Part 1: Overview of Ordinary Dierential Equations 1

Chapter 1 Basic Concepts and Problems 1.1 Problems Leading to Ordinary Dierential Equations Many scientic and engineering problems are modeled by systems of ordinary dierential equations (ODEs). Some examples [] follow. 1. Mechanical Vibrations. Let y(t) denote the displacement at time t of a block of mass m that is connected to a spring of stiness k, a damper of resistance c, and an oscillator f(t) (Figure 1.1.1). If the system is released from position y 0 velocity y 0 then its subsequent motion satises 0 with a m d y dt + c dy dt + ky = f(t) t>0 y(0) = y 0 dy(0) dt = y 0 0 : This is an example of an initial value problem (IVP) for a second-order linear ODE. The variable y(t) is called: The variable t is called: The ODE is second order because: It's linear because: This is an IVP because:

k m c f(t) y(t) Figure 1.1.1: Motion of a spring-mass-damper system.. Ecology. Consider a population of predators and prey living in an ecological niche. The predators survive by eating the prey and the prey exist on an independent source of food. A classical situation involves foxes and rabbits. Let P (t) and p(t), respectively, denote the populations of predators and prey at time t. Given the initial populations P 0 and p 0 of predators and prey, their subsequent populations satisfy the Lotka-Volterra equations dp dt = p(a ; P ) t>0 p(0) = p 0 dp dt = P (;c + p) t>0 P (0) = P 0 where a, c,, and are positive constants corresponding to the prey's natural growth rate, the predator's natural death rate, the prey's death rate upon coming into contact with predators, and the predator's growth rate upon coming into contact with prey. This example involves an IVP for a system of two rst-order nonlinear ODEs. The ODEs are nonlinear because: The system is rst order because:. Column Buckling. The lateral displacement y(t) at position t of a clamped-hinged bar of length l that is subjected to a load P (Figure 1.1.) may be approximated by the Euler-Bernoulli equations d 4 y dt 4 with the boundary conditions y(0) = dy(0) dt + d y =0 0 <t<l dt =0 y(l) = d y(l) dt =0: 4

The parameter = P=EI, where EI is the exural rigidity ofthe bar. This is an example of a boundary value problem (BVP) for a fourth-order linear ODE. This is a boundary value problem because: Observe that y(t) = 0 is a solution of this problem for all values of and l. A more interesting problem is to determine y and those values of and l for which non-trivial (y(t) 6= 0) solutions exist. As such, this BVP is also a dierential eigenvalue problem. Nontrivial solutions y(t) are called eigen functions and their corresponding values of are eigenvalues. P t y(t) P Figure 1.1.: Buckling of an elastic column. 4. Pendelum Oscillations ([1], Section 1.). The position (x(t) y(t)) at time t of a particle of mass m oscillating on a pendelum of length l (Figure 1.1.) is m d x dt = ;T sin = ; T l y x md = mg ; T cos = mg ; T y t>0 dt l where g is the acceleration of gravity, T (t) is the tension in the string, and (t) is the angle of the pendelum relative to the vertical at time t (Figure 1.1.). These equations are, however, insucient to guarantee that the particle stays on the string. To ensure that this is so, we must supplement the ODEs by the algebraic constraint x + y = l : The two second-order dierential equations and the constraint comprise a system of ve dierential algebraic equations (DAEs) for the unknowns x(t), y(t), and T (t). Initial conditions specify x(0), dx(0)=dt, y(0), and dy(0)=dt, but not T (0). 5

Of course, for this problem, it's easy to eliminate the constraint byintroducing the change of variables x = l sin, y = l cos. This would reduce the DAEs to the second-order ODE d = ; g sin : dt l Such simplications would not be possible with more dicult systems. θ y T x mg Figure 1.1.: Oscillations of a simple Pendelum. IVPs will comprise our initial study (Part of these notes). We'll take up BVPs next (Part ) and conclude with a study of DAEs (Part 4). In almost all cases, it will suce to develop and analyze numerical methods for IVPs and BVPs that are written as rst-order vector systems of ODEs in the explicit form y 0 (t) =f(t y) t>0 (1.1.1a) where y(t) = 6 4 y 1 (t) y (t). y m (t) f(t y) = 6 5 4 f 1 (t y 1 ::: y m ) f (t y 1 ::: y m ). f m (t y 1 ::: y m ) 5 (1.1.1b) and y 0 := dy=dt. Data for IVPs would is specied as y(0) = y 0 = 6 4 y 10 y 0. 5 : (1.1.1c) y m0 6

