Families of Solutions to Bernoulli ODEs

Similar documents
1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.

1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation

(9) P (x)u + Q(x)u + R(x)u =0

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

1 1D heat and wave equations on a finite interval

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

The Wave Equation I. MA 436 Kurt Bryan

The Regulated and Riemann Integrals

Chapter 0. What is the Lebesgue integral about?

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Consequently, the temperature must be the same at each point in the cross section at x. Let:

7.3 Problem 7.3. ~B(~x) = ~ k ~ E(~x)=! but we also have a reected wave. ~E(~x) = ~ E 2 e i~ k 2 ~x i!t. ~B R (~x) = ~ k R ~ E R (~x)=!

Review of Calculus, cont d

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

New Expansion and Infinite Series

Math 360: A primitive integral and elementary functions

20 MATHEMATICS POLYNOMIALS

7.2 The Definite Integral

Lecture 1. Functional series. Pointwise and uniform convergence.

Calculus of Variations

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Section 6.1 INTRO to LAPLACE TRANSFORMS

Theoretical foundations of Gaussian quadrature

Summary: Method of Separation of Variables

Math 135-2, Homework 1

AM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h

Conservation Law. Chapter Goal. 5.2 Theory

USA Mathematical Talent Search Round 1 Solutions Year 25 Academic Year

4.4 Areas, Integrals and Antiderivatives

Indefinite Integral. Chapter Integration - reverse of differentiation

HW3, Math 307. CSUF. Spring 2007.

Stuff You Need to Know From Calculus

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Math 100 Review Sheet

Chapter 6 Notes, Larson/Hostetler 3e

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

In Section 5.3 we considered initial value problems for the linear second order equation. y.a/ C ˇy 0.a/ D k 1 (13.1.4)

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

k and v = v 1 j + u 3 i + v 2

Identify graphs of linear inequalities on a number line.

Numerical Integration

a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution

LECTURE 10: JACOBI SYMBOL

ODE: Existence and Uniqueness of a Solution

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

INTRODUCTION TO INTEGRATION

1 The Riemann Integral

Chapter 6 Techniques of Integration

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions

2.4 Linear Inequalities and Interval Notation

Lumpability and Absorbing Markov Chains

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.

The Basic Functional 2 1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

Math 1B, lecture 4: Error bounds for numerical methods

Properties of Lorenz Curves for Transformed Income Distributions

Improper Integrals, and Differential Equations

APPM 1360 Exam 2 Spring 2016

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

Polynomials and Division Theory

Bernoulli Numbers Jeff Morton

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

Sturm-Liouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

Orthogonal Polynomials

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

Math& 152 Section Integration by Parts

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

Jim Lambers MAT 169 Fall Semester Lecture 4 Notes

5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Generalized Coordinates. The Kepler-Coulomb Problem

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Numerical quadrature based on interpolating functions: A MATLAB implementation

Energy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon

Math Calculus with Analytic Geometry II

21.6 Green Functions for First Order Equations

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

38 Riemann sums and existence of the definite integral.

13: Diffusion in 2 Energy Groups

Infinite Geometric Series

Section 14.3 Arc Length and Curvature

STUDY GUIDE FOR BASIC EXAM

Plates on elastic foundation

Recitation 3: More Applications of the Derivative

Pre-Session Review. Part 1: Basic Algebra; Linear Functions and Graphs

MATH 144: Business Calculus Final Review

2008 Mathematical Methods (CAS) GA 3: Examination 2

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

Line Integrals. Partitioning the Curve. Estimating the Mass

Review of basic calculus

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

u t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx

Transcription:

In the fmily of solutions to the differentil eqution y ry dx + = it is shown tht vrition of the initil condition y( 0 = cuses horizontl shift in the solution curve y = f ( x, rther thn the verticl shift tht one might nticite. The comlete solution is obtined in the cse where the coefficients nd r re both constnt. The behviour of the solutions is exlored in the cse = r s vries for severl vlues of n. n Tble of Contents:. Behviour of exonentil functions. Comlete solution of the first order Bernoulli initil vlue roblem 3. Secil cses 4. Exmle for n = 5. Exmle for n = 3 6. Exmle for n = 7. Exmle for n = with n symtote 8. Exmle for n = without n symtote 9. The Generl Cse

