force F, the stress is the displacement x of the spring from its relaxed position. Thus F = kx where k is a constant which depends on the particular spring involved. It is called the spring constant.

24 Second order differential equations 4 Suppose the weight has mass m. If it is hanging at rest then the weight mg is exactly balanced by an extension of the spring. If x eq is this equilibrium extension then mg = kx eq : From now on let x be the displacement from this equilibrium position. There will be force on the weight from weight pulling downwards and either a push or a pull from the extended spring. The total from these effects is mg k(x eq + x) = kx. We have seen that friction is approximated by a linear function of velocity. The total amount of force on the weight at displacement x from equilibrium is therefore kx cx 0 (if we are measuring x upwards). According to Newton s Law F = ma we therefore derive the differential equation or mx 00 = kx cx 0 mx 00 + cx 0 + kx =0 describing the evolution of the system. If there is a force of some kind applied to the system externally (by shaking the whole system up and down, for example) we get an equation where F (t) is the applied force. mx 00 + cx 0 + kx = F (t) 0. Simple electric circuits We are going to look into the behaviour of electric circuits assembled from four different types of components: resistors, inductors, capacitors, and independent voltage sources. If V is the voltage drop across an element and I the current flowing through it, then we have defining relations: Type of element Defining property Resistor V = IR (Ohm s Law) Inductor V = LI 0 Capacitor Q = VC I =V 0 C Independent voltage source V = V (t) is specified as a function of time Here R is resistance, L inductance, C capacitance. Also, Q is the charge stored in the capacitor, so that I = Q 0 is the current effectively flowing through it. The last condition means that the voltage across an independent voltage source is independent of the rest of the circuit. In this chapter we shall look at the series circuit with one each of these components in a loop. V (t) Figure 0.1. A simple electric circuit.

25 Second order differential equations 5 Voltage represents potential energy, so the total voltage drop around this circuit must be 0. This means (with the right sign conventions) that LI 0 + RI + Q=C = V (t) : This involves both I and Q, but if we differentiate it we get where the only unknown function is I. LI 00 + RI 0 + I=C = V 0 (t) 1. The pendulum again Suppose a pendulum to be hanging free on a rigid arm. The pendulum we looked at before had no friction, but that is unrealistic. To make things a bit more reasonable, let s assume that there is a friction force opposing motion which is proportional to velocity. Let us also allow for some kind of external force applied to the pendulum. For example, it might be caused by shaking the top of the pendulum back and forth. Thus all in all we have ` = its length m = the mass of the bob at the end y = the displacement of the bob from the vertical position, measured along the arc = the angular displacement: y = ` c = a friction coefficient F (t) = external force The component of gravity exerted along the arc of the path is mg sin. What is new here is the friction term. For reasons explained in the previous section, we assume that the friction force exerted is proportional to velocity. Finally, Newton s second law here asserts that my 00 = cy 0 mg sin + F (t) or my 00 + cy 0 + mg sin(y=`) =F(t): We shall look at this equation later on in detail, but for the moment we are going to consider only motions of a large massive pendulum which does not move far from the vertical position. In this situation, remains small and sin is approximately the same as. So we get as an approximation the equation my 00 + cy 0 + ky = F (t)

26 Second order differential equations 6 where k = mg=`. This is a linear equation, and it is called the linear approximation of the original one, for motions in the neighbourhood of the equilibrium position at dead vertical. Suppose that there is no friction and c =0. The approximating equation is y 00 +(g=`)y = F (t) : The solutions of this are simple periodic functions of period p g=`, and this is related to the fact that as the amplitude of the pendulum s swing becomes small the period of its swing becomes independent of amplitude.. Impulse response and inhomogeneous equations Suppose the pendulum sits at rest until t =0, when it is hit sharply with a hammer, say in the direction of positive displacement (from the left, in the picture above). What determines how the pendulum moves? The details are messy, of course: when the head hits the bob, the left hand side of the bob squishes in a bit, reacting elastically against the head of the hammer but also transmitting the effect of the blow throughout the bob, eventually to the rod of the pendulum, and then up the rod. But in a short time things stabilize and the pendulum begins to swing back and forth in a simple fashion. After this happens, the force F (t) is equal to 0 and the system obeys the homogeneous equation my 00 + cy 0 + ky =0: The period when interesting things are going on will last a fraction of a second. The subsequent motion will depend only on the state of the pendulum at the end of that brief period. So we must answer the question: At the end of that short period, where is the pendulum, and how fast is it moving? To determine the motion from that point on, we just solve the homogeneous equation with appropriate initial conditions. Let s answer the second question first: how fast is the bob moving when the hammer stops contact? Of course this depends on how the force acts. Here, for example, is a plausible graph of F (t): What we shall see is that the effect of the force depends in a very simple way on this graph, at least if the time interval t is small. Of course the effect of the force is mainly to determine the position and velocity at the end of the initial period. The claim is certainly true as far as position is concerned. The effect on position is a second-order effect, which is to say it is proportional to the square of the length of the time interval in which things are going on. This is just what happens when you let an object drop from your hand the distance travelled from your hand is proportional to t. The force causes acceleration, which causes the velocity to build up noticeably, but it takes a relatively long time for this increase in velocity to affect position. If the time interval is small then the position of the bob won t have changed appreciably. As far as velocity is concerned, we can use Newton s law F = ma, which can be reformulated approximately as F = m v=t; F t = m v

