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Section 5.4 Week 5 Activity

Subsection 5.4.1 Proofs for Differential Operators

Activity 5.4.1.

Prove the four statements in Proposition 5.3.8. You can work entirey with vector in \(\RR^3\text{.}\)

Solution.

Each of the linearity properties involves expanding the vector field in components and using the linearity of single-variable derivatives in the components. I'll just show the divergence identity as an example. I expand the components and then apply the divergence operator. I use the rules for vector addition and scalar multiplication.

\begin{align*} \nabla \cdot (aF \pm bG) \amp = \nabla \cdot \left( a (F_1, F_2, F_3) \pm b (G_1, G_2, G_3) \right)\\ \amp = \nabla \cdot \left( (aF_1,aF_2,aF_3) \pm (bG_1,bG_2,bG_3) \right) \\ \amp = \nabla \cdot \left( aF_1 \pm bG_1,aF_2 \pm bG_2,aF_3 \pm bG_3 \right) \\ \amp = \frac{\del}{\del x} (aF_1 \pm bG_1) + \frac{\del}{\del y} (aF_2 \pm bG_2) + \frac{\del}{\del z} (aF_3 \pm bG_2) \end{align*}

I use the linearity of the single variable derivatives.

\begin{equation*} = a \frac{\del}{\del x} F_1 \pm b \frac{\del}{\del x} G_1 + = a \frac{\del}{\del y} F_2 \pm b \frac{\del}{\del y} G_2 + = a \frac{\del}{\del z} F_3 \pm b \frac{\del}{\del z} G_3 a \end{equation*}

Then I reoder the terms.

\begin{align*} \amp= a \left( \frac{\del}{\del x} F_1 + \frac{\del}{\del y} F_2 + \frac{\del}{\del z} F_3 \right) \pm b \left( \frac{\del}{\del x} G_1 + \frac{\del}{\del y} G_2 + \frac{\del}{\del z} G_3 \right) \\ \amp = a \nabla \cdot F \pm b \nabla \cdot G \end{align*}

The other three solutions follow this pattern almost exactly, just with different use of the specific rules for vector algebra. The calculations for the cross product are a bit lengthy, but no more conceptually challenging than this.

Activity 5.4.2.

Prove this identity for vector fields in \(\RR^3\text{.}\)

\begin{equation*} \nabla \cdot (F \times G) = (\nabla \times F) \cdot G - F \cdot (\nabla \times G) \end{equation*}
Solution.

I expand in components and apply definition of the cross product.

\begin{align*} \amp \nabla \cdot (F \times G) = \nabla \cdot \left( (F_1,F_2,F_3) \times (G_1,G_2,G_3) \right) \\ \amp = \nabla \cdot \left( (F_2G_3 - F_3G_2), (F_3G_1 - F_1G_3), (F_1G_2 - F_2G_1) \right) \\ \amp = \frac{\del}{\del x} (F_2G_3 - F_3G_2)) + \frac{\del}{\del y} (F_3G_1 - F_1G_3)) + \frac{\del}{\del z} (F_1G_2 - F_2G_1)) \end{align*}

I use linearity and the power rule for the partial derivatives to expand this further.

\begin{align*} \amp = G_3 \frac{\del}{\del x} F_2 + F_2 \frac{\del}{\del x} G_3 - (G_2 \frac{\del}{\del x} F_3 + F_3 \frac{\del}{\del x} G_2)\\ \amp + G_1 \frac{\del}{\del y} F_3 + F_3 \frac{\del}{\del y} G_1 - (G_3 \frac{\del}{\del y} F_1 + F_1 \frac{\del}{\del y} G_3) \\ \amp + G_2 \frac{\del}{\del z} F_1 + F_1 \frac{\del}{\del z} G_2 - (G_1 \frac{\del}{\del z} F_2 + F_2 \frac{\del}{\del z} G_1) \end{align*}

I have twelve terms here. I reorder them in the following way.

