TOPIC:Product rules & laplacian 1 #NUMANUSAMA

1. PRESENTED BY
PRODUCT RULES & LAPLACIAN

2. PRODUCT
RULES
𝟏. 𝛁 × (𝛗𝐀)=𝛗(𝛁×𝐀) + 𝐀×( 𝛁𝛗)
Proof:
𝛻 × (φ𝐀)=𝛻 × (φ𝐀1i + φ𝐀2j+φ𝐀3k)
=
i j k
𝜕
𝜕x
𝜕
𝜕y
𝜕
𝜕z
φ𝐀1 φ𝐀2 φ𝐀3

3. =[
𝜕
𝜕y
(φ𝐀3) −
𝜕
𝜕z
(φ𝐀2)]i+[
𝜕
𝜕z
(φ𝐀1)
𝜕
𝜕x
(φ𝐀3)]j +[
𝜕
𝜕x
φ𝐀2 −
𝜕
𝜕y
(φ𝐀1)]k
=[φ
𝜕𝐀3
𝜕y
+
𝜕φ
𝜕𝑦
𝐀3 − φ
𝜕𝐀2
𝜕z
− (
𝜕φ
𝜕𝑧
)𝐀2]i+[φ
𝜕𝐀1
𝜕z
+ (
𝜕φ
𝜕𝑧
)𝐀1 − φ
𝜕𝐀3
𝜕x
−
(
𝜕φ
𝜕𝑥
)𝐀3]j +[φ
𝜕𝐀2
𝜕x
+
𝜕φ
𝜕𝑥
𝐀2 − φ
𝜕𝐀1
𝜕y
− (
𝜕φ
𝜕𝑦
)𝐀1]k
=𝜑 [(
𝜕𝐀3
𝜕y
−
𝜕𝐀2
𝜕y
)i+(
𝜕𝐀1
𝜕z
−
𝜕𝐀3
𝜕x
)j+(
𝜕𝐀2
𝜕x
−
𝜕𝐀1
𝜕y
)k] +[
𝜕𝜑
𝜕𝑦
𝐀3 −
𝜕𝜑
𝜕𝑧
𝐀2 i +
𝜕𝜑
𝜕𝑧
𝐀1 −
𝜕𝜑
𝜕𝑥
𝐀3 +
𝜕𝜑
𝜕𝑥
𝐀2 −
𝜕𝜑
𝜕𝑦
𝐀1 k]

4. =φ 𝛁 × 𝐀 +
i j k
𝜕φ
𝜕x
𝜕φ
𝜕y
𝜕φ
𝜕z
𝐀1 𝐀2 𝐀3
=φ(𝛻×A) + A×( 𝛻φ)

5. 𝟐. 𝛁 × (𝛁𝐟)=𝟎
Proof:
𝛻 × (𝛻f)=𝛻 × (
𝜕f
𝜕𝑥
𝑖 +
𝜕f
𝜕𝑥
j+
𝜕f
𝜕𝑥
𝑘)
=
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕f
𝜕𝑥
𝜕f
𝜕𝑦
𝜕f
𝜕𝑧

6. =[
𝜕
𝜕𝑦
(
𝜕f
𝜕𝑧
) −
𝜕
𝜕𝑧
(
𝜕f
𝜕𝑦
)]𝑖+[
𝜕
𝜕𝑧
(
𝜕f
𝜕𝑥
) −
𝜕
𝜕𝑥
(
𝜕f
𝜕𝑧
)]𝑗 +[
𝜕
𝜕𝑥
(
𝜕f
𝜕𝑦
) −
𝜕
𝜕𝑦
(
𝜕f
𝜕𝑥
)]𝑘
= 0
𝛻 × 𝛻𝑓 = 0, the reason is that 𝛻𝑓 gives a single vector
but curl always operate on vector field.

7. 𝟑. 𝛁 × 𝛁 × 𝑨 = 𝛁 𝛁. 𝑨 − 𝛁 𝟐
𝑨
Proof:
𝛻 × 𝛻 × 𝐴
= 𝛁 ×
i j k
𝜕
𝜕x
𝜕
𝜕y
𝜕
𝜕z
𝐀1 𝐀2 𝐀3
= 𝛁 × [(
𝜕𝐀3
𝜕y
−
𝜕𝐀2
𝜕y
)i+(
𝜕𝐀1
𝜕z
−
𝜕𝐀3
𝜕x
)j+(
𝜕𝐀2
𝜕x
−
𝜕𝐀1
𝜕y
)k]

8. =
i j k
𝜕
𝜕x
𝜕
𝜕y
𝜕
𝜕z
𝜕𝐀3
𝜕y
−
𝜕𝐀2
𝜕y
𝜕𝐀1
𝜕z
−
𝜕𝐀3
𝜕x
𝜕𝐀2
𝜕x
−
𝜕𝐀1
𝜕y
=[
𝜕
𝜕𝑦
𝜕𝐀2
𝜕x
−
𝜕𝐀1
𝜕y
−
𝜕
𝜕𝑧
(
𝜕𝐀1
𝜕𝑧
−
𝜕𝐀3
𝜕𝑥
)]i+[
𝜕
𝜕𝑧
𝜕𝐀3
𝜕𝑦
−
𝜕𝐀2
𝜕𝑧
−
𝜕
𝜕𝑥
(
𝜕𝐀2
𝜕x
−
𝜕𝐀1
𝜕y
)]j+[
𝜕
𝜕𝑥
𝜕𝐀1
𝜕𝑧
−
𝜕𝐀3
𝜕𝑥
−
𝜕
𝜕𝑦
(
𝜕𝐀3
𝜕𝑦
−
𝜕𝐀2
𝜕𝑧
)]k
= (−
𝜕2 𝐀1
𝜕𝑦2 −
𝜕2 𝐀1
𝜕𝑧2 )i+ −
𝜕2 𝐀2
𝜕𝑧2 −
𝜕2 𝐀2
𝜕𝑥2 j + −
𝜕2 𝐀3
𝜕𝑥2 −
𝜕2 𝐀3
𝜕𝑦2 𝑘 +
(
𝜕2 𝐀2
𝜕𝑦𝜕𝑥
+
𝜕2 𝐀3
𝜕𝑧𝜕𝑥
)i+
𝜕2 𝐀3
𝜕𝑧𝜕𝑦
+
𝜕2 𝐀1
𝜕𝑦𝜕𝑥
j+
𝜕2 𝐀1
𝜕𝑧𝜕𝑥
+
𝜕2 𝐀2
𝜕𝑦𝜕𝑧
k

