In this simple case, the solution u = ax + b is found by integrating twice. n = 2: u = 2 u

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1 The Laplace Equatio Ei Pease I. Itoductio Defiitio. Amog the most impotat ad ubiquitous of all patial diffeetial equatios is Laplace s Equatio: u = 0, whee the Laplacia opeato acts o the fuctio u : R ( is ope i R ) by takig the sum of the umixed patial deivatives. Fo example: = : u = 2 u = u = 0 x 2 I this simple case, the solutio u = ax + b is foud by itegatig twice. = 2: u = 2 u + 2 u = 0 x 2 x 2 2 Hee u = u(x, x 2 ) ad the solutio is aleady much moe difficult to obtai.. = N: u = 2 u + 2 u u = N 2 u x 2 x 2 2 x 2 i= = 0 N x 2 i A solutio to Laplace s equatio is a fuctio u satisfyig u = 0. Defiitio 2. Whe combied with bouday coditios such as u = g o, fidig a solutio to Laplace s equatio is efeed to as the Diichlet Poblem. Defiitio 3. I geeal, a C 2 fuctio u satisfyig u = 0 is called a hamoic fuctio. Defiitio 4. The closely elated ohomogeeous case is kow as Poisso s Equatio: u = f. II. Physical Itepetatios Laplace s equatio i 2 ad 3 dimesios occus i time-idepedet poblems ivolvig potetials (e.g. electostatic, gavitatioal) ad velocity i fluid mechaics. A solutio to Laplace s equatio ca also be itepeted as a steady-state tempeatue distibutio. I a typical itepetatio, u deotes the desity o cocetatio of some quatity i equilibium: u deotes chemical cocetatio: Fick s law of diffusio: u = 0 u deotes tempeatue: Fouie s law of heat coductio: u = 0 u deotes electostatic potetial: Ohm s law of electical coductio: u = 0 The (-) hee is fo cosistecy with otatio fo highe-ode elliptic opeatos.

2 2 Jauay 25, 2005 The Laplace Equatio Ei Pease III. The Fudametal Solutio to Laplace s Equatio The basic idea fo deivig the fudametal solutio is to exploit symmety by obsevig that Laplace s equatio is otatio ivaiat: Lemma. Suppose u = 0. If A is a othogoal matix ad we defie v(x) := u(ax) fo (x R ), the v = 0. Poof. sig summatio otatio, wite A = a ij ad coside u(y) whee y i = a ij x j. The ( ) = yi y i x k x k = y i a ik = a ik y i, so takig patials agai gives 2 x 2 k = a ik = a ik ] y i x k y i y j a jk ] = a ik a jk 2 y i y j = 2 y 2 k Now that we have established the adial symmety of Laplace s equatio, we kow to look fo solutios which ae fuctios of = x, i.e., of the fom u(x) = v(), whee = x ad v is to be chose so that u = 0 holds. Fist, ote that we have ) /2 = x = ( x 2 + x x2 ( ) = x i = 2 x 2 + x x 2 /2 (2xi ) = x i fo i =, 2,..., ad x 0. So the fom u(x) = v() we have u xi = v () x i ad usig the poduct ule, we also have u xi x i = v () ( ) x i 2 + v () ( = v () ( x i ) ( 2 + v x 2 () i / = v () ( x i x i (x x2 ) /2 ) 2 ) ( ) 2 + v () x2 i 3 )

3 Jauay 25, 2005 The Laplace Equatio Ei Pease 3 so summig ove i gives ( u = v () x2 i + v () 2 i= = v () i= x 2 i 2 + v () ( x2 i 3 )) i= ( ) x2 i 3 = v () x2 + x ( x2 + v () 2 (x2 + x x2 ) ) ( ) 3 = v () v () 2 3 = v () + v () ( ) Hece, u = 0 v () + v () = 0. Fo v 0, this gives v () = v () = v v The Hece v v = (log v ). =. Now ote that (log v ) = = log v = ( ) log + c v = e ( ) log e c v log ( ) = ae v = a ( ) v = a ( ) = 2 = v = a = v() = b log + c fo > 0. Note that the assumptio ( > 0) is justified because deotes adius. Similaly, 3 = v = a = v() = b + c. ( ) ( 2) Thus we have obtaied { b log = 2 v() = b 3 2 fo some costats b, c. At this poit, we itoduce some coveiet shothad otatio: α() = π/2 Γ(+/2) vol of uit ball i R α() = vol of -ball i R α() = aea of uit sphee i R α() = aea of -sphee i R

