Question

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \operatorname{div}(\operatorname{curl} \mathbf{F})=0 \quad \text { (Theorem } 13.3) $$

Short answer

The proof shows that the divergence of the curl of any vector field is invariably zero, based on the definitions of divergence and curl, and the properties of partial derivatives.

Step-by-step solution

Understanding Definitions

Start by understanding the definition of divergence and curl. The divergence of a vector field \(\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}\) is \(\operatorname{div} \mathbf{F} = \frac{\partial F_1}{\partial x}+ \frac{\partial F_2}{\partial y}+ \frac{\partial F_3}{\partial z}\). The curl of vector field \(\mathbf{F}\) is \(\operatorname{curl} \mathbf{F}= \nabla \times \mathbf{F}\) where \(\nabla = \frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k}\).

Compute \(\operatorname{curl} \mathbf{F}\)

Compute \(\operatorname{curl} \mathbf{F}\) using the determinant form of the tripple product which gives: \(\operatorname{curl} \mathbf{F} =(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z})\mathbf{i} -(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z})\mathbf{j} +(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y})\mathbf{k}\).

Apply the divergence operator

Apply the divergence operator to \(\operatorname{curl} \mathbf{F}\). This is given by: \(\operatorname{div}(\operatorname{curl} \mathbf{F}) = \frac{\partial^2 F_3}{\partial x \partial y}- \frac{\partial^2 F_2}{\partial x\partial z} - \frac{\partial^2 F_3}{\partial y\partial x} + \frac{\partial^2 F_1}{\partial y\partial z} + \frac{\partial^2 F_2}{\partial z\partial x} - \frac{\partial^2 F_1}{\partial z\partial y}\).

Apply the property of mixed partial derivatives

Apply the property of mixed partial derivatives i.e., \(\frac{\partial^2 F_i}{\partial x\partial y} = \frac{\partial^2 F_i}{\partial y\partial x}\) to each pair of mixed partial derivatives in the expression for \(\operatorname{div}(\operatorname{curl} \mathbf{F})\). This shows that the second pairs in each instance cancel each other out, so \(\operatorname{div}(\operatorname{curl} \mathbf{F}) = 0\).