10 Elliptic equations

Transcription

1 1 Elliptic equtions Sections 7.1, 7.2, 7.3, 7.7.1, An Introduction to Prtil Differentil Equtions, Pinchover nd Ruinstein We consider the two-dimensionl Lplce eqution on the domin D, More generl eqution u =, (x, y) D. u = F, (x, y) D is clled the Poisson eqution. Definition 1.1. (1) The prolem defined y the Poisson eqution u = F in D together with the Dirichlet oundry condition u(x, y) =g(x, y) (x, y) D, is clled the Dirichlet prolem. (2) The prolem defined y the Poisson eqution u = F in D together with the Neumnn oundry condition n u(x, y) =g(x, y) (x, y) D, where g is given function on the oundry of D, isclledtheneumnn prolem. ( n u stnds for the norml derivtive of u, i.e.,the directionl derivtive of u in the direction of unit outwrd pointing norml vector n, n u = u n =(u x,u y ) (n 1,n 2 )=u x n 1 + u y n 2.) (3) The prolem defined y the Poisson eqution u = F in D together with the Roin oundry condition u(x, y)+α(x, y) n u(x, y) =g(x, y) (x, y) D, where g nd α re given function defined on the oundry of D, is clled the Roin prolem. Proposition 1.2. Consider the Neumnn prolem u = F (x, y), n u(x, y) =g(x, y), (x, y) D (x, y) D (1.1) Then gds= D D F dxdy. 93

2 Proof. Note tht u = ( x, ) ( u y x, u ) y so tht we cn write Poisson s eqution s u = F, (x, y) D. = 2 2 u + 2 u x y 2 = u Then, in view of the Guss divergence theorem Fdxdy= udxdy= n uds= s climed. D D D D gds, Theorem 1.3 (Wek mximum principle). Let D e ounded plnr domin nd let u C 2 (D) C(D) e hrmonic function in D. Then u ttins its mximum t the oundry point. Proof. Strt with continuous function v in the domin D which stisfies v >. We clim tht v does not hve locl mximum in D. Indeed, ssume tht v ttins locl mximum t the point (x,y )ind. Then v(x,y )=. Thenwemusthve v(x,y ). If v xx (x,y ) >, then v x (x + t, y ) v x (x,y ) v x (x + t, y ) v xx (x,y )=lim =lim >. t t t t Hence for t>ndsmllv x (x + t, y ) > showingthtt v(x + t, y ) is incresing for t>smll. Hencev(x + t, y ) >v(x,y )fort > smll. But this contrdict the fct tht v hs locl mximum t (x,y ). Hence v xx (x,y ) <. Similrly, v yy (x,y ) <. Therefore, v(x,y ) <, contrdicting v > in D. Now given hrmonic function u note tht u ttins mximum t some point in D. Tkeε>ndsetv(x, y) =u(x, y)+ε(x 2 + y 2 ). Then v(x, y) = u(x, y)+4ε =4ε>. The the ove discussion, v ttins its mximum t some oundry point. Set M := mx{u(x, y) (x, y) D} nd L := mx{x 2 + y 2 (x, y) D}. Then v(x, y) M + εl for ll (x, y) D 94

3 nd since v ttins it mximum t the oundry point we hve v(x, y) M + εl for ll (x, y) D. Hence u(x, y) =v(x, y) ε(x 2 + y 2 ) v(x, y) M + εl for ll (x, y) D. Tkingε +,wegetu(x, y) M for ll (x, y) D. Theorem 1.4. Consider u = F (x, y), u(x, y) =g(x, y), (x, y) D (x, y) D (1.2) Then the prolem hs t most one solution. Proof. Assume tht u 1 nd u 2 re solution of (1.3). Set v = u 1 u 2.Then v stisfies v =, (x, y) D (1.3) v(x, y) =, (x, y) D Since v longtheoundry,thewekmximumprincipleimpliestht v(x, y) forll(x, y) D. Next note tht w = v solves w =ind nd w(x, y) =longtheoundry.applyingthewekmximumprinciple to w we get tht v(x, y) =w(x, y) forll(x, y) D. So, v(x, y) for ll (x, y) D. Consequently,v, i.e., u 1 u 2 s climed. 1.1 Lplce eqution on ounded domins Rectngles We study the two-dimensionl Lplce eqution on rectngle R = {(x, y) R 2 <x<, <y<}, u = u xx + u yy = (x, y) R, (1.4) suject to the Dirichlet oundry condition u(,y)=f(y), u(, y) =h(y), u(x, ) = g(x), u(x, ) =k(x), <y<, <x<. (1.5) To find the solution u we write u = u 1 + u 2 where u 1 nd u 2 re hrmonic functions in R nd stisfy the following oundry conditions u 1 (,y)=f(y), u 1 (, y) =h(y), u 1 (x, ) =, u 1 (x, ) =, <y<, <x<. 95

