General Theory Of Differential Equations - ACU Blogs
ABILENE CHRISTIAN UNIVERSITYDepartment of MathematicsGeneral Theory of Differential EquationsSections 2.8, 3.1-3.2, 4.1Dr. John EhrkeDepartment of MathematicsFall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSQuestions Of Existence and UniquenessPart of the difficulty in working with differential equations is understanding whatconstitutes a solution and knowing when a solution even exists. In this lecture, wewill try to address questions of existence and uniqueness as they relate to solutionsof linear differential equations. This will allow us to build up a general theorysupporting our study of differential equations throughout the semester. We willbegin with a small example to illustrate what can go wrong.ExampleSolve the differential equation y dy 2.dxxSolution: This equation is separable and so we proceed as follows:dy2 dx ln y 2 ln x cyx eln y e2 ln x c 2 y ec · eln x y A · x2Slide 2/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSSpecifying a SolutionThe solutions of this equation are parabolas of the form y A · x2 , but we have acouple of questions to still be resolved. Are there some initial conditions for which no solutions exist? If a solution does exist, over what interval(s) are they uniquely defined?In answering the second question, consider what would happen if I asked you tosolve the IVP corresponding to y(0) 0. This is bad for two reasons: There is no rate of change defined at the origin, since f (x, y) is undefined there. A solution could come in on one branch and leave on a completely differentbranch, i.e. a solution of the IVP corresponding to y(1) 2 is the parabolay 2x2 , but this parabola passes through the origin and so cannot be extendeduniquely for negative x. The solution is unique only from [0, ).Even for the first question above, no solution exists for any initial value along they-axis other than the origin, because all the parabolas must pass through the origin.This example highlights just a few of the things that can go wrong when solvingdifferential equations.Slide 3/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSExistence Uniqueness TheoremAt this point one might wonder if there is any way to predict the behavior ofsolutions of a given differential equation without solving the ODE. The answer tothis question is yes, and is summarized by the existence uniqueness theorem (EUT).First Order Existence Uniqueness TheoremSuppose that both the function f (x, y) and its partial derivative f / y are continuouson some rectangle R in the xy-plane containing the point (a, b) in its interior. Then,for some open interval I R containing the point a, the initial value problemdy f (x, y),dxy(a) b(1)has one and only one solution that is defined on the interval I.We will not do a complete proof of this result, but will introduce the machinerybehind the proof and a make a few remarks. Our text does not have a completeproof of this result but many undergraduate texts include such a proof in theappendices. If you are interested in reading through the complete proof let me knowand I would be happy to provide the proof for you.Slide 4/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSMethod of Successive ApproximationsThe approach we employ to establish the existence of solutions is called themethod of successive approximations and utilizes a series of approximationscalled Picard iterates. This method is based on the fact that a function y(x)satisfies the initial value problem, (1) on the open interval I containing x aif and only if it satisfies the integral equationZ xy(x) b f (t, y(t)) dt(2)afor all x I. In particular, if y(x) satisfies (2), then clearly y(a) b, anddifferentiating both sides gives y0 (x) f (x, y(x)) as desired. It is importantto note that we traded solving a differential equation for an integralequation. This is almost always the case when one studies differentialequations, as we will see.Slide 5/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSPicard Iterates (Demonstrating Existence)To attempt to solve (2), we begin with the initial function y0 (x) b, andthen define iteratively a sequence y1 , y2 , y3 , . . . of functions that we hopewill converge