Question

If \(F = \frac{r}{{{r^p}}}\), find div F. is there a value of p for which div \({\bf{F = 0}}\)?

Short answer

The solution is

\(divF = \frac{{3 - p}}{{{r^p}}}\), \(divF = 0\) if and only if \(p = 3.\)

Step-by-step solution

Given information

The given value is:

\(\begin{array}{l}div F = \frac{r}{{{r^p}}}\\div F = 0\end{array}\)

calculate div F

We need to calculate div F where

\(F = \frac{r}{{{r^p}}},\;\;\;r = |r| = \sqrt {{x^2} + {y^2} + {z^2}} \)

Also, we need to find if there exists a value of p such that \(divF = 0\)

The divergence of F is defined as

\(div\,F = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}\)

expand F

\(F = \frac{r}{{{r^p}}} = \frac{{xi + yj + zk}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{p}{2}}}}}\)

Therefore, let

\(\begin{array}{l}P = x{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}}\\Q = y{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}}\\R = z{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}}\end{array}\)

Calculate div F, differentiate using the product rule and then chain rule

\(\begin{array}{c}divF = \frac{\partial }{{\partial x}}\left( {x{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{p}{2}}}} \right) + \frac{\partial }{{\partial y}}\left( {y{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{p}{2}}}} \right) + \frac{\partial }{{\partial z}}\left( {z{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{p}{2}}}} \right)\\ = \left( \begin{array}{l}1 \cdot {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}} + x \cdot \left( { - \frac{p}{2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2} - 1}} \cdot 2x\\ + 1 \cdot {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}} + y \cdot \left( { - \frac{p}{2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2} - 1}} \cdot 2y\\ + 1 \cdot {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}} + z \cdot \left( { - \frac{p}{2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2} - 1}} \cdot 2z\end{array} \right)\end{array}\)

\(\begin{array}{c}divF = 3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}} - p\left( {{x^2} + {y^2} + {z^2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{2}{2} - 1}}\\ = 3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}} - p{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}}\\ = (3 - p){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{p}{2}}}\\ = \frac{{3 - p}}{{{r^p}}}\end{array}\)

find the value of p

For \(divF = 0\),

\(\begin{array}{*{20}{r}}{\frac{{3 - p}}{{{r^p}}} = 0}\\{3 - p = 0}\\{p = 3}\end{array}\)

Conclusion

The solution is

\(divF = \frac{{3 - p}}{{{r^p}}}\), \(divF = 0\) if and only if \(p = 3.\)