# Conservation Law. Chapter Goal. 5.2 Theory

Size: px
Start display at page:

## Transcription

1 Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very importnt conservtion lw used in mny lodels. We use it to derive more specific models. 5.2 Theory Mny PDE models involve the study of how certin quntity chnges with time nd spce. This chnge follows bsic lw clled the conservtion lw. Simply put, this lw sys tht the rte t which quntity chnges in given domin must equl the rte t which the quntity flows cross the boundry of tht domin plus the rte t which the quntity is creted or destroyed, inside the domin. For exmple, consider the study of the popultion of certin niml species within fixed geogrphic re (our domin). The popultion in this geogrphic re will be determined by how mny nimls re born, how mny die, how mny migrte in nd out. The conservtion lw pplied to this exmple sys tht the rte of chnge of the niml popultion is equl to the rte t which nimls migrte in the region minus the rte t which they migrte out plus the birth rte minus the deth rte. Similr sttements cn be mde bout mny other quntities such s het energy, the mss of chemicl, the number of utomobiles on freewy,... Now, we trnsform such sttement into equtions, tht is we quntify it. For this, let u (x, t) denote the density of certin quntity (mss, energy, nimls,...). Recll tht density is mesured in mount of quntity per unit volume or per unit length. So tht if we know the density of the quntity nd the volume of the region where it is contined, then we lso know the mount of the quntity. Let us ssume for now tht ny vrition in the density be restricted to one sptil dimension we will cll x. Tht is, we ssume one-dimensionl 25

2 26 CHAPTER 5. CONSERVATION LAW Figure 5.1: Tube with cross-sectionl re A domin, ech cross section being lbeled by the sptil vrible x. Figure 5.1 illustrtes this ide in the cse of tube s our domin. It s cross-sectionl re is clled A. We ssume tht the lterl sides re insulted so tht the quntity being studied only vries in the x-direction nd in time. For ech vlue of x, u (x, t) does not vry within the cross section t x. Remrk 30 Let us mke some remrks nd introduce further nottion: 1. The domin described here hs constnt cross-sectionl re A. In more complex domin, the cross-sectionl re might depend on x (see problems). 2. The mount of the quntity t time t in smll section of width dx will be u (x, t) Adx for ech x. It follows tht the mount of the quntity in n rbitrry section x b will be u (x, t) Adx. 3. Let φ = φ (x, t) denote the flux of the quntity t x, ttimet. Itmesures the mount of the quntity crossing the section t x, ttimet. Its units re mount of quntity per unit re, per unit time. So, the ctul mount of the quntity crossing the section t x, ttimet is given by Aφ (x, t). By convention, flux is positive if the flow is to the right, nd negtive if the flow is to the left. 4. Let f (x, t) be the given rte t which the quntity is creted or destroyed within the section t x, ttimet. It is mesured in mount of quntity per unit volume, per unit time. f is clled source if it is positive nd sink if it is negtive. So, the mount of the quntity being creted in smll section of width dx for ech x is f (x, t) Adx per unit time. It follows tht the mount of the quntity being creted in n rbitrry section x b will be f (x, t) Adx. We re now redy to formulte the conservtion lw in smll section of the tube, of re A for ech x such tht x b. The conservtion lw sys tht

