Lecture notes on Ordinary Differential Equations

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Ordinary Differential Equation Alexander Grigorian University of Bielefeld Lecture Notes, April - July 2008 Contents 1 Introduction: the notion of ODEs and examples 3 1.1 Separable ODE ................................. 5 1.2 LinearODEof1storder ............................ 8 1.3 Quasi-linear ODEs and differentialforms ................... 11 1.4 Integrating factor ................................ 17 1.5 Second order ODE ............................... 18 1.5.1 Newtons’secondlaw .......................... 18 1.5.2 Electricalcircuit ............................ 18 2 Existence and uniqueness theorems 19 2.1 1storderODE.................................. 19 2.2 Dependenceontheinitialvalue ........................ 27 2.3 Higher order ODE and reduction to the firstordersystem ......... 30 n 2.4 Norms inR ................................... 32 2.5 ExistenceanduniquenessforasystemofODEs ............... 34 2.6 Maximal solutions................................ 40 2.7 Continuity of solutions with respect tof (t,x) ................ 44 2.8 Continuityofsolutionswithrespecttoaparameter ............. 50 2.9 Global existence................................. 53 2.10 Differentiabilityofsolutionsinparameter................... 55 3 Linear equations and systems 66 3.1 Spaceofsolutionsofhomogeneoussystems.................. 66 3.2 Linear homogeneous ODEs with constant coefficients ............ 68 3.3 Spaceofsolutionsofinhomogeneoussystems................. 75 3.4 Linear inhomogeneous ODEs with constant coefficients ........... 76 3.5 SecondorderODEwithperiodicrighthandside............... 82 3.6 Themethodofvariationofparameters .................... 87 3.6.1 Asystemofthe1storder ....................... 87 3.6.2 A scalar ODE ofn-thorder ...................... 90 3.7 WronskianandtheLiouvilleformula ..................... 94 3.8 Linear homogeneous systems with constant coefficients ........... 98 3.8.1 Functionsofoperatorsandmatrices..................101 13.8.2 Jordan cells ...............................106 3.8.3 Jordannormalform...........................108 3.8.4 TransformationofanoperatortoaJordannormalform.......110 4 Qualitative analysis of ODEs 117 4.1 Autonomous systems ..............................117 4.2 Stabilityforalinearsystem ..........................119 4.3 Lyapunov’s theorem ..............................125 21 Introduction: the notion of ODEs and examples Adifferential equation (Differentialgleichung) is an equation for an unknown function that contains not only the function but also its derivatives (Ableitung). In general, the unknown function may depend on several variables and the equation may include various partial derivatives. However, in this course we consider only the differential equations for a function of a single real variable. Such equations are called ordinary differential 1 equations —shortlyODE(diegewohnl ¨icheDifferentialgleichungen). A most general ODE has the form ¡ ¢ 0 (n) Fx,y,y,...,y =0, (1.1) whereF is a given function ofn+2 variables andy =y(x) is an unknown function of a (n) real variablex.Themaximalordern of the derivativey in (1.1) is called the order of the ODE. TheODEsariseinmanyareasofMathematics,aswellasinSciencesandEngineering. In most applications, one needs to find explicitly or numerically a solutiony(x) of (1.1) satisfying some additional conditions. There are only a few types of the ODEs when one can find all the solutions. In Introduction we will be concerned with various examples and specific classes of ODEs of the first and second order, postponing the general theory to the next Chapters. Consider the differential equation of the first order 0 y =f (x, y), (1.2) wherey =y(x) is the unknown real-valued function of a real argumentx,andf (x,y) is a given function of two real variables. 