5.1 Second-Order linear PDE

Transcription

1 5.1 Second-Order liner PDE Consider second-order liner PDE L[u] = u xx + 2bu xy + cu yy + du x + eu y + fu = g, (x,y) U (5.1) for n unknown function u of two vribles x nd y. The functions,b nd c re ssumed to be of clss C 1 nd stisfying 2 +b 2 +c 2 0. The opertor L 0 [u] := u xx + 2bu xy + cu yy consisting of the second order terms of L is clled the principl prt of L. Mny of the fundmentl properties of the solutions of (5.1) re determined by the sign of the discriminnt of L. The discriminnt (L)(x,y) is defined by [ ] b (L)(x,y) = det = b 2 c c b where,b nd c re evluted t the point (x,y). We re interested in how the PDE is trnsformed under chnges of coordintes. We consider C 1 - mp F(x,y) = (ξ(x,y),η(x,y)) whose Jcobin stisfies [ ] ξx ξ detj(x,y) = det y 0 t ech point of (x 0,y 0 ) U. The inverse function theorem implies tht ner the point (x 0,y 0 ) the mp F hs n inverse F 1 (ξ,η) = (x(ξ,η),y(ξ,η)). The inverse is of clss C 1. Now, ssuming tht u is solution of (5.1), define w(ξ,η) = u(x(ξ,η),y(ξ,η)). Then u(x,y) = w(ξ(x,y),η(x,y)) nd, by the chin rule, u x = w ξ ξ x + w η η x u y = w ξ ξ y + w η η y u xx = w ξξ (ξ x ) 2 + 2w ξη ξ x η x + w ηη (η x ) 2 + w ξ ξ xx + w η η xx u yy = w ξξ (ξ y ) 2 + 2w ξη ξ y η y + w ηη (η y ) 2 + w ξ ξ yy + w η η yy u xy = u yx = w ξξ ξ x ξ y + w ξη (ξ x η y + η x ξ y ) + w ηη ξ x η y + w ξ ξ xy + w η η xy Substituting into (5.1), we find tht η x L[w] = Aw ξξ + 2Bw ξη + Cw ηη + Dw ξ + Ew η + Fw = G (5.2) η y 32

2 with the coefficients of the principl prt L 0 [w] = Aw ξξ + 2Bw ξη + Cw ηη given by A(ξ,η) = ξx 2 + 2bξ xξ y + cξy 2 B(ξ,η) = ξ x η x + b(ξ x η y + ξ y η x ) + cξ y η y (5.3) Observe tht C(ξ,η) = η 2 x + 2bη xη y + cη 2 y [ ] A B = B C [ ξx ξ y η y η y ] [ ] b b c [ ] t ξx η x, ξ y η y where t denotes the trnspose of the mtrix. Reclling tht the determinnt of the product of mtrices is equl to the product of the determinnts of mtrices nd tht the determinnt of trnspose of mtrix is equl to the determinnt of mtrix, we get [ ] [ ] A B b det = det (J(x,y)) 2. (5.4) B A b c This shows tht the discriminnt of L hs the sme sign s the discriminnt of the trnsformed eqution nd so it is n invrint of the chnge of coordintes. Consequently, we cn clssify equtions (5.1) ccording to the sign of the discriminnt. Definition 5.1. The eqution (5.1) is clled hyperbolic t (x,y) if (L)(x,y) > 0. prbolic t (x,y) if (L)(x,y) = 0. elliptic t (x, y) if (L)(x, y) < 0. (i) The wve eqution u tt u xx = 0 is hyperbolic equ- Exmple 5.2. tion. (ii) The het eqution u t u xx = 0 is prbolic eqution. (iii) The Lplce eqution u xx = u yy = 0 is n elliptic eqution. Exmple 5.3. Consider the Tricomi eqution yu xx + u yy = 0. (5.5) Here = y,b = 0,c = 1 nd d = e = f = g = 0. Its discriminnt is equl to [ ] [ ] b 0 y det = det = y. c b

3 Hence the eqution (5.5) is hyperbolic for y < 0, prbolic when y = 0, nd elliptic for y > 0. Next we shll show tht we cn find chnges of coordintes in which the (5.1) tkes simple form. Hyperbolic equtions. Suppose tht eqution (5.1) is hyperbolic on the domin U. This mens tht b 2 c > 0 t ech point of U. We shll show tht in this we cn choose (x,y) (ξ(x,y),η(x,y)) so tht A(ξ,η) = ξ 2 x + 2bξ xξ y + cξ 2 y = 0 (5.6) C(ξ,η) = ηx 2 + 2bη xη y + cηy 2 = 0. (5.7) Under such chnge of coordintes nd dividing by 2B the hyperbolic eqution (5.1) tkes its cnonicl form L[w] = w ξη + l[w] = G, (5.8) where l is first-order liner opertor nd G is function. Note tht if nd c re equl to 0, then the eqution (5.1) is lre in its cnonicl form (just divide by 2b). Hence without loss of generlity we my ssume tht 0. Note lso tht the eqution (5.7) for η is the sme s (5.6) for ξ. Hence it suffices to consider only one of the equtions, sy (5.6). It cn be written s product [ ξ x b ] b 2 c ξ y [ ξ x b + ] b 2 c ξ y = 0 nd so, we need to solve the following two liner equtions ξ x µ 1 ξ y = 0 (5.9) nd where we hve bbrevited ξ x µ 2 ξ y = 0, (5.10) µ 1 = b b 2 c nd µ 2 = b b 2 c. Note tht µ 1 nd µ 2 re the rel solutions of the eqution µ 2 + 2bµ + c = 0. (5.11) 34

