Question

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

Short answer

The property for vector fields \[div(\mathbf{F}+\mathbf{G}) = div \mathbf{F} + div \mathbf{G}\] has been proved, assuming that the required partial derivatives are continuous.

Step-by-step solution

Understanding Divergence

The divergence of a vector field is a scalar field representing the quantity of a field flowing out of a given point. If you consider a vector field \(\mathbf{F}\) = \(F_1i + F_2j + F_3k\) where \(F_1\), \(F_2\), and \(F_3\) are component functions of the vector, its divergence is given by: \[ div \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\]

Apply the property on divergence operator

Next, apply the property on the divergence operator that the div of the sum of two vector fields \(\mathbf{F}\) and \(\mathbf{G}\) equals the sum of their individual divergences. From this property, you get : \[div(\mathbf{F}+\mathbf{G}) = div(\mathbf{F} + \mathbf{G}) = \frac{\partial (F_1+G_1)}{\partial x} + \frac{\partial (F_2+G_2)}{\partial y} + \frac{\partial (F_3+G_3)}{\partial z}\]

Simplify the expression

Now, proceed to simplify this expression under the assumption that the required partial derivatives are continuous. This allows the partial derivatives to be redistributed over the terms (since partial derivatives are linear) to yield : \[\frac{\partial F_1}{\partial x} + \frac{\partial G_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial G_2}{\partial y} + \frac{\partial F_3}{\partial z} + \frac{\partial G_3}{\partial z}\] which simplifies to \[div\mathbf{F} + div\mathbf{G} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} + \frac{\partial G_1}{\partial x} + \frac{\partial G_2}{\partial y} + \frac{\partial G_3}{\partial z}\]

Conclusion

From step 3, it's clear that \[div(\mathbf{F}+\mathbf{G}) = div \mathbf{F} + div \mathbf{G}\] Hence, the property we wanted to prove for vector fields is true.