3 5.2. THEORY 27 the rte of chnge of the mount of the quntity in tht section must be equl to the rte t which the quntity flows in t x = minus the rte t which it flows out t b plus the rte t which it is creted within the section x b. Using the remrks nd nottion bove, the lw becomes d dt u (x, t) Adx = Aφ (, t) Aφ (b, t)+ f (x, t) Adx (5.1) LetusfirstnoticethtsinceA is constnt, it cn be tken out of the integrls nd cnceled from the formul to obtin d dt u (x, t) dx = φ (, t) φ (b, t)+ f (x, t) dx (5.2) This eqution is the fundmentl conservtion lw. It simply indictes blnce between how much goes in, how much goes out nd how much is chnged. Eqution 5.1 is n integrl eqution. We cn reformulte it s PDE if we mke further ssumptions. We begin by reminding the reder of theorem known s Leibniz rule, lso known s "differentiting under the integrl". Theorem 31 (Leibniz Rule) If (t), b (t), ndf (x, t) re continuously differentible then d dt (t) (t) F (y, t) dy = This is often known s " (t) (t) F t (y, t) dy+f (b (t),t) b (t) F ( (t),t) (t) (5.3) If we ssume tht u hs continuous prtil derivtives, then using Leibniz rule, the left side of eqution 5.2 becomes d dt u (x, t) dx = u t (x, t) dx If we lso ssume tht φ hs continuous prtil derivtives, then using the fundmentl theorem of clculus, we cn write φ (, t) φ (b, t) = Therefore, eqution 5.2 cn be rewritten φ x (x, t) dx [u t (x, t)+φ x (x, t) f (x, t)] dx =0 It is possible for n integrl to be 0 without the integrnd being equl to 0. However, in our cse, the intervl of integrtion ws rbitrry. In other words,

4 28 CHAPTER 5. CONSERVATION LAW we re sying tht no mtter wht nd b re, this integrl must be 0. Since the integrnd is continuous, it follows tht it must be 0 in other words or u t (x, t)+φ x (x, t) f (x, t) =0 u t (x, t)+φ x (x, t) =f (x, t) (5.4) It is importnt to remember tht this eqution ws obtined under the ssumption tht u nd φ re continuously differentible. Eqution 5.4 will be clled the fundmentl conservtion lw. Summry 32 We study how certin quntity chnges with time in given region. We mke the following ssumptions: 1. u (x, t) denotes the density of the quntity being studied. u is ssumed to be continuously differentible. 2. φ (x, t) is the flux of the quntity t time t t x. It mesures the mount of the quntity crossing cross section of our region t x. φ is ssumed to be continuously differentible. 3. f (x, t) is the rte t which the quntity is creted or destroyed withing our region. f is ssumed to be continuous. 4. We ssume the quntity being studied only vries in the x direction. 5. Then, the eqution describing how our quntity chnges with time in the given region is u t (x, t)+φ x (x, t) =f (x, t) Remrk 33 Let us mke few remrks before looking specific exmples. 1. Eqution 5.4 is often written s for simplicity. u t + φ x = f 2. The functions φ nd f re functions of x nd t. Tht dependence my be through the function u. For exmple, we my hve f = f (u). Similrly, we my hve φ = φ (u). These dependencies my led to nonliner model. We will see one in the exmples. 3. Eqution 5.4 involves two unknown functions: u nd φ, usully the source f is ssumed to be given. This mens tht nother eqution relting u nd φ is needed. Such n eqution usully rises from physicl ssumptions on the medium.

5 5.3. ADVECTION EQUATION Eqution 5.4 is in its most generl form. As we look t specific models, it will tke on different forms. For exmple, when we tlked bout flow in this section, we did not specify how the quntity trveled. In the next sections we will consider vrious possibilities including dvection nd diffusion. Also, the source term cn tke on fifferent forms. We now look t specific exmples. 5.3 Advection Eqution Advection refers to trnsport of certin substnce in fluid (wter, ny liquid, ir,...). An exmple of dvection is trnsport of pollutnt in river. The flow of the river crries the pollutnt. A model where the flux is proportionl to the density is clled n dvection model. It is esy to understnd why. Thinking of the exmple of the river crrying pollutnt, the mount of pollutnt which crosses the boundry of given region in the river clerly depends on the density of the pollutnt. In this cse,wehveforsomeconstntc: φ = cu The constnt c is the speed of the fluid. In our exmple bove, it will be how fst the river flows. Eqution 5.4 becomes: We look t specific exmples. u t + cu x = f (x, t) Advection When f (x, t) =0(No Source) In the bsence of sources, Eqution 5.4 becomes: u t + cu x =0 (5.5) Eqution 5.5 is clled the dvection eqution. Note tht c must hve velocity units (length per time). As noted bove, c is the speed t which the fluid is flowing. Proposition 34 If F is ny differentible function, then F (x ct) is solution of 5.5 for ny vlue of t. Proof. Let u (x, t) =F (x ct) nd ssume F is differentible. Then u t = F (x ct) (x ct) t = cf (x ct)