2 Consider a couple (x,y) as a point inR and assume that functionf is defined on a 2 setD ½ R , which is called the domain (Def initionsber eich)ofthefunctionf and of the equation (1.2). Then the expressionf (x,y) makes sense whenever (x,y)2D. 2 Definition. A real valued functiony(x) definedonanintervalI ½ R,iscalleda (particular) solution of (1.2) ify(x) is differentiable at anyx 2I,thepoint (x,y(x)) 0 belongs toD for anyx2I and the identityy (x)=f (x,y(x)) holds for allx2I. The family of all particular solutions of (1.2) is called the general solution. The graph of a particular solutionis calledan integral curve of the equation. Obviously, anyintegral curveiscontainedinthedomainD. Usually a given ODE cannot be solved explicitly. We will consider some classes of f (x, y) when one find the general solution to (1.2) in terms of indefinite integration. 1 The theory of partial differential equations, that is, the equations containing partial derivatives, is a topic of another lecture course. 2 Here and below by an interval we mean any set of the form (a;b)= fx2R :axbg a;b= fx2R :a·x·bg a;b)= fx2R :a·xbg (a;b= fx2R :ax·bg; where a;b are real or§1 andab. 3Example. Assume that the functionf does not depend ony so that (1.2) becomes 0 3 y =f (x). Hence,y must be a primitive function off. Assuming thatf is a continuous (stetig)functiononanintervalI, we obtain the general solution onI by means of the indefinite integration: Z y =f (x)dx =F (x)+ C, whereF (x) is a primitive off (x) onI andC is an arbitrary constant. Example. Consider the ODE 0 y = y. Let us first find all positive solutions, that is, assume thaty(x) 0. Dividing the ODE byy and noticing that 0 y 0 =(lny), y we obtain the equivalent equation 0 (lny) =1. Solving this as in the previous example, we obtain Z lny =dx =x+ C, whence Cxx y =ee =Ce, 1 CC whereC =e .SinceC 2R is arbitrary,C =e is any positive number. Hence, any 1 1 positive solutiony has the form x y =Ce,C 0. 1 1 Ify(x) 0 for allx,thenuse 0 y 0 =(ln(¡y)) y and obtain in the same way x y =¡Ce, 1 whereC 0. Combine these two cases together, we obtain that any solutiony(x) that 1 remainspositiveornegative,hastheform x y(x)= Ce, where C 0 or C 0.Clearly,C=0 suits as well sincey=0 is a solution. The next plot contains the integrals curves of such solutions: 3 By definition, a primitive function of f is any function whose derivative is equal to f. 4y y 25 12.5 0 -2 -1 0 1 2 x x -12.5 -25 x Let us show that the family of solutionsy = Ce ,C 2 R, is the general solution. Indeed, ify(x) is a solution that takes positive value somewhere then it is positive in x some open interval, sayI. By the above argument,y(x)= Ce inI,where C 0. x Sincee 6=0, this solution does not vanish also at the endpoints ofI. This implies that thesolutionmustbepositiveonthewholeintervalwhereitisdefined. It follows that x y(x)= Ce in the domain ofy(x). The same applies ify(x) 0 for somex. 0x Hence, the general solution of the ODEy =y isy(x)= Ce whereC 2 R.The constantC is referred to as a parameter. It is clear that the particular solutions are distinguished by the values of the parameter. 1.1 Separable ODE Consider a separable ODE, that is, an ODE of the form 0 y =f (x)g(y). (1.3) Any separable equation can be solved by means of the following theorem. Theorem 1.1 (The method of separation of variables) Letf (x) andg(y) be continuous functionsonopenintervalsI andJ, respectively, and assume thatg(y)=0 6 onJ.Let 1 F (x) be a primitive function off (x) onI andG(y) be a primitive function of onJ. g(y) Then a functiony defined on some subinterval ofI, solves the differential equation (1.3) if and only if it satisfies the identity G(y(x)) =F (x)+ C, (1.4) for allx in the domain ofy,whereC is a real constant. 