First Order Partial Differential Equations
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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 Fx,u,u 0, 1.1 where he unknown is he funcion u ux ux 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 ux, 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 ux, is a smooh soluion of (1.2) and le S,x,u R 3 : u ux,. 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., ux, u 0 is he equaion of S and u x udx du 0 is he equaion of he angen plane. Now consider a curve C s, x xs, u us, s I in 3-space defined as a soluion curve for he sysem a,x,u, dx b,x,u, du f,x,u, 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, whichisosay, 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). Noe ha if C is a soluion curve for (1.3) hen a, x,u ux, b,x,u x ux,,u ux, xux, dx du f,x,u, so he pde reduces o an ode along C. In general, he equaions for C mus be solved as a 1
2 sysem. 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 xs, u us, 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 P,x,u, dx or, equivalenly, P dx Q Q,x,u, du 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 aresaiobefuncionally 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 dx x du u or dx x and dx x du u Then dx x lea o x C 1, 2
3 and dx x du u implies x u C 2. 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 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 dx u du x or du x and dx u du x From dx u du x i follows ha x 2 u 2 C 1. Then du x du u 2 C 1 and C 2 u u 2 C 1 C 2u 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 ux, gx, 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. 3
4 Example Consider he following Cauchy problem ux, 4 x ux, 0, ux,0 1 1 x 2. Le C 0 denoe a soluion curve for, 1, dx 4. This pair of equaions is equivalen o he single equaion, dx 4, for which he soluion is, x 4 x 0. The soluion curves, C 0, are a family of sraigh lines all having he same slope. Then along any soluion curve, ux, 4 x ux, ux, x ux, dx du 0, from which i follows ha u is a consan on each such curve. A general soluion for he parial differenial equaion is given by ux, fx 4 where f fz denoes an arbirary smooh funcion of one variable. Then ux,0 fx and his, combined wih he Cauchy iniial condiion, lea o he soluion ux, 1 1 x 4 2 for he Cauchy problem. Noe ha he iniial value u 0 ux 0,0 of he soluion a he poin x 0 propagaes along he line x 4 x 0 ; i.e., ux, u 0 a all poins x, such ha x 4 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, :4 x 4 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, we can solve he equaion x 4 separaely from he equaion u Now consider a Cauchy problem for he variable coefficien equaion ux, x x ux, 0, ux,0 sinx. The coefficiens in his equaion are funcions of he independen variables in he problem 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 soluion curves for he characerisic ode, dx x are given by, lnx 2 /2 c 0, or x c 1 e 2 /2. 2 /2 Evidenly, he soluion curves are he level curves of x, xe and since he pde reduces o he ode u s 0 along level curves of, he soluion u of he parial differenial equaion is consan along hese curves. The mos general such soluion has he form ux, f xe 2 /2 for an arbirary smooh funcion of one variable f. Using his in he iniial condiion lea o, ux,0 fx sinx and ux, sin xe 2 /2. 4
5 Poins worh noing abou hese wo examples: he soluion curves C 0 consruced for hese wo examples are no characerisic curves for he pde s in he examples, since hey are plane curves lying in he x- plane. In fac, hey are he projecions ino he x- plane of he characerisic curves for he pde s. Since he coefficiens in he pde s in hese linear examples do no depend on he soluion u, he characerisic sysem 1.3 or 1.4 can be uncoupled and we can solve for x xs, s wihou solving for u. The soluion curves are curves in he x- plane. In order o disinguish hese plane curves from he previously discussed characerisic curves in 3-space, we will refer o he plane soluion curves as base characerisics and use he erm space characerisics when referring o characerisic curves in 3-space. he domain of influence for an inerval I a,b is he srip x, : 0, ae 2 /2 x be 2 /2 bounded by he base characerisics ha originae a he endpoins of he iniial inerval. he space characerisic curves in hese 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. consan coefficien equaions have sraigh line base characerisics while linear equaions wih variable coefficiens have base characerisics which are curves. 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. To see why characerisics for linear equaions are coheren, le C 1 and C 2 denoe disinc base characerisics of he equaion a,x ux, 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 implies ha ax 0, 0 dx C 1 bx 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. 3. Consider a Cauchy problem for a linear bu inhomogeneous equaion x ux, 2x x ux, 2u, ux,0 x 3. The characerisics are he soluions of 5
6 x, dx 2x, du 2u, 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. The general soluion for he parial differenial equaion can be expressed as and hen u C 1 x Fx 2 x ux,0 Fx x x 3. I follows ha Fx x 4 hence ux, x 2 4 x. When he parial differenial equaion is inhomogeneous, as in his example, he soluion u is no longer consan along he characerisics, bu in fac varies in a manner prescribed by he differenial equaion, u s 2u, and he iniial condiion. Then he space characerisics are no longer plane curves bu are ruly space curves. Noice ha he parial differenial equaion reduces o an ordinary differenial equaion along he characerisics. 4. Now consider he following quasilinear problem, ux, 4 x ux, ux, 2, ux,0 1 1 x. 2 The characerisic sysem can be wrien as 1 dx 4 du or u 2 1 dx 4 and 1 du 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 he base characerisics are he family of sraigh lines x 4 x 0. The second characerisic equaion u u 2 has he soluion u C 1 1 and i follows ha he mos general soluion for he pde can be wrien as, ux, 1 fx 4. The iniial condiion implies ux,0 1 fx 1 1 x, or fx 1 2 x2. Then he soluion of he Cauchy iniial value problem is 6
7 ux, 1 x Noe ha in spie of he smoohness of he iniial daa, he soluion develops a singulariy a x 4, 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 4, as approaches 1. We compue x ux, u(x,) vs x for 0,.2,.6,.9 2x 4 1 x from which i is eviden ha he gradien, x ux, 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 of u(x,) vs x for 0,.2,.6,.9 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 ux, and hen he PDE reduces o a nonlinear ODE along he base characerisics. 5. Consider he quasilinear Cauchy problem, 7
8 ux, u x ux, 0, ux,0 1 1 x 2 fx. Here he coefficiens of ux, and x ux, 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 cons along soluions of he oher equaion, dx/ u. This is o say, u u 0 along he sraigh lines, x u 0 x 0. I follows from he iniial condiion ha he iniial value u 0 originaing a x 0,0 is given by ux 0,0 u 0 fx 0. The soluion u can hen be expressed implicily by wriing ux, 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 fx 0 along x u 0 x 0 and u u 1 fx 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, ux, 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 ux, fx u lea o x ux, f x u1 x ux,; i.e. x ux, f x u 1 f x u. Evidenly he gradien x ux, becomes undefined a any poin where, 1 f x u 0, anoher indicaion ha he soluion breaks down a some finie ime. By ploing he soluion profiles versus x for a sequence of increasing imes, i can be seen ha he iniial wave form moves o he righ, deforming as i propagaes. I is apparen ha a some poin, he angen line o he profile becomes verical and he soluion is no longer single valued beyond his poin. 8
9 Soluion Profiles a 0, 1/2, 1, 3/2 The sponaneously occuring singulariy is also eviden from ploing he gradien versus x a increasing imes. We can see ha as ime increases, he gradien begins o develop a negaively infinie singulariy as he angen line en oward he verical. Here, as in he previous example, he singulariy is due o he nonlineariy of he equaion (which in his case lea o colliding characerisics). Gradien Profiles a 0, 1/2, 1, 3/2 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. Problem 1.3 Consider he Cauchy iniial value problem ux, x x ux, xux,, ux,0 1 1 x. 2 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. 9
10 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 ux, 9 x ux, x u, ux,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 ux, x x ux,, u1, 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
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