Boundary data is more complex. The most general boundary conditions specify a nonlinear relationship between y(0) and y(l) oftheform g 1 (y(0) y(l)) g (y(0) y(l)) g(y(0) y(l)) = 6 = 0: (1.1.1d) 4. 5 g m (y(0) y(l)) In many cases the extension of a scalar to a vector IVP will be obvious and we can study methods for a scalar IVP y 0 (t) =f(t y) t>0 y(0) = y 0 (1.1.) Example 1.1.1. Higher-order ODEs can be written as rst-order systems. Consider the m th order equation z (m) = g(t z z 0 ::: z (m;1) ) where z (i) := d i z=dt i, and introduce the new variables y 1 = z y = z 0 y = z 00 ::: y m = z (m;1) : Then, we have a system in the form (1.1.1) with y 0 1 = y y 0 = y ::: y 0 m = g(t y 1 y ::: y m ): For an IVP, that data would prescribe as z (i) (0) = c i i =0 1 ::: m; 1 with the understanding that z (0) = z. Written in terms of y, we have y 1 (0) = c 0 y (0) = c 1 ::: y m (0) = c m;1 : With DAEs and sometimes with IVPs and BVPs, it will be necessary to consider implicit dierential systems F(t y y 0 )=0: (1.1.) If the Jacobian @F=@y 0 is nonsingular then, by the implicit function theorem, (1.1.) is equivalent to (1.1.1) however, it may be natural to solve (1.1.) in its implicit form to maintain, e.g., sparsity. DAEs involve systems where @F=@y 0 is singular.

Example 1.1.. Let us write the oscillating pendelum problem as a rst-order system by letting y 1 = x y = x 0 y = y y 4 = y 0 : Then, we have y 0 1 = y y 0 = ; T ml y 1 y 0 = y 4 y 0 4 = g ; T ml y y 5 = T with the constraint y 1 + y = l : This can be written in the form (1.1.) with y 0 ; y 1 y 0 + y 1y 5 F(t y y 0 ml )= 6 y 0 ; y 4 4 y 0 ; g + y y 5 4 ml y + y ; l 1 5 = 0: The Jacobian is clearly singular. F y = 6 4 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 Problems 5 1. Letting y 1 = p and y = P, write the predator-prey model dp dt = p(a ; P ) t>0 p(0) = p 0 dp dt = P (;c + p) t>0 P (0) = P 0 in the vector form (1.1.1).. Write the mechanical vibration problem m d y dt in the form (1.1.1a-c) + c dy dt + ky = f(t) t>0 y(0) = y 0 8 dy(0) dt = y 0 0

. Write the column buckling problem d 4 y dt 4 + d y =0 0 <t<l y(0) = dy(0) y(l) =0 y(l) = d =0 dt dt dt in the form (1.1.1a,b,d) 1. Existence, Uniqueness, and Stability of IVPs Before considering numerical procedures for (1.1.1, 1.1.), let us review some results from ODE theory that ensure the existence of unique solutions which depend continuously on the initial data. For simplicity, letusfocus on a scalar IVP having the form (1.1.). Denition 1..1. A function f(t y) satises a Lipschitz condition in a domain D if there exists a non-negative constant L such that jf(t y) ; f(t z)j Ljy ; zj 8(t y) (t z) D: (1..4) Satisfying a Lipschitz condition guarantees that solutions y(t) of (1.1.) are unique as expressed by the following theorem. Theorem 1..1. Let f(t y) be continuous and satisfy a Lipschitz condition on f(t y) j 0 t T ;1 <y<1g: Then the IVP (1.1.) has a unique continuously dierentiable solution on [0 T] for all y 0 (;1 1). Proof. The proof appears in most books on ODE theory, e.g., []. Remark 1. A Lipschitz conditions guarantees unique solutions and is not needed for esistence. For example, the IVP y 0 =y = y(0) = 0 has the two solution y(t) = 0 and y(t) = t. This f(t y) = y = Lipschitz condition. does not satisfy a 9