to the solution. Specifically, we letZxy1 (x) b f (t, y0 (t)) dtaZxy2 (x) b f (t, y1 (t)) dta.Zyn 1 (x) b xf (t, yn (t)) dtaSuppose we know that each of these functions {yn (x)} 0 is defined on someopen interval (the same for each n) containing x a, and that the limity(x) lim yn (x) exists at each point of this interval.n Slide 6/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSContinuing.Continuing from the previous slide, it follows thaty(x) lim yn 1 (x)n Z x lim b f (t, yn (t)) dtn aZ xf (t, yn (t)) dt b limn aZ x f t, lim yn (t) dt b n Za x b f (t, y(t)) dtaThis solution method works provided that: we can validate the interchange of limit operations above. we can show there exists only this one solution.Slide 7/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSDemonstrating UniquenessIf we assume at this point that we have established the existence of a solution viasuccessive approximations (there still is some work to be done to make thatargument solid), then we only need to show that the hypotheses of the theorem leadto a unique solution. When proving uniqueness you generally assume there existsanother solution to the IVP different from y(x), say φ(x) and look at the difference y(x) φ(x) and try to reach a contradiction by showing this difference must bezero. In this case, that is done by proving thatZ x y(t) φ(t) dt. y(x) φ(x) A0To do this requires the use of some heavy duty analysis and the determination of aLipschitz condition. A Lipschitz condition for two functions exists if there is apositive constant A such that f (x, y1 ) f (x, y2 ) A y1 y2 .Arguments of this type are generally considered in a first semester ODE course ingraduate school. We will not pursue this problem beyond this point.Slide 8/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSApplying the Existence Uniqueness TheoremExampleConsider the first order ODE, xy0 y 1. Identify for what values of (a, b), theinitial value problem xy0 y 1, y(a) b has a unique solution, no solution, ormore than one solution.Solution: Consider the point(s), (0, b) for all b 6 1. At these points, x 0, and sof (x, y) (y 1)/x fails to be continuous at x 0 and so by the EUT there can be nosolution to the IVP,y 1dy , y(0) b.dxxConsider the point, (0, 1). When we consider the point (0, 1), f (x, y) 0/0 which isindeterminate. So the EUT says nothing about this point in regards to existence. Wemay have solutions, or we may not have solutions. In the case, a solution existsmight it be unique? We will check the partial derivative, y 1 1 yxxwhich is discontinuous at x 0. This means, that solutions to the IVP y(0) a whenthey exist are not guaranteed to be unique and in particular we see this for (0, 1).Slide 9/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSWorking With Complex NumbersTo achieve our goals this semester we will need to learn how to handlecomplex numbers and operations involving complex variables. A complexnumber is of the form z a bi where a, b R. The complex conjugate of z,denoted z is given byz a bi.Remember, that multiplying a number by its complex conjugate turns thenumber to a real number. In particular,z · z a2 b2 .The absolute value of z, denoted z , involves conjugation sincep z z · z a2 b2 .Moreover, using z and z we obtainz zz zand b (z) .22iRecall that when working with complex numbers you will often need tomultiply or divide by i. Recall that i 1 and so i2 1, i3 i andi4 1.a (z) Slide 10/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSPolar Representation of a Complex Numberz a bi r cos θ i(r sin θ) r (cos θ i sin θ) reiθEuler’s Formula: eiθ cos θ i sin θThe form z reiθ is called the polar form of the complex number a bi.Under this description, a is called the real part, b the imaginary part when z iswritten in rectangular form. In polar form, r is called the modulus of z(denoted z ), and θ is called the argument of z (denoted arg(z)).Slide 11/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSThe Complex ExponentialTypically, when we call something an exponential in mathematics, we havein mind two very specific properties that it must satisfy. An exponential should satisfy the exponential rule. That is, if e is anexponential, then ex · ey ex y . An