9. = -(
𝜕2
𝜕𝑥2 +
𝜕2
𝜕𝑦2 +
𝜕2
𝜕𝑧2)(𝐀1i+𝐀2j+𝐀3k) + i
𝜕
𝜕x
(
𝜕𝐀1
𝜕x
+
𝜕𝐀2
𝜕y
+
𝜕𝐀3
𝜕z
)
+j
𝜕
𝜕y
(
𝜕𝐀1
𝜕x
+
𝜕𝐀2
𝜕y
+
𝜕𝐀3
𝜕z
) +k
𝜕
𝜕z
(
𝜕𝐀1
𝜕x
+
𝜕𝐀2
𝜕y
+
𝜕𝐀3
𝜕z
)
= −𝛻2 𝐀 +𝛻 (
𝜕𝐀1
𝜕x
+
𝜕𝐀2
𝜕y
+
𝜕𝐀3
𝜕z
)
=𝛻 𝛻. 𝐀 −𝛻2 𝑨 (proved)

10. 𝛁 × 𝐀 × 𝐁 = 𝐁. 𝛁 𝐀 − 𝐁 𝛁. 𝐀 − 𝐀. 𝛁 𝐁 + 𝐀 𝛁. 𝐁
Proof:
Let A=A1i+A2j+A3k , B=B1i+B2j+B3k
A × B =
𝑖 𝑗 𝑘
A1 A2 A3
B1 B2 B3
= (A2 𝐵3 − A2 𝐵3)i + (A1 𝐵3 − A3 𝐵1)j + (A1 𝐵2 − A2 𝐵1)k

11. =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
A2 𝐵3 − A2 𝐵3 A1 𝐵3 − A3 𝐵1 A1 𝐵2 − A2 𝐵1
= i((
𝜕
𝜕𝑦
(A1 𝐵2 − A2 𝐵1) -
𝜕
𝜕𝑧
(A1 𝐵3 − A3 𝐵1)) - j(
𝜕
𝜕𝑥
(A1 𝐵2 −
A2 𝐵1) −
𝜕
𝜕𝑧
(A2 𝐵3 − A3 𝐵2)) +k(
𝜕
𝜕𝑦
(A2 𝐵3 −
A3 𝐵2)−
𝜕
𝜕𝑥
(A3 𝐵1 − A1 𝐵3))

12. after solving for i, j , k we will get:
Solving for i:
= 𝐀 𝟏 𝛁. 𝐁 - 𝐀. 𝛁 𝐁 𝟏 − 𝐁 𝟏 𝛁. 𝐀 − 𝐁. 𝛁 𝐀 𝟏
Similarly for j and k:
= 𝐀 𝟐 𝛁. 𝐁 - 𝐀. 𝛁 𝐁 𝟐 − 𝐁 𝟐 𝛁. 𝐀 − 𝐁. 𝛁 𝐀 𝟐
= 𝐀 𝟑 𝛁. 𝐁 - 𝐀. 𝛁 𝐁 𝟑 − 𝐁 𝟑 𝛁. 𝐀 − 𝐁. 𝛁 𝐀 𝟑
Now by adding these three equations ,we will get the desired
result i.e.
= 𝐁. 𝛁 𝐀 − 𝐁 𝛁. 𝐀 − 𝐀. 𝛁 𝐁 + 𝐀 𝛁. 𝐁
(proved)

13. 𝛻 × 𝐴=0
Proof:
Let A=A1i+A2j+A3k
𝛻 × 𝐴=
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐴1 𝐴2 𝐴3
=i
𝜕𝐴3
𝜕𝑦
−
𝜕𝐴2
𝜕𝑧
− 𝑗
𝜕𝐴3
𝜕𝑥
−
𝜕𝐴2
𝜕𝑧
+
k
𝜕𝐴2
𝜕𝑥
−
𝜕𝐴1
𝜕𝑦
=0

14. 𝛻 × 𝐴=0 , means field is irrotational.

15. LAPLACIAN 𝛁 𝟐
The laplacian is a differential operator given by the
divergence of the gradient of a scalar function V , written as
𝛻2
𝑉 = 𝛻. (𝛻𝑉)
 The laplacian of a scalar field is scalar.

16. Physical Significance
 As a second derivative, one dimensional laplacian
operator is related to minima and maxima.
 when the second derivative is negative(positive), the
curvature is concave (convex).
 If the laplacian is zero , the function is harmonic.
MCQ’S:
1.The curl of a gradient is
a. Zero
b. Divergence
c. None of these

17. 2. If 𝛻 × 𝐴 = 0, then the field
a. Rotational
b. Irrtotational
c. solenoidal