Elliptic Equations. Laplace equation on bounded domains Circular Domains

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1 Elliptic Equtions Lplce eqution on bounded domins Sections 7.7.2, 7.7.3, An Introduction to Prtil Differentil Equtions, Pinchover nd Rubinstein 1.2 Circulr Domins We study the two-dimensionl Lplce eqution u = u xx + u yy = (x, y) B (1.18) subject to the Dirichlet boundry condition u(x, y) =g(x, y) (x, y) B, (1.19) where B stnds for the open disk of rdius nd center t (, ). We first write this eqution in polr coordintes, x = r cos ϑ, y = r sin ϑ. Then r = x r x + y r y =cosϑ x +sinϑ y ϑ = x ϑ x + y = r sin ϑ ϑ y x + r cos ϑ y. Solving for x nd y,weget so tht = 2 x y 2 = x =cosϑ r sin ϑ r ϑ y =sinϑ r cos ϑ r ϑ ( = 2 r r r r 2 ϑ 2. cos ϑ r sin ϑ r 14 )( ϑ sin ϑ r cos ϑ r ) ϑ

2 Consequently, if u is solution of (1.18), then setting w defined stisfies the eqution w(r, ϑ) =u(r cos ϑ, r sin ϑ) w rr + 1 r w r + 1 r 2 w ϑϑ = <r<, ϑ<2π. (1.2) subject to the condition where w(, ϑ) = h(ϑ), < ϑ 2π, (1.21) h(ϑ) :=u( cos ϑ, sin ϑ) =g( cos ϑ, sin ϑ). We seek solution of the form w(r, θ) = Θ(ϑ)R(r). Substituting into the eqution (1.2) we find tht Θ(ϑ)R (r)+ 1 r Θ(ϑ)R (r)+ 1 Θ(ϑ)R(r) =. r2 Multiplying by r 2 nd dividing by Θ(ϑ)R(r), we get r 2 R (r) R(r) + r R (r) R(r) = (ϑ) Θ Θ(ϑ) Since the left-hnd side depends only on r nd the right-hnd side depends only on ϑ, botmustbeconstnt,syλ. Thisledstotwoequtions nd Θ + λθ =, <ϑ<2π (1.22) r 2 R + rr λr =, <r<. (1.23) In order to tht the solution is of clss C 2 we hve to impose two periodicity conditions: Θ() = Θ(2π) Θ () = Θ (2π). (1.24) The periodic boundry problem (1.22) (1.24) doesn t hve negtiveeigenvlues. Indeed, if λ<is negtive eigenvlue of (1.22) (1.24) with corresponding eigenfunction Θ, then multiplying (1.22) by Θ nd integrting over [, 2π] weget Θ Θdϑ = λ Θ 2 dϑ. 15

3 Integrting by prts Θ Θ dϑ = [ Θ Θ ] 2π Θ 2 dϑ = Θ 2 dϑ = λ Θ 2 dϑ. Consequently, Θ 2 dϑ =sothtθ. If λ =,thenthegenerl solution of (1.23) is of the form Θ(ϑ) = Aϑ + B nd the periodic boundry conditions (1.24) led to Θ(ϑ) =B. This implies tht λ =isn eigenvlue nd the corresponding eigenfunction Θ (ϑ) =1. Ifϑ>, the generl solution is Θ(ϑ) =A cos λϑ + B sin λϑ. The periodic boundry conditions imply tht A = A cos 2π λ + B sin 2π λ, B = A sin 2π λ + B cos 2π λ. Multiplying the first eqution by A, the second by B, nd dding leds to A 2 + B 2 =(A 2 + B 2 )cos2π λ. If A 2 + B 2 =,wegettrivilsolutionθ ndifa 2 + B 2,then cos 2π λ =1sotht2πλ =2nπ for n 1. Hence the eigenvlues re λ n = n 2, n 1 with the corresponding eigenfunctions Θ n (ϑ) =A n cos nϑ + B n sin nϑ, n 1. Now for ech eigenvlue λ n = n, n, we need to solve r 2 R n + rr n n 2 R n =. The indicil eqution of this eqution is s(s 1) + s n 2 = s 2 n 2 =. (See pges 8-81 of the Notes-Week 9.) If n =,thenthesolutionr is of the form R = C + D ln r nd if n 1, then R n (r) =Cr n + Dr n. Consequently, the product solutions w n re of the form {[ C + D ln r ], n = w n (r, ϑ) =R n (r)θ n (r) = [ Cn r n + D n r n ]A n cos nϑ + B n sin n], n 1. 16