exponential should satisfy the differential equationdy ay.dtThis says that an exponential has the property that it is self-replicatingunder differentiation (and integration).Question?Does our definition of the complex exponential, eiθ as described by Euler’sformula define an exponential as stated above? eiθ cos θ i sin θ is called acomplex-valued function of a real variableSlide 12/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSPower Series RepresentationsThe final motivation for our description of cos θ i sin θ as being thecomplex exponential is provided by our knowledge of the Taylor series forthe exponential, which is given byex 1 x2x3x4x ··· .1!2!3!4!Inserting x iθ we obtainiθ (iθ)2(iθ)3(iθ)4eiθ 1 ···1!2!3!4! θ2θ4θ3 1 ··· i θ ··· .2!4!3!This points to the fact that the real part of eiθ should be cos θ and theimaginary part is the Taylor series for sin θ.Slide 13/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSThe Art of ComplexifyingTo see how truly amazing and wondrous operations with complex variablescan be we will demonstrate two examples which take advantage of the artof complexifying (yes that’s a word).ExampleObtain a solution to the integralRe x cos x dx by complexifying the integral.Solution: Looking at the integrand, we can rewrite it as the real part of acomplex variable, e x cos(x) e x · eix .By substituting this in for the real integrand we complexify the integral, Z ( 1 i)x Ze x( 1 i)xe cos(x) dx edx . 1 iTaking the real part of this expression gives 1 1 i xe x( 1 i)x e e (cos x i sin x) (sin x cos x) . 1 i22Slide 14/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSConstant Coefficient EquationsIn this example, we will introduce a technique that we will lean on later inthe semester called exponential response.ExampleObtain a general solution to the first order linear equation with constantcoefficients given byx0 (t) 2x(t) cos(t).Solution: Complexifying the ODE leads to z0 (t) 2z(t) eit . We change toz(t) to represent the fact that we expect the solution to such an equation tobe complex. We guess that the solution to such an equation is of the formz(t) Aeit (i.e. matches the exponential response term up to a constant)where A is complex valued. Substituting our guess into the ODE givesA 1/(2 i) soz(t) 1 ite2 iSlide 15/27 — Dr. John Ehrke — Lecture 3 — Fall 2012 (z(t)) x(t) 21cos t sin t.55
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSThe Exponential Response TheoremThis method is so useful that we will establish the general case, that is, solvex0 kx Aert . The results of this are summarized in the theorem below:Exponential Response TheoremThe exponential response to the first order linear constant coefficient ODEx0 kx Aertis given byx(t) A rter k(3)as long as r k 6 0.This formula is valid for complex values of r, but depending on the specificnature of the original oscillation you might have to obtain the real orimaginary part of (3).Slide 16/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSSecond Order Linear EquationsThe general form of a second order differential equation is given byy00 f (t, y, y0 ). In this lecture, we will consider the second order, linearhomogeneous ODE with constant coefficients, given byay00 by0 cy 0(4)We are going to make some assumptions about solutions to (4) which wewill come back and verify later.1A general solution (4) is of the form y(x) c1 y1 c2 y2 .2y1 and y2 are linearly independent solutions of (4). (y1 6 cy2 , for someconstant c 6 0)34Initial conditions involve both y and y0 which help to determine c1 andc2 above.The solutions, y1 and y2 are called the fundamental set of solutions for theequation (4).Slide 17/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSMaking a guess.A natural question to ask at this point is, “What is the nature of thesolutions y1 and y2 ?” This question is best answered by making a guess (aswe saw in the last lecture) that such solutions are exponential in nature. So,to try and solve the equationay00 by0 cy 0,rtwe guess that y(t) e is a solution. Substituting into (2) givesar2 ert brert cert 0 (r2 pr q)ert 0where p b/a and q c/a and a 6 0. This implies that y(t) ert is asolution if and only ifr2 pr q 0.Slide 18/27 — Dr. John Ehrke — Lecture 3 — Fall 2012(5)