Remark. Theorem 1..1 applies when f satisfys a Lipschitz condition on a compact, rather than an unbounded, domain D as long as the solution y(t) remains in D []. In addition to existence and uniqueness, we will want to know something about the stability of solutions of the IVP. In particular, we will usually be interested in the sensitivity of the solution to small changes in the data. Perturbations arise naturally in numerical computation due to discretization and round o errors. A formal study of sensitivity would lead us to the following notion of a well-posed system. Denition 1... The IVP (1.1.) is well-posed if there exists positive constants k and ^ such that, for any ^, the perturbed IVP z 0 = f(t z)+(t) z(0) = y 0 + 0 (1..5) satises jy(t) ; z(t)j k (1..6) whenever j 0 j <and j(t)j <for t [0 T]. Again, a Lipschitz condition ensures that we are dealing with a well-posed IVP. Theorem 1... If f(t y) satises a Lipschitz condition on f(t y) j 0 t T ;1 <y<1g then y 0 = f(t y) is well posed on [0 T] with respect to all initial data. Proof. Let (t) =z(t) ; y(t) and subtract (1.1.) from (1..5) to obtain 0 (t) =f(t z) ; f(t y)+(t) (0) = 0 : Taking an absolute value and using the Lipschitz condition (1..4) j 0 (t)j Lj(t)j + j(t)j j(0)j = 0 : 10

We easily see that j(t)j 0 j 0 (t)j provided that j(t)j 0 exists and it is fairly easy to show that j(t)j 0 exists. Additionally, by assumption, max j(t)j < 0tT max j 0 j < 0tT so j(t)j 0 Lj(t)j + j(0)j <: Multiply by the integrating factor e ;Lt to obtain (e ;Lt j(t)j) 0 e ;Lt j(0)j <: Integrating j(t)j Since the denominator is smallest when t = 0 jz(t) ; y(t)j thus, y 0 = f(t y) iswell posed on [0 T]. L[(L +1)e Lt ; 1] : L = k 8t [0 T] The notion of awell-posed system is related to the more common notion of stability as indicated by the following denition. Denition 1... Consider the dierential equation y 0 = f(t y) and (without loss of generality) let the origin (0 0) be anequilibrium point, i.e., f(0 0) = 0. Then, the origin is: stable if a perturbation of the initial condition jy(0)j <grows no larger than for subsequent times, i.e., if jy(t)j <for t>0. asymptotically stable if it is stable and jy(0)j <implies that lim t!1 jy(t)j =0. unstable if it is not stable. Remark. This denition could, like Denition 1.., also involve perturbations of f(t y). We have omitted these for simplicity. 11

An autonomous system is one where f(t y) does not explicitly depend on t, i.e., f(t y) =f(y). If (0 0) is an equilibrium point then, in this case, f(0) = 0. Expanding the solution in a Taylor's series in y, we have y 0 (t) =f(y) =f(0) + f y (0)y(t)+O(y ): Since f(0) = 0, y 0 (t) =f y (0)y(t)+O(y ) where f y = @f =@y. Recall that a function g(y) iso(y p ) if there exists a constant C > 0 such that jg(y)j Cy p as y! 0. Letting = f y (0), we see that the stability of an autonomous system is related to that of the simple linear IVP y 0 = y t>0 y(0) = : (1..) Additional details on autonomous and non-autonomous systems appear in Birkho and Rota [] or Hairer et al. [5], Section 1.1. The solution of the linear IVP (1..) is y(t) =e t hence, (1..) is stable when Re() 0, asymptotically stable when Re() < 0, and unstable when Re() > 0. These conclusions remain true for the original nonlinear autonomous problem when Re() 6= 0 and y is small enough for the O(y ) term to be negligible relative toy. This cannot happen in the stable case (Re() = 0) hence, it requires a more careful analysis. Remark 4. The analysis of non-autonomous systems is similar [1]. Analyzing the stability of autonomous vector systems y 0 = f(y(t)) (1..8) 1

