LECTURE 12. Special Solutions of Laplace's Equation. 1. Separation of Variables with Respect to Cartesian Coordinates. Suppose.

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2 LECTURE 12 Special Solutions of Laplace's Equation 1. Sepaation of Vaiables with Respect to Catesian Coodinates Suppose è12.1è èx; yè =XèxèY èyè is a solution of è12.2è Then we must have è12.3è 2 x 2 2 y =0 : 2 Y d2 X dx 2 X d2 Y dy 2 =0 : Applying the usual sepaation of vaiables agument, we ænd è12.4è d 2 X dx 2 2 X = 0 è Xèxè =A sinèèbcosèè d 2 Y dy, 2 2 Y = 0 è Y èyè =Ce,y De y : when the sepaation constant K is a positive eal numbe 2 é 0. The possibility that the K = 0 should not be excluded, howeve. Fo this special case we would have è12.5è No should we exclude the case when K =, 2 é 0: d 2 X dx 2 = 0 è Xèxè =ax b ; d 2 Y dy 2 = 0 è Y èyè =cy d : è12.6è d 2 X dx 2, 2 X = 0 è Xèxè =Ae,x Be x d 2 Y dy 2 2 Y = 0 è Y èyè =C sinèyèd cosèyè : Thus sepaation of vaiables yields thee 4-paamete families of solutions. è12.7è 0 èx; yè = axy bx cy d èx; yè = Ae y sinèxèbe,y sinèxèce y cosèxède,y cosèxè i èx; yè = Ae x sinèyè Be,x sinèyèce x cosèyède,x cosèyè 2. Sepaation of Vaiables with Respect to Pola Coodinates If we make a change of vaiables to pola coodinates è12.8è x = cosèè y = sinèè æ p = x 2, y 2 = tan,1 y x 51

3 SPECIAL SOLUTIONS OF LAPLACE'S EQUATION then and è12.9è x = x x = cosèè, sinèè = y y y = sinèè cosèè 2 x 2 = 2 y 2 = cosèè, sinèè = cos 2 2, 2 2 cosèè sinèè 2 sinèè cosèè = sin 2 èè 2, cosèè, sinèè sinèè cosèè sin2 èè sin2 èè sinèè cosèè sinèè cosèè 2 cosèè sinèè 2 cos2 èè cos2 èè cosèè sinèè, 2 sinèè cosèè 2 2 x 2 2 y = : 2 2 Thus, in pola coodinates Laplace's equation takes the fom è12.10è If we set è12.11è =0 : 2 è;è=rèèæèè : and plug è12.11è into è12.10è and then divide by Rèèæèè then we obtain R 00 R R0 R æ00 2 æ =0 o 2 R00 R R0 R =,æ00 æ : Applying the sepaation of vaiables agument we now look fo solutions of è12.12è If 2 6= 0, then the second equation has as solutions è12.13è 2 R 00 R 0, 2 R = 0 æ 00 2 æ = 0 æèè =A cosèèb sinèè : In ode fo such solutions to be continuous acoss the ay = 0 èi.e. so that æè2è = æè0èè we must demand that = n 2 N = f1; 2; 3;:::g. Fo such the æst equation in è12.12è is an Eule-type equation which has as solutions è12.14è If 2 = 0, then è12.12è educes to è12.15è Rèè =A n B,n : R 00 R 0 = 0 æ 00 = 0 The geneal solution of the second equation is obviously è12.16è æèè =a b : To solve the æst, we set W = R 0 so that we can educe it to the following æst ode ODE è12.17è, 1 = W 0 W = d èlnèw èè d

4 Integating both sides of this equation yields è12.18è o 3. POLYNOMIAL SOLUTIONS 53, ln jj =lnèw èc 0 è12.19è W = C Replacing the left hand side of è12.19è R 0 and then integating both sides of yields è12.20è Rèè =C ln jj D We thus aive at the following two families of solutions è12.21è è12.22è n è;è = A n cosènèb n sinènèc,n cosènèd,n sinènè 0 è;è = a ln jj b ln jj c d Fom the solution 0 è;è with 0 = a = c = d, we can infe the existence of a solution of the fom èx; yè =ln æ æ æèx, x o è 2 èy, y o è ææ 2 It can be easily checked that this is the only solution èup to a multiplicative andèo additive constantè of Laplace's equation that depends only the distance fom the point èx o ;y o è. It is called the fundamental solution of Laplace's equation. 3. Polynomial Solutions We have aleady seen that èx; yè =axy bx cy d is a solution of Laplace's equation. Let us now look to see if thee ae othe polynomial solutions. Suppose then Thus, is a solution of Laplace's equation. Similaly, if we set then and so èx; yè =Ax 2 By 2 0= 2 x 2 2 y 2 = A B è B =,A : èx; yè =A, x 2, y 2æ èx; yè =Ax 3 Bx 2 y Cxy 2 Dy 3 0= 2 x 2 C =,3A =6Ax 2By 2Cx6Dy è 2 y2 B =,3D will be a solution of Laplace's equation. èx; yè =A, x 3, 2xy 2æ B, y 3, 3x 2 y æ Now let èx; yè be an abitay homogeneous polynomial of degee n: è12.23è èx; yè = nx a i x n,i y i :