4 4 Jauay 25, 2005 The Laplace Equatio Ei Pease Defiitio 5. Now to omalize o the uit ball (ad fo the sake of tidy statemets of late esults) we defie { log x = 2 2π Φ(x) = 3 ( 2)α() x 2 fo x R, x 0 to be the fudametal solutio of Laplace s equatio. IV. Solvig Poisso s Equatio Now we geealize ad exted ou solutio fo u = 0 u = f. to solve Poisso s equatio Remak. Note that Φ(x) is hamoic by costuctio fo x 0, so Φ(x y) is hamoic fo x y. Next, coside that fo ay f : R R, x Φ(x y)f(y) will also be hamoic fo ay y x, as will ay sum of such expessios. This would suggest the followig easoig: u(x) = 0 = x Φ (x y) = 0 = x Φ (x y) f (y) = 0 = x Φ (x y) f (y) dy = 0 R so that we ae essetially just slippig the Laplacia iside the itegal to obtai as solutio the covolutio u(x) = Φ (x y) f (y) dy R fotuately, the sigulaity at y = x pevets this simple appoach ad moe cae must be take. Theoem. Suppose that f Cc 2 (R ) ad { log ( x y ) f (y) dy = 2 2π R u(x) = Φ (x y) f (y) dy = 2 f(y) R dy 3 ( 2)α() R. x y 2 The u povides a solutio of Poisso s equatio, i.e., (i) u C 2 (R ), ad (ii) u = f i R. Poof. (i) We have u (x) = Φ (x y) f (y) dy = Φ (y) f (x y) dy. R R

5 Jauay 25, 2005 The Laplace Equatio Ei Pease 5 Thus, we make the substitutio u (x + he i ) u (x) = Φ (y) f (x + he i y) dy Φ (y) f (x y) dy R R = Φ (y) f (x + he i y) f (x y)] dy R to obtai the diffeece quotiet u(x+he i ) u(x) = Φ (y) h R so that we may calculate the patial deivative f(x+hei y) f(x y) h ] dy x i (x) = lim u(x+he i ) u(x) h 0 h f(x+hei y) f(x y) h 0 h = lim Φ (y) R = Φ (y) lim R = f(x+hei y) f(x y) h 0 h R Φ (y) f x i (x y) dy ] dy ] dy Note: we ca justify movig the limit ude the itegal sig (thid equality i the pevious computatio) because f Cc 2(R ) gives us that f is uifomly cotiuous, ad hece that this covegece is uifom. Similaly, f Cc 2 (R ) gives us that x i is uifomly cotiuous ad we ca make a ealy idetical epetitio of this agumet to obtai Sice R Φ (y) 2 u x j x i (x) = lim h 0 h ( ) x i (x + he j ) x i (x) = lim Φ (y) h 0 h R = Φ (y) lim R h 0 h = Φ (y) 2 f x i x j (x y) dy R f x i (x + he j y) f 2 f x i x j (x y) dy is cotiuous i x, 2 u x 2 i ] x i (x y) dy f x i (x + he j y) f x i (x y) ] dy = u C 2 (R ).

6 6 Jauay 25, 2005 The Laplace Equatio Ei Pease Poof. (ii) By pat (i) we have u (x) = i= 2 u x 2 i (x) = Φ (y) 2f (x y) dy x 2 i= R i ] = Φ (y) 2 f (x y) dy x 2 R i i= = Φ (y) x f (x y) dy R Now we eed to isolate the sigulaity of Φ at 0, so we fix some ε > 0 ad split the itegal Φ (y) x f (x y) dy = Φ (y) x f (x y) dy + Φ (y) x f (x y) dy. R B(0,ε) R }{{} B(0,ε) }{{} I ε J ε Now deotig B := B(0, ε), we have I ε = Φ (y) x f (x y) dy B Φ (y) x f (x y) dy B Φ (y) x f (x y) dy B = x f (x y) Φ (y) dy B = C D 2 f Φ (y) dy B { Cε 2 log ε = 2 Cε 2 3. Thus lim I ε = 0. Tuig to the othe itegal ad deotig B := R B(0, ε), J ε = ε 0 B Φ (y) xf (x y) dy, so we use itegatio by pats applyig the fomula u xi v dx = uv xi dx + uvν i ds with u xi = x f (x y) dy ad v = Φ (y), so that J ε = DΦ (y) D y f (x y) dy + Φ (y) f (x y) ds (y) B }{{}}{{} K ε L ε