4 Since we look for solutions which don t blow up s r +,wehveto reject the solution ln r nd r n,i.e.,wetked n =forlln. Hence we consider the solution w of (1.2) of the form At r =, wehve w(r, ϑ) = α 2 + r n [α n cos nϑ + β n sin nϑ]. h(ϑ) =w(, ϑ) = α 2 + n [α n cos nϑ + β n sin nϑ]. Using tht sequences sin nϑ nd cos nϑ re orthogonl on [, 2] we find the coefficients α n nd β n s follows. First integrte the equlity over [, 2π] to get Multiplying by cos nϑ gives α = 1 π nd multiplying by sin nϑ gives h(ϑ) dϑ. α n = 1 π n h(ϑ)cosnϑ dϑ n 1, β n = 1 π n h(ϑ)sinnϑ dϑ n 1. Summing up, solution of the Lplce eqution on the disc B in polr coordintes is given by where w(r, ϑ) = α 2 + r n [α n cos nϑ + β n sin nϑ], α = 1 π α n = 1 π n β n = 1 π n h(ϑ) dϑ h(ϑ)cosnϑ dϑ h(ϑ)sinnϑ dϑ 17

5 Next substituting the formule for α n,β n into the the series solution w we find tht w(r, ϑ) = 1 2π r n π n h(ϕ) dϕ [ + = 1 h(ϕ) 2π = 1 h(ϕ) 2π ] h(ϑ)cosnϕ cos nϑ dϕ + h(ϑ)sinnϕ sin nϑ dϕ ] [ 1+2 r n ( r [ 1+2 ( ) cos nϕ cos nϑ +sinnϕ sin nϑ n ) n cos n(ϕ ϑ)] dϕ Recll tht e iα =cosα + i sin α so tht cos α = eiα +e iα 2. Hence [ 1+2 ( ) r n cos n(ϕ ϑ)] =1+ ( ) r n [ e i(ϕ ϑ) + e i(ϕ ϑ)] dϕ =1+ ( re i(ϕ ϑ) ) n + ( re i(ϕ ϑ) ) n Therefore, w(r, ϑ) = 1 2π re i(ϕ ϑ) =1+ rei(ϕ ϑ) re i(ϕ ϑ) + re i(ϕ ϑ) 2 r 2 = 2 2r cos(ϕ ϑ)+r 2. 2 r 2 h(ϕ) 2 2r cos(ϕ ϑ)+r 2 dϕ. This formul cn be written in the (x, y)-coordintes s follows. Let p = (x, y) bepointinthediscb nd let (x, y) =(r cos ϑ, r sin ϑ). Let q = (x,y )bethepointoftheboundryb of the disc B hving the polr coordintes (x,y )=(cos ϕ, r sin ϕ). By the lw of cosines, p p 2 := (x x ) 2 +(y y ) 2 = 2 + r 2 2r cos(ϕ ϑ). Therefore, using tht ds = dϕ, u(x, y) = 1 u(x,y 2 (x, y) 2 ) 2π B (x x ) 2 +(y y ) 2 ds = 2 (x, y) 2 2π u(x,y ) B (x x ) 2 +(y y ) 2 ds, 18