. Behviour of exonentil functions kx Among the fetures of the exonentil function y = e is tht stretch of its grh by fctor of constnt ( > 0 in the y direction is identicl to trnsltion by c = ln in the x k direction. Put nother wy, the horizontl distnce between ny two oints on the grhs of kx kx y = e nd y= e ( > 0 tht shre the sme y coordinte is constnt c = ln. k Figure Grhs of the exonentil fmily y= e kx The roof is simle. Setting the vlues of the scled nd trnslted functions equl t ech x, kx kx ( c kx kc kc e e e e = e kc = ln Therefore the horizontl shift c is relted to the verticl stretch fctor by c = ln. k If k > 0 nd > (s illustrted, then c < 0 nd the trnsltion is to the left. If < 0, then there is reflection in the x-xis, together with horizontl shift of c = ln. k This roerty of the exonentil function exlins the behviour of the grhs of fmilies of solutions to first order Bernoulli initil vlue roblems with constnt coefficients. - -

. Comlete solution of the first order Bernoulli initil vlue roblem with constnt coefficients: The ordinry differentil eqution (ODE dx + x y x = r x y x ( ( ( ( ( n is the generl first order Bernoulli ODE, where n is some rel constnt nd ( x nd r( x re integrble rel functions of the rel vrible x. In this er we restrict our ttention to the cse of constnt coefficients. Let, n, nd r ll be indeendent rel constnts. Consider the initil vlue roblem (IVP n y ry, y( 0 dx + = = If n 0 then y 0 is solution to the ordinry differentil eqution. y 0 my be rt of nother fmily of solutions, deending on the vlues of, r nd n. In most cses, y 0 is singulr solution to the ordinry differentil eqution. If in ddition = 0, then the comlete solution of the IVP includes ( 0 y x. If n = then the IVP is liner nd homogeneous: + ( r y = 0, y( 0 =, for which dx ( r x the comlete solution is quickly found to be y= e. y 0 is clerly the member of this fmily of solutions for which = 0. There is therefore no singulr solution in this cse, which is no surrise: liner ordinry differentil equtions never hve singulr solutions. If in ddition = r, then the solution reduces to y. Otherwise, using the result from the revious section, chnge in the vlue of the initil condition from y( 0 = to y( 0 = (where nd hve the sme sign results in horizontl trnsltion of the solution curve by c = ln. r y If n then the chnge of vribles w = trnsforms the ODE into liner form. n n n y dw d y n n dw w = = = y = y n dx n dx dx dx dx n n dw n dw n + y = ry y + y = ry + y = r dx dx dx dw ( n w r dx + = n - 3 -

Solving this liner ODE, ( ( h = P dx = n dx = n x h ( r e R dx = e r dx = e n ( h ( n x e e = (integrting fctor ( n x n x h (unless = 0, in which cse e R dx = r dx = r x The generl solution of the ODE for 0 nd n is n y n h h ( = w = e e R dx + C ( ( n x r n x = e e + C ( n n r ( ( n x r y = + n Ce y ( x = + ( Imosing the initil condition, n Ce ( n x ( n n r n r y( 0 = = + ( nc ( nc = Therefore the comlete solution to the generl Bernoulli initil vlue roblem with constnt coefficients in the cse n nd 0 is together with the singulr solution 0 r n r y( x = + e ( n x ( n / y in the cse ( n 0 nd 0 > =. / - 4 -

3. Secil cses: If n > 0 then y 0 is singulr solution of The generl solution of dx Imosing the initil condition, / n y0 nc ( ( n ry dx =. n / n = ry is y( x = ( n( rx+ C ( = ( = ( = ( ( nc n The comlete solution in the cse = 0 is therefore n / ( ( n ( ( y x = n rx+ / ( ( n This cn be re-written s y( x ( n r( x c =, where n c= r ( n ( r 0 The cse = r = 0 hs the simle solution y. In ll other cses for which = 0, ech member of the fmily cn be found from nother solution curve by horizontl shift. There is n no solution curve when is not rel. One cn deduce this excetion s the limit s 0 of the more generl 0 cse: / ( ( n r n r n x lim + e 0 / ( ( ( n n n x r n x = lim e + e 0 / ( n n ( n x r ( ( ( n x n x = lim e + + + + 0!! (using the Mclurin series for the exonentil function / ( n n ( n x ( ( nx = lim e + r + ( nx 0 n / n = + r nx ( (( (. - 5 -

The comlete solution y(x is constnt ( y In the ordinry solution the term only if n = nd/or ( = 0 nd n > 0. r n is well-defined t 0 = only if n> 0 n<. When the initil condition is y ( 0 = 0, the ordinry solution exists only if n < nd the singulr solution exists only if n > 0. Therefore the comlete solution includes both the ordinry nd singulr solutions when = 0 nd 0< n <. In the following sections we will look t two such cses: n = nd 3 n =, then t one cse ( nd finlly t one cse ( n = where the singulr solution does not exist. n = where only the singulr solution exists t = 0 Furthermore, if ( n is frction tht, in its lowest terms, hs n even denomintor, then there is no rel solution when < 0 nd there my be distinct ir of rel solutions when > 0. The inhomogeneous liner IVP is secil cse (n = 0, y r, y( 0 dx + = = for which the comlete solution is r r y( x = + e x. n From the differentil eqution + y = ry one cn deduce the existence of horizontl dx symtote shred by mny ordinry solutions: As 0, dx n n y ry y 0 or y r / ( n Therefore, whenever is rel number, ll non-singulr solutions shre the horizontl r / ( n / ( n symtote y = (nd, for some vlues of n, y =. r r - 6 -

