The Transport Equation

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1 The Transpor Equaion Consider a fluid, flowing wih velociy, V, in a hin sraigh ube whose cross secion will be denoed by A. Suppose he fluid conains a conaminan whose concenraion a posiion a ime will be denoed by u(,). Then a ime, he amoun of conaminan in a secion of he ube beween posiions, 1 and is given by he epression u, Ad amoun of conaminan in, a ime Similarly, we can wrie an epression for he amoun of conaminan ha flows hrough a plane locaed a posiion,, during he ime inerval from 1 o u, AVd amoun of conaminan flowing hrough a plane a posiion,, during he inerval 1, Then an equaion epressing a maerial balance for he conaminan can be wrien as follows, u, Ad u, Ad u, AVd u, AVd i.e., he amoun of conaminan in he secion 1, a ime equals he amoun of conaminan in he secion 1, a ime 1, plus he amoun of conaminan ha flowed hrough he plane a posiion 1 during he ime inerval 1, minus he amoun of conaminan ha flowed hrough he plane a posiion during he ime inerval 1,. Of course his equaion is based on he assumpion ha here are no oher sources of conaminan in he ube and here is no loss of conaminan hrough he walls of he ube. Now he fundamenal heorem of calculus implies ha u, Ad u, Ad u, Add u, AVd u, AVd u, AVdd and, combining hese resuls wih he balance equaion leads o u, A d d u, AV d d 0. If we assume ha his equaliy holds for every segmen 1, in he ube and for each ime inerval 1,, and if he funcion u, and is parial derivaives of order one are coninuous funcions of and, hen i follows from an elemenary propery of coninuous funcions ha u, A u, AV 0 for all,. If he fluid velociy V and he cross secion of he ube, A, are consans, hen his equaion reduces o u, V u, 0 for all,. This is he so called ranspor equaion in one dimension. Since his equaion conains parial derivaives of order a mos equal o one, i is called a firs order parial differenial equaion. Iis,moreover,alinear parial differenial equaion. This erminology relaes o he fac ha if we define an operaor, L, as follows Lu, u, V u, 1

2 hen i is a simple maer o verify ha LC 1 u 1, C u, C 1 Lu 1, C Lu,. Any operaor wih his propery is called a linear operaor (any funcion of one variable f wih he propery ha fc 1 1 c c 1 f 1 c f is a funcion whose graph is a sraigh line), and any parial differenial equaion (PDE) epressing an equaliy for a linear parial differenial operaor is called a linear equaion. Problem 1 Show ha any PDE ha conains erms involving producs of he unknown funcion wih is derivaives is no a linear PDE. Wha is he form of he mos general linear firs order PDE for a funcion of variables, u,? Wha is i for a funcion of 4 variables, u, y,z,? Problem Show ha for any smooh funcion of one variable, F, we have ha u, F V solves u, V u, 0 Devise a similar such soluion for he equaion, u,y,z, V 1 u,y,z, V y u,y,z, V 3 z u,y,z, 0. Consider now he more general siuaion involving he flow of a fluid, flowing wih velociy, v, in a region U in R n. Suppose he fluid conains a conaminan whose concenraion a posiion a ime will be denoed by u,. Then a ime, he amoun of conaminan in an arbirary ball B in U is given by he epression B u, d amoun of conaminan in B a ime Similarly, he ouflow hrough he boundary of he ball during he ime inerval 1, is given by u, v n dsd ouflow hrough he boundary of he ball during B heimeinerval 1, where n denoes he ouward uni normal o B, he boundary he surface of he ball. Now he maerial balance equaion becomes B u, d B u, 1 d B u, v n dsd epressing he fac ha he amoun of conaminan in he ball a ime equals he amoun of conaminan in he ball a ime 1, less ha amoun ha has flowed ou of he ball hrough he boundary. As before, we can use he fundamenal heorem o wrie B u, d B u, 1 d B u, dd Recall now ha he divergence heorem assers ha for an arbirary smooh vecor field, G, Then for G u, v we ge B G n ds B div G d. u, dd B B divu, v d d 0.

3 Problem 3 Show ha for any smooh scalar funcion, u,, and any consan vecor v, divu, v v grad u, I follows from he resul of he problem ha since B is an arbirary ball in U, and 1, is similarly arbirary, hen if u and is derivaives of order one are all coninuous in U, u, v grad u, 0, in U for all. This is he ranspor equaion in n-dimensions. Soluions for Firs Order Equaions Consider firs he problem of finding he general soluion for he equaion u, V u, 0 for all,. By a soluion o he equaion, we mean a funcion, u,, ha is coninuous and has coninuous firs derivaives a all poins (,), and in addiion is such ha he coninuous funcion u, V u, is equal o zero a all poins. If u, has all hese properies, we say ha u, is a classical soluion for he PDE. Suppose ha s s s is he parameric descripion of some curve C in he - plane. Then for u u, a smooh funcion of and, i is clear ha du/ds u, s u, s epresses he derivaive of u along he curve C. In paricular, if he curve C is such ha s V and s 1, i.e., s Vs 0, and s s 0 Then du/ds u, V u, 0, assers ha u is consan along C from which i is apparen ha u u, solves he PDE if and only if u is consan along C. I is also clear ha if V is a consan, hen C is a sraigh line. The parameric represenaion for his sraigh line is hen s Vs 0 s s 0 s Eliminaing he parameer, s, leads o he implici equaion V 0 V 0 : C 0, for he sraigh line passing hrough he poin 0, 0 in he - plane. Clearly, for F any smooh funcion of one variable, u, F V is a smooh funcion of and which is consan along C. Then by our previous observaion, u u, solves he PDE if and only if u, F V for F any smooh funcion of one variable. Noe ha he general soluion o a linear firs order parial differenial equaion conains an arbirary funcion, in conras o he general soluion for a linear firs order ordinary differenial equaion which conains an arbirary consan. Now consider he more general problem of finding he mos general soluion for he equaion u, v u, 0 for all,. 3