7 Jauay 25, 2005 The Laplace Equatio Ei Pease 7 Now fo L ε, we see L ε = Φ (y) f (x y) ds (y) Φ (y) f (x y) ds (y) Φ (y) f (x y) ds (y) = f (x y) Φ (y) ds (y) = Df Φ (y) ds (y) { Cε log ε = 2 Cε 3 Thus lim ε 0 L ε = 0. Now fo K ε = DΦ (y) D B y f (x y) dy, we use itegatio by pats agai with u xi = D y f(x y)dy ad v = DΦ(y), to obtai Φ K ε = Φ (y) f (x y) dy (y) f (x y) ds (y) B Φ = (y) f (x y) ds (y) whee the fist itegal vaishes because we costucted Φ to be hamoic, i the fist place (hece Φ = 0 = Φ = 0. Moe pecisely, Φ(y) is hamoic fo y 0, but this itegatio takes place ove the potio of R which is at least ε fom 0, so we ae okay. Now we have by defiitio that Φ (y) := ν DΦ (y), so which becomes ν = y ad DΦ(y) = y fo y 0, y α() y ν = y ε ad DΦ(y) = α() o the sphee (0, ε). Puttig these togethe, ( Φ y (y) = ν DΦ (y) = y ε α() = y y α()ε + = y 2 α()ε + ε = 2 α()ε + = α()ε ε ) y ε ( y = ε o (0, ε))

8 8 Jauay 25, 2005 The Laplace Equatio Ei Pease We substitute this expessio ito ou fomula fo K ε to obtai K ε = f (y) ds (y). α()ε (x,ε) Now we make the coveiet defiitio f ds := α() (x,) (x,) f ds fo the aveage value of f o the sphee (x, ), so that we ca ewite this esult as K ε = f (y) ds (y). Thus Combiig all tems, lim K ε = lim ε 0 ε 0 (x,ε) (x,ε) ] f (y) ds (y) = f(x). u (x) = lim ε 0 (I ε + K ε + L ε ) = 0 f(x) + 0 = f(x) Remak. Φ is costucted to be hamoic away fom the oigi, i.e., Φ(x) = 0 fo x 0, but what is Φ(0)? Although this questio is pehaps ot quite well-fomed, we ca povide a faily pecise aswe by defiig Φ(x) = δ 0 i R, the Diac delta fuctio (distibutio) i R. Retuig to ou pevious ituitio, we ca ow make the fomal computatio u (x) = x Φ (x y) f (y) dy R = δ x f (y) dy R = f (y) dµ x R = f (x) whee dµ x is the measue defied by dµ x (E) = { x E 0 x / E. Remak. The pevious theoem ca be pove ude less estictive coditios. itegable fuctio f o a domai Ω (ad Φ as above), let u be defied o R by u(x) = Φ(x y)f(y)dy. Ω Fo a The the pevious esult ca be exteded as follows: let f be bouded ad locally Hölde cotiuous (with expoet α ) i Ω. The u C 2 (Ω) ad u = f i Ω.

9 Jauay 25, 2005 The Laplace Equatio Ei Pease 9 V. Mea-value Fomulas Theoem 2. If u C 2 () satisfies u = 0 fo a ope set i R, the B(x, ), u (x) = u ds = u dy. (x,) Poof. Fo the fist equality, defie φ () := u (y) ds (y) = so that φ () = = = = = = 0 (0,) (x,) (x,) (x,) B(x,) (x,) Du (x + z) z ds (z) Du (y) y x ds (y) Du (y) νds (y) ds (y) u(y)dy B(x,) (0,) u (x + z) ds (z) ( ds = ) udx Hece φ is costat, which implies φ () = lim φ (t) = lim t 0 B(x,) t 0 (x,t) 0 u (y) ds (y) = u (x) Fo the secod equality, usig pola coodiates gives ( ) u dy = uds ds = 0 (x,s) u (x) α () s ds = u (x) α () 0 s ds So B(x,) = u (x) α () = u (x) α () udy = u (x) α () = u (x) = α() B(x,) udy = udy B(x,)

