Question

Prove the identity where \(R\) is a simply connected region with boundary \(C .\) Assume that the required partial derivatives of the scalar functions \(f\) and \(g\) are continuous. The expressions \(D_{\mathrm{N}} f\) and \(D_{\mathrm{N}} g\) are the derivatives in the direction of the outward normal vector \(\mathrm{N}\) of \(C,\) and are defined by \(D_{\mathrm{N}} f=\nabla f \cdot \mathbf{N}\) and \(D_{\mathrm{N}} g=\nabla g \cdot \mathrm{N}\). Green's first identity: \(\iint_{R} \int\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d A=\int_{C} f D_{\mathrm{N}} g d s\) [Hint: Use the second alternative form of Green's Theorem and the property \(\operatorname{div}(f \mathbf{G})=f \operatorname{div} \mathbf{G}+\nabla f \cdot \mathbf{G}\).]

Short answer

By applying the divergence theorem after rewriting the left-hand side of Green's first identity using the given property, we obtain \(\int_{C} f D_{\mathrm{N}} g ds\), proving Green's first identity.

Step-by-step solution

Rewrite the Left Hand Side

Start by rewriting the left-hand side of Green's first identity using the property given in the hint. We have \[ \iint_{R} \left( f \nabla^{2}g + \nabla f \cdot \nabla g \right)dA = \iint_{R} \operatorname{div}(f\nabla g) dA. \] Here, \( \nabla g \) is the gradient vector field of the function 'g' and 'f' multiplies this vector field. The 'div' operator calculates the divergence of this new vector field.

Apply the Divergence Theorem

Next, apply the divergence theorem, also known as Gauss's theorem or the second form of Green's theorem, which states that the volume integral of the divergence of a vector field is equal to the surface integral of the vector field. This gives us \[ \iint_{R} \operatorname{div}(f\nabla g) dA = \int_{C} f \nabla g \cdot d\mathbf{s}, \] where \(d\mathbf{s}\) is the outward-pointing normal vector on the boundary of the region 'R'.

Compute the Dot Product

We now compute the dot product and obtain the final expression as \[ \int_{C} f \nabla g \cdot d\mathbf{s} = \int_{C} f D_{\mathrm{N}} g ds. \] Here, \(D_{\mathrm{N}} g = \nabla g \cdot \mathbf{N}\) is the derivative in the direction of the outward normal vector on \(C\), which completes the proof.