5 SPECIAL SOLUTIONS OF LAPLACE'S EQUATION Then if this is to be a solution of Laplace's equation we must have è12.24è è12.25è è12.26è è12.27è 0 = 2 x 2 2 y 2 = = = nx n,2 X n,2 X èn, ièèn, i, 1èa i x n,i,2 y i nx X n,2 èn, ièèn, i, 1è a i x n,i,2 y i i èi, 1è a i x n,i y i,2 èi 2èèi 1èa i2 x n,2,i y i èèn, ièèn, i, 1èa i èi 2èèi 1èa i2 è x n,2,i y i and so è12.23è will be a solution of Laplace's equation if the coeæcients a i satisfy the following ecusion elation è12.28è a i2 = èn, ièèn, i, 1è a i : èi 1èèi 2è Note that these ecusion elations imply that fo each n 2 N thee ae pecisely two linealy independent homogeneous polynomial solutions of Laplace's equation èsince è12.28è tells that all the coeæcients a i ae completely detemined by a 0 and a 1 è. 4. Seies Solutions Conside the following PDEèBVP è12.29è = 0 èr; è = fèè which is just Laplace's equation in pola coodinates with Diichlet bounday conditions imposed on the bounday of the cicle = R. To constuct a solution of è12.29è we shall æst expand the solution è;è in a seies of -dependent eigenfuctions: è12.30è è;è= 1 2 a 0èè 1X èa n èè cosènèb n èè sinènèè : We note that the tigonometic functions cosènè and sinènè ae eigenfunctions of the Stum-Liouville poblem with diæeential equation ècompae with è12.12è and bounday conditions æ 00 2 æ=0 æè0è = æè2è : Inseting è12.30è into è12.29è we obtain è12.31è 0 = a a0 P P 1 0 1,, a 00, æ,n 2 a n cosènè, n 2 b n sinènè, æ æ n 1 næ a0 cosènè b 00 n 1 b0 n sinènè

6 4. SERIES SOLUTIONS 55 Multiplying both sides by cosènè o sinènè, integating with espect to fom 0 to 2, and employing the identities R 2 2 ; n =0 cosènèd = 0 0 ; n 6= 0 è12.32è we obtain R 2 sinènè cosèmèd = 0 ; 8 n; m 2 R Z 0 2 cosènè cosèmèd = æ 0 nm R 2 0 sinènè sinèmèd = æ nm a 00 n 1 a0 n, n2 2 a n = 0 ; n =0; 1; 2; 3;::: è12.33è b 00 n 1 b0 n, n2 2 b n = 0 ; n =1; 2; 3;::: These ae all Eule type ODEs which have as thei geneal solution è12.34è Rèè = A n B,n ; n 6= 0 Rèè = A B ln jj ; n =0 In ode fo ou solution to be egula at the oigin we must exclude solutions popotional to,n and ln jj. We theefoe take a n èè = A n n ; n =0; 1; 2; æææ è12.35è b n èè = B n n ; n =1; 2; 3; æææ to be the appopiate solutions of è12.33è. Hence, è12.25è becomes è12.36è è;è= 1 2 A 0 1X èa n n cosènèb n n sinènèè : To æx the constants A n ;B n wenow impose the bounday condition at = R; è12.37è fèè = 1 2 A 0 1X èa n R n cosènèb n R n sinènèè : Multiplying both sides of è12.37è by cosèmè o sinèmè and integating fom 0 to 2 then yields è12.38è R 1 2 A n = R n fèè cosènèd ; n =0; 1; 2; 3;::: R0 1 2 B n = R n fèè sinènèd ; n =1; 2; 3;::: 0 Homewok: Poblem 4.5.4