6 30 CHAPTER 5. CONSERVATION LAW Similrly Thus,weseetht u x = F (x ct) = F (x ct) (x ct) x u t + cu x = cf (x ct)+cf (x ct) = 0 Such solutions re clled right-trveling wves. The grph of F (x ct) is the grph of F (x) shifted ct sptil units to the right. This shows tht whtever the shpe of the initil solution is, it will trvel to the right s time increses. Once we know the initil condition u (x, 0), welsoknowthesolutionu (x, t), s indicted below. The pure initil vlue problem for the dvection eqution is { ut + cu x =0 x R t>0 u (x, 0) = u 0 (x) x R (5.6) Where u 0 (x) is given initil density. From proposition 34, it follows tht u (x, t) =u 0 (x ct) is the solution to the problem given by equtions 5.6. Exmple 35 Consider the problem { ut +2u x =0 x R t>0 u (x, 0) = sin x x R Then u (x, t) =sin(x 2t) is the solution to this problem Exmple: Advection nd Decy Recll from elementry differentil equtions tht decy is modeled by the lw du dt = λu where λ is the decy rte. For exmple, substnce dvecting through tube t velocity c, nd decying t rte of decy λ would be modeled by the dvectiondecy eqution u t + cu x = λu (5.7) Here, λu corresponds to the source term (the function f in eqution 5.4.

7 5.4. METHODS OF CHARACTERISTICS Generl Advection eqution In its most generl form, the dvection eqution is where: nd c re constnts. u t + cu x + u = f (x, t) (5.8) cu x is the term which corresponds to the flux. Recll, in the dvection model, the flux φ is φ = cu, c being the speed of the flow of the fluid in the dvection model. u nd f (x, t) correspond to the source. u sys tht the rte of chnge of the quntity within the domin is proportionl to the quntity. Exmples include rdioctive decy, popultion growth. f (x, t) sys tht the rte of chnge of the quntity within the domin is some given function. An exmple of this would be pollutnts flowing in river, some pollutnts being dumped withing the domin. f (x, t) would give the rte t which the pollutnt is being dumped. Exmple 36 If the eqution is u t +cu x +u = f (x, t), then it mens tht there is dvection (cu x ), decy or growth (u) swellsquntitybeingddedt rte given by f (x, y). Exmple 37 If the eqution is u t + cu x + u =0, then it mens tht there is dvection nd growth or decy. Exmple 38 If the eqution is u t + cu x =0, then then it mens tht there is only dvection. 5.4 Methods of Chrcteristics Eqution of the Form u t + cu x + u = f (x, t) We now look t n importnt method used to solve PDE s, the method of chrcteristics. We see how it is used to solve generl dvection eqution of the form u t + cu x + u = f (x, t) The purpose of this technique is to write this eqution s first-order liner ODE nd use the integrting fctor technique to solve such n eqution. The first step is to chnge eqution 5.8 into first-order liner ODE. For this, we perform chnge of vribles. We define new independent vribles ξ nd τ, clled chrcteristic coordintes (hence the nme of this technique) s follows: ξ = x ct (5.9) τ = t