4 (hence u 1 stisfies homogeneous oundry conditions long the sides [,] {} nd [,] {}), nd u 2 (,y)=,u 2 (, y) =, u 2 (x, ) = g(x), u 2 (x, ) =k(x), <y<, <x< (hence u 2 stisfies homogeneous oundry conditions long the sides {} [,]nd{} [,]). Exmple 1.5. Consider u = u xx + u yy = (x, y) R, (1.6) suject to the Dirichlet oundry condition u(,y)=,u(, y) =, u(x, ) = g(x), u(x, ) =k(x), <y<, <x<. (1.7) Note tht u stisfies homogoneous oundry conditions long the sides {} [, 1] nd {} [, 1]. We use seprtion of vriles method nd look for solutions of the form u(x, y) =X(x)Y (y). Inserting into the equtions we get X (x)y (y)+x(x)y (y) =. After dividing y X(x)Y (y), we rrive t X (x) X(x) = Y (y) Y (y). Since the left-hnd sides depends only on x nd the right-hnd side on y, ot sides re equl to constnt, sy From this we otin two equtions X (x) X(x) = Y (y) Y (y) = λ. X + λx = (1.8) nd Y λy =. (1.9) 96

5 Note tht = u(, y) =X()Y (y) =X()Y (y) =u(, y) forlly which implies tht X() = X() =. Hencewehvetosolvethefollowingeigenvlue prolem, X + λx =, <x<, (1.1) X() = X() =. The eigenvlues re positive. Indeed, if λ isneigenvluendx is corresponding eigenfunction, then multiplying the eqution y X nd integrting over [, ], we get λ X 2 dx = X Xdx= [ X X ] (X ) 2 dx = (X ) 2 dx. The left-hnd side is greter or equl to wheres the right-hnd side is less or equl to. Consequently, oth sides re equl to. This implies tht X =,i.e.,x =constntndsincex() = this constnt hs to e equl to. So, the eigenvlues re positive. For λ>, the generl solution of the eqution is given y X(x) =A cos λx + B sin λx. Using oundry conditions we find tht A = nd B sin λ =. So for nontrivil solution we must hve sin λ =whichimpliestht the eigenvlues re ( ) nπ 2 λ n =, n 1 nd the corresponding eigenvlues re X n (x) =sin nπx, n 1. Next we look t the eqution (11.7) with λ = λ n. The generl solution is given y Y n (y) =A n cosh nπy + B n sinh nπy. (1.11) Hence the product solution u n (x, y) =X n (x)y n (y) isequlto u n (x, y) =sin nπx [ A n cosh nπy + B n sinh nπy ], 97

6 nd proposed solution u of Lplce eqution is liner comintion of these product solution, u(x, y) = u n (x, y) = sin nπx [ A n cosh nπy + B n sinh nπy ]. The solution u stisfy Lpce eqution nd the homogeneous oundry conditions, u(,y)=u(, y) =. Aty =, g(x) =u(x, ) = The coefficients A n re given y A n = 2 A n sin nπx. g(x)sin nπx. At y =, k(x) =u(x, ) = [ A n cosh nπ + B n sinh nπ ] sin nπx from which we find tht nd A n cosh nπ + B n sinh nπ = 2 B n = A n tnh nπ + 2 cosh nπ k(x)sin nπx k(x)sin nπx. To simplify the clcultion one cn tke for the generl solution Y n (insted of liner comintion of cosh nπy nd sinh nπy s in (11.8)) liner comintion of sinh nπy nd sinh nπ (y ) (othrelinerlyindependent solutions of (11.7)), nmely Then u(x, y) = Y n (y) =C n sinh nπy + D n sinh nπ (y ). sin nπx [ C n sinh nπy + D n sinh nπ ] (y ). 98

7 At y =, g(x) =u(x, ) = ( D n )sinh nπ nπx sin from which we find tht At y =, D n = from which we find tht 2 sinh nπ k(x) =u(x, ) = C n = Exmple 1.6. Consider 2 sinh nπ g(x)sin nπx dx. C n sinh nπ g(x)sin nπx dx. u = u xx + u yy = <x<π, <y<π, (1.12) suject to the Neumnn t oundry condition u y (x, ) =, u y (x, π) =x π/2, u x (,y)=u(π, y) =, <x<π, <y<π. (1.13) Note tht long the oundry of the rectngle R = {(x, y) <x,y<π}, n uds =. R Indeed, n u := u u =(u x,u y ) (n 1,n 2 ). Along [, 1] {}, wehve (u x,u y )=(u x, ) nd (n 1,n 2 )=(, 1) so tht n u =. Along{π} [, 1], (u x,u y )=(,u y )nd(n 1,n 2 )=(1, ), nd n u =. Along[, 1] {π}, (u x,u y )=(u x,x π/2) nd (n 1,n 2 )=(, 1), nd n u = x π/2. Finlly, Along {} [, 1], (u x,u y )=(,u y )nd(n 1,n 2 )=( 1, ) so tht n u =. Consequently, π n uds= (x π/2) dx =. R Hence the necessry condition for solvility is stisfied. Next we seek solution u of the form u(x, y) =X(x)Y (y). 99