10 0 Jauay 25, 2005 The Laplace Equatio Ei Pease Coollay. u C 2 () satisfies u (x) = (x,) u ds, B (x, ) = u is hamoic. Poof. If u / 0, the B(x, ) such that u > 0 (o u < 0) o B(x, ). But the fo φ as above, 0 = φ () = < u (y) dy > 0 B(x,) Theoem 3. R is ope ad bouded, u C 2 () C(Ū) is hamoic withi. (i) Suppose is coected. The x 0 such that u (x 0 ) = max Ū {u} = u is costat o. (ii) max Ū {u} = max {u} Poof. (i) Suppose x 0 such that u (x 0 ) = M := max Ū {u}. The B(x 0, ) wheeve 0 < < dist (x 0, ), so we ca apply the mea-value theoem to get M = u (x 0 ) = udy. B(x,) Sice M is the maximum of u o the ball, equality ca oly hold if u M o the ball; i.e., it must be that u(y) = M, y B(x 0, ). This also shows {u = M} is a ope set: u (x 0 ) = M = u(y) = M, y B(x 0, ). Moeove, {u = M} is also elatively closed i : {x. u (x) = M} = u ({M}) shows that {u = M} is the peimage of a closed set (ude a cotiuous map), ad hece is closed. Thus coected = = {x. u (x) = M}. Poof. (ii) Ū, so m = max {u} max Ū {u} = M. To see that m M, ote that Ū is compact, u attais its maximum at some poit x 0 Ū. The we have case i) x 0. The m M follows immediately. case ii) x 0 Ū. The x 0 = u is costat o by (i), so m = M. Remak. This theoem has a aalogous vesio fo miima (poved aalogously).

11 Jauay 25, 2005 The Laplace Equatio Ei Pease Remak. Suppose is coected ad fo some g 0, u C 2 () C(Ū) satisfies { u = 0 i u = g o. The g(x) > 0 somewhee o implies u(x) > 0 eveywhee o. The easoig behid this is as follows: g 0, so the lowest mi Ū {u} ca be is 0, because the miimum is attaied o the bouday, whee u = g. The u(x 0 ) = 0 fo x 0 implies u 0 by (i). But the the assumptio that g(x) > 0 fo some x gives 0 u = g > 0 o <. The maximum piciple also povides a easy poof fo the uiqueess of the solutio to Poisso s equatio. Theoem 4. Suppose g C( ), f C(). The thee is at most oe solutio u C 2 () C(Ū) to the bouday-value poblem { u = f i u = g o. Poof. Suppose u, v satisfy u = f, u = g ad v = f, v = g. The w = u v ad w = v u ae hamoic. Hece w = w = g g 0 o, so max Ū w = max Ū w = mi w = 0 = u v. Ū Theoem 5. If u C() satisfies u (x) = u ds = u dy, B (x, ), (x,) B(x,) the u C (). Cosequetly, u hamoic = u C (). Poof. Let η C (R ) be the stadad mollifie defied by { ( ) C exp x < x η(x) = 2 0 x, whee the costat C > 0 is selected such that η dx =. Note that spt(η) = B(0, ), R ad that η is a adial fuctio. Fix ε > 0 ad defie η ε (x) = η ( ) x ε ε so that η e C, η R ε dx =, ad spt (η ε ) B (0, ε). Now set { } u ε = η ε u i ε = x. dist (x, ) > ε so that u ε = η ε (x y) u (y) dy = B(0,ε) η ε (y) u (x y) dy fo x ε.

12 2 Jauay 25, 2005 The Laplace Equatio Ei Pease We show u u ε o ε. Fo x ε, u ε (x) = η ε (x y) u (y) dy = η ( ) x y ε ε u (y) dy = = B(x,ε) ε 0 = u (x) η ε η ( ε ε ε 0 ε = u (x) = u (x) = u (x) 0 ( ) x y u (y) dy ε ) ( (x,ε) ) u ds d η ( ε ε) α () d η ε () α () d B(0,ε) η ε dy Thus u ε u i ε, ad sice u ε C by costuctio, this shows u C ( ε ), ε > 0. Thus u C (). Remak. Somehow, the algebaic stuctue of Laplace s equatio u = i= u x i x i = 0 leads to the aalytic deductio that all the patial deivatives of u exist, eve those which do ot appea i the PDE. Theoem 6. (Estimates o deivatives). u is hamoic i = D α u (x 0 ) C k +k u L (B(x 0,)), B (x 0, ) ad fo evey multiidex α with α = k. C k is give hee by C 0 = α (), C k = (2+ k) k α () (k =, 2,...) Theoem 7. (Liouville). Let u : R R be hamoic. The u bouded = u costat. Poof. Fix x 0 R ad > 0. Now by the pevious theoem, u hamoic o B(x 0, ) gives Du (x 0 ) C + u L (B(x 0,)) C α () u L (B(x 0,)) 0 Thus Du (x 0 ) 0, ad hece u is costat.

