MATH 220: SECOND ORDER CONSTANT COEFFICIENT PDE. We consider second order constant coefficient scalar linear PDEs on R n. These have the form

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1 MATH 220: SECOND ORDER CONSTANT COEFFICIENT PDE ANDRAS VASY We conside second ode constant coefficient scala linea PDEs on R n. These have the fom Lu = f L = a ij xi xj + b i xi + c i whee a ij b i and c ae (complex) constants and f is given. As a geneal pinciple the leading i.e. second ode tems ae the most impotant in analyzing the PDE. The second ode tems then take the fom L = a ij xi xj ; i hee we ae inteested in the case when the a ij ae eal constants. Since xi xj = xj xi we might as well assume that a ij = a ji (othewise eplace both by 1 2 (a ij + a ji ) which does not change L) i.e. that A is symmetic. We ewite L in a diffeent fom. With and we can wite whee x is the tanspose of x : x1 i=1 x = x2... xn a a 1n A =... a n1... a nn L = xa x x = [ x1... xn ]. Technically thee is no need fo the x in the subscipts and one should wite e.g. = [ ] 1... n. Howeve the subscipts ae vey useful below as they enable us to compess (and abuse) the notation in the change of vaiables fomula. We would like to bing it to a simple fom; the simple fom we wish fo may depend on the intended use. As a fist step we would like to have L bought into a diagonal fom i.e. A made diagonal. We can achieve this by changing coodinates. Namely if we let ξ = Ξ(x) i.e. ξ j = Ξ j (x 1... x n ) j = 1... n with invese function X i.e. x = X(ξ) meaning that X Ξ and Ξ X ae both the identity maps (both X and Ξ ae assumed to be C 1 ) then by the chain ule fo any C 1 function u we have Ξ l xk (u Ξ) = ( ξl u) Ξ. x k l=1 1

2 2 ANDRAS VASY We can ewite this in a matix fom: (1) Ξ 1 x 1... x1... =... Ξ 1 xn xn... Ξ n x 1 Ξ n x n so in paticula if ξ is a linea function of x i.e. ξ i = n B ijx j then (2) x = B ξ. Thus ξ 1... ξn L = xa x = ξ BAB ξ. ξ = Bx B = [ b ij ] n i so Hence in ode to make L be given by a diagonal matix in the ξ coodinates we need that BAB is diagonal. Recalling that A is symmetic we poceed as follows. Any eal symmetic matix such as A can be diagonalized by conjugating it by an othogonal matix i.e. thee is an othogonal matix O (ecall that othogonal means that OO = O O = Id i.e. O = O 1 ) and a diagonal matix Λ such that A = O ΛO whee the diagonal enties λ j of Λ ae the eigenvalues of A and the ows of O i.e. the columns of O ae coesponding (unit length) eigenvectos e j. (Thus the othogonality of O is a consequence of the othogonality of the e j.) Choosing B = O we deduce that BAB = OO ΛOO = Λ is indeed diagonal. In fact we can go a bit futhe. We wite sgn t fo the sign of a eal numbe t so sgn t = 1 if t > 0 sgn t = 1 if t < 0 and sgn t = 0 if t = 0. One can facto a diagonal matix with eal enties like Λ by witing its diagonal enties as λ j = λ j 1/2 (sgn λ j ) λ j 1/2. It tuns out that if λ j = 0 fo some j it is convenient to simply wite λ j = 1 sgn λ j 1. Thus if D is the diagonal matix with diagonal enties d j = λ j 1/2 when λ j 0 d j = 1 if λ j = 0 and P is the diagonal matix with diagonal enties p j = sgn λ j then Λ = DP D so A = O DP DO. Now D is invetible since it is diagonal and none of its diagonal enties vanish. We let B = D 1 O hence BAB = (D 1 O)(O DP DO)(O D 1 ) = P. We deduce that fo any L thee is a change of coodinates ξ = Bx such that in the ξ coodinates L = ξ P ξ = p j ξ 2 j. Now the oveall sign of L is not impotant so e.g. if all the p j < 0 the behavio of solutions of Lu = f is the same as those of Lu = g (simply let g = f) and the coesponding coefficients of L ae p j > 0. Also the case when some p j vanish is degeneate and fist ode tems influence the behavio (and existence) of solutions of Lu = f temendously. As an example the heat opeato 2 x t behaves vey diffeently fom the backwad heat opeato 2 x + t. We thus make the following classification of the non-degeneate setting i.e. when det A 0 so each p j is 1 o 1: (i) If all p j (i.e. λ j ) have the same sign L (and L) is called elliptic.

