REUNotes08-O DE s May 30, 2008Chapter TwoDifferential EquationsREUNotes08-O DE s May 30, 20084 CHAPTER 22.1 LINEAR EQUATIONSWe consider or dinary differential equations of the following formdxdt= f(t, x).where x is either a real variable or a vector of several variables, t is a re al,independent variable which is often conveniently thought as a time, and fis a given function of both variables. The dimension of the equation isreferred to the dimensio n of the dependent variable x.In some cases, f does not depend on both variables. If f does not dependon the time variable, the equationx′= f(x)is called an autonomous equation.For simplicity we denote the derivative either by ‘prime’ x′= dx/dt or by‘dot’ ˙x = dx/dt. Thus the equation can also be written asx′= f(t, x) o r ˙x = f(t, x).One Dimensional EquationsWe begin by considering one of the simplest types of equations – 1 dimen-sional, linear, with constant coefficientx′= f(x) = ax, (2.1)where a is a cons tant.A so lution of the equation is a function x(t) of time t so tha t the equationholds when x(t) is substituted in. For example, x(t) = 3e2tis a s olution tothe equationx′= 2xbec ause by the chain rule and the derivative o f exponential functions,x′(t) = 3e2t· 2 = 2x(t).The equation thus holds and 3e2tis a solution. In contrast, x(t) = sin t isnot a solution beca usex′(t) = cos t 6= 2x(t) = 2 sin(t).Method of Trial Solutions. Finding solutions to an arbitrarily given equa-tion is never trivial. Different equation types require different approaches,which usually are not transferable from one ty pe to another. Fortunately forthe equatio n (2.1) under c onsideration, what takes to solve it is an educatedinsight not beyond the realm of elementary functions.The equation x′= ax says that a solution x(t) is such a function thatits derivative x′(t) is essentially itself ≈ x(t). What types of elementaryfunctions fit this description? There are many. But a simple candidate isthe family of exponential functions of the following formCertREUNotes08-O DE s May 30, 2008DIFFERENTIAL EQUATIONS 5with C, r being arbitrary parameters. The question then beco mes for whatparameter values o f C and r is Certa solution? To this extent, the equationx′= ax is the only constraint and there is only one way to find it out – plugit in and see if it fits. First,[Cert]′= Cert· r.Setting it to the right hand side of the equation,[Cert]′= Cert· r = a[Cert]which is true if r = a. Hence, we conclude thatx(t) = Ceatis a solution to the equation for any arbitrary constant C.General Solutions: The appeara nce of one arbitrary constant in a solutionis expected for any first order differential e quation of one variable beca useof the following reasons. To solve differe ntial equations is to find antideriva-tives, almost always indire ctly and implicitly, and each differentiation of onevariable genera tes one arbitrary ‘integration’ co nstant upon the completionof such an antiderivative process. Thus, as a rule of thumb, we anticipatetwo arbitrary constants for the solutions of a second o rder equation of onevariable, or a first order equation of two variables, and so on. In general,we anticipate n ×m many arbitrary constants for an n th order equations ofm many varia bles. Such a s olution containing the right amount of arbitraryconstants is called a general solution. Hence, x(t) = Ceatis a generalsolution to the equation x′= ax.The arbitrary constant C can be determined by s pecifying one additionalcondition to the equation. The case we consider almost exclusively is theinitial condition which in general takes the formx(t0) = x0with t0, x0prescribed. In most cases we take t0= 0. For example, using theinitial conditionx(0) = x0the arbitrary constant C is determined uniquely by the followingx0= x(0) = Cea0= C.Hence, x(t) = x0eatis the solution to the initial value problem(x′= f(x) = axx(0) = x0Example 2.1.1 Solve the initial value problem(x′= −2xx(1) = 3REUNotes08-O DE s May 30, 20086 CHAPTER 2−1 −0.5 0 0.5 1 1.5 2−1−0.8−0.6−0.4−0.200.20.40.60.81Figure 2.1 Solution portrait of Example 1.Solution: Since a = −2, x(t) = Ce−2tis the general solution. By the initialcondition3 = x(1) = Ce−2·1=⇒ C = 3e2and the solution to the initial value problem isx(t) = 3e2e−2t= 3e2(1−t).daTypical solutions of the example are sketched in Fig.2.1, as well as itsvector field. Such a portrait is c alled a solution portrait.Important properties of this type of equations are the following•x(t) ≡ 0 is a constant solution, called an equilibr ium solution.•If a < 0, all solutions exponentially converge to the equilibrium s olutionas t tends to infinitylimt→∞Ceat= 0.The equilibrium solution x = 0 is said to be asymptotically stable.•If a > 0, all solutions, except for the equilibrium solution, exponentiallydiverge to infinity as t tends to infinitylimt→∞Ceat=(∞, if C > 0−∞, if C < 0Such an equilibrium solution x = 0 is said to b e asymptotically un-stable.Two Dimensional EquationsLetx =x1x2REUNotes08-O DE s May 30, 2008DIFFERENTIAL EQUATIONS 7be a vector of two real variables x1, x2, andA =a11a12a21a22be a 2 × 2 matrix. The system of two linear equations of the following form(x′1= a11x1+ a12x2x′2= a21x1+ a22x2can also be written in its matrix formx′= Ax.Method of Trial Solutions. Similar to the rationale for the scalar case,we try it out to see if exponential functions of the following for mx = ξeλt=u1u2eλtare solutions. Using the derivative conventionx′=x′1x′2and its resulting algebraic rules , we have(ξeλt)′= ξeλt· λ.Setting it to the right hand side of the equation, we have(ξeλt)′= ξeλt· λ = A(ξeλt) = (Aξ)eλt.Cancelling out eλtwe derive the following relationAξ = λξ ⇐⇒ (A − λI)ξ = 0.Since we seek solutions other than the trivial equilibr ium so lution x = 0,we want ξ 6= 0 for the solution candidate ξeλt. Hence the condition aboveimplies that λ must be an eigenvalue of the matrix A and ξ must be an eigen-vector of the eigenvalue λ. By a reason of matrix theory, λ is an eigenvalueif and only if it is the root the characteristic equation|A − λI| = λ2− (a11+ a22)λ + a11a22− a12a21= 0.Depending on the na tur e of the roots, the method follows up one of threepossibilities, each goes through a distinct set of steps.Case I: Distinct Real Eigenvalues.In this case, both λ1, λ2are real and unequal. We first find