## Finding a Vector Field With Divergence Equal to a Power of r

To solve the equation
$\mathbf{\nabla} \cdot \mathbf{F} =r^n$
in spherical polar coordinates, use the facet that fore a radial vector field
$F=F_r \mathbf{e_r}$
,
$\mathbf{\nabla} \cdot \mathbf{F} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 F_r ) =r^n$

Hence
$\frac{\partial}{\partial r} (r^2 F_r ) =r^{n+2}$

Integrate both sides to get
$r^2 F_r = \frac{r^{n+3}}{n+3}+A$

Then
$F_r = \frac{r^{n+1}}{n+3} + \frac{A}{r^2}$
if
$n \neq 0$
and
$F_r = \frac{r^{n+1}}{n+3}$
if
$r=0$
is possible, since
$\mathbf{\nabla} \cdot \mathbf{F}=0 \rightarrow F_r =\frac{1}{r^2}$
using this result.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://astarmathsandphysics.com/university-maths-notes/vector-calculus/4194-finding-a-vector-field-with-divergence-equal-to-a-power-of-r.html?tmpl=component&print=1&page=", "fetch_time": 1519269583000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00183.warc.gz", "warc_record_offset": 618862321, "warc_record_length": 3016, "token_count": 281, "char_count": 653, "score": 3.953125, "int_score": 4, "crawl": "CC-MAIN-2018-09", "snapshot_type": "longest", "language": "en", "language_score": 0.5538523197174072, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}