13 Jauay 25, 2005 The Laplace Equatio Ei Pease 3 Theoem 8. Suppose f Cc 2 (R ), 3. The ay bouded solutio of u = f i R has the fom u (x) = Φ (x y) f (y) dy + C (x R ) R fo some costat C. x Poof. Fo 3, we have Φ (x) = 0. ( 2) α () x 2 This shows that u (x) = R Φ (x y) f (y) dy is a bouded solutio of u = f i R. Now if v is some othe solutio, defie w := v u. The w = (v u) = v u = ( f) ( f) = 0 shows w is hamoic. Hece by Liouville s theoem, w is costat. Thus w = v u = C = v = u + C = Φ (x y) f (y) dy + C R Theoem 9. (Aalyticity) u is hamoic i = u aalytic i. That is, u may be epeseted at x 0 by the Taylo seies D α u(x 0 ) (x x α! 0 ) α, which coveges fo x x 0 < dist(x 0, ) e α Theoem 0. (Haack s Iequality) Fo each coected ope set V, C > 0 such that sup V u C if V u, fo all oegative hamoic fuctios u i. Poof. Let = dist (V, ). Choose x, y V with x y. The 4 u (x) = u dz B(x,2) = α()(2) B(x,2) 2 α() = u dz 2 B(y,) = 2 u (y) B(y,) u dz u dz = 2 u (y) u (x) 2 u (y) if x, y V ad x y. Sice V is coected ad V is compact, we ca cove V by a chai of fiitely may balls {B j } N j=, each of which has adius 2, ad such that B i B i+, i =,... N. The u(x) 2 N u(y), x, y V.

14 4 Jauay 25, 2005 The Laplace Equatio Ei Pease VI. Deivatio of Gee s Fuctio I this sectio we develop a solutio of Poisso s equatio u = f i, subject to the Diichlet coditio u = g o. Fo coveiece, we also make the assumptio that R is ope ad bouded, ad that is C. 2 Stat with a abitay fuctio u C (Ū). Fix a x, ad the fix ε > 0 such that B(x, ε). Defie V ε := B(x, e). Coside u(y) ad Φ(y x) o V ε. Recall (oe of may) Gee s fomula: u v v u dx = u v v ds Applyig this fomula to u(y) ad Φ(y x) o V ε gives u (y) Φ (y x) dy Φ (y x) u (y) dy ( ) V ε V ε = u (y) Φ (y x) Φ (y x) (y) ds (y) V ε whee ν deotes the outwad uit omal vecto o V ε. Next obseve that V ε = (disjoit uio), so we may sepaate the itegal ove V ε ito two pieces accodigly. Now o we have the estimates Φ (y x) (y) ds (y) Cε max Φ ε 0 0 ad u (y) Φ (y x) ds (y) = u (y) ds (y) ε 0 u (x). Theefoe, whe we let ε 0, the ight-had side of ( )) becomes u (y) Φ (y x) Φ (y x) (y) ds (y) + u (x). Moeove, Φ(y x) = 0 fo y x, so V ε = B(x, e) implies u (y) Φ (y x) ds (y) = 0 ds = 0, ε > 0. V ε V ε