8 Inserting into the eqution nd dividing y XY we get X (x) X = Y Y. Since the left-hnd side depends only on x nd the right-hnd side depends only on y, X (x) = Y X Y = λ for some constnt λ. HenceX nd Y stisfy nd X (x)+λx = (1.14) Y (x) λy =. (1.15) Moreover, if u x (,y)=x ()Y (y) =ndu x (π, y) =X (π)y (y) =. Since we look for nontrivil solution, we must hve X () = X (π) =. Sohve to study eigenvlue prolem X (x)+λx =, X () = X (π) =, <x<π (1.16) The eigenvlues re λ n = n 2, n, nd the corresponding eigenfunctions re X n (x) =cosnx. Next we solve (11.12) with λ = λ n = n 2, Y (x) n 2 Y =. If n 1, then Y n (y) =A n cosh ny + B n sinh ny, nd if n =,then Y = A + B y. Note tht long the side [, 1] {}, u y (x, ) = so tht X(x)Y () = showing tht we must hve Y n () = for every n. Since Y (y) =B nd Y n(y) =na n sinh ny +nb n cosh ny. HenceY () = B =ndy n() = nb n =. Consequently, Y n = A n cosh ny, n. 1

9 The product solution u n (x, y) =X n (x)y n (y) hstheform u n (x, y) =A n A n cos nx cosh ny, nd solution u is infinite series of product solutions u(x, y) = u n (x, y) = A n cosh ny cos nx. n n Differentiting (formlly) with respect to y we get u y (x, y) = na n sinh ny cos nx. At y = π, x π/2 =u y (x, π) = na n sinh nπ cos nx. Thus, na n sinh nπ = 2 π (x π/2) cos nx dx = 2 π π x cos nx dx cos nx dx π π = 2 [ ] x sin nx π 2 π π sin nx dx cos nx dx π n nπ = 2 [ ] [ ] π sin nx π cos nx n 2 π n = 2 n 2 π [( 1)n 1] = { 4 n 2 π, n is odd, n is even. This mens tht A n =isn is even nd if n is odd, then so tht u(x, y) =A + 4 π A n = 4 n 3 π sinh nπ cosh((2n 1)y)cos((2n 1)x) (2n 1) 3. sinh(2n 1)π Finlly, we consider Lplce eqution with mixed oundry conditions. 11

10 Exmple 1.7. Consider u =, u(,y) u x (,y)=,u(, y) =f(y), u(x, ) = u(x, ) =, Using seprtion of vriles, we hve <x<, <y< <y<, <x<. (1.17) I.e., X X = Y Y = λ, X λx = Y + λy = Next note tht u(x, ) = u(x, ) =sothtx(x)y () = X(x)Y () =. This implies tht Y () = Y () = nd we we consider the eigenvlue prolem Y + λy =, <y< The eigenvlues re Y () = Y () =. λ n = with the corresponding eigenfunctions Now we need to solve ( ) nπ 2, n 1 Y n (y) =sin nπ, n 1. y X λ n X =. Along the side {} [, 1], u(,y) u x (,y)=sotht X()Y (y) X ()Y (y) = ( ) 2 implying tht X() = X nπ () =. For λ = λ n =,thegenerlsolution of X λ n X =isgiveny X n (x) =A n cosh nπx + B n sinh nπx. 12

11 Since X n(x) nπ nπx =A n sinh +B n nπ impliestht Consequently, X n (x) =B n nπ cosh nπx,theconditionx() = X () = A n = B n nπ. cosh nπx + B n sinh nπx. Therefore the we look for the solution of (11.14) s series of product solutions u(x, y) = B n sin nπy [ nπ nπx cosh +sinh nπx ]. Finlly, long {} [, 1], f(y) =u(, y) = B n [ nπ cosh nπ +sinh nπ ] sin nπy. Therefore, [ nπ B n i.e., B n = cosh nπ [ nπ +sinh nπ ] = 2 2 nπ cosh +sinh nπ ] f(y)sin nπy f(y)sin nπy dy, dy. 13