0 Forexample,consideragaintheODEy =y in the domainx 2 R, y 0.Then f (x)=1 andg(y)=y6=0 so that Theorem 1.1 applies. We have Z Z F (x)=f (x)dx =dx =x 5and Z Z dydy G(y)= = =lny g(y)y where we do not write the constant of integration because we need only one primitive function. The equation (1.4) becomes lny =x+ C, x whence we obtainy =Ce asinthepreviousexample.NotethatTheorem1.1doesnot 1 cover the case wheng(y) may vanish, which must be analyzed separately when needed. Proof. Lety(x)solve(1.3). Sinceg(y)6=0,wecandivide(1.3)byg(y),whichyields 0 y =f (x). (1.5) g(y) 1 0 0 Observe that by the hypothesisf (x)=F (x) and =G (y), which implies by the 0 g (y) chain rule 0 y 0 0 0 =G (y)y =(G(y(x))). g(y) Hence, the equation (1.3) is equivalent to 0 0 G(y(x)) =F (x), (1.6) which implies (1.4). Conversely, if functiony satisfies (1.4) and is known to be differentiable in its domain then differentiating (1.4) inx, we obtain (1.6); arguing backwards, we arrive at (1.3). The only question that remains to be answered is whyy(x) is differentiable. Since the functiong(y) does not vanish, it is either positive or negative in the whole domain. 1 Then the functionG(y),whosederivativeis , is either strictly increasing or strictly g(y) −1 decreasing in the whole domain. In the both cases, the inverse functionG is defined and is differentiable. It follows from (1.4) that −1 y(x)=G (F (x)+C). (1.7) =1 Since bothF andG are differentiable, we conclude by the chain rule thaty is also differentiable, which finishes the proof. Corollary. Under the conditions of Theorem 1.1,forallx 2I andy 2J there exists 0 0 a unique value of the constantC such that the solutiony(x) defined by (1.7) satisfies the conditiony(x )=y. 0 0 The conditiony(x )=y is called the initial condition (Anfangsbedingung). 0 0 Proof. Settingin(1.4)x =x andy =y,weobtainG(y )=F (x )+C,whichallows 0 0 0 0 to uniquely determine the value ofC,thatis,C =G(y )¡F (x ). Conversely, assume 0 0 thatC is given by this formula and prove that it determines by (1.7) a solutiony(x).If therighthandside of (1.7)is definedonanintervalcontainingx , thenbyTheorem1.1it 0 defines a solutiony(x), and this solution satisfiesy(x )=y bythechoiceofC.Weonly 0 0 have to make sure that the domain of the right hand side of (1.7) contains an interval −1 aroundx (a priori it may happen so that the the composite functionG (F (x)+C) 0 has empty domain). Forx =x the right hand side of (1.7) is 0 −1−1 G (F (x )+C)=G (G(y )) =y 0 0 0 6−1 so that the functiony(x) is defined atx =x . Since both functionsG andF +C are 0 continuous and defined on open intervals, their composition is definedonanopenset. Since this set containsx , it contains also an interval aroundx . Hence, the functiony is 0 0 definedonanintervalaroundx,which finishes the proof. 0 One can rephrase the statement of Corollary as follows: for allx 2I andy 2J 0 0 there exists a unique solutiony(x)of(1.3)thatsatisfies in addition the initial condition y(x)=y;thatis,foreverypoint (x ,y ) 2I£J thereisexactlyoneintegralcurve 0 0 0 0 of the ODE that goes through this point. However, the meaning of the uniqueness claim in this form is a bit ambiguous because out of any solutiony(x), one can make another solution just by slightly reducing the domain, and if the reduced domain still containsx 0 thentheinitialconditionwillbesatisfiedalsobythenewsolution. Thepreciseuniqueness claim means that any two solutions satisfying the same initial condition, coincide on the intersection of their domains; also, such solutions correspond to the same value of the parameterC. In applications of Theorem 1.1, it is necessary to find the functionsF andG.Techni- cally it is convenient to combine the evaluation ofF andG with other computations as follows. The first step is always dividing (1.3) byg to obtain (1.5). Then integrate the both sides inx to obtain Z Z 0 ydx =f (x)dx. (1.8) g(y) Then we need to evaluate the integral in the right hand side. IfF (x) is a primitive off then we write Z f (x)dx =F (x)+ C. 0 Inthelefthandsideof(1.8),wehaveydx =dy. Hence, we can change variables in the integral replacing functiony(x) by an independent variabley.Weobtain Z Z 0 ydxdy = =G(y)+ C. g(y)g(y) Combining the above lines, we obtain the identity (1.4). 