In the examples that follow, show that the given vector field
is lamellar everywhere in
and determine its scalar potential
.\\
Example 1. \, Given
![{\displaystyle {\begin{matrix}{\vec {U}}\,:=\,y\,{\vec {i}}+(x+\sin {z})\,{\vec {j}}+y\cos {z}\,{\vec {k}}.\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0122f4e1de96e71d3a6d28c3c1fc47a3d5ca04b)
For the rotor (curl) of the field we obtain
</math>\nabla\!\times\!\vec{U} = \left|\begin{matrix}
\vec{i} & \vec{j} & \vec{k}\\
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\
y & x\!+\!\sin{z} & y\cos{z}
\end{matrix}\right| \\=
\left(\frac{\partial(y\cos{z})}{\partial{y}}-\frac{\partial(x\!+\!\sin{z})}{\partial{z}}\right)\vec{i}
+\left(\frac{\partial{y}}{\partial{z}}-\frac{\partial(y\cos{z})}{\partial{x}}\right)\vec{j}
+\left(\frac{\partial(x\!+\!\sin{z})}{\partial{x}}-\frac{\partial{y}}{\partial{y}}\right)\vec{k}Failed to parse (syntax error): {\displaystyle ,\\ which is identically <math>\vec{0}}
for all
,
,
.\, Thus, by the definition given in the parent entry,
is lamellar.\\
Since \,
,\, the scalar potential \,
\, must satisfy the conditions
Thus we can write
where
may depend on
or
. Differentiating this result with respect to
and comparing to the second
condition, we get
Accordingly,
where
may depend on
.\, So
Differentiating this result with respect to
and comparing to the third condition yields
This means that
is an arbitrary constant. Thus the form
expresses the required potential function.\\
Example 2. \, This is a particular case in
:
![{\displaystyle {\begin{matrix}{\vec {U}}(x,\,y,\,0)\,:=\,\omega y\,{\vec {i}}+\omega x\,{\vec {j}},\quad \omega ={\mbox{constant}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/008353139141e842316d844b6d98fa0f22d3c26c)
Now,\; </math>\nabla\!\times\!\vec{U} = \left|\begin{matrix}
\vec{i} & \vec{j} & \vec{k}\\
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\
\omega y & \omega x & 0
\end{matrix}\right| =
\left(\frac{\partial(\omega x)}{\partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}
\vec{U}
with\,
.\, We deduce successively:
Thus we get the result
which corresponds to a particular case in
.\\
Example 3. \, Given
![{\displaystyle {\begin{matrix}{\vec {U}}\,:=\,ax{\vec {i}}+by{\vec {j}}-(a+b)z){\vec {k}}.\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f87e910426a75099f4f8ff7a6bcb8fdd956a3e5f)
The rotor is now\, </math>\nabla\!\times\!\vec{U} =
\left|\begin{matrix}
\vec{i} & \vec{j} & \vec{k}\\
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\
ax & by & -(a+b)z
\end{matrix}\right|= \vec{0}.
\nabla u=\vec{U}
Differentiating (1) and (2) with respect to
and using (3) give
We substitute
and
again into (1) and (2) and deduce as follows:
putting
,
into (1), (2) then gives us
whence, by comparing,\,
,\, so that by (3), the expression
and
itself have been found, that is,
Unlike Example 1, the last two examples are also solenoidal, i.e.\,
,\, which physically may be interpreted as the continuity equation of an incompressible fluid flow.\\
Example 4. \, An additional example of a lamellar field would be
with a differentiable function \,
;\, if
is a constant, then
is also solenoidal.