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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8854ffd-431b-44a2-b47e-daec534c3a26%2Fc060e1b6-5e44-4a40-a2cc-c1500342f1f9%2F82tbt5y_processed.jpeg&w=3840&q=75)
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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