8 32 CHAPTER 5. CONSERVATION LAW Remrk 39 In cse you forgot your Greek lphbet, ξ is xi nd τ is tu. One cn think of this new coordinte system s moving coordinte system. We need to rewrite the eqution in terms of the new vrible. Using equtions 5.9, we hve u (x, t) =u (ξ + ct, t) This gives function in ξ nd τ. WecllthisfunctionU, thtisu (x, t) = u (ξ + ct, τ) =U (ξ,τ) or u (x, t) =U (x ct, t). We lso need to express the originl derivtives in terms of the new vribles. By the chin rule, we hve nd Eqution 5.8 becomes u t = U ξ ξ t + U τ τ t = cu ξ + Uτ u x = U ξ ξ x + U τ τ x = U ξ cu ξ + Uτ + cu ξ + U = F (ξ,τ) where F (ξ, τ) =f (x, t). Simplifying it gives us U τ + U = F (ξ, τ) The next step is to solve this new eqution. It only hs one derivtive. We cn tret it s n ODE. This is liner ODE. You will recll from your ODE clss tht such n eqution cn be solved by multiplying it by the integrting fctor e τ.weobtin U τ e τ + Ue τ = e τ F (ξ,τ) The left side is τ (Ueτ ). So, our eqution becomes τ (Ueτ )=e τ F (ξ, τ) which we cn solve by integrtion. Integrting ech side with respect to τ gives tht is τ (Ueτ ) dτ = e τ F (ξ,τ) dτ Ue τ = e τ F (ξ, τ) dτ Solving for U gives U = e τ e τ F (ξ,τ) dτ

9 5.4. METHODS OF CHARACTERISTICS 33 Of course, in the generl cse, this is how fr we go. In specific cses, we integrte. The lst step is to rewrite the solution in terms of the originl vribles x nd t. We outline these steps below. Algorithm 40 (Methods of Chrcteristics) Tosolvenequtionofthe form u t + cu x + u = f (x, t) where f is given, nd c re constnts, following the steps below: Step 1 Chnge eqution 5.8 into first order liner ODE by performing the chnge of vribles indicted in equtions 5.9 Step 2 The new eqution becomes U τ + U = F (ξ,τ). Step 3 Solve the new eqution using the integrting fctor technique (multiply by e τ.) Step 4 Rewrite the solution in terms of the originl vribles. We illustrte this with n exmple. Exmple 41 Find the generl solution of u t +2u x u = t. If we let ξ = x 2t nd τ = t, then our eqution becomes Multiplying by e τ gives U τ U = τ e τ U τ e τ U = τe τ Which is Therefore ( e τ U ) = τe τ τ Ue τ = τe τ dτ Integrting the right side by prts gives Ue τ = (1 + τ) e τ + C Since U is function of both ξ nd τ, nd we were integrting with respect to τ, constnt is ny function of ξ. Therefore Ue τ = (1 + τ) e τ + g (ξ) where g is n rbitrry function of ξ. Solving for U nd rewriting the nswer in terms of x nd t gives u (x, t) = (1 + t)+g (x 2t) e t To find g, we need more informtion such s the initil condition.

10 34 CHAPTER 5. CONSERVATION LAW Exmple 42 Solve the pure initil vlue problem { ut +2u x u = t x R t>0 u (x, 0) = sin x x R This is the sme problem s bove. The generl solution is u (x, t) = (1 + t)+ g (x 2t) e t.wewntu (x, 0) = sin x. But u (x, 0) = 1+g (x) (simply replce t by 0 in the generl solution. Therefore, we must hve tht is g (x) 1=sinx g (x) =sinx +1 It follows tht u (x, t) = (1 + t)+e t (sin (x 2t)+1) The reder will verify tht this solves the initil vlue problem given Eqution of the Form u t + c (x, t) u x = f (x, t, u) A similr method s the one described bove cn be pplied. Here, we do the following chnge of vribles: { ξ = v (x, t) τ = t Where v (x, t) =C is the generl solution of the ODE dx = c (x, t). We illustrte dt this with n exmple. Exmple 43 Find the generl solution of u t +2tu x =0. Here, c (x, t) =2t. To find v (x, t), wesolve dx =2t. The generl solution is dt x = t 2 + C or x t 2 = C. Thus, v (x, t) =x t 2. We perform the chnge of coordintes So, using U (ξ, τ) =u (x, t), wehve nd ξ = x t 2 τ = t u t = U ξ ξ t + U τ τ t = 2tU ξ + U τ u x = U ξ ξ x + U τ τ x = U ξ