3 3 (ii) If all but one p j have the same sign L is called hypebolic. (iii) If thee ae at least two p j of each sign then L is called ultahypebolic. In cetain ways ultahypebolic equations ae like hypebolic equations (though not as fa as initial conditions on hypesufaces ae concened!); the biggest diffeence is between elliptic equations and the othe two classes. We conclude that by a change of coodinates and possible multiplication by 1 evey non-degeneate second ode linea constant coefficient PDE with eal coefficients in its leading pat can be bought to the following fom: (i) If L is elliptic then L = ξ 2 j + b j ξj + c (ii) If L is hypebolic then n 1 L = ξ 2 j + ξ 2 n + b j ξj + c (iii) If L is ultahypebolic then fo some 2 k n 2 n k L = ξ 2 j + ξ 2 j + b j ξj + c. j=n k+1 The model examples when b j and c vanish ae: (i) If L is elliptic then = ξ 2 j the Laplacian (ii) If L is hypebolic then n 1 = ξ 2 j + ξ 2 n the wave opeato o d Alembetian. We can of couse change ou notation and wok with = 2 x j and = n 1 2 x j + x 2 n. Fo one often wants to think of x n as time and wite it as t and wite x fo (x 1... x n 1 ); then R n = R n 1 + t 2. We stat ou study of second ode equations by studying hypebolic equations in R 2. The advantage of these is that they can be factoed (disegading the fist ode tems) into the poduct of two fist ode opeatos which we aleady know how to analyze. Thus with the slightly moe geneal d Alembetian = 2 t c 2 2 x we facto it as = ( t c x )( t + c x ). With such a factoization it is easy to solve u = 0. Namely we fist let v = ( t + c x )u so the PDE states ( t c x )v = 0. This is a fist ode linea PDE which we solve easily (indeed we solved this in Lectue 2 with slightly diffeent notation) to obtain v(x t) = h(x + ct)

4 4 ANDRAS VASY fo some h. (Explicitly we could impose initial conditions say at t = 0 namely v(x 0) = h(x) and then we get v(x t) = h(x + ct).) Next we need to solve ( t + c x )u = v with v(x t) = h(x + ct). We use the method of chaacteistics again so we have chaacteistic equations with initial data e.g. at the x axis (3) dx = c x( 0) = dt = 1 t( 0) = 0 dz = h(x + ct) u( 0) = φ(). We obtain that t = s x = cs + hence x + ct = 2cs + and = x ct so the ODE fo z is dz = h(2cs + ) z( 0) = φ(). Integating fom s = 0 we deduce that z = 0 h(2cs + ) + φ(). Letting H(s) = h(s ) (so H = h and H() = 0) we deduce that (4) z = 1 H(2cs + ) + φ() 2c and (5) u(x t) = 1 H(x + ct) + φ(x ct) 2c so u is of the fom (6) u(x t) = f(x + ct) + g(x ct). Convesely it is easy to check that evey u of this fom with f g in C 2 solves the PDE so this is ou geneal solution of the wave equation. Note that x + ct = C and x ct = C C a constant ae the chaacteistics of the two factos t c x and t + c x espectively (ecall that is constant along the chaacteistics in ou method to solve fist ode PDEs). Thus the geneal solution is the sum of two waves each of which is constant along the chaacteistics coesponding to one of the factos. We now find the solution with initial conditions: Diectly fom (6) u(x 0) = φ(x) u t (x 0) = ψ(x). f(x) + g(x) = φ(x) cf (x) cg (x) = ψ(x). Diffeentiating the fist line gives f (x)+g (x) = φ (x). Combining with the second line and solving fo f and g yiel f (x) = 1 2 (φ (x) + c 1 ψ(x)) g (x) = 1 2 (φ (x) c 1 ψ(x)).