15 Jauay 25, 2005 The Laplace Equatio Ei Pease 5 Thus lettig ε 0 tasfoms ( ) fom u (y) Φ (y x) dy Φ (y x) u (y) dy = V ε V ε ito the simple expessio Φ (y x) u (y) dy = V ε u (y) Φ (y x) Φ (y x) (y) ds (y) u (y) Φ (y x) Φ (y x) (y) ds (y) + u(x). Reaagig ad substitutig, we obtai the followig expessio fo u: u(x) = Φ (y x) Φ (y) u (y) (y x)] ds (y) Φ (y x) u (y) dy = Φ (y x) Φ (y) g (y) (y x)] ds (y) + Φ (y x) f(y)dy. 2 This idetity is valid x, u C (Ū), so we could solve fo u if we kew o. fotuately, we do t kow what the omal deivative is alog. Theefoe, we itoduce (fo fixed x) a coecto fuctio φ x = φ x (y) that solves the bouday value poblem { φ x = 0 i φ x = Φ(y x) o We apply Gee s fomula u v v u dx = u v v ds to φx ad obtai u (y) φ x (y) φ x (y) u (y)] dy = u (y) φ x (y) φx (y) (y)] ds (y). Now by defiitio of φ x, φ x = 0 i ad φ x = Φ(y x) o, so this expessio is eally φ x (y) u (y) dy = u (y) φ x (y) Φ (y x) (y)] ds (y) = Φ (y x) (y) ds (y) = φ x (y) u (y) dy + u (y) φx (y) ds (y). With the tem thus isolated, we substitute back ito the expessio fo u(x): u (x) = φ x (y) u (y) dy + u (y) φx (y) ds (y) u (y) Φ (y x) ds (y) Φ (y x) u (y) dy = φ x (y) Φ (y x)] u (y) dy + u (y) φ x Φ (y) (y x)] ds (y) Defiitio 6. The Gee s fuctio fo the egio is Note that G is hamoic fo x y. G(x, y) := Φ(y x) φ x (y), x, y, x y.

16 6 Jauay 25, 2005 The Laplace Equatio Ei Pease Cotiuig fom above, this temiology yields u (x) = G (x, y) u (y) dy u (y) G (x, y) ds (y), whee G (x, y) := D yg(x, y) ν(y) is the oute omal deivative of G with espect to y. We summaize these esults to obtai 2 Theoem. Suppose that u C (Ū) solves the bouday value poblem { u = f i u = g o fo give f, g. The u (x) = G (x, y) f (y) dy g (y) G (x, y) ds (y) (x ). (*) Lemma 2. (Symmety of Gee s fuctio). x, y, x y, we have G(y, x) = G(x, y). Poof. Fix x, y, x y. Fo z, let us deote (x,ε) v(z) := G(x, z) ad w(z) := G(y, z). The (i) v(z) = 0 fo z y, (ii) w(z) = 0 fo z y, ad (iii) v = w = 0 o. Choose ε > 0 so small that (x, ε) (y, ε) =. Now apply Gee s Idetity o V := (B (x, ε) B (y, ε)) to get v w w v] ds (z) = w v v w] ds (z) (y,ε) whee ν is the iwad-poitig uit vecto field o (x, ε) (y, ε). Now w is smooth ea x, so (x,ε) (x,ε) w v ds (z) Cε sup (x,ε) v ε 0 0. O the othe had, v(z) = Φ(y x) φ x (z) whee φ x is smooth i, so v Φ lim w ds (z) = lim (x y) w (z) ds (z) = w (x), ε 0 ε 0 (x,ε) by calculatios as i the deivatio of the solutio fo Poisso s equatio. Now v (x,ε) w ds (z) (x,ε) w v ds (z) = (y,ε) w v ds (z) (y,ε) w(x) 0 = v(y) 0 v w ds (z) shows G(y, x) = w(x) = v(y) = G(x, y).

17 Jauay 25, 2005 The Laplace Equatio Ei Pease 7 VII. Gee s fuctio fo a half-space Defiitio 7. We will use the followig otatio fo the half-space i R : R + := {(x,..., x ) R. x > 0}. It is the goal of this sectio to poduce a explicit expessio fo Gee s fuctio fo = R +. Note: Some of the pevious esults do ot diectly apply hee, as this egio is ot bouded. Nevetheless, we will be able to fid some solutios usig these methods with the caveat that we must check thei validity aftewads. Defiitio 8. If x = (x,..., x ) R +, the its eflectio i the plae R + x := (x,..., x ) / R +. is the poit Ituitively, the idea is to use the coecto fuctio φ x to eflect the sigulaity out of the domai R +. To that ed, we defie φ x (y) := Φ(y x) = Φ(y x, y 2 x 2,..., y x, y + x ) (x, y R + ). The fo y R +, y is equidistat fom x ad x. Theefoe, y x = y x = Φ(y x) = Φ(y x) = Φ(y x) = φ x (y) (y R +) because Φ is adial. Ad sice x / R +, φ x (y) = Φ(y x) = 0 (y R + ). Thus we have satisfied the equiemets fo φ x ad ae justified i makig the followig defiitio: Defiitio 9. Gee s fuctio fo R + is G(x, y) := Φ(y x) Φ(y x) (x, y R +, x y) At this poit, we pause to compute Φ y (y x) = ( ) ( 2)α() y y x 2 = ( 2)α() = α() 2 = y x α() y x so that G y (x, y) = Φ y (y x) Φ y (y x) = (y y x ) 2 + (y 2 x 2 ) (y x ) 2] 2 (y x ) 2 + (y 2 x 2 ) (y x ) 2] 2 2 (y x ) G α() G (x, y) = y (x, y) = 2x α() ] y x y x y+x, ad hece y x. y x

