First Order Partial Differential Equations

Transcription

1 Firs Order Parial Differenial Equaions 1. The Mehod of Characerisics A parial differenial equaion of order one in is mos general form is an equaion of he form F x,u, u 0, 1.1 where he unknown is he funcion u u x u x 1,...,x n of n real variables. Here, we will no consider problems of such generaliy bu will focus insead on a smaller class of problems. For example, he equaion (1.1) is said o be a quasilinear equaion in wo variables if i is of he form a,x,u u x, b,x,u x u,x f,x,u, 1.2 i.e., he equaion is linear in he derivaives u and x u bu is nonlinear in u. If f,x,u 0, he equaion is said o be homogeneous. In order o make he noaion more convenien laer, we are choosing o call he independen variables, and x. Suppose u u x, is a smooh soluion of (1.2) and le S,x,u R 3 : u u x,. Then S is said o be a soluion surface for (1.2). The smoohness of he soluion u means ha S has a angen plane a each poin,x,u S. The normal vecor n o he angen plane has he direcion numbers u, x u, 1 ; i.e., u x, u 0 is he equaion of S and u x udx 0 is he equaion of he angen plane. Now consider a curve C s, x x s, u u s, s I soluion curve for he sysem a,x,u, dx b,x,u, du f,x,u, in 3-space defined as a 1.3 If T denoes a vecor angen o C a,x,u hen he direcion numbers of T mus be a,b,f. Bu hen (1.2) implies ha T n, which is o say, T lies in he angen plane o he surface S. Bu if T lies in he angen plane, hen C mus lie in S. Evidenly, soluion curves of (1.2) lie in he soluion surface S associaed wih (1.2). Such curves are called characerisic curves for (1.2). We could also consider a plane curve C B defined by he pair of equaions, a,x,u, dx b,x,u, 1.3 Noe ha if u u x, is a soluion for (1.2) hen u x, a,b a,x,u u x, b,x,u x u x,,u f,x,u, Bu u x, a,b is jus a direcional derivaive of u x, along he direcion angen o C B. Then we can view he PDE (1.2) as an ordinary differenial equaion along he special curve C B ; i.e., a,x,u u x, b,x,u x u x,,u u x, xu x, dx f,x,u. 1

2 We refer o he curves C as "space characerisics" and o he plane curves C B as "base characerisics" for he PDE (1.2). In general, he PDE is solved by solving he ordinary differenial equaions (1.3) for C as a sysem. In cerain special cases, he soluion process can be accomplished by solving he pair of equaions (1.3 ) firs and hen solving he ODE for u separaely. We will recall now some noions from differenial geomery ha will clarify he procedure for solving he sysem (1.3). Inegral Curves for Vecor Fiel A vecor valued funcion, V P,x,u,Q,x,u,R,x,u is called a vecor field if P,Q,R are all smooh funcions and if P 2 Q 2 R 2 is never zero. A space curve, C s, x x s, u u s, s I is said o be an inegral curve or rajecory for V if V is angen o C a every poin; i.e., if or, equivalenly, P,x,u, dx P dx Q Q,x,u, du R R,x,u, 1.4a 1.4b A funcion,x,u is said o be a firs inegral for he vecor field V P,Q,R if P,x,u Q,x,u x R,x,u u The rajecories C for V will be found by represening C as he inersecion of level surfaces of firs inegrals. The level surfaces S j,x,u : j,x,u C j j 1,2 inersec ransversally a each poin if heir normals, n 1 and n 2 are never parallel. This siuaion occurs if 1 and 2 are such ha he expression 1 2 is differen from zero a each poin. In his case he funcions 1 and 2 are said o be funcionally independen and heir level surfaces S 1 and S 2 inersec in a curve C. Since C hen lies in boh of he surfaces, S 1 and S 2, he angen o C is normal o boh n 1 and n 2, ha is o boh 1 and 2. This is he same hing as saying boh 1 and 2 saisfy (1.5). We will illusrae wih examples. Example Consider he radial vecor field V,x,u. A firs inegral mus saisfy,x,u x x,x,u u u,x,u 0. To obain a soluion, we consider he following sysem of ode s Then and dx x u or dx x lea o x C 1, dx x u implies x u C 2. dx x and dx x Tha is, 1,x,u x and 2,x,u u x are a pair of firs inegrals for V,x,u. We can show ha for any smooh funcion F of wo variables, 3,x,u F 1,x,u, 2,x,u u 2