0 Ifintheequationy =f (x)g(y)thefunctiong(y)vanishesatasequenceofpoints,say y ,y,..., enumerated in the increasing order, then we have a family of constant solutions 1 2 y(x)=y . The method of separation of variables provides solutions in any domain k y y y . The integral curves in the domainsy y y can in general touch kk+1kk+1 the constant solutions, as will be shown in the next example. Example. Consider the equation p 0 y = jyj, which is defined for ally 2R. Since the right hand side vanish fory=0, the constant functiony´ 0 is a solution. In the domains y 0 and y 0, the equation can be solved using separation of variables. For example, in the domain y 0,weobtain Z Z dy =dx p y whence p 2y =x+C 7and 1 2 y = (x+C) ,x¡C 4 (the restriction x¡C comes from the previous line). Similarly, in the domain y 0, we obtain Z Z dy p =dx ¡y whence p ¡2 ¡y =x+C and 1 2 y =¡ (x+C) ,x ¡C. 4 We obtain the following integrals curves: y y 4 3 2 1 0 -2 -1 0 1 2 3 4 5 x x -1 -2 -3 -4 Weseethattheintegralcurvesinthedomain y 0 touch the curvey=0 and so do the integral curves in the domain y 0. This allows us to construct more solution as follows: take a solutiony (x) 0 that vanishes atx =a and a solutiony (x) 0 that vanishes 1 2 atx =b where ab are arbitrary reals. Then define a new solution: 8 y (x),xa 1 y(x)= 0,a·x·b, : y (x),xb. 2 Note that such solutions are not obtained automatically by the method of separation of 2 variables. Itfollowsthatthroughanypoint (x ,y )2R thereareinfinitelymanyintegral 0 0 curves of the given equation. 1.2 Linear ODE of 1st order Consider the ODE of the form 0 y +a(x)y =b(x) (1.9) wherea andb are given functions ofx,definedonacertainintervalI.Thisequationis 0 called linear because it depends linearly ony andy. A linear ODE can be solved as follows. 8Theorem 1.2 (The method of variation of parameter) Let functionsa(x) andb(x) be continuous in an intervalI. Then the general solution of the linear ODE (1.9) has the form Z −A(x)A(x) y(x)=eb(x)edx, (1.10) whereA(x) is a primitive ofa(x) onI. Note that the functiony(x) given by (1.10) is definedonthefullintervalI. A(x) Proof. Let us make the change of the unknown functionu(x)=y(x)e ,thatis, −A(x) y(x)=u(x)e. (1.11) Substituting this to the equation (1.9) we obtain ¡ ¢ 0 −A−A ue +aue =b, 0−A−A 0−A ue ¡ueA +aue =b. 0 SinceA =a, we see that the two terms in the left hand side cancel out, and we end up with a very simple equation foru(x): 0−A ue =b 0A whenceu =be and Z A u =bedx. Substituting into (1.11), we finish the proof. Onemaywonderhowonecanguesstomakethechange(1.11). Hereisthemotivation. Consider first the case whenb(x)´ 0. In this case, the equation (1.9) becomes 0 y +a(x)y=0 and it is called homogeneous. Clearly, the homogeneous linear equation is separable. In the domains y 0 and y 0 we have 0 y =¡a(x) y and Z Z dy =¡a(x)dx =¡A(x)+ C. y Then lnjyj =¡A(x)+C and −A(x) y(x)= Ce whereC can be any real (includingC=0 that corresponds to the solutiony´ 0). For a general equation (1.9) take the above solution to the homogeneous equation and replace a constantC by a functionC (x) (or which was denoted byu(x) in the proof), which will result in the above change. Since we have replaced a constant parameter by a function, this method is called the method of variation of parameter. It applies to the linear equations of higher order as well. 9Example. Consider the equation 1 2 0x y +y =e (1.12) x in the domain x 0.Then Z Z dx A(x)=a(x)dx = =lnx x (we do not add a constantC sinceA(x) is one of the primitives ofa(x)), Z Z ³ ´ 1 2 1 2 1 2 xx 2x y(x)=exdx =edx =e +C, x 2x 2x whereC is an arbitrary constant. Alternatively, one can solve first the homogeneous equation 1 0 y +y=0, x using the separable of variables: 0 y 1 = ¡ yx 0 0 (lny) = ¡(lnx) lny = ¡lnx+C 1 C y =. x Next, replace the constantC by a functionC(x) and substitute into (1.12): µ ¶ 0 C(x) 1C 2 x + =e, xxx 0 Cx¡CC 2 x + =e 2 2 xx 0 C 2 x =e x 2 0x C =ex Z ³ ´ 1 2 2 xx C(x)=exdx =e +C. 0 2 Hence, ³ ´ C(x) 1 2 x y = =e +C, 0 x 2x whereC is an arbitrary constant. 0 Corollary. Under the conditions of Theorem 1.2,foranyx 2I and anyy 2R there 0 0 is exists exactly one solutiony(x) defined onIandsuchthaty(x )=y . 0 0 That is, though any point (x ,y )2I£Rtheregoesexactlyoneintegralcurveofthe 0 0 equation. 10−A Proof. LetB(x) be a primitive ofbe so that the general solution can be written in the form −A(x) y =e (B(x)+C) withanarbitraryconstantC. Obviously, anysuchsolutionisdefinedonI. Thecondition y(x )=y allows to uniquely determineC from the equation: 0 0 A(x ) 0 C =ye ¡B(x ), 0 0 whence the claim follows.‘ 1.3 Quasi-linear ODEs and differential forms 2 LetF (x,y) be a real valued function definedinanopensetΩ ½ R . Recall thatF is differentiable at a point (x,y)2Ω if there exist real numbersa,b such that F (x+dx,y +dy)¡F (x,y)=adx+bdy +o(jdxj+jdyj), as jdxj +jdyj 0. Heredx anddy the increments ofx andy, respectively, which are considered as newindependent variables (the differentials). The linear functionadx+bdy of the variablesdx,dy is called the differential ofF at (x, y) and is denoted bydF,that is, dF =adx+bdy. (1.13) In general,a andb are functions of (x,y). Recall also the following relations between the notion of a differential and partial derivatives: 1. IfF is differentiable at some point (x,y) and its differential is given by (1.13) then ∂F ∂F the partial derivativesF = andF = exist at this point and xy ∂x ∂y F =a,F =b. xy 2. IfF is continuously differentiable inΩ, that is, the partial derivativesF andF xy exist inΩ and are continuous functions thenF is differentiable at any point inΩ. Definition. Given two functionsa(x,y) andb(x, y) inΩ, consider the expression a(x,y)dx+b(x,y)dy, which is called a differential form. The differential form is called exact inΩ if there is a differentiable functionF inΩ such that dF =adx+bdy, (1.14) and inexact otherwise. If the form is exact then the functionF from (1.14) is called the integral of the form. Observe that not every differential form is exact as one can see from the following statement. 11Lemma 1.3 If functionsa,b are continuously differentiable inΩ then the necessary con- dition for the formadx+bdy to be exact is the identity a =b. yx Proof. Indeed, if there isF is an integral of the formadx +bdy thenF =a and x F =b, whence it follows that the derivativesF andF are continuously differentiable. yxy By a well-know fact from Analysis, this implies thatF =F whencea =b . xy yxyx Example. The form ydx¡ xdy is inexact becausea =1 whileb =¡1. yx The form ydx+xdy is exact because it has an integralF (x,y)=xy. 3 y 2 2 2 The form 2xydx+(x +y )dy is exact because it has an integralF (x,y)=xy + 3 (it will be explained later how one can obtain an integral). If the differential formadx+bdy is exact then this allows to solve easily the following differential equation: 0 a(x, y)+b(x,y)y =0. (1.15) 0 This ODE is called quasi-linear because it is linear with respect toy but not neces- dy 0 sarily linear with respect toy.Usingy = , one can write (1.15) in the form dx a(x,y)dx+b(x,y)dy=0, which explains why the equation (1.15) is related to the differential formadx+bdy. We say that the equation (1.15) is exact if the formadx+bdy is exact. 2 Theorem 1.4 LetΩ be an open subset ofR ,a,b be continuous functions onΩ,suchthat the formadx+bdy is exact. LetF be an integral of this form. Consider a differentiable functiony(x) defined on an intervalI ½ R such that the graph ofy is contained inΩ. Theny solves the equation (1.15) if and only if F (x,y(x)) = const on I. Proof. Thehypothesisthatthegraphofy(x) is contained inΩ implies that the composite functionF (x,y(x)) is defined onI. By the chain rule, we have d 0 0 F (x,y(x)) =F +Fy =a+by. xy dx d 0 Hence, the equationa +by =0 is equivalent toF (x, y(x)) = 0,andthelatteris dx equivalent toF (x,y(x)) = const. 0 Example. The equationy +xy =0 is exact and is equivalent toxy =C because ydx+xdy =d(xy).Thesamecanbeobtainedusingthemethodofseparationofvariables. 2 2 0 The equation 2xy+(x +y )y =0 is exact and is equivalent to 3 y 2 xy + = C. 3 Below are some integral curves of this equation: 12y y 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 -7.5 -6.25 -5 -3.75 -2.5 -1.25 0 1.25 2.5 3.75 5 6.25 7.5 x x How to decide whether a given differential form is exact or not? A partial answer is given by the following theorem. 2 We say that a setΩ½R is a rectangle (box) if it has the formI£J whereI andJ are intervals inR. 2 Theorem 1.5 (The Poincaré lemma) LetΩ be an open rectangle in R.Leta,b be continuously differentiable functions onΩ such thata ´b . Then the differential form yx adx+bdy is exact inΩ. Proof of Theorem 1.5. Assume firstthattheintegralF exists andF (x ,y )=0 0 0 for some point (x ,y ) 2Ω (the latter can always be achieved by adding a constant 0 0 toF). For any point (x,y) 2Ω, also the point (x,y ) 2Ω;moreover,theintervals 0 (x ,y ),(x, y ) and (x,y ),(x,y) are contained inΩ becauseΩ is a rectangle. Since 0 0 0 0 F =a andF =b, we obtain by the fundamental theorem of calculus that xy Z Z xx F (x,y )=F (x,y )¡F (x ,y )=F (s,y )ds =a(s,y )ds 0 0 0 0x 0 0 xx 0 0 and Z Z yy F (x,y)¡F (x,y )=F (x,t)dt =b(x,t)dt, 0y yy 0 0 whence y x Z Z F (x;y)= a(s;y )ds+ b(x;t)dt: (1.16) 0 x y 0 0 Now use the formula (1.16) to define functionF (x,y).LetusshowthatF is indeed the integral of the formadx+bdy. Sincea andb are continuous, it suffices to verify that F =a andF =b. xy 13It is easy to see from (1.16) that y Z ∂ F =b(x,t)dt =b(x,y). y ∂y y 0 Next, we have Z Z xy ∂∂ F =a(s,y )ds+b(x,t)dt x 0 ∂x ∂x xy 0 0 Z y ∂ =a(x,y )+b(x,t)dt. (1.17) 0 ∂x y 0 ∂ Thefactthattheintegralandthederivative canbeinterchangedwillbejustifiedbelow ∂x (see Lemma 1.6). Using the hypothesisb =a , we obtain from (1.17) xy Z y F =a(x,y )+a (x,t)dt x 0y y 0 =a(x,y )+(a(x,y)¡a(x,y )) 0 0 =a(x,y), which finishes the proof. Now we prove the lemma, which is needed to justify (1.17). Lemma 1.6 Letg(x,t) be a continuous function onI£J whereI andJ are bounded closed intervals inR. Consider the function Z β f (x)=g(x,t)dt, α where α,β=J,whichisdefined for allx2I. If the partial derivativeg exists and is x continuous onI£J thenf is continuously differentiable onI and, for anyx2I, Z β 0 f (x)=g (x,t)dt. x α Inotherwords,theoperationsofdifferentiationinxandintegrationint,whenapplied tog(x,t), are interchangeable. ProofofLemma1.6. We need to show that, for allx2I, Z β 0 f (x)¡f (x) 0 g (x,t)dt asx x, x 0 x ¡x α which amounts to Z Z ββ 0 g(x ,t)¡g(x,t) 0 dtg (x,t)dt asx x. x 0 x ¡x αα Note that by the definition of a partial derivative, for anyt2 α,β, 0 g(x ,t)¡g(x,t) 0 g (x,t) asx x. (1.18) x 0 x ¡x 140 Consider all parts of (1.18) as functions oft,with fixedx and withx as a parameter. Then we have a convergence of a sequence of functions, and we would like to deduce that their integrals converge as well. By a result from Analysis II, this is the case, if the convergence is uniform (gl eichmassig ¨ )inthewholeinterval α,β, that is, if ¯ ¯ 0 ¯ ¯ g(x ,t)¡g(x,t) 0 ¯ ¯ sup ¡g (x,t) 0 asx x. (1.19) x ¯ 0 ¯ x ¡x t∈α,β 0 By the mean value theorem, for anyt2 α,β,thereisξ2 x,x such that 0 g(x ,t)¡g(x,t) =g (ξ, t). x 0 x ¡x Hence, the difference quotient in (1.19) can be replaced byg (ξ, t). To proceed further, x recall that a continuous function on a compact set is uniformly continuous. In particular, the functiong (x,t) is uniformlycontinuous onI£J,thatis,forany ε 0 there is δ 0 x such that x,ξ2 I,jx¡ξj δ andt,s2 J,jt¡sj δ)jg (x,t)¡g (ξ, s)j ε. (1.20) xx 0 Ifjx¡xj δ then alsojx¡ξj δ and, by (1.20) withs =t, jg (ξ, t)¡g (x,t)j ε for allt2 J. xx 0 In other words,jx¡xj δ implies that ¯ ¯ 0 ¯ ¯ g(x ,t)¡g(x,t) ¯ ¯ sup ¡g (x,t) ·ε, x ¯ ¯ 0 x ¡x t∈J whence (1.19) follows. Consider some examples to Theorem 1.5. 2 2 2 Example. Consider again the differential form 2xydx+(x +y )dy inΩ =R.Since ¡ ¢ 2 2 a =(2xy) =2x =x +y =b, yx y x we conclude by Theorem 1.5 that the given form is exact. The integralF can be found by (1.16) takingx =y =0: 0 0 Z Z xy 3 ¡ ¢ y 2 2 2 F (x,y)= 2s0ds+x +tdt =xy +, 3 0 0 as it was observed above. Example. Consider the differential form ¡ydx+xdy (1.21) 2 2 x +y 2 inΩ =R nf0g.Thisformsatisfies the conditiona =b because yx µ ¶ 2 2 2 2 2 y (x +y )¡2yy ¡x a =¡ =¡ = y 2 2 2 2 2 2 2 2 x +y (x +y ) (x +y ) y 15and µ ¶ 2 2 2 2 2 x (x +y )¡2xy ¡x b = = =. x 2 2 2 2 2 2 2 2 x +y (x +y ) (x +y ) x By Theorem 1.5 we conclude that the given form is exact in any rectangular domain in Ω. However, let us show that the form is inexact inΩ. Consider the functionθ(x,y) which is the polar angle that is defined in the domain 0 2 Ω =R nf(x,0) :x· 0g by the conditions yx sinθ =, cosθ = ,θ2 (¡π, π), rr p 0 2 2 wherer =x +y . Let us show that inΩ ¡ydx+xdy dθ =. (1.22) 2 2 x +y y In the half-planefx 0g we have tanθ = andθ2 (¡π/2,π/2) whence x y θ = arctan. x Then (1.22) follows by differentiation of the arctan: 1xdy¡ ydx ¡ydx+xdy dθ = =. 2 2 2 2 xx +y 1+(y/x) x In the half-planefy 0g we have cotθ = andθ2 (0,π) whence y x θ = arccot y x and (1.22) follows again. Finally, in the half-plane fy 0g we have cotθ = andθ 2 y (¡π,0) whence µ ¶ x θ =¡arccot ¡, y 0 is the union of the three half-planesfx 0g,fy 0g, and (1.22) follows again. SinceΩ 0 0 fy 0g,weconcludethat(1.22)holdsinΩ and, hence, the form (1.21) is exact inΩ. Why the form (1.21) is inexact inΩ? Assume from the contrary that the form (1.21) is exact inΩ and thatF is its integral inΩ,thatis, ¡ydx+xdy dF =. 2 2 x +y 0 4 ThendF =dθ inΩ whence it follows thatd(F ¡θ)=0 and, henceF =θ+const in 0 0 Ω. It follows from this identity that functionθcanbeextendedfromΩ to a continuous 4 WeusethefollowingfactfromAnalysisII:ifthedifferential of a function is identical zero in a n connected open set U ½ R then the function is constant in this set. Recall that the set U is called connected if any two points from U can be connected by a polygonal line that is contained in U. 0 The set− is obviously connected. 16function onΩ, which however is not true, because the limits ofθ when approaching the point (¡1,0)(oranyotherpoint (x,0) with x 0) from above and below are different. The moral of this example is that the statement of Theorem 1.5 is not true for an arbitrary open setΩ. It is possible to show that the statement of Theorem 1.5 is true if and only if the setΩ is simply connected, that is, if any closed curve inΩ can be continuouslydeformedtoapoint while stayinginΩ. Obviously, therectangles aresimply 0 2 connected (as well asΩ), while the setΩ =R nf0g is not simply connected. 1.4 Integrating factor Consider again the quasilinear equation 0 a(x,y)+b(x,y)y =0 (1.23) and assume that it is inexact. Write this equation in the form adx+bdy=0. After multiplying by a non-zero functionM (x,y),weobtainanequivalentequation Madx+ Mbdy=0, which may become exact, provided functionM is suitably chosen. Definition. AfunctionM (x,y) is called the integrating factor for the differential equa- tion (1.23) inΩ ifM isanon-zerofunctioninΩ such that the form Madx + Mbdy is exact inΩ. If one has found an integrating factor then multiplying (1.23) byM the problem amounts to the case of Theorem 1.4. Example. Consider the ODE y 0 y =, 2 4xy +x in the domainfx 0,y 0g and write it in the form ¡ ¢ 2 ydx¡ 4xy +xdy=0. 2 Clearly, this equation is not exact. However, dividing it byx , we obtain the equation µ ¶ y 1 dx¡ 4y +dy=0, 2 xx which is already exact in any rectangular domain because µ ¶ ³ ´ y 1 1 = =¡ 4y +. 2 2 xxx y x Taking in (1.16)x =y =1, we obtain the integral of the form as follows: 0 0 Z Z µ ¶ xy 1 1y 2 F (x,y)=ds¡ 4t+dt=3¡2y ¡. 2 sxx 1 1 By Theorem 1.4, the general solution is given by the identity y 2 2y + = C. x 171.5 Second order ODE 00 A general second order ODE, resolved with respect toy has the form 00 0 y =f (x, y , y ), wheref is a given function of three variables andy =y(x) is an unknown function. We consider here some problems that amount to a second order ODE. 1.5.1 Newtons’ second law Consider movement of a point particle along a straight line and let its coordinate at 0 timet bex(t).Thevelocity(Geschwindigkeit) of the particle isv(t)=x (t) and the 00 acceleration (Beschleunig ung) isa(t)=x (t). The Newton’s second law says that at any time 00 mx = F, (1.24) wherem is the mass of the particle andF is the force (Kraf t) acting on the particle. In 0 general,F is a function oft,x,x so that (1.24) can be regarded as a second order ODE forx(t). The forceF is called conservative ifF depends only on the positionx. For example, conservative are gravitation force, spring force, electrostatic force, while friction and the airresistancearenon-conservativeastheydependinthevelocityv. AssumingF =F (x), denote byU (x)aprimitivefunctionof¡F (x). The functionU is called the potential of 0 the forceF. Multiplying the equation (1.24) byx and integrating int,weobtain Z Z 00 0 0 mxxdt =F (x)xdt, Z Z md 2 0 (x)dt =F (x)dx, 2dt 2 mv =¡U (x)+C 2 and 2 mv +U (x)= C. 2 2 mv The sum +U (x) is called the total energy of the particle (which is the sum of the 2 kinetic energyandthe potential energy). Hence,wehaveobtainedthelaw of conservation of energy: the total energy of the particle in a conservative field remains constant. 1.5.2 Electrical circuit Consider anRLC-circuit that is, an electrical circuit (Schaltung) where a resistor, an inductor and a capacitor are connected in a series: 18 R + L _ V(t) C DenotebyRtheresistance(Widerstand)oftheresistor,byLtheinductance(Induktivitat ¨ ) of the inductor, and byC the capacitance (Kapaz itat ¨ ) of the capacitor. Let the circuit contain a power source with the voltageV (t) (Spannung)wheret is time. Denote by I (t) the current (Strom)inthecircuitattimet. Using the laws of electromagnetism, we obtain that the potential differencev on the resistorR is equal to R v =RI R (Ohm’s law), and the potential differencev on the inductor is equal to L dI v =L L dt (Faraday’s law). The potential differencev on the capacitor is equal to C Q v =, C C 0 whereQisthecharge(Ladungsmenge) of the capacitor; also we haveQ =I.By Kirchhoff’s law, we have v +v +v =V (t) RLC whence Q 0 RI + LI + =V (t). C Differentiating int,weobtain I 00 0 0 LI +RI + =V, (1.25) C which is a second order ODE with respect toI (t). Wewillcomebacktothisequation after having developed the theory of linear ODEs. 2 Existence and uniqueness theorems 2.1 1st order ODE We change notation, denoting the independent variable byt and the unknown function byx(t). Hence, we write an ODE in the form 0 x =f (t,x), 192 wheref is a real value function on an open setΩ½R and a pair (t,x) is considered as 2 apointinR . LetusassociatewiththegivenODEtheinitial value problem (Anfangswertproblem) -shortly,IVP,whichistheproblemoffindingasolutionthatsatisfiesinadditiontheinitial conditionx(t)=x where (t ,x ) is a given point inΩ.WewriteIVPinacompact 0 0 0 0 form as follows: ½ 0 x =f (t,x), (2.1) x(t )=x. 0 0 A solution to IVP is a differentiable functionx(t):I R whereI is an open interval containingt , such that (t,x(t))2Ω for allt2I,whichsatisfies the ODE inI and the 0 initial condition. Geometrically, the graph of functionx(t) is contained inΩ and goes through the point (t ,x ). 0 0 In order to state the main result, we need the following definitions. Definition. We say that a functionf :ΩR is Lipschitz inxifthereisaconstantL such that jf (t,x)¡f (t,y)j·Ljx¡yj for allt,x,y such that (t,x)2Ω and (t,y)2Ω.TheconstantL is called the Lipschitz constant off inΩ. We say that a functionf :ΩR is locally Lipschitz inx if, for any point (t ,x )2Ω 0 0 there exist positive constantsε,δ such that the rectangle R=t ¡ δ, t +δ£x ¡ε,x +ε (2.2) 0 0 0 0 is contained inΩ and the functionf is Lipschitz inR; that is, there is a constantL such that for allt2 t ¡ δ, t +δ andx,y2 x ¡ε,x +ε, 0 0 0 0 jf (t,x)¡f (t,y)j·Ljx¡yj. Note that in the latter case the constantL may be different for different rectangles. 2 Lemma 2.1 (a) If the partial derivativef exists and is bounded in a rectangleR½R x thenf is Lipschitz inx inR. 2 (b) If the partial derivativef exists and is continuous in an open setΩ½R thenf x is locally Lipschitz inx inΩ. Proof. (a)If (t,x)and (t,y)belongtoRthenthewholeintervalbetweenthesepoints is also inR, and we have by the mean value theorem f (t,x)¡f (t,y)=f (t,ξ)(x¡y), x for someξ2 x,y. By hypothesis,f is bounded inR,thatis, x L := supjf j1, (2.3) x R whence we obtain jf (t,x)¡f (t,y)j·Ljx¡yj. 20

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