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSThe Characteristic EquationSo we have determined that y ert if and only if r2 pr q 0. Thisequation is called the characteristic equation for the ODE, (2). There are threecases to consider, each of which leads to different solution behaviors.1distinct real roots r1 , r2 y1 er1 t , y2 er2 t2complex conjugate pair r a bi y1 e(a bi)t , y2 e(a bi)t3repeated real root, r gives you y1 ert , but how do we find y2 ?We will explore the physical interpretations of these cases, as well assolution techniques in the next lecture, but for the remainder of this lecture,we will discuss the general theory which underpins these techniques. Thequestions we need to answer are:1Why is c1 y1 c2 y2 a solution to the equation, (2)?2Why are all solutions of this form?Slide 19/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSSuperposition PrincipleSuperposition PrincipleIf y1 and y2 are solutions to a linear homogeneous ODE, then y c1 y1 c2 y2is also a solution.To make our lives easier in manipulating differential equations we oftenwant to think about differential equations as linear operators acting on y.We define the linear operator L corresponding to the ODE y00 py0 qy 0byy00 py0 qy 0 D2 y pDy qy 0 (D2 pD q)y 0 Ly 0We denote by D, the differential operator, D d/dy. L D2 pD q is alinear operator if it satisfies the following properties: L(u1 u2 ) L(u1 ) L(u2 ) L(αu) αL(u), for some constant α.Slide 20/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSLinear OperatorsUnder the definition of a linear operator on the previous slide, we show thatL D2 pD q is linear.L(u1 u2 ) (D2 pD q)(u1 u2 ) D2 (u1 u2 ) pD(u1 u2 ) q(u1 u2 ) u001 u002 p(u01 u02 ) qu1 qu2 (u001 pu01 qu1 ) (u002 pu02 qu2 ) Lu1 Lu2We claim it is obvious that L(αu) αL(u) by properties of the derivative. We arenow prepared to answer our first question: “If we suppose that y1 and y2 aresolutions to y00 py0 qy 0, does that automatically mean c1 y1 c2 y2 is asolution?”L(c1 y1 c2 y2 ) L(c1 y1 ) L(c2 y2 ) c1 L(y1 ) c2 L(y2 ) 0(since y1 solves the ODE, then L(y1 ) 0)Thus, c1 y1 c2 y2 solves y00 py0 qy 0 if y1 and y2 do.Slide 21/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSTwo solutions are better than one.In answering the second of our questions for this lecture, “Why do all solutions looklike c1 y1 c2 y2 ?” we will show that this solution form is necessary to satisfy IVPs.TheoremThe set of solutions {c1 y1 c2 y2 } is enough to satisfy any initial values imposed onthe equation y00 py0 qy 0. This means there does not exist some initialcondition y(x0 ) a, y0 (x0 ) b such that a solution not of the form c1 y1 c2 y2 is theonly solution that satisfies this IVP.Proof: Let y(x0 ) a, y0 (x0 ) b. Then applying these initial conditions to the solutionc1 y2 c2 y2 we obtain the system of linear equations with unknowns c1 , c2 given byc1 y1 (x0 ) c2 y2 (x0 ) ac2 y01 (x0 ) c2 y02 (x0 ) bThis system of equations is solvable if and only if the determinant,Slide 22/27 — Dr. John Ehrke — Lecture 3 — Fall 2012y1y01y26 0.y02
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSThe WronskianGiven y1 , y2 , the Wronskian of y1 , y2 , denoted W(y1 , y2 ) is given byW(y1 , y2 ) y1y01y2y02 y1 y02 y2 y01The Wronskian explains why linearly independent solutions are requiredsince if y2 cy1 , thenW(y1 , y2 ) y1y01y2y02 y1y01cy1cy01 cy1 y01 cy1 y01 0In this case, because W(y1 , y2 ) 0, then the resulting system of equationsfrom the initial conditions is not solvable. This leads to a very importanttheorem,Abel’s TheoremIf y1 , y2 are solutions to the ODE y00 py0 qy 0, then either1W(y1 , y2 ) 0 for all x2or W(y1 , y2 ) 6 0 for all xSlide 23/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSAbel’s TheoremProof: To prove Abel’s theorem we note that y1 and y2 satisfyy001 p(t)y01 q(t)y1 0(6)y002 p(t)y02 q(t)y2 0(7)If we multiply the first equation by y2 , the second by y1 and add the results, weobtain(y1 y002 y001 y2 ) p(t)(y1 y02 y01 y2 ) 0.