3 is also a firs inegral for V and 3 is hen viewed as an implici represenaion for he mos general soluion of he firs inegral pde. Problem 1.1 Show ha if 1,x,u and 2,x,u are a pair of firs inegrals for V hen 3,x,u F 1,x,u, 2,x,u, where F denoes an arbirary smooh funcion of wo variables, is also a firs inegral for V.. Problem 1.2 Show ha 1,x,u x and 2,x,u u x are funcionally independen. 2. Consider he vecor field V,u,x. The rajecories are soluion curves for he following sysem of ode s From dx u dx u x or x and dx u x i follows ha x 2 u 2 C 1. Then x x u 2 C 1 and C 2 u u 2 C 1 C 2 u x. Evidenly, 1 x 2 u 2 and 2 u x are a pair of firs inegrals for he vecor field V,u,x. Each of hese funcions saisfies,x,u u x,x,u x u,x,u 0, and he mos general soluion of his equaion can be wrien implicily as F x 2 u 2, u x 0, where F denoes an arbirary smooh funcion of wo variables. Characerisics for Quasilinear PDE s of Order 1 We are aware now ha C is a characerisic curve for he quasilinear pde (1.2) if C is a rajecory for he vecor field V a,b, f. Then soluions for he pde can be obained from firs inegrals for he vecor field. However, we are no usually ineresed in finding he mos general soluion for he pde bu are insead ineresed in finding cerain paricular soluions. For example, we shall be ineresed in finding a soluion for (1.2) ha saisfies he addiional condiion ha u x, g x, on C I : x x,. where he curve C I and he funcion g are given. A condiion of his form is called a Cauchy condiion, and he problem of finding a soluion for (1.2) ha saisfies he Cauchy condiion is called a Cauchy iniial value problem. Example For consan, consider he following Cauchy problem u x, x u x, 0, u x,0 1 3

4 The characerisic sysem (1.3) becomes, x s, s 1, u s 0 The base characerisics, C B are soluion curves for, 1, dx. This pair of equaions is equivalen o he single equaion, dx, for which he soluion is, x x 0. The base characerisic curves, C B, are a family of sraigh lines all having he same slope. Then along any base characerisic curve, u x, x u x, u x, x u x, dx from which i follows ha u is a consan on each such curve. To express he fac ha u is consan along base characerisics bu need no equal he same consan on every base characerisic, we can wrie, u x, f x where f f z denoes an arbirary smooh funcion of one variable. Then u x, f x is a general soluion for he parial differenial equaion. Now u x,0 f x and his, combined wih he Cauchy iniial condiion, lea o he paricular soluion u x, 1 1 x 2 for he Cauchy problem. Noe ha he iniial value u 0 u x 0,0 of he soluion a he poin x 0 propagaes along he line x x 0 ; i.e., u x, u 0 a all poins x, such ha x x 0. As a resul, if he iniial daa is specified only on he inerval, say 0 x 10, hen he soluion is deermined only in he srip, x, : x 10, 0. The srip is he domain of influence of he iniial inerval I 0 x 10. The pde in his example is linear which lea o he resul ha he characerisic sysem of ode s uncouples. Tha is, he firs wo equaions are independen of u which means we can solve he equaion x separaely from he equaion u 0. 0, 2. Now consider a Cauchy problem for he variable coefficien equaion u x, x u x, 0, u x,0 1 Here we replaced he consan coefficien wih he variable coefficien,. The coefficiens in his equaion are funcions of he independen variables, x, bu do no depend on he unknown funcion u. Hence he equaion is a linear parial differenial equaion as was he equaion in he previous example. The base characerisics are soluion curves for he sysem s 1, and x s. This is equivalen o he single ode, dx whose soluion is given by, x 2 /2 c 0, or x 2 /2 c 1.Then he base characerisics in his problem are a family of parabolas. The remaining equaion in he characerisic sysem, u s 0, implies ha u is consan along he base characerisic curves. Since u need no have he same consan value on every base characerisic, he general soluion has he form u x, F x 2 /2 for an arbirary smooh funcion of one variable F. Using his in he iniial condiion lea o, u x,0 F x 1 and hen he paricular soluion ha solves he iniial value problem is given by, 4

