\[\mathbf{v}\]
satisfying \[\mathbf{\nabla} \cdot \mathbf{v} =0\]
for which we want to find a vector field \[\mathbf{w}\]
such that \[\mathbf{\nabla} \times \mathbf{w} = \mathbf{v}\]
. (1)For any vector
\[\mathbf{u}\]
, \[\mathbf{\nabla} \cdot (\mathbf{\nabla}\times \mathbf{u}) =0\]
so all solutions of (1) are given by \[\mathbf{u} = \mathbf{u_o} + \mathbf{\nabla} f \]
where \[f=f(x,y,z)\]
is an arbitrary scalar and \[\mathbf{w_0}\]
is any vector satisfying nbsp;\[\mathbf{\nabla} \times \mathbf{w} = \mathbf{v}\]
.Example
\[\mathbf{v}=3x \mathbf{i} +2y \mathbf{j} - 5z \mathbf{k}\]
\[\mathbf{\nabla} \cdot \mathbf{v} = \frac{\partial (3x) }{\partial x} + \frac{\partial (2y) }{\partial y} + \frac{\partial (-5x) }{\partial z}=0\]
Because
\[\mathbf{w_0}\]
is arbitrary we can choose \[\mathbf{w_0} = w_1 \mathbf{i} + w_2 \mathbf{j}\]
Then \[\begin{equation} \begin{aligned} \mathbf{\nabla} \times \nabla{w_0} &= (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k}) \times (w_1 \mathbf{i} + w_2 \mathbf{j}) \\ &=- \frac{\partial w_2}{\partial z} \mathbf{i} + \frac{\partial w_1}{\partial z} \mathbf{j} + (\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y})\mathbf{k} = 3x \mathbf{i} + 2y \mathbf{j} -5z \mathbf{k} \end{aligned} \end{equation}\]
Equating components gives
\[- \frac{\partial w_2}{\partial z} = 3x\]
, \[- \frac{\partial w_1}{\partial z} = 2y\]
, \[\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y} = -5z\]
Integration gives
\[w_2 = -3xz+ g(x,y)\]
and \[w_1 = 2yz+ h(x,y), \]
Substituting thse into
\[\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y} = -5z\]
gives \[-3z + \frac{\partial g}{\partial z} -2z - \frac{\partial h}{\partial y} = -5z\]
This equation is satisfied with
\[g(x,y)=h(x,y)=0\]
Hence
\[\mathbf{w} = \mathbf{w_0} + \mathbf{\nabla} f = 2yz \mathbf{i} -3xz{j} + \mathbf{\nabla} f\]
where \[f\]
is arbitrary.