\begin{align*} \amp = G_1 \left( \frac{\del}{\del y} F_3 - \frac{\del}{\del z} F_2 \right) + G_2 \left( \frac{\del}{\del z} F_1 - \frac{\del}{\del x} F_3 \right) + G_3 \left( \frac{\del}{\del x} F_2 - \frac{\del}{\del y} F_1 \right) \\ \amp - \left( F_1 \left( \frac{\del}{\del y} G_3 - \frac{\del}{\del z} G_2 \right) + F_2 \left( \frac{\del}{\del z} G_1 - \frac{\del}{\del x} G_3 \right) + F_3 \left( \frac{\del}{\del x} G_2 - \frac{\del}{\del y} G_1 \right) \right) \end{align*}

From here, I just need to recognize that the result is precisely the expanded form of the right side that I'm trying to work towards.

\begin{equation*} = G \cdot ( \nabla \times F) - F \cdot ( \nabla \times G) \end{equation*}

Subsection 5.4.2 Vector Field Differential Operators

Activity 5.4.3.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = (yz,xz,xy) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} yz + \frac{\del}{\del y} xz + \frac{\del}{\del z} xy = 0+0+0 = 0 \\ \nabla \times F \amp = \left( \frac{\del}{\del y} xy - \frac{\del}{\del z} xz, \frac{\del}{\del z} yz - \frac{\del}{\del x} xz, \frac{\del}{\del x} xz - \frac{\del}{\del y} yz \right) \\ \amp = (x-x,y-y,z-z) = (0,0,0) \end{align*}

The field is both irrotational and incompressible.

Activity 5.4.4.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = \left( \ln(xy),\frac{1}{x}, \frac{-x}{z} \right) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} \ln (xy) + \frac{\del}{\del y} \frac{1}{x} + \frac{\del}{\del z} \frac{-x}{z} = \frac{1}{y} + 0 + \frac{x}{z^2} = \frac{1}{y} - \frac{x}{z^2}\\ \nabla \times F \amp = \left( \frac{\del}{\del y} \frac{-x}{z} - \frac{\del}{\del z} \frac{1}{x}, \frac{\del}{\del z} \ln (xy) - \frac{\del}{\del x} \frac{-x}{z}, \frac{\del}{\del x} \frac{1}{x} - \frac{\del}{\del y} \ln (xy)\right)\\ \amp = \left( 0-0, 0 + \frac{1}{z}, \frac{-1}{x^2} - 0 \right) = \left( 0, \frac{1}{z}, \frac{-1}{x^2} \right) \end{align*}

The field is neither irrotational nor incompressible.

Activity 5.4.5.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = (x^2,yz,xz) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} x^2 + \frac{\del}{\del y} yz + \frac{\del}{\del z} xz = 2x + z + z = 2x + 2z\\ \nabla \times F \amp = \left( \frac{\del}{\del y} xz - \frac{\del}{\del z} yz, \frac{\del}{\del z} x^2 - \frac{\del}{\del x} xz, \frac{\del}{\del x} yz - \frac{\del}{\del y} x^2 \right)\\ \amp = ( 0 - 0, 0 - z, 0 - 0) = (0,-z,0) \end{align*}

The field neither irroational or incompressible

Activity 5.4.6.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = (4,9,-12) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} 4 + \frac{\del}{\del y} 9 + \frac{\del}{\del z} -12 = 0 + 0 + 0 = 0 \\ \nabla \times F \amp = \left( \frac{\del}{\del y} -12 - \frac{\del}{\del z} 9, \frac{\del}{\del z} 4 - \frac{\del}{\del x} -12, \frac{\del}{\del x} 9 - \frac{\del}{\del y} 4 \right) = (0,0,0) \end{align*}

The field is irrotational and incompressible.

