Question

If \( f \) and \( g \) are the solutions of Laplace equation then \( \nabla \cdot \{(f \nabla g) - (g \nabla f)\} = \)

• A. \( \nabla g \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( \nabla f \)

Hint:

This expression is the integrand used in Green's Second Identity: \( \int_V (f\nabla^2 g - g\nabla^2 f) dV = \oint_S (f\nabla g - g\nabla f) \cdot d\mathbf{\hat{n}} \). Since both functions satisfy Laplace's equation, both sides of the identity are identically zero.

Answer 1

The Correct Option is B

Concept: A scalar function \( \phi \) satisfies the Laplace equation if its Laplacian equals zero: \[ \nabla^2 \phi = 0 \] Therefore, since \( f \) and \( g \) are explicitly stated to be solutions of the Laplace equation, we have: \[ \nabla^2 f = 0 \quad \text{and} \quad \nabla^2 g = 0 \] We will also make use of the standard vector identity for the divergence of a scalar multiplied by a vector field, \( \nabla \cdot (\phi \mathbf{A}) = \phi (\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla \phi) \). When applied to a gradient field \( \mathbf{A} = \nabla \psi \), this identity becomes: \[ \nabla \cdot (\phi \nabla \psi) = \phi \nabla^2 \psi + \nabla \phi \cdot \nabla \psi \]

Step 1: Expanding the given divergence expression using vector identities.
The objective expression to expand and calculate is: \[ W = \nabla \cdot \{(f \nabla g) - (g \nabla f)\} \] By the linearity property of the divergence operator, we can split this into two separate divergence terms: \[ W = \nabla \cdot (f \nabla g) - \nabla \cdot (g \nabla f) \quad \cdots (1) \]

Step 2: Evaluating the first divergence term \( \nabla \cdot (f \nabla g) \).
Applying our vector identity directly where the scalar function is \( f \) and the gradient field belongs to \( g \): \[ \nabla \cdot (f \nabla g) = f (\nabla \cdot \nabla g) + \nabla f \cdot \nabla g = f \nabla^2 g + \nabla f \cdot \nabla g \]

Step 3: Evaluating the second divergence term \( \nabla \cdot (g \nabla f) \).
Applying our vector identity where the roles are reversed (the scalar function is \( g \) and the gradient field belongs to \( f \)): \[ \nabla \cdot (g \nabla f) = g (\nabla \cdot \nabla f) + \nabla g \cdot \nabla f = g \nabla^2 f + \nabla g \cdot \nabla f \]

Step 4: Combining both expanded terms back into the original equation.
Substitute these two expanded formulas back into equation (1): \[ W = (f \nabla^2 g + \nabla f \cdot \nabla g) - (g \nabla^2 f + \nabla g \cdot \nabla f) \] Distributing the negative sign through the second expression: \[ W = f \nabla^2 g + \nabla f \cdot \nabla g - g \nabla^2 f - \nabla g \cdot \nabla f \] Since the scalar dot product is commutative, \( \nabla f \cdot \nabla g = \nabla g \cdot \nabla f \). Therefore, these two terms cancel each other out completely: \[ W = f \nabla^2 g - g \nabla^2 f \]

Step 5: Applying the Laplace equation conditions.
Since \( f \) and \( g \) satisfy the Laplace equation, we substitute \( \nabla^2 g = 0 \) and \( \nabla^2 f = 0 \) into our expression: \[ W = f(0) - g(0) = 0 - 0 = 0 \] Thus, the expression simplifies cleanly to \( 0 \), which matches Option (B).