27 Second order differential equations 7 for small intervals t. Several smaller intervals will add their effects together, which means that roughly speaking the change in momentum mv is the integral of F (t) dt, which is the area under the graph of F (t). We are assuming that the effects of the force of gravity are small over the small interval, compared to the force F. This is not badly inaccurate. So the effect of the hammer blow on position is practically negligible and that on velocity depends primarily on the area under its graph. It is in fact proportional to this area, and the constant of proportionality is the mass m. The integral of Fdtis called impulse or action. It measures the transfer of momentum. An ideal hammer would transfer the momentum instantaneously, and we can measure even such ideal hammers according to the amount of momentum transferred. The ideal hammer of unit size can be thought of as the limit of ones acting over a very small but still finite interval, each with unit area under the graph of F (t). We can now ask a simpler equation than the original one: What is the effect of a unit ideal hammer? It changes the momentum of the pendulum from 0 to 1 in an instant, and does not in that instant affect position. After that instant, the pendulum motion follows the homogeneous differential equation. So up to t =0the graph of position is dead horizontal, and at t =0it starts to rise suddenly with a slope equal to 1=m. In other words, for t>0the position is given by the solution y(t) of the homogeneous equation with y(0) = 0, y 0 (0) = 1=m. This is what I have called, in the case where m =1, the second fundamental solution y 1;0 (t) of the homogeneous equation. So we see that there is a simple interpretation of this solution in terms of a transfer of unit momentum to a physical system at t =0. Here is the graph of the solution of the equation y 00 +0:4y 0 +y=f(t); y(0) = 0; y 0 (0) = 0 where F (t) = 5 if 0 t 1=5 0 otherwise. I have chosen the friction coefficient low enough to allow some oscillation, and chosen t =1=5large enough to show what happens in that small interval. The light gray line is the actual function y 1;0 (t). If force acts on the linear system over a long period of time, we can think of it as being broken up into a succession of forces of short duration, displaced in time. What happens during the interval [s; s +s]is added to the accumulated effect. This new addition is equal to F (s)y 1;s (t)s for t s. If the system is at rest and the force does not act until t =0, we get an approximation y(t) : = 1 m X 0st y 1;0 (t s)f (s)s and in the limit y(t) = 1 m Z t 0 y 1;0 (t s)f(s)ds for the solution of the inhomogeneous system with vanishing initial conditions. significance of this formula. This tells us the physical

28 Second order differential equations 8 Part III. Vibrations and resonance In realistic situations, solutions of second order linear equations will often correspond to physical vibration. Understanding the nature of vibrations is a major part of the subject. Vibration is concerned with periodic phenomena. A function f (t) is said to have period T if f (t + T ) = f(t) for all t, or equivalently if f (t) is equal to itself shifted by T. 3. Simple periodic functions A simple periodic function is one of the form A cos!t + B sin!t. In these notes we shall characterize such functions in a uniform manner, and see in particular how to draw their graphs. The simplest functions of this sort are y = cos t y = sin t: They both have period, and their graphs look like this: Figure 3.1. The graph of y = cos t. Figure 3.. The graph of y = sin t. These two graphs are in fact essentially the same they differ only in that the graph of sin t is obtained from that of cos t by shifting it to the right by an interval of =, which is one quarter of a period. Now if we are given a function f (t) and shift its graph to the right by an interval of h, we obtain the graph of f (t h) (for example, at t = h the shifted graph will have the same height as that of f at 0). Therefore the shift in the graphs of cos t and sin t is a consequence of the trigonometrical identity sin t = cos(t =) : Of course it is also possible to write cos t = sin(t + =) : However, it is conventional to use the function cos t as the basic periodic function, and express all others in terms of it. Thus we shall say that cos t has a phase shift of 0 and that sin t has a phase shift of =. Now consider the functions y = cos t y = sin t