11 5.5. CONCLUSION 35 Thus u t +2tu x =0becomes Tht is 2tU ξ + U τ +2tU ξ =0 U τ =0 Hence U (ξ,τ) =g (ξ) for some rbitrry function g. If we rewrite the solution in terms of the originl vribles, we hve u (x, t) =g ( x t 2) To find g, we need more informtion such s the initil condition. 5.5 Conclusion In this chpter, we derived the eqution for the very importnt conservtion lw. We pplied it to derive the eqution corresponding to the dvection model. We lso looked how to solve some simple equtions nd in the process introduced the method of chrcteristics. 5.6 Problems 1. How does the conservtion lw in chnge if the tube hs vrible crosssectionl re A = A (x) insted of constnt one? 2. Solve the initil vlue problem { ut + cu x =0 x R t>0 u (x, 0) = e x2 x R Pick c =2nd sketch the solution surfce using your fvorite softwre t severltimesnpshots. 3. Find the generl solution of the trnsport eqution u t +cu x =0.Useyour generl solution to solve the initil vlue trnsport eqution { ut 2u x =0 x R t>0 u (x, 0) = sin x x R 4. Find the generl solution of eqution 5.7 by using the chnge of coordintes ξ = x ct nd τ = t nd proceeding s we did for the methods of chrcteristics. 5. Show tht the decy term in eqution 5.7 cn be removed by mking chnge of the dependent vrible to w = ue λt

12 36 CHAPTER 5. CONSERVATION LAW 6. Using the methods of chrcteristics, solve the pure initil vlue problem { ut + u x 3u = t x R t>0 u (x, 0) = x 2 x R

### Conservation Law. Chapter Goal. 6.2 Theory

Chpter 6 Conservtion Lw 6.1 Gol Our long term gol is to unerstn how mthemticl moels re erive. Here, we will stuy how certin quntity chnges with time in given region (sptil omin). We then first erive the

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

HW 2 Comments: L1-3. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the

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

1 1.1. Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem. 1.1.1. Overview

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

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

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

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

### 7.2 The Definite Integral

7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

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

Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

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

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

### The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### Chapters 4 & 5 Integrals & Applications

Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

### MA 124 January 18, Derivatives are. Integrals are.

MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

### 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 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt

### 3 Conservation Laws, Constitutive Relations, and Some Classical PDEs

3 Conservtion Lws, Constitutive Reltions, nd Some Clssicl PDEs As topic between the introduction of PDEs nd strting to consider wys to solve them, this section introduces conservtion of mss nd its differentil

### The Wave Equation I. MA 436 Kurt Bryan

1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

### Math 113 Exam 1-Review

Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

### 10 Vector Integral Calculus

Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

### 1 The fundamental theorems of calculus.

The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

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

7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

### Math Calculus with Analytic Geometry II

orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

### PDE Notes. Paul Carnig. January ODE s vs PDE s 1

PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................

### Section 14.3 Arc Length and Curvature

Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

### Heat flux and total heat

Het flux nd totl het John McCun Mrch 14, 2017 1 Introduction Yesterdy (if I remember correctly) Ms. Prsd sked me question bout the condition of insulted boundry for the 1D het eqution, nd (bsed on glnce

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

### Math 5440 Problem Set 3 Solutions

Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 213 1: (Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping

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

MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

### ODE: Existence and Uniqueness of a Solution

Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

### 1 The fundamental theorems of calculus.

The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Theorem Suppose f is continuous

### The Regulated and Riemann Integrals

Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

### Overview of Calculus I

Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

### We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn

### Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

### MATH , Calculus 2, Fall 2018

MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

### The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

### Review of Riemann Integral

1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

### 1.9 C 2 inner variations

46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for

### 4.4 Areas, Integrals and Antiderivatives

. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

### 1 1D heat and wave equations on a finite interval

1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion

### 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?

Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

### Math 8 Winter 2015 Applications of Integration

Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

### Math 5440 Problem Set 3 Solutions

Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 25 1: Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping

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

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

### The practical version

Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

### The Basic Functional 2 1

2 The Bsic Functionl 2 1 Chpter 2: THE BASIC FUNCTIONAL TABLE OF CONTENTS Pge 2.1 Introduction..................... 2 3 2.2 The First Vrition.................. 2 3 2.3 The Euler Eqution..................