6 where (x, y) 2 = x 2 + y 2. The is the Poisson s formul for the solution of Lplce eqution on the disc B. 11 The mximum principle for the het eqution The mximum principle lso holds for the het eqution. Consider the het eqution for the function u of three vribles u(x, y, t) where t > nd(x, y) belongs to two-dimensionl bounded domin D, u t = k u, (x, y) D, t >. (11.1) Here u = u xx + u yy. To formulte the mximum principle set let T> nd set Q T = {(x, y, t) (x, y) D, <t T }. Define the prbolic boundry of Q T to be P Q T =(D {}) (D [,T]). Note tht the boundry Q T is equl to Q T =(D {}) (D {T }) (D [,T]), tht is, the prbolic boundry P Q T does not contin the top D {T }. By C T bbrevite the set of functions which re twice differentible in x nd y nd once differentible in t in Q T,ndcontinuousinQ T. Theorem 11.1 (Wek mximum principle for the het eqution). Let u C T be solution of (11.1). Then u ttins its mximum on the prbolic boundry. Proof. Strt with function v in C T for which v t k v <inq T.Thenwe clim tht it ttins its mximum on the prbolic boundry P Q T.Indeed, ssume tht v ttins its mximum t point (x,y,t ) Q T.Then v t (x,y,t )=v x (x,y,t )=v y (x,y,t )=. Moreover, v(x,y,t ). Otherwise t lest one the second prtil derivtives of v is strictly greter thn. Sy 2 v x (x 2,y ) >. Since 2 v v x 2 (x x,y,t )=lim (x + h, y,t ) v x (x,y,t ) h h v =lim t x (x + t, y ) t. 19

7 This mens, in prticulr, tht v x (x + h, y,t ) > forh>ndh smll. Hence v(x + h, y,t ) <v(x,y,t )forh>ndh smll, contrdicting tht v hs mximum t (x,y,t ). So, v(x,y,t ) sclimed. Hence v t (x,y,t ) k v(x,y,t ) contrdictingv t k v <in Q T.Sincevhs to ttin its mximum in Q T,thiscnonlyhppenedthe boundry Q T. Assume tht v ttins its mximum t q =(x,y,t ) Q T.Itsufficestoshowthtt T since then q P Q T. If t = T,then since v(x,y,t) v(x,y,t)fort<t nd close to T.Then v(x,y,t) v(x,y,t) v t (q) = lim nd v(q). t T t T But then v t (q) k v(q) contrdictingv t k v <inq T. So, v hs to ttin its mximum on the prbolic boundry P Q T. Now consider u stisfying the ssumptions nd for ε>, let Clerly, v(x, y, t) =u(x, y, t) εt. mx{v(x, y, t) (x, y, t) P Q T } M := mx{u(x, y, t) (x, y, t) P Q T }. Moreover, v t v = ε +(u t u) = ε <. So, v ttins its mximum on P Q T. Hence v M t ll points of Q T. Since u = v + εt, wehve u M + εt t ll points of Q T. Tking ε, u M t ll points of Q T. As consequence of the mximum principle we hve the uniqueness nd the stbility of the solution to the Dirichlet problem for the het eqution. Theorem Let u 1 nd u 2 be two solutions of the het eqution u t k u = F (x, t), (x, y) D, <t<t stisfying the initil conditions u i (x, y, ) = f i (x, y), (x, y) D, i =1, 2 nd the boundry conditions u i (x, y, t) =h i (x, y, t) (x, y) D, <t<t, i=1, 2 Then mx{ u 1 (x, y, t) u 2 (x, y, t) (x, y, t) Q T } δ 11

8 where δ = mx f 1(x, y) f 2 (x, y) + mx h 1(x, y, t) h 2 (x, y, t). (x,y) D (x,y) D,t> In prticulr, the problem the Dirichlet problem u t k u = F (x, t), u(x, y, ) = f(x, y), u(x, y, t) =h(x, y, t), (x, y) D, <t<t (x, y) D (x, y) D, <t<t hs t most one solution. Proof. Set v = u 1 u 2.Thenv t = k v for (x, y) D nd <t<t nd v(x, y, ) = f(x, y) :=f 1 (x, y) f 2 (x, y), (x, y) D, nd v(x, y, t) =h(x, y, t) :=h 1 (x, y),t h 2 (x, y, t), (x, y) D, <t<t. By the mximum principle the mximum of v is ttin t some point on the prbolic boundry P Q T.So, v(x, y, t) δ, for ll (x, y, t) Q T where δ is defined bove. Replcing v by v we get tht v(x, y, t) δ, for ll (x, y, t) Q T. Consequently, δ v(x, y, t) δfor ll (x, y, t) Q T, tht is, mx{ v(x, y, t) (x, y, t) Q T } δ. 111