4 In order to obtin nonsingulr mp (x,y) (ξ(x,y),η(x,y)), we choose ξ to be the solution of (5.9) nd η to be the solution of (5.10). To solve (5.9), we use the method of chrcteristics (except tht we don t specify the initil condition). The chrcteristic equtions re dx dt = 1, dt = µ 1, dz dt = 0. The lst eqution sys tht the solution ξ is constnt long ech of the chrcteristics (x(t), y(t)). In view of the first two equtions, the chrcteristics cn be obtin s curves y = y(x) solving dx = dt = µ 1, (5.12) dx dt nd then the solution ξ is constnt t points (x,y(x)). Similrly, one solves (5.10) to obtin η. In summry, to choose ξ nd η one solves the (5.11) to obtin two rel roots µ 1 nd µ 2. Then, denoting by f(x,y) = C 1 nd g(x,y) = C 2 the solutions of chrcteristics equtions dx = µ 1(x,y nd the vribles ξ nd η re defined by dx = µ 2(x,y), (5.13) ξ(x,y) = f(x,y), η(x,y) = g(x,y). The solutions of both equtions in (5.13) re clled the two fmilies of chrcteristics of (5.1). Exmple 5.4. Consider yu xx + u yy = 0 In the region where y < 0, the eqution is hyperbolic. Solving yµ = 0, one finds two rel solutions µ 1 = 1 ( y) 1/2 nd µ 2 = 1 ( y) 1/2 We look for two rel fmilies of chrcteristics, dx +µ 1 = 0 nd dx +µ 2 = 0, dx 1 ( y) 1/2 = 0 nd dx + 1 = 0. ( y) 1/2 35

5 The solutions of the equtions re Therefore, we set 2 3 ( y)3/2 + x = C 1 nd 2 3 ( y)3/2 + x = C 2. ξ = 2 3 ( y)3/2 + x nd η = 2 3 ( y)3/2 + x. The derivtives of ξ nd η re, ξ x = 1, ξ y = ( y) 1/2, η x = 1, η y = ( y) 1/2. With u(x,y) = v(ξ(x,y),η(x,y), one gets u x = v ξ + v η u y = ( y) 1/2 v ξ + ( y) 1/2 v η u xx = v ξξ + 2v ξη + v ηη u yy = yv ξξ + 2yv ξη yv ηη ( y) 1/2 [v ξ v η ]. Substituting into the eqution, one obtins 0 = yu xx + u yy = 4yv ξη ( y) 1/2 [v ξ v η ] = 4y [ v ξη 1 ] 8 ( y) 3/2 (v ξ v η ). Since ξ η = 4 3 ( y)3/2, one concludes tht v ξη Prbolic equtions. 1 6(ξ η) (v ξ v η ) = 0. Suppose tht (5.1) is prbolic on the domin U. Hence b 2 c = 0 t ech point of U. As before ssume tht 0 on U. We find mp (x,y) (ξx,y),η(x,y)) so tht B(ξ,η) = A(ξ,η) = 0. It suffices to mke A = 0 since 0 = B 2 AC = B 2 implies tht B(ξ,η) = 0. Under such chnge of coordintes the prbolic eqution (5.1) cn be brought to its cnonicl form L[w] = w ξξ + l[w] = G(ξ,η) where l is first-order liner opertor nd G is function. 36

6 To do this we look for (x,y) ξ(x,y) so tht Since b 2 = c, we hve ξ 2 x + 2bξ x ξ y + cξ 2 y = A(ξ,η) = ξ 2 x + 2bξ x ξ y + cξ 2 y = 0. (ξ 2x + 2 b ) ξ xξ y + b2 2ξ2 y = (ξ x µξ y ) 2, where µ = b is the double root of µ 2 + 2bµ + c = 0. So, we look for the solution ξ of the first-order liner eqution, ξ x µξ y = 0. The solution ξ is constnt long ech chrcteristic which is determined by the eqution dx = µ = b. (5.14) For the mp (x,y) η(x,y) we cn tke ny mp so tht ξ x ξ ξ y ξ x 0. In summry, in the prbolic cse to choose ξ nd η one solves the (5.11) to obtin double root µ = b/. Then, denoting by f(x,y) = C the solution of chrcteristics eqution the vrible ξ is defined by = µ (5.15) dx ξ(x,y) = f(x,y) nd the vrible η is chosen so tht ξ x ξ ξ y ξ x 0. Exmple 5.5. Reduce the following eqution to its cnonicl form nd then find the generl solution, for x > 0. x 2 u xx 2xyu xy + y 2 u yy + xu x + yu y = 0 37

