Exempelsamling Vektoranalys
Exempelsamling Vektoranalys Exempelsamling Vektoranalys
57c) ∇ · (∇ × A) = ∂ i ǫ ijk ∂ j A k = 0d)[(B × C) · (∇ × A)] i = (ǫ ijk B j C k )(ǫ ilm ∂ l A m ) == (δ jl δ km − δ jm δ kl )B j C k (∂ l A m ) == B l C k (∂ l A k ) − B m C l (∂ l A m ) == [C · (B · ∇)A − B · (C · ∇)A] ie)[(B · ∇)(φA)] i = B j ∂ j (φA i ) == B j (∂ j φ)A i + φB j (∂ j A i ) == [A(B · ∇φ) + φ(B · ∇)A] if)[(B · ∇)(A × B)] i = B j ∂ j (ǫ ikl A k B l ) == ǫ ikl B j {(∂ j A k )B l + A k (∂ j B l )} == −ǫ ilk B l B j ∂ j A k + ǫ ikl A k B j ∂ j B l == [−B × (B · ∇)A + A × (B · ∇)B] ig)[A × (∇ × A)] i = ǫ ijk A j (∇ × A) k = ǫ ijk ǫ klm A j ∂ l A m == (δ il δ jm − δ im δ jl )A j ∂ l A m == A m ∂ i A m − A j ∂ j A i = 1 2 ∂ i(A m A m ) − A j ∂ j A i =h)= [ 1 2 gradA2 − (A · ∇)A] i[(A × ∇) × A] i = = ǫ ijk (∇ × A) j A k= ǫ ijk ǫ jlm A l ∂ m A k = (δ kl δ im − δ km δ il )A l ∂ m A k == A k ∂ i A k − A i ∂ k A k == [ 1 2 gradA2 − A divA] i (jfr g)65. 65 a-g saknash) alt. I:[rot((a × r) × b)] i = ǫ ijk ∂ j ((a × r) × b) k == ǫ ijk ǫ klm ∂ j (a × r) l b m == ǫ ijk ǫ klm ǫ lpq a p b m ∂ j r q == (δ il δ jm − δ im δ jl )ǫ lpq a p b m ∂ j r q == ǫ ipq a p b m ∂ m r q − ǫ jpq a p b i ∂ j r q == ǫ ipq a p b m δ mq − ǫ jpq a p b i δ jq = [a × b] i
58h) alt. II:tyochh) alt. III:rot((a × r) × b) = {(8.24)} = (b · ∇)(a × r) − b(∇ · (a × r)) == {ex. 64 f) och (8.23)} == a × (b · ∇)r + b(a · (∇ × r)) = a × b()∂(b · ∇)r = b x∂x + b ∂y∂y + b ∂z (x, y, z) = (b x , b y , b z ) = b∂z∇ × r =∣e x e y e z∂ ∂ ∂∂x ∂y ∂zx y z= 0∣rot((a × r) × b) = rot(b × (r × a)) = rot((b · a)r − (b · r)a) == {(8.22)} = (b · a)(∇ × r) −(∇(b · r)) × a =} {{ }=0= −b × atyeller∇(b · r) =( ∂∂x , ∂ ∂y , ∂ )(b x x + b y y + b z z) = (b x , b y , b z ) = b∂z∇(b · r) = {(8.25)} = (b · ∇)r +b × (∇ × r)} {{ } } {{ }=b =066. 66a)[grad(a · gradφ)] i = ∂ i (a j ∂ j φ) = a j ∂ i ∂ j φ = [(a · ∇)∇φ] ib)[rot(a × gradφ)] i = ǫ ijk ∂ j (a × ∇φ) k =Alltså: A = −B = (a · ∇)∇φc) Medblir= ǫ ijk ǫ klm ∂ j (a l ∂ m φ) == (δ il δ jm − δ im δ jl )a l ∂ j ∂ m φ == a i ∂ m ∂ m φ − a l ∂ l ∂ i φ = [a ∇ 2 φ −(a · ∇)∇φ]}{{}i=0∇φ = e x yz + e y xz + e z xyA = e x (a y z + a z y) + e y (a x z + a z x) + e z (a x y + a y x)
- Page 7 and 8: 629. 29 Beräkna linjeintegralen av
- Page 9 and 10: 8Beräkning av divergens och rotati
- Page 11 and 12: 1050. 50 Använd Gauss’ sats för
- Page 13 and 14: 1260. 60 Beräkna integralen ∮dä
- Page 15 and 16: 1466. 66 Låt φ vara ett skalärf
- Page 17 and 18: 1678. 78 Beräkna integralendär S
- Page 19 and 20: 18längs S:s slutna randkurva C. ˆ
- Page 21 and 22: 2093. 93 Vektorfältetsatisfierar e
- Page 23 and 24: 22102. 102 Temperaturfördelningen
- Page 25 and 26: 24110. 110 Låt a vara en konstant
- Page 27 and 28: 26Vilket svar hade man fått om fä
- Page 29 and 30: 28Kontinuitetsekvationen, Greens sa
- Page 31 and 32: 30Den specifika värmeledningsförm
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- Page 35 and 36: 34a) Transformera Laplaces ekvation
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- Page 39 and 40: 38167. 167 En tensor har i det kart
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- Page 43 and 44: 42a) rotA.b) (B · ∇)A.c) (A ·
- Page 45 and 46: 44189. 189 Använd Stokes’ sats f
- Page 47 and 48: 46198. 198 Vektorfältet A har kont
- Page 49 and 50: 48Svar och lösningsanvisningar1. 1
- Page 51 and 52: 5020. 20 Betrakta nivåytan φ = x
- Page 53 and 54: 52e) 12π/534. 34 4πR 335. 35 3233
- Page 55 and 56: 5450. 50 Låt S ∗ vara en cirkel
- Page 57: 5661. 61 Eftersom rotA = (1, 1, 1),
- Page 61 and 62: 6072. 7212 ○ ∫∫SdS × (a × r
- Page 63 and 64: 6283. 83a) På C ärr = a(cosϕ, si
- Page 65 and 66: 6487. 87 Linjeintegralens i:e kompo
- Page 67 and 68: 66(4, 5) → (3) ger slutligen:dvs.