### Math 31S. Rumbos Fall Solutions to Assignment #16

Mth 31S. Rumbos Fll 2016 1 Solutions to Assignment #16 1. Logistic Growth 1. Suppose tht the growth of certin niml popultion is governed by the differentil eqution 1000 dn N dt = 100 N, (1) where N(t)

### 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

The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

### MATH 144: Business Calculus Final Review

MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

### Indefinite Integral. Chapter Integration - reverse of differentiation

Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

### Improper Integrals, and Differential Equations

Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

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

Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

### Math 124A October 04, 2011

Mth 4A October 04, 0 Viktor Grigoryn 4 Vibrtions nd het flow In this lecture we will derive the wve nd het equtions from physicl principles. These re second order constnt coefficient liner PEs, which model

### x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

### Math 100 Review Sheet

Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

### Final Exam - Review MATH Spring 2017

Finl Exm - Review MATH 5 - Spring 7 Chpter, 3, nd Sections 5.-5.5, 5.7 Finl Exm: Tuesdy 5/9, :3-7:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.

### Week 10: Line Integrals

Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.

### df dt f () b f () a dt

Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem

### Interpreting Integrals and the Fundamental Theorem

Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

### Line and Surface Integrals: An Intuitive Understanding

Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of

### Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles

### 4. Calculus of Variations

4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the

### Introduction to the Calculus of Variations

Introduction to the Clculus of Vritions Jim Fischer Mrch 20, 1999 Abstrct This is self-contined pper which introduces fundmentl problem in the clculus of vritions, the problem of finding extreme vlues

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

Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

### Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

### 1 Probability Density Functions

Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

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

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t

### f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

### MA Handout 2: Notation and Background Concepts from Analysis

MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,

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

Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

### Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b

Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite

### Calculus of Variations

Clculus of Vritions Com S 477/577 Notes) Yn-Bin Ji Dec 4, 2017 1 Introduction A functionl ssigns rel number to ech function or curve) in some clss. One might sy tht functionl is function of nother function

### Main topics for the First Midterm

Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

### SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

### 1.3 The Lemma of DuBois-Reymond

28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBois-Reymond We needed extr regulrity to integrte by prts nd obtin the Euler- Lgrnge eqution. The following result shows tht, t lest sometimes, the extr

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

Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5. - 5.3) Remrks on the course. Slide Review: Sec. 5.-5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description

### MAT187H1F Lec0101 Burbulla

Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

### INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

### Section 4: Integration ECO4112F 2011

Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

### Definite integral. Mathematics FRDIS MENDELU

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

### The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

### p(t) dt + i 1 re it ireit dt =

Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

### Math 116 Calculus II

Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

### 20 MATHEMATICS POLYNOMIALS

0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

### Chapter 6 Techniques of Integration

MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

### Mathematics Tutorial I: Fundamentals of Calculus

Mthemtics Tutoril I: Fundmentls of Clculus Kristofer Bouchrd September 21, 2006 1 Why Clculus? You ve probbly tken course in clculus before nd forgotten most of wht ws tught. Good. These notes were put

### Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

### Math& 152 Section Integration by Parts

Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

### Topic 1 Notes Jeremy Orloff

Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble

### P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)

1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

### potentials A z, F z TE z Modes We use the e j z z =0 we can simply say that the x dependence of E y (1)

3e. Introduction Lecture 3e Rectngulr wveguide So fr in rectngulr coordintes we hve delt with plne wves propgting in simple nd inhomogeneous medi. The power density of plne wve extends over ll spce. Therefore

### 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

Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since

### 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

Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

### Chapter 0. What is the Lebesgue integral about?

Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

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

REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

### Chapter 6 Notes, Larson/Hostetler 3e

Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

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

Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

### MAT 168: Calculus II with Analytic Geometry. James V. Lambers

MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd