Differential Operators Theorems
Theorem1. If the vector f is expressible in terms of its Cartesian components f1,f2,f3 as f= f1i+f2j+f3k, Then ∇.f\(=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial v}+\frac{\partial f_3}{\partial z}\)
Proof:
Given f=f1i+f2j+f3k
∴ div f =∇ .f\(=\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{f}}{\partial v}+\mathbf{k} \cdot \frac{\partial \mathbf{f}}{\partial \mathbf{z}}\)
⇒ \(=\mathbf{i} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial x}+\mathbf{j} \frac{\partial f_2}{\partial x}+\mathbf{k} \frac{\partial f_3}{\partial x}\right)+\mathbf{j} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial y}+\mathbf{j} \frac{\partial f_2}{\partial y}+\mathbf{k} \frac{\partial f_3}{\partial y}\right)\)\(+\mathbf{k} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial z}+\mathbf{j} \frac{\partial f_2}{\partial z}+\mathbf{k} \cdot \frac{\partial f_3}{\partial z}\right)\)
=\(\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}\)
Theorem 2. Prove that \(\text{div}(\mathbf{f} \pm \mathbf{g})=\text{div} . \mathbf{f} \pm \text{div} . \mathbf{g}\)
Proof:
⇒ \(\text{div}(\mathbf{f} \pm \mathbf{g})=\Sigma \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{f} \pm \mathbf{g})=\Sigma \mathbf{i} \cdot\left(\frac{\partial \mathbf{f}}{\partial x} \pm \frac{\partial \mathbf{g}}{\partial x}\right)\)
= \(\Sigma \mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x} \pm \Sigma \mathbf{i} \cdot \frac{\partial \mathbf{g}}{\partial x}=\text{div} \mathbf{f} \pm \text{div} \mathbf{g}\)
Differential Operator’s Theorems And Proofs
Differential Operators The Laplacian Operator∇2
⇒ \(\nabla . \nabla \phi=\Sigma i \cdot \frac{\partial}{\partial x}\left(i \frac{\partial \phi}{\partial x}+j \frac{\partial \phi}{\partial y}+k \frac{\partial \phi}{\partial z}\right)=\sum \frac{\partial^2 \phi}{\partial x^2}\)
= \(\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}=\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right) \phi=\nabla^2 \phi\)
Thus the operator \(\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\) is called the Laplacian operator.
Note: \(\nabla^2 \phi=0\) is called Laplacian equation.
Differential Operators Curl Of A Vector
Definition: Let \(\mathbf{f}\) be any continuously differentiable vector point function. Then the vector function defined by \(\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{f}}{\partial z}\) is called the Curl of \(\mathbf{f}\) and is denoted by \(\nabla \times \mathbf{f}\)
∴ Curl \(\mathbf{f}=\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{f}}{\partial z}=\left(\mathbf{i} \times \frac{\partial}{\partial x}+\mathbf{j} \times \frac{\partial}{\partial y}+\mathbf{k} \times \frac{\partial}{\partial z}\right) \mathbf{f}\)
= \(\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \times \mathbf{f}=\nabla \times \mathbf{f}\)
Theorem: If f is a differentiable vector point function given by f=f1i+f2j+f3k then, curl f
Proof:
Curl \(\left.\mathbf{f}=\nabla \times \mathbf{f}=\sum \mathbf{i} \times \frac{\partial \mathbf{t}}{\partial x}=\sum \mathbf{i} \times \frac{\partial}{\partial x} f_1 \mathbf{i}+f_2 \mathbf{j}+f_3 \mathbf{k}\right)\)
= \(\sum\left(\frac{\partial f_2}{\partial x} \mathbf{k}-\frac{\partial f_3 \mathbf{j}}{\partial x}\right)=\left(\frac{\partial f_2}{\partial x} \mathbf{k}-\frac{\partial f_3}{\partial x} \mathbf{j}\right)+\left(\frac{\partial f_3}{\partial y} \mathbf{i}-\frac{\partial f_1}{\partial y} \mathbf{k}\right)+\left(\frac{\partial f_1}{\partial z} \mathbf{j}-\frac{\partial f_2}{\partial z} \mathbf{i}\right)\)
= \(\mathbf{i}\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right)+\mathbf{j}\left(\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x}\right)+\mathbf{k}\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)\)
Note 1. The expression for Curl f can be well remembered if we treat \(\nabla\) as an operative vector quantity that is Curl \(\mathbf{f}=\left|\begin{array}{lll}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_1 & f_2 & f_3\end{array}\right|\)
2. If f is a constant vector then \(\text{Curl} \mathbf{f}=\mathbf{0}\)
Laplacian Operator ∇² Vector Identities
Differential Operators Irrotational Vector
A vector point function f is said to be irrotational if. Curl f = 0 i.e. ∇x f = 0
Physical Interpretation of curl If w is the angular velocity of a rigid body rotating about a fixed axis and v is the velocity of any point P(x,y,z) on the body, then w =1/2 curl v . Thus the angular velocity of rotation at any point is equal to half the curl of the velocity vector. This justifies the use of the word “curl of a vector”.’
