Proof. Write \(\mathbf U = U_1\,\mathbf i + U_2\,\mathbf j + U_3\,\mathbf k\) and \(\mathbf V = V_1\,\mathbf i + V_2\,\mathbf j + V_3\,\mathbf k\). Properties 1 and 3 hold because differentiation is linear. For property 2, \[\nabla \cdot (f\mathbf U) = \sum _{i}\frac {\partial (fU_i)}{\partial x_i} = \sum _i f\frac {\partial U_i}{\partial x_i} + \sum _i U_i\frac {\partial f}{\partial x_i} = f\,\nabla \cdot \mathbf U + \mathbf U\cdot \nabla f ,\] writing \(x_1, x_2, x_3\) for \(x, y, z\). For property 4, the \(\mathbf i\)-component of \(\nabla \times (f\mathbf V)\) is \[\frac {\partial (fV_3)}{\partial y} - \frac {\partial (fV_2)}{\partial z} = f\left (\frac {\partial V_3}{\partial y} - \frac {\partial V_2}{\partial z}\right ) + \left (\frac {\partial f}{\partial y}V_3 - \frac {\partial f}{\partial z}V_2\right ),\] which is the \(\mathbf i\)-component of \(f\,\nabla \times \mathbf V + \nabla f\times \mathbf V\); the other components are similar. For property 5, the \(\mathbf i\)-component of \(\nabla \times \nabla f\) is \(f_{zy} - f_{yz} = 0\), and \[\nabla \cdot (\nabla \times \mathbf V) = (V_{3,yx} - V_{2,zx}) - (V_{3,xy} - V_{1,zy}) + (V_{2,xz} - V_{1,yz}) = 0 ,\] where \(V_{3,yx}\) denotes \(\frac {\partial ^2V_3}{\partial x\partial y}\); the terms cancel in pairs because mixed partial derivatives agree. □

Solution.

1.

The divergence is \[\nabla \cdot \mathbf F = \frac {\partial }{\partial x}(x^2y) + \frac {\partial }{\partial y}(-xyz) + \frac {\partial }{\partial z}(yz^2) = 2xy - xz + 2yz ,\] and the curl is \begin {align*} \nabla \times \mathbf F & = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k\\[1ex] \frac {\partial }{\partial x} & \frac {\partial }{\partial y} & \frac {\partial } {\partial z}\\[1ex] x^2y & -xyz & yz^2 \end {vmatrix}\\[1ex] & = \left (\frac {\partial }{\partial y}(yz^2) - \frac {\partial }{\partial z}(-xyz)\right )\mathbf i - \left (\frac {\partial }{\partial x}(yz^2) - \frac {\partial }{\partial z}(x^2y) \right )\mathbf j\\ & \qquad + \left (\frac {\partial } {\partial x}(-xyz) - \frac {\partial }{\partial y}(x^2y)\right )\mathbf k\\[1ex] & = (z^2+xy)\,\mathbf i - 0\,\mathbf j - (yz + x^2)\,\mathbf k . \end {align*}

2.

By Example 4.3.1 with \(n = -1\), \(\nabla \left (\frac 1r\right ) = -r^{-3}\mathbf r\). Then by property 2, with \(f = -r^{-3}\) and \(\mathbf U = \mathbf r\), \[\nabla ^2\left (\frac 1r\right ) = \nabla \cdot \left (-r^{-3}\mathbf r\right ) = -r^{-3}\,\nabla \cdot \mathbf r - \mathbf r\cdot \nabla (r^{-3}) .\] Here \(\nabla \cdot \mathbf r = 1 + 1 + 1 = 3\) and, again by Example 4.3.1, \(\nabla (r^{-3}) = -3r^{-5}\mathbf r\), so \[\nabla ^2\left (\frac 1r\right ) = -3r^{-3} + 3r^{-5}(\mathbf r\cdot \mathbf r) = -3r^{-3} + 3r^{-5}r^2 = 0 .\] So \(\frac 1r\) is harmonic away from the origin: the gravitational and electric potentials satisfy Laplace’s equation in empty space.

Solution. \[\nabla \times \mathbf F = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k\\[1ex] \frac {\partial }{\partial x} & \frac {\partial }{\partial y} & \frac {\partial } {\partial z}\\[1ex] 2xy & x^2+ z^2 & 2zy \end {vmatrix} = (2z - 2z)\,\mathbf i - (0 - 0)\,\mathbf j + (2x - 2x)\,\mathbf k = \mathbf 0 ,\] and \(\mathbf F\) is defined on all of \(\mathbb {R}^3\), so by Theorem 4.4.3 it is conservative. A potential \(f\) satisfies \[\frac {\partial f}{\partial x} = 2xy ,\qquad \frac {\partial f}{\partial y}= x^2+z^2 ,\qquad \frac {\partial f}{\partial z} = 2yz .\] Integrating the first with respect to \(x\), \(f = x^2y + h(y,z)\), where the constant of integration may depend on \(y\) and \(z\). The second then gives \(x^2 + h_y = x^2 + z^2\), so \(h_y = z^2\) and \(h = yz^2 + g(z)\). Hence \(f = x^2y + yz^2 + g(z)\), and the third equation gives \(2yz + g'(z) = 2yz\), so \(g\) is a constant. Therefore \[f(x,y,z) = x^2y + yz^2 + C .\]

Solution. Regarding both as fields in space with zero \(\mathbf k\)-component, \(\nabla \cdot \mathbf F = 0\) and \(\nabla \times \mathbf F = (1 - (-1))\,\mathbf k = 2\,\mathbf k\), while \(\nabla \cdot \mathbf G = 2\) and \(\nabla \times \mathbf G = \mathbf 0\). The rotation spreads nothing out but rotates everything, with angular velocity \(1\) about the \(z\)-axis (half the curl); the radial field rotates nothing but spreads out from every point.