5 u x, 1 1 x 2 / Finally, consider a Cauchy problem for an inhomogeneous equaion u x, x u x, 4u, u x,0 1 Since he lef side of his equaion is he same as he previous example, his problem will have he same base characerisics, x 2 /2 c 1.The hird equaion in he characerisic sysem rea u 4u, and he soluion is given by u C 2 e 4 along each characerisic. Since he parameer C 2 is consan on each base characerisic, bu no necessarily he same consan on all base characerisics, we wrie u x, F x 2 /2 e 4 for he general soluion of he PDE. The paricular soluion o he iniial value problem is easily found o be u x, e 4 1 x 2 /2 2. Poins worh noing abou hese examples: he soluion curves of he sysem s a x, and x s b x, can be uncoupled from he equaion u s f x,,u and we can solve for x x s, s wihou solving for u. The soluion curves are curves in he x- plane. In order o disinguish hese plane curves from he characerisic curves in 3-space, we refer o he plane soluion curves as base characerisics and use he erm space characerisics when referring o characerisic curves in 3-space. The base characerisics are he projecions ino he x- plane of he characerisic curves for he pde s. Since he pde s in hese examples are linear, (i.e.,he coefficiens a and b do no depend on u, he base characerisics can be deermined wihou finding u. he consan coefficiens in example 1 led o sraigh line base characerisics while he variable coefficien in example 2 led o base characerisics which were parabolas. However even when he base characerisics are curved, he family of curves is coheren; i.e., base characerisics which originae a disinc poins can never cross.(see he remark below) he space characerisic curves in he firs wo examples are curves of he form x x, u cons ; i.e., hey are plane curves lying in planes parallel o he x- plane. This is due o he fac ha in boh examples, he parial differenial equaion is homogeneous and in such cases, he pde reduces o du/ 0 along base characerisics wih he obvious resul ha soluions are consan along base characerisics and along space characerisics. In example 3, he inhomogeneous pde lea o space characerisics ha are no plane curves since in his case du/ was no equal o zero. To see why characerisics for linear equaions are coheren, le C 1 and C 2 denoe disinc base characerisics of he equaion a,x u x, b,x x u,x,u f,x,u, which cross a some poin x 0, 0. Inersecion a a poin requires ha he curves have noncoinciden angens a he poin of inersecion. Bu his is inconsisen wih (1.2) which 5

6 implies ha a x 0, 0 dx C 1 b x 0, 0 dx C 2 i.e., he angens o C 1 and C 2 have equal slopes a he poin of inersecion. In he case of quasilinear equaions, where he coefficiens can depend on u, we will see ha his coherence can fail. 4. Consider a Cauchy problem for a linear bu inhomogeneous equaion x u x, 2x x u x, 2 u, u x,0 x 3. The characerisics are he soluions of x, dx 2x, du 2 u, or x dx 2x, and du 2u dx 2x. Since he equaion is linear, he base characerisic curves can be obained independen of he soluion u. The soluions for he characerisic equaions are given by x 2 C 0 and x u C 1. Since he parameer C 1 is consan on any base characerisic bu may have disinc values on differen characerisics, he general soluion for he parial differenial equaion can be expressed as Then u C 1 x F x 2 x. u x,0 F x x x 3. implies ha F x x 4 hence he paricular soluion for he iniial value problem is, u x, x 2 4 x. In each of hese linear examples, he soluion procedure was he same: firs find he base characerisics and hen find he soluion of he parial differenial equaion. Noe ha in each case, he soluion obained was a global soluion in he sense ha i exiss and saisfies he pde for all 0.We shall see ha in quasilinear problems, he soluion may no be global in. 4. Consider he following quasilinear problem, u x, x u x, u x, 2, u x,0 1 The characerisic sysem can be wrien as 1 dx or u 2 1 dx and 1 u 2. The pde is linear in he leading erms (in fac, his special subclass of quasilinear problems is referred o as he class of semilinear problems) so ha he characerisic sysem uncouples and he base characerisics can be found wihou knowing he soluion u. Clearly 6

7 he base characerisics are he family of sraigh lines x x 0. The remaining characerisic equaion u u 2 has he soluion u C 1 1, where C 1 is consan along he base characerisics. Since C 1 need no be he same for all base characerisics, i follows ha he mos general soluion for he pde can be wrien as, u x, 1 F x. The iniial condiion implies u x,0 1 F x 1 1 x, 2 of he Cauchy iniial value problem is u x, 1 x 2 1 or F x 1 x2. Then he soluion Noe ha in spie of he smoohness of he iniial daa, he soluion develops a singulariy a x, 1. Ploing soluion profiles a imes approaching 1 shows ha he soluion behaves like a wave ha propagaes from lef o righ and sharpens o a spike a x, as approaches 1. u(x,) versus x for 0,.5,.9 We compue x u x, 2 x 4 1 x from which i is eviden ha he gradien, x u x, has an even sronger singulariy a x 4, 1. The developing gradien singulariy can be seen in he following plo of gradien profiles for increasing. gradien u x, versus x for 0,.5,.9 7

