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

Step 1: Recognise the expression as Green's second identity.
The combination \( \nabla \cdot (f\nabla g - g\nabla f) \) is exactly the divergence form that appears in Green's second identity, which states \[ \nabla \cdot (f\nabla g - g\nabla f) = f\nabla^2 g - g\nabla^2 f \] This comes from expanding each divergence term with the product rule and cancelling the cross terms \( \nabla f \cdot \nabla g \) and \( \nabla g \cdot \nabla f \), which are identical since the dot product is commutative.

Step 2: Apply the given condition.
We are told that both \( f \) and \( g \) solve Laplace's equation, so \[ \nabla^2 f = 0, \qquad \nabla^2 g = 0 \] Substituting these into the identity from Step 1: \[ \nabla \cdot (f\nabla g - g\nabla f) = f(0) - g(0) = 0 \]
So the whole expression collapses to \( \boxed{0} \), matching option (B). This is also why the flux integral \( \oint (f\nabla g - g\nabla f)\cdot \hat{n}\, dS \) vanishes for harmonic functions, since the divergence being zero everywhere means there is nothing left to integrate.