4 If s 1s,..., Ns s s is he parameric descripion of some curve C in R N1 hen for u u, a smooh funcion of and, i is clear ha du/ds 1 u, 1 s... N u, N s u, s epresses he derivaive of u along he curve C. If he curve C is such ha s v and s 1, i.e., s vs 0, and s s 0 hen du/ds v u u, 0, assers ha u is consan along C, (obviously C is a sraigh line in R N1. By he previous argumen, he following saemens are equivalen: 1 u u, is a soluion of u, v u, 0 u u, is consan along he sraigh line s vs 0, and s s 0 3 u, f v for f any smooh scalar valued funcion of N variables Problem 4 Show ha he hree saemens are, in fac, equivalen and ha C is a sraigh line if v is a consan. Saemen 3 here assers ha he general soluion for he equaion given in saemen 1 is any funcion of he form u, f v where f is any smooh funcion from R N ino R 1. Evidenly he soluion o his equaion is a very long way from being unique. On he oher hand, consider he problem of finding a funcion u, which saisfies he condiions a) u, v u, F, for all, and all 0, b) u,0 g, for all The condiion a) is an inhomogeneous version of he equaion in saemen 1). The erm inhomogeneous is used o epress he fac ha he righ side of he equaion is no zero and, in fac, conains he forcing funcion, F,. The equaion in saemen 1) is said o be a homogeneous equaion, since he righ side of he equaion is zero. The condiion b) is referred o as an iniial condiion since i specifies he sae variable, u,, a he iniial ime 0. The funcions F, and g are called he daa for he problem and are assumed o be given. We will show ha for each pair of daa funcions, F, and g, we canfindafuncion u, which saisfies boh of he condiions a) and b). Then we say, u, isasoluionoheiniial value problem. Noe ha his consrucion does no gauranee ha he soluion is unique (alhough we may be able o show laer ha he soluion is unique). We will describe he consrucion in he simples case ha N 1. Then he iniial value problem (IVP) assumes he form u, V u, F,,, 0, u,0 g, Firs, le u 1, denoe he soluion of his IVP in he case ha F, 0. The general 4

5 soluion of he homogeneous PDE is any funcion of he form, u 1, f V, where f denoes a smooh funcion of one variable. In order o saisfy he iniial condiion we mus have, u 1,0 f g, from which i follows ha u 1, g V. Here we see ha in order for he homogeneous iniial value problem o have a soluion, he given iniial funcion mus be coninuously differeniable. Now le u, denoe he soluion of he IVP in he case ha g 0 bu F, is no zero. Then u, V u, F,,, 0, u,0 0, The inhomogeneous equaion is equivalen o d/dsu s,s Fs,s, where s V and s 1; i.e., s Vs 0, and s s 0 Then for arbirary parameer values s s 1, s u s,s u s 1,s 1 d/dsus,sds s1 s u s 1,s 1 Fs,s ds s1 If we choose s 0 s 1, hen 0 s, 0 s. Tha is, s V0, and s 0 s 1 V, and s 1 and 0 u, u V,0 F Vs, s ds 0 0 F V, d where we made he change of variable, s in he inegral. Now i is easy o show ha u, u 1, u, g V F V, d 0 saisfies boh of he condiions a) and b); i.e., his is a soluion of he iniial value problem. Problem 5 Verify ha he soluion consruced here does, in fac, saisfy boh condiions in he IVP. Wha are he condiions on F(,) in order for his soluion o be valid? Verify ha in he case of general N, he soluion of he IVP is given by u, g V F V 0, d Suppose N 1, F 0 and ha he iniial daa funcion, g, is such ha Then g 1 if 1 if 1 5

6 u, 1 if V 1 if V 1 formally saisfies he IVP bu since he firs derivaives of u, fail o eis a poins on he line V 1, i canno be said o be a soluion in he classical sense. In he coming weeks we will ry o deermine wheher here is some weaker sense in which his can be said o be a soluion of he IVP. Mehod of Characerisics The mehod of finding a soluion for he ranspor equaion in he previous eamples is a special case of he so called mehod of characerisics. If we consider a more general firs order equaion in variables, A, u, B, u, C, u, F,, hen a curve C in he - plane is said o be a characerisic curve for his PDE if C is a soluion curve for he following sysem of ordinary differenial equaions, s As,s, s Bs,s. Noe ha along any characerisic curve C, he PDE reduces o an ordinary differenial equaion d/dsus,s Cs,sus,s Fs,s. Noe ha alhough he linear PDE reduces o a linear ODE along characerisic curves, he ordinary differenial equaions which produce he characerisics may be nonlinear. In he eamples we have considered, he coefficiens A, B and C were consans and correspondingly, he characerisics urned ou o be sraigh lines. In general, when he coefficiens in he PDE are no consan, hen he characerisics urn ou o be a family of curves. We will reurn o he mehod of characerisics laer when considering conservaion laws. 6

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