Activity 5.4.7.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = (\sin(x),\cos(y),\sin(z)) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} \sin x + \frac{\del}{\del y} \cos y + \frac{\del}{\del z} \sin z = \cos x - \sin y + \cos z \\ \nabla \times F \amp = \left( \frac{\del}{\del y} \sin z - \frac{\del}{\del z} \cos y, \frac{\del}{\del z} \sin x - \frac{\del}{\del x} \sin z, \frac{\del}{\del x} \cos y - \frac{\del}{\del y} \sin x \right)\\ \amp = (0,0,0) \end{align*}

The field is irrotational but not incompressible.

Activity 5.4.8.

Determine if the following vector field is irrotational and/or incompressible.

\begin{equation*} F(x,y,z) = (xy^2,z^2+y^2,x^2-z^2) \end{equation*}
Solution.

I need to calculate both the curl and the divergence.

\begin{align*} \nabla \cdot F \amp = \frac{\del}{\del x} xy^2 + \frac{\del}{\del y} z^2 + y^2 + \frac{\del}{\del z} x^2 - z^2 = y^2 + 2y - 2z\\ \nabla \times F \amp = \left( \frac{\del}{\del y} x^2 - z^2 - \frac{\del}{\del z} z^2 + y^2, \frac{\del}{\del z} xy^2 - \frac{\del}{\del x} z^2 + y^2, \frac{\del}{\del x} z^2 + y^2 - \frac{\del}{\del y} xy^2 \right)\\ \amp = ( 0 - 2z, 0 - 0, 0 - 2xy) \end{align*}

The field is neither irrotational nor incompressible.

Subsection 5.4.3 Harmonic Scalar Fields

Activity 5.4.9.

Determine if this scalar field is harmonic.

\begin{equation*} f(x,y,z) = x^2yz \end{equation*}
Solution.

I calculate the Laplacian (the divergence of the gradient).

\begin{align*} \nabla^2 f \amp = \frac{\del^2}{\del x^2} x^2 y z + \frac{\del^2}{\del y^2} x^2 y z + \frac{\del^2}{\del z^2} x^2 y z = 2yz + x^2 z + x^2 y \end{align*}

The field is not harmonic.

Activity 5.4.10.

Determine if this scalar field is harmonic.

\begin{equation*} f(x,y,z) = x^2+3y^2-4z^2 \end{equation*}
Solution.

I calculate the Laplacian (the divergence of the gradient).

\begin{align*} \nabla^2 f \amp = \frac{\del^2}{\del x^2} x^2 + 3y^2 - 4z^2 + \frac{\del^2}{\del y^2} x^2 + 3y^2 - 4z^2 + \frac{\del^2}{\del z^2} x^2 + 3y^2 - 4z^2 \\ \amp = 2 + 6 - 8 = 0 \end{align*}

The field is harmonic.

Activity 5.4.11.

Determine if this scalar field is harmonic.

\begin{equation*} f(x,y,z) = e^{x+y+z} \end{equation*}
Solution.

I calculate the Laplacian (the divergence of the gradient).

\begin{align*} \nabla^2 f \amp = \frac{\del^2}{\del x^2} e^{x+y+z} + \frac{\del^2}{\del y^2} e^{x+y+z} + \frac{\del^2}{\del z^2} e^{x+y+z} \\ \amp = e^{x+y+z} + e^{x+y+z} + e^{x+y+z} = 3 e^{x+y+z} \end{align*}

The field is not harmonic.

Activity 5.4.12.

Determine if this scalar field is harmonic.

\begin{equation*} f(x,y,z) = 7x+8y-4z \end{equation*}
Solution.

I calculate the Laplacian (the divergence of the gradient).

\begin{align*} \nabla^2 f \amp = \frac{\del^2}{\del x^2} 7x+8y-4z + \frac{\del^2}{\del y^2} 7x+8y-4z + \frac{\del^2}{\del z^2} 7x+8y-4z \\ = 0+0+0 = 0 \end{align*}

The field is harmonic.

Subsection 5.4.4 Conceptual Review Questions

  • What is a vector field?

  • What is an integral curve?

  • What is curl?

  • What is divergence?

  • How does the Leibniz rule extend?