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4.18. Prove: (a) V• (A + B) = V• A + V• B, (b) V• (ФА) = (Vф)• A + ф(V• A). Solution (a) Let A = Aii+Azj+Azk, B = Bii + B2 j + Bzk. Then V. (A + B) = ( 11: - = = = = әф Ә Ә = ( i+-+-k). (Aii + Azj+ A3k) + ax = V. A + V. B (b) V. (ФА) = V• (фAji + ФAzj +3k ax аф Ә i+ j+ -А1 + Ф Ә Ә (A1 + Bi) + (Az + B2) + = (A3 + B3) ду ax дz ax 2A1 242 aA3 AB aBz JB3 + + + + + ax Əy az ax ду дz -A1 + 2A1 ax аф ду Ә K) - (A, . [(A1 + Bili + (A2 + B2)j + (A3+B3)k] az + A2 + әф -А2 + ф Əy аф әф -A3 + ф дz аф аф i+=j+ Əy az ax = (Vф). A+ ф(V.A) даг ду а (ФА)) + - (ф2) + ax ду + аф - A3 + ф- az 2A3 az i+ i+k). (B₁i ax 2A 242 243 + + ax ду az Ә (ФА3) az quation. k). (A,4- k) (A₁i+A2j+A3k) + 6( -i+=j+-k).(1 +2+3) a _K). (A1+ A2 + ax дz (Bii+Bzj+B3k)

4.18. Prove: (a) V• (A + B) = V• A + V• B, (b) V• (ФА) = (Vф)• A + ф(V• A). Solution (a) Let A = Aii+Azj+Azk, B = Bii + B2 j + Bzk. Then V. (A + B) = ( 11: - = = = = әф Ә Ә = ( i+-+-k). (Aii + Azj+ A3k) + ax = V. A + V. B (b) V. (ФА) = V• (фAji + ФAzj +3k ax аф Ә i+ j+ -А1 + Ф Ә Ә (A1 + Bi) + (Az + B2) + = (A3 + B3) ду ax дz ax 2A1 242 aA3 AB aBz JB3 + + + + + ax Əy az ax ду дz -A1 + 2A1 ax аф ду Ә K) - (A, . [(A1 + Bili + (A2 + B2)j + (A3+B3)k] az + A2 + әф -А2 + ф Əy аф әф -A3 + ф дz аф аф i+=j+ Əy az ax = (Vф). A+ ф(V.A) даг ду а (ФА)) + - (ф2) + ax ду + аф - A3 + ф- az 2A3 az i+ i+k). (B₁i ax 2A 242 243 + + ax ду az Ә (ФА3) az quation. k). (A,4- k) (A₁i+A2j+A3k) + 6( -i+=j+-k).(1 +2+3) a _K). (A1+ A2 + ax дz (Bii+Bzj+B3k)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Is there any chance that someone can explain what b. In 4.18 is showing? Is this divergence? Thank you!
2 ду2 Then, by addition, Then 1 √x² + y² + 2², = = = аф Ә Əx ах == (A1 + B1) + аф ax The equation V²ф = 0 is called Laplace's equation. It follows that 4 = 1/r is a solution of this equation. 4.18. Prove: (a) V• (A + B) = V• A + V • B, (b) V. (ФА) = (VФ). A + ф(V• A). ivnd Dimas tra Solution (a) Let A = Aji+Az j+Azk, B = Byi + B2 j + Bzk. Then V. (A + B) = a i+ j+ = -А1 + ф A1 + 2y²-2²-x² (x² + y² +2²)5/2 аф ду ax² a i+ ax Əy az 2A1 A2 2A3 aB aB2 + + + + + ду ax Ә а + A2 + i+ i+ 22 22 дуг 2A1 аф ax -j+-k Ә + Ә -k ) . [(A1 + By)i + (A2 + B2)j+ (A3+B3)k] az ау = V• A + V. B Ә (b) V• (ФА) = V• (ФА,i + ФАzj + Ф4,k) = ax (ФА) + 7 (ФА2) + 7 (ФА3) JB3 az ar ду az аф + Ә - (A2 + B2) + - (A3 + B3) taZA ду Az ( Əz² and А2 + ф -A3 + ф (Aii+2+3+ 22 3A2 аф ду Az 2A + Аз + ф ²+y²+z = 0. 2A3 az аA2 2A3 + + ду дz dx ду To Ә + + 2z? - x² - y² (x² + y² +22)5/2 а -k) . (Bji + Bzj + Bzk) dz аф әф аф, Ә a -k). (Aii + Azj+ Azk) + =i+_j+-k). (Aii + Azj+A3k) ax ду az dy az = (VФ). А + фV. A) nothile
Transcribed Image Text:2 ду2 Then, by addition, Then 1 √x² + y² + 2², = = = аф Ә Əx ах == (A1 + B1) + аф ax The equation V²ф = 0 is called Laplace's equation. It follows that 4 = 1/r is a solution of this equation. 4.18. Prove: (a) V• (A + B) = V• A + V • B, (b) V. (ФА) = (VФ). A + ф(V• A). ivnd Dimas tra Solution (a) Let A = Aji+Az j+Azk, B = Byi + B2 j + Bzk. Then V. (A + B) = a i+ j+ = -А1 + ф A1 + 2y²-2²-x² (x² + y² +2²)5/2 аф ду ax² a i+ ax Əy az 2A1 A2 2A3 aB aB2 + + + + + ду ax Ә а + A2 + i+ i+ 22 22 дуг 2A1 аф ax -j+-k Ә + Ә -k ) . [(A1 + By)i + (A2 + B2)j+ (A3+B3)k] az ау = V• A + V. B Ә (b) V• (ФА) = V• (ФА,i + ФАzj + Ф4,k) = ax (ФА) + 7 (ФА2) + 7 (ФА3) JB3 az ar ду az аф + Ә - (A2 + B2) + - (A3 + B3) taZA ду Az ( Əz² and А2 + ф -A3 + ф (Aii+2+3+ 22 3A2 аф ду Az 2A + Аз + ф ²+y²+z = 0. 2A3 az аA2 2A3 + + ду дz dx ду To Ә + + 2z? - x² - y² (x² + y² +22)5/2 а -k) . (Bji + Bzj + Bzk) dz аф әф аф, Ә a -k). (Aii + Azj+ Azk) + =i+_j+-k). (Aii + Azj+A3k) ax ду az dy az = (VФ). А + фV. A) nothile
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