7 The discriminnt is equl to [ ] xy x 2 det y 2 = 0 xy so tht the eqution is prbolic. The qudrtic eqution x 2 µ 2 2xyµ+y 2 = 0 hs exctly one solution, µ = y x. Next we look for chrcteristics. These re solutions of the eqution dx = µ, i.e., dx = y x. The fmily of solutions is given by xy = C. Therefore, we define ξ(x,y) = xy nd we tke s the second independent vrible η(x,y) = x. Then ξ x = y, ξ y = x, η x = 1, η y = 0. The Jcobin of the mp (x,y) (ξ(x,y),η(x,y)) is nonzero. Let v(ξ,η) = u(x(ξ, η), y(ξ, η)), tht is, u(x, y) = v(ξ(x, y), η(x, y)). Using the chin rule, u x = yv ξ + v η u y = xv ξ u xx = y 2 v ξξ + 2yv ξη + v ηη u xy = xyv ξξ + xv ξη + v ξ u yy = x 2 v ξξ. Substituting into the eqution, one gets x 2 v ηη + xv η = 0, so tht, using x = η, v ηη + 1 η v η = 0. To solve this eqution introduce the function w = v η. Then w η = 1 η w which hs the generl solution w = 1 ηa(ξ). So, v η = 1 η A(ξ) 38

8 which fter integrtion gives v(ξ,η) = A(ξ)ln η + B(ξ). Since ξ(x,y) = xy nd η(x,y) = x nd u(x,y) = v(ξ(x,y),η(x,y)), the generl solution of the eqution hs the form u(x,y) = A(xy)ln x + B(xy) where A,B re two rbitrry functions of clss C 2. Elliptic equtions. Suppose tht (5.1) is elliptic on the domin U. Then b 2 c < 0 t ech point of U. This time we look for the mp (x,y) (xξ(x,y),η(x,y)) so tht A(ξ,η) = ξ 2 x + 2bξ xξ y + cξ 2 y = C(ξ,η) = η2 x + 2bη xη y + cη 2 y B(ξ,η) = ξ x η x + b(ξ x η y + ξ y η x ) + cξ y η y = 0. Under such chnge of coordintes nd dividing by A, the elliptic eqution (5.1) cn be brought to its cnonicl form L[w] = w ξξ + w ηη + l[w] = G(ξ,η) where l is first-order liner opertor. The bove system consists of two nonliner first-order equtions. Subtrcting C(ξ,η) form A(ξ,η) nd multiplying B(ξ,η) by 2i leds to the following system, (ξ 2 x η2 x ) + 2b(ξ xξ y η x η y ) + c(ξ 2 y η2 y ) = 0 ξ x (2iη x ) + b(ξ x (2iη y ) + ξ y (2iη x )) + cξ y (2iη y ) = 0. Consequently, setting φ = ξ + iη, one finds tht the bove system is equivlent to φ 2 x + 2bφ x φ y + cφ 2 y = 0. This cn be written s product, where we hve bbrevited [φ x µ 1 φ y ] [φ x µ 2 φ y ] = 0, µ 1 = b i c b 2 nd µ 2 = b i c b 2. 39

9 These re complex roots of µ 2 + 2bµ + c = 0. Note tht µ 1 nd µ 2 re conjugted, i.e., µ 1 = µ 2. As in the hyperbolic cse we solve the chrcteristics equtions dx = µ 1 nd dx = µ 2. (5.16) This time the solutions re complex. If f(x,y) = C 1 nd g(x,y) = C 2 re complex solutions of (5.16), then Then set φ(x,y) = f(x,y) ξ = 1 (φ + ψ) 2 nd nd ψ(x,y) = g(x,y). η = 1 (φ ψ). 2i Exmple 5.6. Consider yu xx + u yy = 0 In the region where y > 0, the eqution is elliptic. Solving yµ = 0, one finds two complex solutions µ 1 = i y 1/2 nd µ 2 = i y 1/2 We look for two complex fmilies of chrcteristics, dx +µ 1 = 0 nd dx +µ 2 = 0, dx + i y 1/2 = 0 nd dx i = 0. y1/2 The solutions of the equtions re Therefore, we set 2 3 y3/2 + ix = C 1 nd 2 3 y3/2 ix = C 2. φ = 2 3 y3/2 + x nd ψ = 2 3 y3/2 x, nd then ξ = 1 2 (φ + ψ) = 2 3 y3/2 nd η = 1 (φ ψ) = x. 2i 40

10 The derivtives of ξ nd η re, ξ x = 0, ξ y = y 1/2, η x = 1, η y = 0. With u(x,y) = v(ξ(x,y),η(x,y), one gets u x = v η u y = y 1/2 v ξ u xx = v ηη u yy = yv ξξ y 1/2 v ξ. Substituting into the eqution, one obtins v ξξ + v ηη + 1 2y 3/2v ξ = 0. Finlly, since ξ = 2 3 y3/2, the eqution becomes v ξξ + v ηη + 3 ξ v ξ = 0. 41