Theorem Prove that curl (A ± B) = curl A ± curl B
Proof:
Curl \((\mathrm{A} \pm \mathrm{B})=\nabla \times(\mathrm{A} \pm \mathrm{B})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathrm{~A} \pm \mathrm{B})=\sum \mathbf{i} \times\left(\frac{\partial \mathrm{A}}{\partial x} \pm \frac{\partial \mathrm{B}}{\partial x}\right)\)
= \(\sum \mathbf{i} \times \frac{\partial \mathrm{A}}{\partial x} \pm \sum \mathbf{i} \times \frac{\partial \mathrm{B}}{\partial x}=\nabla \times \mathrm{A} \pm \nabla \times \mathrm{B}=\text{curl} \mathrm{A} \pm \text{curl} \mathrm{B}\)
Theorem: The differential operators’ grad, div, and curl are invariant with respect to any Cartesian coordinate frame.
Proof: Let P (x, y, z) be a point in the coordinate frame OXYZ and i, j, k be the right-handed unit vectors along the axes. Let (x’, y’,z’ ) be the coordinates of P.w.r. to the new coordinate frame O’ X’ Y’ Z’.
If \(\mathbf{i}^{\prime}, \mathbf{j}^{\prime}, \mathbf{k}^{\prime}\) are the right handed unit vectors along \(\overrightarrow{\mathrm{OX}^{\prime}}, \overrightarrow{\mathrm{OY}^{\prime}}, \overrightarrow{\mathrm{OZ}^{\prime}}\), then \(\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OO}^{\prime}+} \overrightarrow{\mathrm{O}^{\prime} \mathrm{P}^{\prime}}\)
⇒ \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=\overrightarrow{\mathrm{OO}^{\prime}}+x^{\prime} \mathbf{i}^{\prime}+y^{\prime} \mathbf{j}^{\prime}+z^{\prime} \mathbf{k}^{\prime}\)
As \(x^{\prime}, y^{\prime}, z^{\prime}\) are the functions of x y z differentiating partially with respect to x, y, z we get \(\mathbf{i}=\mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial x}+\mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial x}+\mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial x}\)…..(1)
⇒ \(\mathbf{j}=\mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial y}+\mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial y}+\mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial y}\)
⇒ \(\mathbf{k}=\mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial z}+\mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial z}+\mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial z}\)
Now taking the dot product of (1), (2), (3) with \(\mathbf{i}^{\prime}, \mathbf{j}^{\prime}, \mathbf{k}^{\prime}\) we get \(\mathbf{i}. \mathbf{i}^{\prime}=\frac{\partial x^{\prime}}{\partial x}, \mathbf{i} . \mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial x}, \mathbf{i} \cdot \mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial x}\)
⇒ \(\mathbf{j} \cdot \mathbf{i}^{\prime}=\frac{\partial x^{\prime}}{\partial y}, \mathbf{j} \cdot \mathbf{j} \frac{\partial y^{\prime}}{\partial y}, \mathbf{j} \cdot \mathbf{k} \frac{\partial z^{\prime}}{\partial y}\)
⇒ \(\mathbf{k} \cdot \mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial z}, \mathbf{k} \cdot \mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial z}, \mathbf{k} \cdot \mathbf{k} \frac{\partial z^{\prime}}{\partial z}\)
Now by the principle of partial differentiation
1. \(\frac{\partial}{\partial y}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial x}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial x}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial x}=(i.i)\frac{\partial}{\partial x^{\prime}}+ (i. \left.\mathbf{j}^{\prime}\right) \frac{\partial}{\partial y^{\prime}}+(i. \left.\mathbf{k}^{\prime}\right) \frac{\partial}{\partial z^{\prime}} \quad\)……. from (4)
2. \(\frac{\partial}{\partial y}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial z}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial z}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \frac{\partial}{\partial x^2}+\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right) \frac{\partial}{\partial y^1}+\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right) \frac{\partial}{\partial z} \quad \cdots\) from (5)