- Page 69 and 70: 68b)c)∇ × A ==1r 2 sinθ1r 2 sin
- Page 71 and 72: 70a) rotF = 0b)divF = cos3θr 4 sin
- Page 73 and 74: 72111. 111 − 4 cosθr 4 e r112. 1
- Page 75 and 76: 74a)b)∫∫1○S r e r · ˆn dS =
- Page 77 and 78: 76125. 125 Differentialekvationerna
- Page 79 and 80: 78Då ε → 0 blir4πhφ(0, −) =
- Page 81 and 82: 80139. 139 Integralen ==∮C 1∮
- Page 83 and 84: 82d)e)f)gradφ =(1 ∂φ√u2 + v 2
- Page 85 and 86: a) Ur ⎧⎪ ⎨⎪ ⎩ξ 2 = ρ
- Page 87 and 88: 86148. 148r = (r cosϕsin θ, r sin
- Page 89 and 90: 88b) δ ij ε ijk = ε iik = 0c)ε
- Page 91 and 92: 90161. 161∮− φA · drC162. 162
- Page 93 and 94: 92172. 172∫∫∫(A div A − A
- Page 95 and 96: 94177. 177∇ · (ω × r) = r · (
- Page 97 and 98: 96184. 184 ∫∫∫∫(∇φ ×
- Page 99 and 100: 98Här har vi utnyttjat att∇φ
- Page 101 and 102: 100a)∫∫e ϕ × ˆndS =S∫∫=
- Page 103 and 104: 102b)∇ 2 ∇ 2 u =∇ 2 u =u =1 (
57c) ∇ · (∇ × A) = ∂ i ǫ ijk ∂ j A k = 0d)[(B × C) · (∇ × A)] i = (ǫ ijk B j C k )(ǫ ilm ∂ l A m ) == (δ jl δ km − δ jm δ kl )B j C k (∂ l A m ) == B l C k (∂ l A k ) − B m C l (∂ l A m ) == [C · (B · ∇)A − B · (C · ∇)A] ie)[(B · ∇)(φA)] i = B j ∂ j (φA i ) == B j (∂ j φ)A i + φB j (∂ j A i ) == [A(B · ∇φ) + φ(B · ∇)A] if)[(B · ∇)(A × B)] i = B j ∂ j (ǫ ikl A k B l ) == ǫ ikl B j {(∂ j A k )B l + A k (∂ j B l )} == −ǫ ilk B l B j ∂ j A k + ǫ ikl A k B j ∂ j B l == [−B × (B · ∇)A + A × (B · ∇)B] ig)[A × (∇ × A)] i = ǫ ijk A j (∇ × A) k = ǫ ijk ǫ klm A j ∂ l A m == (δ il δ jm − δ im δ jl )A j ∂ l A m == A m ∂ i A m − A j ∂ j A i = 1 2 ∂ i(A m A m ) − A j ∂ j A i =h)= [ 1 2 gradA2 − (A · ∇)A] i[(A × ∇) × A] i = = ǫ ijk (∇ × A) j A k= ǫ ijk ǫ jlm A l ∂ m A k = (δ kl δ im − δ km δ il )A l ∂ m A k == A k ∂ i A k − A i ∂ k A k == [ 1 2 gradA2 − A divA] i (jfr g)65. 65 a-g saknash) alt. I:[rot((a × r) × b)] i = ǫ ijk ∂ j ((a × r) × b) k == ǫ ijk ǫ klm ∂ j (a × r) l b m == ǫ ijk ǫ klm ǫ lpq a p b m ∂ j r q == (δ il δ jm − δ im δ jl )ǫ lpq a p b m ∂ j r q == ǫ ipq a p b m ∂ m r q − ǫ jpq a p b i ∂ j r q == ǫ ipq a p b m δ mq − ǫ jpq a p b i δ jq = [a × b] i
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