∆𝑢 = 0 1 < 𝑟 < 2
1
{ 𝑢|𝑟=1 = cos 𝜃
2
|
𝑢 𝑟=2 = 1 + cos 2𝜃
∞
𝑢(𝑟, 𝜃) = ∑(𝐴𝑛 𝑟 𝑛 + 𝐵𝑛 𝑟 −𝑛−1 )𝑃𝑛 (cos 𝜃)
∞
𝑛=0
𝑢|𝑟=1 = ∑(𝐴𝑛 + 𝐵𝑛 )𝑃𝑛 (cos 𝜃) = (𝐴0 + 𝐵0 )𝑃0 (cos 𝜃 ) + (𝐴1 + 𝐵1 )𝑃1 (cos 𝜃 ) + (𝐴2 + 𝐵2 )𝑃2 (cos 𝜃) + ⋯ =
∞
𝑛=0
1
1
cos 𝜃 = 𝑃1 (cos 𝜃)
2
2
1
1
1
𝑢|𝑟=2 = ∑(𝐴𝑛 2𝑛 + 𝐵𝑛 2−𝑛−1 )𝑃𝑛 (cos 𝜃) = (𝐴0 + 𝐵0 ) 𝑃0 (cos 𝜃) + (2𝐴1 + 𝐵1 ) 𝑃1 (cos 𝜃) + (4𝐴2 + 𝐵2 ) 𝑃2 (cos 𝜃) + ⋯ = 1 + cos 2𝜃 =
2
4
8
𝑛=0
2
4
= 𝑃0 (cos 𝜃) + 𝑃2 (cos 𝜃)
3
3
4
𝐴0 + 𝐵0 = 0
𝐵0 = −𝐴0
𝐵0 = −𝐴0
𝐴0 =
3
1
2⇔{
1
2⇔{1
2 ⇔{
{
4
𝐴0 + 𝐵0 =
𝐴0 − 𝐴0 =
𝐴 =
𝐵0 = −
2
3
2
3
2 0 3
3
1
1
1
1
1
𝐵1 = − 𝐴1
𝐴1 + 𝐵1 =
𝐵1 = − 𝐴1
𝐵1 = − 𝐴1
𝐴1 = −
2
2 ⇔{
2
2
14
{
⇔{
⇔{
⇔{
1 1
1
1 1
7
1
4
2𝐴1 + ( − 𝐴1 ) = 0
2𝐴1 + 𝐵1 = 0
2𝐴1 + − 𝐴1 = 0
𝐴1 = −
𝐵1 =
4 2
4
8 4
4
8
7
32
𝐴2 + 𝐵2 = 0
𝐵2 = −𝐴2
𝐵2 = −𝐴2
𝐴2 =
93
1
4
1
4
31
4
{
⇔{
⇔{
⇔{
32
4𝐴2 + 𝐵2 =
4𝐴2 − 𝐴2 =
𝐴 =
𝐵2 = −
8
3
8
3
8 2 3
93
𝐴𝑛 + 𝐵𝑛 = 0
{
𝑛 ≥ 3 ⇒ 𝐴0 = 𝐵0 = 0 ∀𝑛 ≥ 3
𝐴𝑛 2𝑛 + 𝐵𝑛 2−𝑛−1 = 0
4 4
𝑟
4
32𝑟 2
32 3 cos2 𝜃 − 1
)
𝑢(𝑟, 𝜃) = ( − ) 𝑃0 (cos 𝜃) + (−
+ 2 ) cos 𝜃 + (
−
3 3𝑟
14 7𝑟
93
93𝑟 3
2
2
4
4
𝑧
4𝑧
16𝑧
16 2
16𝑧 2
16
2
2
(
)
𝑢(𝑥, 𝑦, 𝑧) = −
−
+
+
−
𝑥
+
𝑦
+
𝑧
−
+
3
5
3 3√𝑥 2 + 𝑦 2 + 𝑧 2 14 7(√𝑥 2 + 𝑦 2 + 𝑧 2 )3
31
93
93(√𝑥 2 + 𝑦 2 + 𝑧 2 )
31(√𝑥 2 + 𝑦 2 + 𝑧 2 )