$\nabla \cdot f + w \cdot f = 0$

Question

Let $w(x,y,z)$ be a fixed vector field on $\mathbb{R}^3$. What are the solutions of the equation $$ \nabla \cdot f + w \cdot f = 0 \, ? $$ Note that if $w = \nabla \phi $, then the above equation is equivalent to $$ \nabla \cdot (e^\phi f) = 0, $$ for which the solutions are of the form $f = e^{-\phi} \nabla \times g$ for some arbitrary $g$.

Answer 1

You can try to proceed along the following lines: \begin{equation*} (\partial _{\mathbf{x}}+\mathbf{w})\cdot \mathbf{f}=0 \end{equation*} Special case $\mathbf{w}$ is constant. Then \begin{equation*} \exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x} ]=\partial _{\mathbf{x}}+\mathbf{w} \end{equation*} so \begin{eqnarray*} \exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}% ]\cdot \mathbf{f} &=&0 \\ \partial _{\mathbf{x}}\cdot \{\exp [+\mathbf{w\cdot x}]\mathbf{f}\} &=&0 \\ \exp [+\mathbf{w\cdot x}]\mathbf{f} &\mathbf{=a}&+\partial _{\mathbf{x}% }\times \mathbf{b(x)} \\ \mathbf{f(x)} &=&\exp [-\mathbf{w\cdot x}]\{\mathbf{a}+\partial _{\mathbf{x}% }\times \mathbf{b(x)}\} \end{eqnarray*} In general you have to find $\mathbf{U}(\mathbf{x})$ ($3\times 3$ matrix) such that \begin{equation*} \mathbf{U}^{-1}(\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \mathbf{U}(% \mathbf{x})=\partial _{\mathbf{x}}+\mathbf{w(x}) \end{equation*} Let \begin{eqnarray*} \mathbf{U}(\mathbf{x}) &=&\exp [\mathbf{A(x)}] \\ \partial _{\mathbf{x}}\cdot \mathbf{U}(\mathbf{x}) &=&\mathbf{U}(\mathbf{x}% )[\partial _{\mathbf{x}}+\{\partial _{\mathbf{x}}\cdot \mathbf{A(x)\}]} \\ \partial _{\mathbf{x}}\cdot \mathbf{A(x)} &=&\mathbf{w(x}) \end{eqnarray*} Then \begin{eqnarray*} (\partial _{\mathbf{x}}+\mathbf{w(x)})\cdot \mathbf{f} &=&\mathbf{U}^{-1}(% \mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\}=0 \\ \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\} &=&0 \\ \mathbf{U}(\mathbf{x})\cdot \mathbf{f(x)} &=&\{\mathbf{a}+\partial _{\mathbf{ x}}\times \mathbf{b(x)}\} \\ \mathbf{f(x)} &=&\mathbf{U}^{-1}(\mathbf{x})\cdot \{\mathbf{a}+\partial _{ \mathbf{x}}\times \mathbf{b(x)}\} \end{eqnarray*}