Give an example of a conservative vector field whose divergence is uniformly equal to zero in${\mathrm{ℝ}}^3$.

The goal is to demonstrate a conservative vector field whose divergence is uniformly equal to zero in all directions.${\mathrm{ℝ}}^3$

If $\mathbf{F}\left(x_y,y,z\right)$is a conservative vector field, then.

$\mathbf{F}(x,y,z)=\mathbb{V}f(x,y,z)$

The divergence of a vector field $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows $divF=\nabla .F$

Then the divergence of conservative vector field$\mathbf{F}(x,y,z)=\nabla f(x,y,z)$will be,

$divF=\nabla .F$

$$\begin{gathered} =\nabla .\nabla f \\ =(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k) \\ =\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)+\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)+\frac{\partial }{\partial z}\left(\frac{\partial f}{\partial z}\right) \\ =\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2} \end{gathered}$$

Hence, a conservative vector field whose divergence is uniformly equal to zero in ${\mathrm{ℝ}}^3$satisfies the following conditions,

$\frac{{\partial }^{2}f}{\partial x^2}=0,\frac{{\partial }^{2}f}{\partial y^2}=0,\frac{{\partial }^{2}f}{\partial z^2}=0$

Suppose that , $f=x+y+z$, then

$$\begin{gathered} \nabla f=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k)(x+y+z) \\ =\frac{\partial }{\partial x}(x+y+z)i+\frac{\partial }{\partial y}(x+y+z)j+\frac{\partial }{\partial z}(x+y+z)k \\ =1i+1j+1k \\ =i+j+k \end{gathered}$$

Then, the vector field $F(x,y,z)=i+j+k$is a conservative vector field, such that $F=\nabla f$

Where $F=x+y+z$

Notice that,

$$\begin{gathered} \frac{{\partial }^{2}f}{\partial x^2}=\frac{{\partial }^2}{\partial x^2}(x+y+z) \\ =\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\right(x+y+z\left)\right) \\ =\frac{\partial }{\partial x}\left(1\right) \\ =0 \\ \frac{{\partial }^{2}f}{\partial y^2}==\frac{{\partial }^2}{\partial y^2}(x+y+z) \\ =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}\right(x+y+z\left)\right) \\ =\frac{\partial }{\partial y}\left(1\right) \\ =0 \\ \frac{{\partial }^{2}f}{\partial z^2}=\frac{{\partial }^2}{\partial z^2}(x+y+z) \\ =\frac{\partial }{\partial z}\left(\frac{\partial }{\partial z}\right(x+y+z\left)\right) \\ =\frac{\partial }{\partial z}\left(1\right) \\ =0 \end{gathered}$$

Hence, the divergence for this conservative vector field $F(x,y,z)=i+j+k$ is uniformly equal to zero in .

Therefore, an example of a conservative vector field whose divergence is uniformly equal to zero in ${\mathrm{ℝ}}^3$is