Conservation Law. Chapter Goal. 5.2 Theory


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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 onedimensionl 25
2 26 CHAPTER 5. CONSERVATION LAW Figure 5.1: Tube with crosssectionl 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 crosssectionl re is clled A. We ssume tht the lterl sides re insulted so tht the quntity being studied only vries in the xdirection 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 crosssectionl re A. In more complex domin, the crosssectionl 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 righttrveling 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 firstorder liner ODE nd use the integrting fctor technique to solve such n eqution. The first step is to chnge eqution 5.8 into firstorder 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
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