(8)Next, we let W(t) W(y1 , y2 )(t) and observe that W 0 (t) y1 y002 y001 y2 . Then we canwrite (3) in the formW 0 p(t)W 0(9)which is a first order linear equation in W, and has solution Z W(t) c · exp p(t) dt(10)where c is a constant. The value of c depends on the fundamental solutions, y1 andy2 . However, since the exponential function is never 0, W(t) is not 0 unless c 0, inwhich case W(t) 0 for all t, which completes the proof.Slide 24/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSNormalized SolutionsNot all solutions y1 and y2 are created equal. There are in fact, “bestsolution”, called normalized solutions which we will introduce below.Normalized SolutionsTwo solutions Y1 and Y2 of the second order linear ODE, y00 py0 qy 0are called normalized if they satisfy the initial conditions,Y1 (0) 1, Y10 (0) 0Y2 (0) 0, Y20 (0) 1ExampleUsing the above definition, find the normalized solutions for y00 y 0 andy00 y 0. Compare your results. Remember to use the characteristicequation to determine your fundamental set of solutions.Slide 25/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
ABILENE CHRISTIAN UNIVERSITYDEPARTMENT OF MATHEMATICSWhy do we care about normalized solutions?If Y1 and Y2 are normalized at x 0, then the solution to the IVPy00 py0 qy 0with initial conditions y(0) a, y0 (0) b isy(x) aY1 bY2 .In example, we can simply read the solutions off straight from the ODE.This is highly desired by those in the engineering fields, and is generally anicer version of linearly independent solutions than what we have beenworking with thus far. The general solutiony(x) c1 Y1 c2 Y2is called the normalized general solution.Slide 26/27 — Dr. John Ehrke — Lecture 3 — Fall 2012
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of linear differential equations. This will allow us to build up a general theory supporting our study of differential equations throughout the semester. We will begin with a small example to illustrate what can go wrong. Example Solve the differential equation dy dx 2 y x : Solution: This equation is separable and so we proceed as follows .
(iii) introductory differential equations. Familiarity with the following topics is especially desirable: From basic differential equations: separable differential equations and separa-tion of variables; and solving linear, constant-coefficient differential equations using characteristic equations.
Andhra Pradesh State Council of Higher Education w.e.f. 2015-16 (Revised in April, 2016) B.A./B.Sc. FIRST YEAR MATHEMATICS SYLLABUS SEMESTER –I, PAPER - 1 DIFFERENTIAL EQUATIONS 60 Hrs UNIT – I (12 Hours), Differential Equations of first order and first degree : Linear Differential Equations; Differential Equations Reducible to Linear Form; Exact Differential Equations; Integrating Factors .
3 Ordinary Differential Equations K. Webb MAE 4020/5020 Differential equations can be categorized as either ordinary or partialdifferential equations Ordinarydifferential equations (ODE's) - functions of a single independent variable Partial differential equations (PDE's) - functions of two or more independent variables
3.1 Theory of Linear Equations 97 HIGHER-ORDER 3 DIFFERENTIAL EQUATIONS 3.1 Theory of Linear Equations 3.1.1 Initial-Value and Boundary-Value Problems 3.1.2 Homogeneous Equations 3.1.3 Nonhomogeneous Equations 3.2 Reduction of Order 3.3 Homogeneous Linear Equations with Constant Coeffi cients 3.4 Undetermined Coeffi cients 3.5 V
Introduction to Advanced Numerical Differential Equation Solving in Mathematica Overview The Mathematica function NDSolve is a general numerical differential equation solver. It can handle a wide range of ordinary differential equations (ODEs) as well as some partial differential equations (PDEs). In a system of ordinary differential equations there can be any number of
Chapter 1 Introduction 1 1.1 ApplicationsLeading to Differential Equations 1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 16 Chapter 2 First Order Equations 30 2.1 Linear First Order Equations 30 2.2 Separable Equations 45 2.3 Existence and Uniqueness of Solutionsof Nonlinear Equations 55
DIFFERENTIAL EQUATIONS FIRST ORDER DIFFERENTIAL EQUATIONS 1 DEFINITION A differential equation is an equation involving a differential coefficient i.e. In this syllabus, we will only learn the first order To solve differential equation , we integrate and find the equation y which
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