POISSON S EQUATION 2 V 0

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Oswin Heath
5 years ago
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1 POISSON S EQUATION We have seen how to solve the equation but geneally we have V V4k We now look at a vey geneal way of attacking this poblem though Geen s Functions. It tuns out that this poblem has applications fa beyond electostatics. Fo example, it is closely elated to the famous popagatos of field theoy. We now take We stat with the obsevation that 3 d' da, G with 3 ' G,' ' then

2 3 3 ' ' G,' d ' d 'G,' ' d ' ' 'G,' G,' ' ' da 1 3 d 'G,' ' G,' ' ' da' ' 'G,' da' We know that Poisson s equation will have a unique solution fo eithe Diichlet o Neumann bounday conditions. In the fome we take on the bounday. G,' s 1 3 d 'G,' ' ' 'G,' da' Hence we need only the value of ϕ on the suface which is povided by the bounday condition. In the case of Neumann conditions we take Whee is the unit nomal to the suface. 'G, ' nˆ 1 3 G,' ' d ' G,' ' ' da' Hee we need only ϕ on the suface, which again is povided by the bounday conditions.

3 In eithe case we have the solution of Poisson s equation povided we can find the appopiate Geen s function. We begin with Diichlet case whee ϕ is known on the bounday. We need 3 ' G,' ' with G,' s whee s is a field point on the suface. Note that the bounday condition G,' Is the same whethe the bounday is a conducto o not. s We note that G is simply the solution fo a point chage of magnitude ε located at. The total voltage is then gotten fom the pinciple of supeposition by integating ove the volume. It is easy to show that To see this conside two functions G,' G ', G,' andg,' 1 3 G,' 1 ' G,' G,' ' G,' 1 d ' G,' 1 'G,' G,' 'G,' 1 da' But ' G,' ', G,' 3 s

4 Hence G is symmetic in its two aguments. G, G, 1 1 Case 1 Specific Cases We can wite down the Geen s function immediately fo cetain simple cases. Conside a point chage with no boundaies save at infinity. G is the potential of a point chage ε we have Case G,' 1 k 1 '4 Conside an infinite, gounded, conducting plane. Thus This will simply be the sum of the Geen s functions fo the two point chage G x,y,z,x',y',z' 1 1/ xx' yy' zz' 4 1 1/ xx' yy' zz' 4 Case 3 Conside a gounded, conducting sphee. as in Case :

5 G,' whee is the adius of the sphee. 1 1 ' 4 ˆ We stat with the eigenvalue equation Geen s Functions fom Eigenfunctions OV V whee the V ae eigenfunctions of the opeato with V nomalized in the usual way V ' V ' d ' 3 ', ' We now assume that the V fom a complete set. any function can be expanded in tems of them F av V F d a V V d ' 3 3 ' a a ', ' a V F d 3 and 3 3 F V ' F ' d 'V V 'V F' d '

6 Since F is itay we find V ' V ' 3 We now take the V to be eigenfunctions of the opeato V ' V V ' V V ' V Conside the sum Hence V ' V S SV ' V ' 3 G,' V ' V Subject to the condition that G = on the bounday (o G = on bounday if Neumann conditions). Fo Dichlet conditions we can simplify this somewhat by educing the sum ove 3 vailes to one ove. We suppose that can be boken into two pats whee with n n F n F n

7 F n' ' F n n n' ' wite G n 'n' ' F n', ' F n, g, ' We need G ' nn' ' Now n Hence n G F n', ' F n, g, ' n F n', ' F n, g g F F n', ' F n, g F g Hence we need If we take This becomes g g F n, F n', ' ' nn' ' g g ' 'F n, F n', ' ' n n' ' which is what we need fo G. We must also make g () = at the bounday.

8 To solve g g ' we note that eveywhee except = it is the equation g g We now solve this and the impose the condition that g be continuous at = ε. Next we make the deivative discontinuous in such a way as to get the function. We conside the pocess fo spheical coodinates. We have Spheical Coodinates 1 ctn sin Let G,' Y ', ' Y, g,' m m m,m G,' Y ', ' Y, g,' m m,m m m g m,'y m ', 'Y m, g m,' 1Y m ', 'Y m, gm 1 gm Y m ', ' Y m,

9 1 G,' g 1 g Y ', ' Y, m m m m m Now let m m g 1 g ' m m,m ' G,' Y ', ' Y, To evaluate the sum we use the fact that the Y m ae a complete set. Thus fo an itay function F(,) we have Hence Now m m m F, A Y, F, Y, sindd A Y, Y, sindd 'm' m 'm' m m A A m m ' mm' 'm' F, F ', ' Y ', ' sin 'd 'd 'Y, m m m m m m F ', ' sin ' Y ' ' Y, d ' d ' m m m sin ' Y ', ' Y, ' '

10 Fx i xx df dx xi i Hence ' ' sin sin cos cos ' Only the fist tem occus in the ange. Hence Hence m m m Y ', ' Y, cos cos ' ' ' G,' 1 coscos ' ' This is actually what we want. Because 1 3 d 'G,' ' ' 'G,' da' Conside the fist tem on the HS 1 ' ' sin ' ', ', ' coscos ' ' d'd 'd ' 1 ' ' ' ' ' ' as equied. To find g m (,N) we must solve with,, m m g 1 g '

11 At all points except = N we have Ty d 1g m d g m A This has solutions A A 1 1 A 1 1 1, 1 We now suppose the bounday condition is specified on a sphee of adius. we conside two egions (1),N < (),N > In both egions we have sub egions: < N, >. Fist conside egion (1). fo N we must have m g,' A since g is finite at the oigin. Fo N < < we can have both tems We need g m continuous at = N. Thus C g,' B m 1

12 C B' 1 A' ' At = we need g m =. Thus C 1 B C B 1 1 A' B' B ' 1 Hence fo < N we have Fo > N we have g m,' A' B 1 ' 1 ' A ' 1 g,' A m 1 A 1 1 A A ' ' ' ' 1 dg ' A d 1

13 dg A 1 ' 1 1 d 1 ' 1 Hence 1 1 ' 1 dg dg A d ' ' d ' ' ' A But ' dg dg d g A 1 d d d ' 1 1 d 1 We want dg d ' To get this we take A 1 1 ' ' 1 1

14 ' ' dg 1 dg ' d d ' d > N g,' ' 1 ' 1 ' ' 1 1 < N 1 1 ' ' g,' 1 1 Thus < 1 1 ' ' G,' Y 1 m ', ' Y m, 1,m,m ' Y m ', ' Y m, 1 ' > N

15 1 1 ' G,' Y 1 m ', ' Y m, 1,m,m ' ' Y m ', ' Y m, 1 o 1 G,' Y 1 1 m ', ' Y m, m 1 Poceeding in the same way fo egion () we find 1 1 G,' Y m ', ' Y m, m 1

f(k) e p 2 (k) e iax 2 (k a) r 2 e a x a a 2 + k 2 e a2 x 1 2 H(x) ik p (k) 4 r 3 cos Y 2 = 4

f(k) e p 2 (k) e iax 2 (k a) r 2 e a x a a 2 + k 2 e a2 x 1 2 H(x) ik p (k) 4 r 3 cos Y 2 = 4 Fouie tansfom pais: f(x) 1 f(k) e p 2 (k) p e iax 2 (k a) 2 e a x a a 2 + k 2 e a2 x 1 2, a > 0 a p k2 /4a2 e 2 1 H(x) ik p 2 + 2 (k) The fist few Y m Y 0 0 = Y 0 1 = Y ±1 1 = l : 1 Y2 0 = 4 3 ±1 cos Y