3. \(\frac{\partial}{\partial z}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial z}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial z}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \frac{\partial}{\partial x^{\prime}}+(\mathbf{k} \cdot \mathbf{j}) \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right) \frac{\partial}{\partial z^{\prime}} \ldots\) from (6)
Also, we know that \(\left.\begin{array}{l}
\mathbf{i}^{\prime}=\left(\mathbf{i}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+\left(\mathbf{i}^{\prime} \cdot \mathbf{j}\right) \mathbf{j}^{\prime}+\left(\mathbf{i}^{\prime} \cdot \mathbf{k}\right) \mathbf{k} \\
\mathbf{j}^{\prime}=\left(\mathbf{j}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+(\mathbf{j} \cdot \mathbf{j}) \mathbf{j}+(\mathbf{j} \cdot \mathbf{k}) \mathbf{k} \\
\mathbf{k}^{\prime}=\left(\mathbf{k}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+\left(\mathbf{k}^{\prime} \cdot \mathbf{j}\right) \mathbf{j}+\left(\mathbf{k}^{\prime} \cdot \mathbf{k}\right) \mathbf{k}
\end{array}\right\} \cdots\)….(A)
1. To Prove grad is invariant
Multiplying (1), (2) by \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and adding we get \(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\)
= \(\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{i}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{i}^{\prime}\right)+\right. \left.\mathbf{k}\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right)\right] \frac{\partial}{\partial x^{\prime}}+\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{j}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right)+\mathbf{k}\left(\mathbf{k} \cdot \mathbf{j}^{\prime}\right)\right] \frac{\partial}{\partial y^{\prime}}\)
+ \(\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{k}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right)+\mathbf{k}\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right)\right] \frac{\partial}{\partial z^{\prime}}=\mathbf{i}^{\prime} \frac{\partial}{\partial x^{\prime}}+\mathbf{j}^{\prime} \frac{\partial}{\partial y^{\prime}}+\mathbf{k}^{\prime} \frac{\partial}{\partial z^{\prime}}\)
Therefore grad is invariant.
2. To prove div is invariant
Taking dot product of (4), (5), (6) with \(\mathbf{i}, \mathbf{j}, \mathbf{k}\)
Taking the cross product of (4), (5), (6) with \(\mathbf{i}, \mathbf{j}, \mathbf{k}\)
⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}=\left(\mathbf{i} . \mathbf{i}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{i} \cdot \mathbf{j}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{i} \cdot \mathbf{k}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial z^{\prime}}\)
⇒ \(\mathbf{j} \times \frac{\partial}{\partial y}=\left(\mathbf{j} \cdot \mathbf{i}^{\prime}\right) \mathbf{j} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right) \mathbf{j} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial z^{\prime}}\)
⇒ \(\mathbf{k} \times \frac{\partial}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{k} \cdot \mathbf{j}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial z^{\prime}}\)
Adding vertically and applying (A) \(\mathbf{i} \times \frac{\partial}{\partial x}+\mathbf{j} \times \frac{\partial}{\partial y}+\mathbf{k} \times \frac{\partial}{\partial z}=\mathbf{i}^{\prime} \times \frac{\partial}{\partial x^{\prime}}+\mathbf{j}^{\prime} \times \frac{\partial}{\partial y^{\prime}}+\mathbf{k}^{\prime} \times \frac{\partial}{\partial z^{\prime}} \text {. }\)
Thus curl is invariant.