5 5 Integating yiel f(x) = 1 2 φ(x) + 1 2c g(x) = 1 2 φ(x) 1 2c x 0 x 0 ψ(σ) dσ + A ψ(σ) dσ + B fo some constants A and B substitution into f + g = φ gives B = A. summay u(x t) = f(x + ct) + g(x ct) = 1 2 (φ(x + ct) + φ(x ct)) + 1 2c x+ct x ct ψ(σ) dσ. This is d Alembet s solution of the wave equation. Notice that if φ is C 2 and ψ is C 1 then this indeed satisfies the wave equation in the classical sense and it also satisfies the initial conditions. This method was somewhat ad hoc in that we ignoed the initial conditions at fist. We could have gone though the whole calculation enfocing the initial conditions. The key issue is to find initial conditions fo v. But this is easy: v(x 0) = u t (x 0) + cu x (x 0) = ψ(x) + cφ (x) i.e. the initial condition is v(x 0) = h(x) h() = ψ() + cφ (). Then ou solution is Then in (4) H(s) = and so hence v(x t) = h(x + ct) = ψ(x + ct) + cφ (x + ct). h(s ) = z( s) = 1 2c u(x t) = 1 2c 2cs+ (ψ(s ) + cφ (s )) = x+ct x ct ψ(s ) + c(φ(s) φ()) ψ(s ) + 1 (φ(2cs + ) φ()) + φ() 2 ψ(s ) + 1 (φ(x + ct) + φ(x ct)) 2 in ageement with the pevious calculation. If we ae given a second ode hypebolic linea PDE a 11 u ξξ + 2a 12 u ξη + a 22 u ηη = 0 by a change of coodinates we could always bing it to the fom 2 t 2 x which we just analyzed hence we can solve this. Instead we can also poceed by diectly factoing this equation. Note that hypebolicity means in this case exactly that det A < 0 i.e. that a 11 a 22 a 2 12 < 0. If a 11 = a 22 = 0 then it is vey easy to facto the equation it is just 2a 12 ξ η u = 0. Othewise eithe a 11 o a 22 is non-zeo; assume it is the fome. Then we facto the quadatic polynomial a 11 µ 2 + 2a 12 µ + a 22 = a 11 (µ µ + )(µ µ ) whee µ ± ae the oots of this polynomial so µ ± = 2a 12 ± 4a a 11a 22 2a 11. Note that hypebolicity is exactly the condition that the quantity whose squae oot we ae taking is positive. Then a 11 α 2 + 2a 12 αβ + a 22 β 2 = a 11 (α µ + β)(α µ β) In

6 6 ANDRAS VASY fo all α and β (note that fo β = 0 this is automatic and fo β 0 divide though by β 2 ) so a 11 2 ξ + 2a 12 ξ η + a 22 2 η = a 11 ( ξ µ + η )( ξ µ η ). Fo instance if a 12 = 0 a 11 = 1 a 22 = c 2 this gives µ ± = ±c and the factoing 2 ξ c 2 2 η = ( ξ + c η )( ξ c η ); exactly the factoing we had fo the wave opeato befoehand. A slightly diffeent method is to change the hypebolic equation to a diffeent model fom i.e. anothe fom athe than t 2 c 2 x 2 namely ξ η. As aleady mentioned L = ξ η is aleady factoed and the geneal solution of Lu = 0 can be ead off immediately: u(ξ η) = f(ξ) + g(η). Note that this change of coodinates can always be done fo a hypebolic second ode opeato on R 2 fo they can all (including ξ η ) be changed to the standad wave opeato hence by composing one of these changes with the invese of anothe one they can be changed into each othe. The coodinates (ξ η) ae called chaacteistic coodinates. To see why note that fo the wave opeato we want ξ = t + c x η = t c x o vice vesa i.e. [ ] [ ] [ ] ξ 1 c t = η 1 c x which in view of (1)-(2) means that (with the ole of ξ and x switched) we need [ ] [ ] [ ] t 1 1 ξ = x c c η i.e. t = ξ + η x = cξ cη o ξ = 1 2c (x + ct) η = 1 (x ct) 2c so the coodinate lines ξ = C and η = C ae exactly the chaacteistics. (If we did not mind a constant in font of ξ η we could have chosen ξ = x + ct η = x ct.) While so fa we woked with classical C 2 solutions of the wave equation one can immediately genealize this using Poblem 2 on Poblem Set 2. Namely fo any distibutions f g D (R) the distibution given by u(x t) = f(x + ct) + g(x ct) (in the same sense as on the poblem set) solves ( 2 t c 2 2 x)u = 0 in the sense of distibutions. This can be ead off since ( 2 t c 2 2 x)u = ( t + c x )( t c x )f(x + ct) + ( t c x )( t + c x )g(x ct) = ( t + c x )0 + ( t c x )0 = 0 whee we used the esult of Poblem 2 on Poblem Set 2. So fo instance the function u(x t) = H(x ct) whee H is now the step function solves the wave equation as does u(x t) = δ 0 (x + ct). The convese of ou claim is also tue i.e. evey distibution u satisfying ( 2 t c 2 2 x)u = 0 is of the fom u(x t) = f(x + ct) + g(x ct). This is paticulaly easy to ead off in chaacteistic coodinates whee one has to solve equations like ξ v = 0 η u = v using a slightly modified vesion of Poblem 1 on Poblem Set 2 namely with the extension of this poblem to R 2.