is more complicated. Once again, assume that y = 0 is an equilibrium point and expand f(0) in a series y 0 = f(0)+f y (0)y + O(kyk )=f y (0)y + O(kyk ): We need a brief digression for a few denitions. Denition 1..4. The Jacobian matrix of a vector-valued function f(y) with respect to y is the matrix f y := @f @y = 6 4 @f 1 @f 1 @f 1 @y 1 @y @ym @f @f @f @y 1 @y @ym @fm @y 1 @fm @y... @fm @ym Denition 1..5. The norm of a vector y is a scalar kyk such that 1. kyk 0 and kyk =0if and only if y = 0,. kyk = jjkyk for any scalar, and. kx + yk kxk + kyk. We'll identify specic vector norms when they are needed. 5 : (1..9) For the moment, let's return to our stability analysis and let A = f y (0). Then, the stability of the nonlinear autonomous system (1..8) is related to that of the linear system The solution of this system is y 0 = Ay y(0) = y 0 ky 0 k: (1..10a) y(t) =e At y 0 (1..10b) where the matrix exponential is dened by the series Often, A can be diagonalized as e At = I + ta + t + t + : (1..10c)! A! A 1 T ;1 AT = = 6 4... m 5 (1..11) 1

where 1 ::: m are the eigenvalues of A and the columns of T are the corresponding eigenvectors. In this case, we may easily verify that the solution of (1..10) is y(t) =Te t T ;1 y 0 : Thus, (1..10) is stable if all of the eigenvalues have non-positive real parts, asymptotically stable if all of the eigenvalues have negative real parts, and unstable otherwise. Unfortunately, not all matrices are diagonalizable. However, A can always be reduced to the Jordan canonical form 1 T ;1 AT = 6 (1..1a) 4... 5 l where each Jordan block has the form i = 6 4 i 1 i... 1 5 (1..1b) i The dimension of the Jordan block i corresponds to the multiplicity ofthe eigenvalue i. Thus, if i is simple, the block is a scalar. With this, it is relatively easy to show [] that y = 0 is stable when either Re( i ) < 0orRe( i )=0and i is simple, i =1 ::: m, asymptotically stable when Re( i ) < 0, i =1 ::: m, and unstable otherwise. Example 1..1. Consider the predator-prey model of Section 1.1 y 0 1 = y 1 (a ; y ) y 0 = y (;c + y 1 ) where a,, c, and are positive constants. This autonomous problem has two equilibrium points: y1 y = 0 0 14 c= a= :

The point [0 0] T corresponds to extinction of predators and prey, whereas [c= a=] T implies a co-existence of both species. The stability of[0 0] T the linear system 0 y1 a 0 = y 0 ;c y1 y : may be analyzed by using Since the system matrix is diagonal, its eigenvalues are 1 = a, = ;c. Thus, 1 > 0 and [0 0] T is unstable. In order to analyze the stability of the second equilibrium point, we introduce the change of variables z 1 = y 1 ; c z = y ; a to move the equilibrium point to the origin. Substituting this transformation into the predator-prey equations gives z 0 1 =(z 1 + c )[a ; (z + a )] = ;z ( c + z 1) z 0 =(z + a )[;c + (z 1 + c )] = z 1( a + z ): The linearized system is now 0 z1 = z 0 ;c= a= 0 z1 z : The eigenvalues of this system are 1 = p i c=a. Since the eigenvalues are imaginary, the linear system is stable however, the stability of the nonlinear system is undetermined without additional analysis. Problems 1. Suppose @f =@y is continuous on a closed convex domain D. Show that f(t y) satises a Lipschitz condition on D with (Hint: use the Mean Value Theorem.) @f(t y) L = max j j: (t y)d @y. Are the following IVPs well posed ([4], p. )? Explain. y 0 = p 1 ; y y(0) = 0 y 0 = p y ; 1 y(0) = : 15

16

Bibliography [1] U.M. Ascher and L.R. Petzold. Computer Methods for Ordinary Dierential Equations and Dierential-Algebraic Equations. SIAM, Philadelphia, 1998. [] G. Birkho and G.-C. Rota. Ordinary Dierential Equations. John Wiley and Sons, New York, third edition, 198. [] W.E. Boyce and R.C. DiPrima. Elementary Dierential Equations and Boundary Value Problems. John Wiley and Sons, New York, third edition, 19. [4] C.W. Gear. Numerical Initial Value Problems in Ordinary Dierential Equations. Prentice Hall, Englewood Clis, 191. [5] E. Hairer, S.P. Norsett, and G. Wanner. Solving Ordinary Dierential Equations I: Nonsti Problems. Springer-Verlag, Berlin, second edition, 199. 1