University Math
Theorem: If f and g are vector fields and φ is scalar field defined in given
region R and C is a scalar constant then prove that:
1- div 𝑓 + 𝑔 = div 𝑓 + div 𝑔
2- div 𝑐 𝑓 = c div 𝑓
3- div φ𝑓 = φ div 𝑓 + gradφ. 𝑓
1)- div 𝑓 + 𝑔 = div 𝑓 + div 𝑔
Proof:
=L.H.S
= div 𝑓 + 𝑔
= ∇. 𝑓 + 𝑔
=
𝜕
𝑖
𝜕𝑥
+
𝜕
𝐽
𝜕𝑦
+
𝜕
𝑘
𝜕𝑧
⋅ 𝑓+𝑔
Let
𝑓 = 𝑓1 𝑖 + 𝑓2 𝐽 + 𝑓3 𝑘
And
So,
𝑔 = 𝑔1 𝑖 + 𝑔2 𝐽 + 𝑔3 𝑘
𝑓 + 𝑔 =(𝑓1 𝑖 + 𝑓2 𝐽 + 𝑓3 𝑘) + (𝑔1 𝑖 + 𝑔2 𝐽 + 𝑔3 𝑘)
𝑓 + 𝑔 = 𝑓1 + 𝑔1 𝑖 + 𝑓2 + 𝑔2 𝑗 + 𝑓3 + 𝑔3 𝑘
𝑑𝑖𝑣 𝑓 + 𝑔 =
𝑑𝑖𝑣 𝑓 + 𝑔 =
𝑑𝑖𝑣 𝑓 + 𝑔 =
𝑑𝑖𝑣 𝑓 + 𝑔 =
𝑑𝑖𝑣 𝑓 + 𝑔 =
𝜕
𝜕
𝜕
𝑖 + 𝐽 + 𝑘 ⋅ [ 𝑓1 + 𝑔1 𝑖 + 𝑓2 + 𝑔2 𝑗 + 𝑓3 + 𝑔3 𝑘]
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
𝑖 . (𝑓1 + 𝑔1 )𝑖 +
𝑗 . (𝑓2 + 𝑔2 )𝑗 +
𝑘 . (𝑓3 + 𝑔3 )𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
(𝑓 + 𝑔1 ) +
(𝑓 + 𝑔2 ) +
(𝑓 + 𝑔3 )
𝜕𝑥 1
𝜕𝑦 2
𝜕𝑧 3
𝜕𝑓1
𝜕𝑔1
𝜕𝑓2
𝜕𝑔2
𝜕𝑓3
𝜕𝑔3
+
+
+
+
+
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑓1
𝜕𝑓2
𝜕𝑓3
𝜕𝑔1
𝜕𝑔2
𝜕𝑔3
+
+
+
+
+
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝑑𝑖𝑣 𝑓 + 𝑔 = div 𝑓 + div 𝑔
Hence proved. It can also be written as;
∇. 𝑓 + 𝑔 = ∇. 𝑓 + ∇. 𝑔
2- div 𝑐 𝑓 = c div 𝑓
Proof:
L.H.S
= div 𝑐 𝑓
= ∇. 𝑐 𝑓
= 𝑐 ∇. 𝑓
We know that;
∇=
𝜕
𝑖
𝜕𝑥
𝜕
+ 𝐽
𝜕𝑦
+
𝜕
𝑘
𝜕𝑧
and 𝑓 = 𝑓1 𝑖 + 𝑓2 𝐽 + 𝑓3 𝑘
div 𝑐 𝑓 = div(𝑐𝑓1 𝑖 + 𝑐𝑓2 𝐽 + 𝑐𝑓3 𝑘)
div 𝑐 𝑓 = ∇.(𝑐𝑓1 𝑖 + 𝑐𝑓2 𝐽 + 𝑐𝑓3 𝑘) =
𝜕
𝑖
𝜕𝑥
+
𝜕
𝐽
𝜕𝑦
+
𝜕
𝑘
𝜕𝑧
. (𝑐𝑓1 𝑖 + 𝑐𝑓2 𝐽 + 𝑐𝑓3 𝑘)
div 𝑐 𝑓
div 𝑐 𝑓
𝜕(𝑐𝑓1 )
𝜕(𝑐𝑓2 )
𝜕(𝑐𝑓3 )
=
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑓1
𝜕𝑓2
𝜕𝑓3
=𝑐
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
OR
div 𝑐 𝑓 = 𝑐
𝜕𝑓
𝜕𝑥
+
𝜕𝑓
𝜕𝑦
𝜕𝑓
+
𝜕𝑧
div 𝑐 𝑓 = 𝑐 𝑑𝑖𝑣 𝑓
Hence proved.
We can also write it as;
∇. 𝑐 𝑓 = 𝑐 ∇. 𝑓
3- div φ𝑓 = φ div 𝑓 + gradφ. 𝑓
Proof:
L.H.S
= div φ𝑓
= ∇. φ𝑓
=
=
=
=
𝜕
𝜕
𝜕
𝑖 + 𝐽 + 𝑘 ⋅ (φ𝑓1 𝑖 + φ𝑓2 𝐽 + φ𝑓3 𝑘)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
(φ𝑓1 ) + (φ𝑓2 ) + (φ𝑓3 )
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑓1
𝜕φ
𝜕𝑓2
𝜕φ
𝜕𝑓3
𝜕φ
φ
+ 𝑓1
+φ
+ 𝑓2 + φ
+ 𝑓3
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑓1
𝜕𝑓2
𝜕𝑓3
𝜕φ
𝜕φ
𝜕φ
φ( +
+ ) + (𝑓1
+ 𝑓2
+ 𝑓3 )
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
=φ
=φ
=φ
𝜕φ
𝜕φ
𝜕φ
+
+
+ (𝑓1
+ 𝑓2
+ 𝑓3 )
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
∇. 𝑓 +
𝑖 + 𝐽 + 𝑘 . (𝑓1 𝑖 + 𝑓2 𝐽 + 𝑓3 𝑘)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
+
𝑖 + 𝐽 + 𝑘 . (𝑓1 𝑖 + 𝑓2 𝐽 + 𝑓3 𝑘)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑓1
(
𝜕𝑥
𝜕𝑓2
𝜕𝑦
𝜕𝑓3
)
𝜕𝑧
= φ div 𝑓 + gradφ. 𝑓
= R.H.S
Hence proved.
It also can be written as;
=φ ∇. 𝑓 + ∇( φ. 𝑓 )