Vector Calculus Identities Involving Laplacian Operator
Differential Operators Vector Identities
Theorem1. If A is a differential vector function and Φ is a differentiable scalar function, then prove that div ( Φ A)=(grade Φ).A+ Φdiv A or ∇. ( Φ A) + Φ (∇.A)
Proof: div\((\phi \mathbf{A})=\nabla \cdot \phi(\mathbf{A})=\Sigma \mathbf{i} \cdot \frac{\partial}{\partial x}(\phi \mathbf{A})=\sum \mathbf{i} \cdot\left(\frac{\partial \phi}{\partial x} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial x}\right)\)
= \(\sum\left(\mathbf{i} \frac{\partial \phi}{\partial x}\right) \cdot \mathbf{A}+\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \phi\)
= \((\nabla \phi) \cdot \mathbf{A}+(\nabla \cdot \mathbf{A}) \phi\)
Theorem 2. Prove that curl (Φ A)=(gradeΦ) × A + Φ curl A or ∇× (Φ A)= (∇Φ) × A +Φ(∇ ×A)
Proof: curl\((\phi \mathbf{A})=\nabla \times(\phi \mathbf{A})=\Sigma \mathbf{i} \times \frac{\partial}{\partial x}(\phi \mathbf{A})\)
= \(\sum \mathbf{i} \times\left(\frac{\partial \phi}{\partial x} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial x}\right)=\sum\left(\mathbf{i} \frac{\partial \phi}{\partial x}\right) \times \mathbf{A}+\sum\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right) \phi=\nabla \phi \times \mathbf{A}+(\nabla \times \mathbf{A}) \phi\)
Theorem 3. Prove that grad(A.B) = (B .∇) A+(A.∇) B+B × curl A+A × curl B
Proof: Now \(\mathbf{A} \times \text{curl} \mathbf{B}=\mathbf{A} \times(\nabla \times \mathbf{B})\)
= \(\mathbf{A} \times \sum \mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}=\sum \mathbf{A} \times\left(\mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)
= \(\sum\left\{\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{i}-(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}\right\}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)-\left(\mathbf{A} \cdot \sum \mathbf{i} \frac{\partial}{\partial x}\right) \mathbf{B}\)
∴ \(\mathbf{A} \times \text{curl} \mathbf{B}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)-(\mathbf{A} \cdot \nabla) \mathbf{B}\)
⇒ \(\mathbf{A} \times \text{curl} \mathbf{B}+(\mathbf{A} \cdot \nabla) \mathbf{B}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)\)
Similarly \(\mathbf{B} \times \text{curl} \mathbf{A}=\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)-(\mathbf{B} \cdot \nabla) \mathbf{A}\)
⇒ \(\mathbf{B} \times \text{curl} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}=\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)\)
Adding (1) and (2) we have
∴ \(\mathbf{A} \times \text{curl} \mathbf{B}+(\mathbf{A} \cdot \nabla) \mathbf{B}+\mathbf{B} \times \text{curl} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}\)
= \(\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)+\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}+\frac{\partial \mathbf{A}}{\partial x} \cdot \mathbf{B}\right)\)
= \(\sum \mathbf{i} \frac{\partial}{\partial x}(\mathbf{A} \cdot \mathbf{B})=\nabla(\mathbf{A} \cdot \mathbf{B})=\text{grad}(\mathbf{A} \cdot \mathbf{B})\)
Theorems For Laplacian Operator In Scalar And Vector Fields
Theorem 4. Prove that div (A×B)= B.curlA-A, curl B ∇ . (A×B) =B.(∇×A) – A . (∇×B)
Proof: \(\text{div}(\mathbf{A} \times \mathbf{B})=\nabla \cdot(\mathbf{A} \times \mathbf{B})\)
= \(\sum \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})=\sum \mathbf{i} \cdot\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)
= \(\sum \mathbf{i} .\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\sum \mathbf{i} .\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)
= \(\sum\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right) \cdot \mathbf{B}-\left(\sum \mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}\right) \cdot \mathbf{A}\)
= \((\nabla \times \mathbf{A}) \cdot \mathbf{B}-(\nabla \times \mathbf{B}) \cdot \mathbf{A}=\mathbf{B} \cdot \text{curl} \mathbf{A}-\mathbf{A} \cdot \text{curl} \mathbf{B}\)
Note. If \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are irrotational then \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}\) is solenoidal.
⇒ \(\overline{\mathrm{A}}, \overline{\mathrm{B}}\) are irrotational.
⇒ \(\nabla \times \overline{\mathrm{A}}=\overline{0}, \nabla \times \overline{\mathrm{B}}=\overline{0}\) substituting in the theorem, \(\nabla(\overline{\mathrm{A}} \times \overline{\mathrm{B}})=\overline{\mathrm{B}} \cdot(\nabla \times \overline{\mathrm{A}})-\overline{\mathrm{A}} \cdot(\nabla \cdot \overline{\mathrm{B}})=0\)
∴ \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}\) is solenoidal.