4. Exmle for n = The initil vlue roblem y y, y( 0 dx + = = is n exmle of first order Bernoulli ODE with = r = nd n=. Let us exlore the evolution of the grhs of the solution s the vlue of the initil condition vries. From the generl cse bove, the comlete solution is x ( + ( ± e ( > 0 x y( x = ( e or 0 ( = 0 no rel solution ( < 0 Note tht the initil vlue roblem hs singulr solution ( y 0 in ddition to the ordinry solution when (nd only when = 0. At = one brnch of the ordinry solution becomes nother constnt solution, y. An interesting result emerges uon investigting the behviour of the solution to ech side of = : Figure Fmily of solutions for = r =, n=, 0 < All of the solution grhs for 0< < re similr to ech other, one brnch shifting further to the left s, the other shifting further to the right. They ll shre the sme limiting lim y x =, with the excetion of the singulr solution y 0. behviour ( x Whenever = r, ll of the ordinry solutions shre the sme horizontl symtote, / ( n y = =. Figure 3 dislys some of the solution grhs for. r - 7 -

Figure 3 Fmily of solutions for = r =, n=, All of the solution grhs for one brnch of > re similr to ech other, shifting further to the + left s nd to the right s. The other brnch is identicl to the grhs for 0< <, shifting further to the right s. Similrity is estblished esily. For one brnch of 0< <, ( x ( ( x y = + e = ( ( e + c x ( x c = ( e e = e, c where e = c= ln (. The sme y-xis intercet (0, is ttined by horizontl trnsltion of c ln ( For 0< <, ln ( < 0 so tht the shift is to the left. c = 0 when = 0. For the other brnch of 0< <, ( =. ( x ( ( x y = + e = ( + e + c x ( x c = ( e e = e, c where e = + c= ln ( +. The sme y-xis intercet (0, is ttined by horizontl trnsltion of c ln ( ln ( + > 0 so tht the shift is to the right. c = 0 when = 0. = +. - 8 -

Therefore ll of the solution curves for 0 shifted to the right by n mount c ln ( brnch c ln ( =. For one brnch, c s. For one brnch of >, y ( ( e x + c x c where e c ln ( = + ( < < re irs of identicl coies of ( y = e x, = ± (which is ctully shift to the left for the ( x c ( = + e e = + e, = =. c = 0 when = = = 4. Therefore ll of the solution curves for this brnch of > re identicl coies of y e x y 0 = = 4, shifted to the right by n mount ( = +, (the solution when ( ( + c= ln. The shift is ctully to the left when < < 4. c s nd c + s. For the other brnch of >, x ( x ( ( ( c where e = + c= ln ( +. = + = + + c x ( y e e ( x c ( = e e = e, Therefore ll of the solution curves for this brnch of > re identicl coies of y e x c= ln +. For this brnch, c + s = (, shifted to the right by n mount (. Therefore the ccommodtion of vrying vlues of the initil condition y ( 0 is chieved by horizontl shifts in the solution grh rther thn verticl shifts, excet tht second distinct she emerges for nd tht there is no solution grh t ll for < 0. - 9 -

5. Exmle for n = 3 If we now tke = r = 3 nd n= 3, then the initil vlue roblem becomes /3 3y 3 y, y( 0 dx + = = with comlete solution 3 3 x ( + ( e ( 0 y( x = x 3 ( e or 0 ( = 0 Agin the behviour differs on ech side of =, (where the solution simlifies to y nd there is singulr solution for = 0 in ddition to the ordinry solution. All of the solution curves for Figure 4 Fmily of solutions for = r = 3, n= 3 < re horizontl trnsltions of ( 3-0 - y = e x, the ordinry c =, by n mount c, where ( 3 3 solution curve for 0 = e c = ln. The trnsltion is to the left when 0< < nd to the right when < 0. All of the solution curves for > re horizontl trnsltions of ( 3 y = + e x, the ordinry c =, by n mount c, where ( 3 3 solution curve for 8 = e c = ln. The trnsltion is to the left when < < 8 nd to the right when > 8.

