## Finding a Vector With a Given Curl

Suppose we have a vector field
$\mathbf{v}$
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{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}

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.