Theorem 5. Prove that curl(A×B) = Adiv B-B div A +(B.∇)A-(A.∇)B
Proof: curl\((\mathbf{A} \times \mathbf{B})=\nabla \times(\mathbf{A} \times \mathbf{B})\)
= \(\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})=\sum \mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)
= \(\sum \mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\sum \mathbf{i} \times\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)
= \(\sum\left\{(\mathbf{i} \cdot \mathbf{B}) \frac{\partial \mathbf{A}}{\partial x}-\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}\right\}+\sum\left\{\left(\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-(\mathbf{i} \cdot \mathbf{A}) \frac{\partial \mathbf{B}}{\partial x}\right\}\)
= \(\sum(\mathbf{B} \cdot \mathbf{i}) \frac{\partial \mathbf{A}}{\partial x}-\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}+\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-\sum(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}\)
= \(\left(\mathbf{B} \cdot \sum i \frac{\partial}{\partial x}\right) \mathbf{A}-\left(\sum i \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}+\left(\sum i \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-\left(\mathbf{A} \cdot \sum i \frac{\partial}{\partial x}\right) \mathbf{B}\)
= \((\mathbf{B} \cdot \nabla) \mathbf{A}-(\nabla \cdot \mathbf{A}) \mathbf{B}+(\nabla \cdot \mathbf{B}) \mathbf{A}-(\mathbf{A} \cdot \nabla) \mathbf{B}\)
= \(\mathbf{A} \text{div} \mathbf{B}-\mathbf{B} \text{div} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}-(\mathbf{A} \cdot \nabla) \mathbf{B}\)
Theorem 6. Prove that curl grad Φ=0
Proof: \(\text{grad} \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\)
curl\((\text{grad} \phi)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}
\end{array}\right|\)
= \(\mathbf{i}\left(\frac{\partial^2 \phi}{\partial y \partial z}-\frac{\partial^2 \phi}{\partial z \partial y}\right)-\mathbf{j}\left(\frac{\partial^2 \phi}{\partial x \partial z}-\frac{\partial^2 \phi}{\partial z \partial x}\right)+\mathbf{k}\left(\frac{\partial^2 \phi}{\partial x \partial y}-\frac{\partial^2 \phi}{\partial y \partial x}\right)=0\)
Note : Since Curl \((\text{grad} \phi)=0\) grad \(\phi\) is always irrotational.
Understanding Theorems Involving ∇² In Differential Operators
Theorem 7. Prove that div. curl f=0
Proof:
Let \(\mathbf{f}=f_1 \mathbf{i}+f_2 \mathbf{j}+f_3 \mathbf{k}\)
∴ \(\text{curl}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
f_1 & f_2 & f_3
\end{array}\right|\)
= \(\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right) \mathbf{i}-\left(\frac{\partial f_3}{\partial x}-\frac{\partial f_1}{\partial z}\right) \mathbf{j}+\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right) \mathbf{k}.\)
∴ \(\text{div}(\text{curl} \mathbf{f})=\nabla \cdot(\nabla \times \mathbf{f})\)
= \(\frac{\partial}{\partial x}\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right)-\frac{\partial}{\partial y}\left(\frac{\partial f_3}{\partial x}-\frac{\partial f_1}{\partial z}\right)+\frac{\partial}{\partial z}\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)\)
= \(\frac{\partial^2 f_3}{\partial x \partial y}-\frac{\partial^2 f_2}{\partial x \partial z}-\frac{\partial^2 f_3}{\partial y \partial x}+\frac{\partial^2 f_1}{\partial y \partial z}+\frac{\partial^2 f_2}{\partial x \partial z}-\frac{\partial^2 f_1}{\partial z \partial y}=0\)
Note. Since div. (curl \(\mathbf{f})=0\), curl \(\mathbf{f}\) is always solenoidal.
Theorem 8. Prove that ∇× (∇×A)=∇(∇.A) − ∇2 A
Proof: \(\nabla \times(\nabla \times \mathbf{A})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\nabla \times \mathbf{A})\)
Now, \(\mathbf{i} \times \frac{\partial}{\partial x}(\nabla \times \mathbf{A})=\mathbf{i} \times \frac{\partial}{\partial x}\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right)\)
= \(\mathbf{i} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{A}}{\partial x^2}+\mathbf{j} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial y}+\mathbf{k} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial z}\right)\)
= \(\mathbf{i} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{A}}{\partial x^2}\right)+\mathbf{i} \times\left(\mathbf{j} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial y}\right)+\mathbf{i} \times\left(\mathbf{k} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial z}\right)\)
= \(\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{A}}{\partial x^2}\right) \mathbf{i}-\frac{\partial^2 \mathbf{A}}{\partial x^2}+\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{A}}{\partial x \partial y}\right) \mathbf{j}+\left(\mathbf{i} \cdot \frac{\partial^2 \mathbf{A}}{\partial x \partial z}\right) \mathbf{k}\)
Differential Operators Scalar Potential Of An Irritational Vector
Definition: a vector f is said to be irrotational if curl f=0 If f is irrotational, there will always exist a scalar function φ(x,y,z) such that f= grade Φ.This Φ is called the Scalar potential of f.
it is easy to prove that, if f= grad Φ, then curl f=0. Hence ∇× f=0 ⇔ there exists a scalar function Φ such that f=∇Φ.