18 8 Jauay 25, 2005 The Laplace Equatio Ei Pease If u solves the bouday value poblem { u = 0 u = g i o, we ca plug i to the peviously deived fomula (*) to obtai This esult is well-kow: u (x) = 2x α() R + g(y) y x dy ( x R + ). Defiitio 0. Poisso s fomula fo R + : Poisso s keel fo R + : u (x) = 2x α() K (x, y) := 2x g(y) R +. α() x y y x dy ( x R + ). Now as pomised, we must validate these esults. Theoem 2. (Poisso s fomula fo R + ). Fo g C(R ) L (R ) ad u solves the Diichlet poblem o R + : (i) u C (R + ) L (R + ), (ii) u = 0 i R +, ad (iii) u (x) u (x) = 2x α() R + g(y) y x dy, x x 0, x R + g (x 0 ), fo each x 0 R +. Poof. (i)&(ii) Fo fixed x, the mappig y G(x, y) is hamoic fo y x. Symmety Lemma, x G(x, y) is hamoic fo y x. Thus The by x G y (x, y) = 2x α() x y is a hamoic mappig fo x R +, y R +. We use the fact that K(x, y)dy = fo each x R + (calculatio omitted). R +

19 Jauay 25, 2005 The Laplace Equatio Ei Pease 9 Thus g L = u bouded, as follows: u (x) 2x α() 2x α() R + g 2x α() g(y) y x dy g dy R y x + R + dy y x = g Sice x K(x, y) is smooth fo x y, u (x) = x K (x, y) g (y) dy = 0 R + shows that u C. Poof. (iii) Fix x 0 R + ad ε > 0. Choose δ > 0 small eough that y x 0 < δ = g (y) g ( x 0 ) < ε, y R +. The if x x 0 < δ 2 fo x R +, ( ) u (x) g x 0 = K (x, y) g (y) dy g ( x 0) R + = K (x, y) g (y) dy K (x, y) g ( x 0) dy R + R + = K (x, y) ( ) g (y) g x 0 dy R + R + B(x0,δ) K (x, y) g (y) g ( x 0 ) dy }{{} I + K (x, y) ( ) g (y) g x 0 dy. R + B(x0,δ) } {{ } J

20 20 Jauay 25, 2005 The Laplace Equatio Ei Pease Now fo I, y x 0 < δ fo y ( R + B (x0, δ) ) gives I = K (x, y) ( ) g (y) g x 0 dy ε = ε. R + B(x0,δ) R + K (x, y) dy Fo J, ote that x x 0 < δ 2, y x 0 δ = y x 0 y x + δ y x = y x 2 y x 0. y x 0 Thus ad theefoe J 2 g L 2+2 x g L α () R + B(x0,δ) K (x, y) dy R + B(x0,δ) y x 0 dy x 0 + 0, x x 0 < δ 2 = ( ) u (x) g x 0 < ε. Beto] BoyDi] Evas] GilT] Refeeces Betoues, D. (999) Patial Diffeetial Equatios fo Computatioal Sciece. Spige- Velag. Boyce, W.E., ad DiPima, R.C. (997) Elemetay Diffeetial Equatios ad Bouday Value Poblems. Joh Wiley. Evas, L.C. (998) Patial Diffeetial Equatios. AMS (GSM v.9). Gilbag, D., ad Tudige, N.S. (983) Elliptic Patial Diffeetial Equatios of Secod Ode (2d ed). Spige-Velag.