The solution curve for = is limiting cse of both fmilies nd rises from the ordinry solution. Only the singulr solution y 0 fils to roch y = s x. These exmles from sections 4 nd 5 re both in the rnge 0< n <, for which both singulr = y 0 = 0. nd ordinry solutions exist for ( 6. Exmle for n = If we now tke = r = nd n=, then the initil vlue roblem becomes y y, y( 0 dx + = = with comlete solution ( 0 x y( x = + e 0 ( = 0 This time only the singulr solution exists for = y( 0 = 0, lthough it is esy to show tht lim = 0 x, identicl to the singulr solution. 0 x + e If nd only if 0 < < 0 < 0 or >, then the solution curve hs n x infinite discontinuity t e = x = ln. We gin obtin two fmilies of solutions, one for 0< < nd the other outside tht rnge. Agin we cn show tht every member of ech fmily is simle horizontl trnsltion of either y( x = x or y( x = x resectively. + e e - -

Figure 5 Fmily of solutions for = r =, n=, < 0 Figure 6 Fmily of solutions for = r =, n=, 0 - -

Figure 7 Fmily of solutions for = r =, n=, 7. Exmle for n = with n symtote If we now tke = r = nd n=, then the initil vlue roblem becomes + y =, y( 0 = dx y with comlete solution x + + ( e ( 0 y( x = x + ( e ( 0 This time there is no singulr solution. However, there re two distinct solutions when (nd only when = 0. The solution is rel for ll x if nd only if. There re two constnt solutions: y when = nd y when =. All of the solution grhs shre the sme limiting behviour y( x lim =. x - 3 -

Figure 8 Fmily of solutions for = r =, n= Agin we hve two fmilies of solution curve: y x = sgn e x (for < <+ nd horizontl trnsltions of ( ( horizontl trnsltions of ( ( y x = sgn + e x (otherwise. In ll of the exmles here I hve set = r. The effect of other ositive choices for r ( / n ( / n r r (or ±, s in the cse n = bove. If there is no horizontl symtote. is to re-scle the limiting vlue of y(x from to r ( / n is not rel number, then - 4 -

8. Exmle for n = without n symtote If we now tke =, r = nd n=, then the initil vlue roblem becomes y =, y( 0 = dx y with comlete solution x + ( + e ( 0 y( x = x ( + e ( 0 Figure 9 Fmily of solutions for =, r =+, n= This time there re no symtotes, becuse r ( n / ( / = is not rel number. There re two fmilies of solution curve: x horizontl trnsltions of y( x = + e (for 0 nd x = (for 0. horizontl trnsltions of y( x e There re two distinct solutions when nd only when = 0. There is no singulr solution. - 5 -

9. The Generl Cse It is esy to show tht this henomenon, (verticl trnsltions in the initil condition cuse horizontl trnsltions in the solution curves, must occur for ll non-singulr solutions of ny first order Bernoulli initil vlue roblem with constnt coefficients (excet for = r = 0 or n =. / In the comlete solution ( ( ( n r n r n x y x = + e, equte the trnsltion ( nx ( c e n r ( n x n r to the term e when > r n ( n x n r (or to e when <. In the first cse, ( nx + ( nc n r ( nx + ( n c n r e e e e n r n r + ( n c = ln c = ln ( n r n A similr clcultion in the second cse leds to c = ln. ( n Therefore ll non-singulr solutions re simle trnsltions of ( y x or of r = + e ( ( n x ( n / ( n / to the right by c n r = ln ( n r n x r n y( x = e to the right by c = ln ( n In ech cse, the trnsltion is ctully to the left if c < 0. when when n r > <. n r The behviour of the exonentil function (s mentioned in section is the fundmentl reson for this feture of the solutions to the Bernoulli initil vlue roblem. The boundry cse between the two fmilies, the horizontl symtote y( x lso the limiting cse of n infinite trnsltion for both fmilies. r ( / n, is - 6 -

All of the non-singulr solutions tend to the sme limiting vlue of cse ( n > 0 (or s x in the cse ( r n < 0, excet when This behviour ersists even for = 0: For = 0, r 0 nd n the generl solution my be re-written s n ( ( n ( ( / / = + =, where ( ( ( ( n y x n rx n r x c c = ( / n r n n r ( s x + in the ( / n is not rel. If r = = 0, then the ODE becomes the trivil = 0, whose solution is just y for ll n. dx If ( n= nd r = then gin the ODE becomes = 0, whose solution is y for ll n. dx Together with the singulr solutions, these re the only excetions to the following conclusion. A verticl trnsltion in the initil vlue y( 0 = results in horizontl trnsltion of nother n solution curve to the Bernoulli ODE with constnt coefficients, y ry dx + =. GLYN GEORGE Fculty of Engineering nd Alied Science, Memoril University of Newfoundlnd, St. John s, NL, Cnd, AC N8 web site: htt://www.engr.mun.c/~ggeorge e-mil: glyn@mun.c Return to the list of ublictions - 7 -