8 Since his pde is semilinear (i.e., linear in is leading erms) and has smooh iniial daa, he sponaneous singular behavior in he soluion mus be due o he nonlinear erm on he righ side of he equaion. Noe also ha semilinear equaions permi he base characerisics (i.e. he soluion curves for s a, x s b ) o be found independen of he soluion u u x, and hen he PDE reduces o a nonlinear ODE along he base characerisics. 5. Consider he quasilinear Cauchy problem, u x, u x u x, 0, u x,0 1 1 x 2 f x. Here he coefficiens of u x, and x u x, depend on u so he characerisic equaions become 1, dx u and du 0, or dx u and du 0. In his case hese equaions do no uncouple. In order o solve he equaions as a sysem, we noe firs he equaion u 0. This implies u C 1 along soluions of he oher equaion, dx/ u. This is o say, u u 0 along base characerisics which are he sraigh lines, x u 0 x 0. Then, as usual, we have a general soluion of he form u x, F x u. I follows from he iniial condiion ha he iniial value u 0 originaing a x 0,0 is given by u x 0,0 F x 0 f x 0. The soluion u can hen be expressed implicily by wriing u x, 1 1 x u 2 and his equaion can be solved for u in erms of x and. The resul, obainable using Maple or Mahemaica, is a complicaed funcion of x,. As in he previous example, he quasilineariy of he pde produces some singular behavior in he soluion. For insance, suppose u u 0 f x 0 along x u 0 x 0 and u u 1 f x 1 along x u 1 x 1 x 0. If u 1 u 0, hen he wo sraigh lines inersec a x, where x 1 x 0 u 1 u 0, x u 1x 0 u 0 x 1 u 1 u 0. Noe ha he ime of inersecion,, is posiive if x 1 x 0 and u 1 u 0. A such a poin of inersecion, u x, has he impossible requiremen of being simulaneously equal o he disinc values u 1 u 0. We conclude ha he soluion breaks down in some way a his poin. Noe furher ha i.e. u x, f x u lea o x u x, f x u 1 f x u. x u x, f x u 1 x u x, ; Evidenly he gradien x u x, becomes undefined a any poin where, 1 f x u 0, anoher indicaion ha he soluion breaks down a some finie ime. 8

9 In each of he las wo examples we have seen equaions wih smooh coefficiens and iniial daa develop sponaneous singulariies due o he nonlineariy of he equaions. The soluions in hese wo examples break down a some finie ime and no classical soluion for he iniial value problems exiss pas his poin of breakdown. I will be necessary o weaken he noion of soluion in order for hese nonlinear problems o be solvable globally. Finally, consider following inhomogeneous example. 6. Consider he inhomogeneous quasilinear Cauchy problem, u x, u x u x, 1, u x,0 x. Here he characerisic equaions become 1, dx so we can rewrie hese as u and du 1, Then we arrive a firs inegrals, 1, 2 dx du u and du 1. 1 x,,u u C 1 and 2 x,,u x 1 2 u2 C 2 which lea o x 1 2 u2 C C 1 2 C C 1 C 3 and hen a hird inegral of he sysem is 3 x,,u x u C 3 Then a general soluion can be wrien as, i.e., The iniial condiion, 1 x,,u F 3 x,,u ; u F x u x u x,0 0 F x 0 lea o he paricular soluion, u x u which can be solved explicily for u as follows u x, x Noe ha he inhomogeneous equaion led o a simpler soluion han he homogeneous problem in he previous example. In addiion, his soluion has no singulariy for 0. Problem 1.3 Consider he Cauchy iniial value problem 9

10 u x, x x u x, x u x,, u x,0 1 Solve he characerisic equaions and find he equaion of he base characerisic ha passes hrough he poin 1,2. Find he mos general soluion of he pde. Find he paricular soluion ha saisfies he iniial condiion and describe he se of poins x, where he soluion is deermined if he iniial values are specified only for 1 x 1. Problem 1.4 Consider he Cauchy iniial value problem 4x u x, 9 x u x, x u, u x,0 x2 Solve he characerisic equaions o find wo firs inegrals of he associaed vecor field. Find he mos general soluion of he pde Find he paricular soluion ha saisfies he iniial condiion. Does he soluion have any singular behavior? Problem 1.5 Consider he Cauchy iniial value problem u u x, x x u x,, u 1, 2 Solve he characerisic equaions o find wo firs inegrals of he associaed vecor field. Find he mos general soluion of he pde. Find he paricular soluion ha saisfies he iniial condiion. Does he soluion have any singular behavior? 10