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[parent] examples of lamellar field (Example)

In the examples that follow, show that the given vector field $\vec{U}$ is lamellar everywhere in $\mathbb{R}^3$ and determine its scalar potential $u$.

Example 1.  Given

$\displaystyle \vec{U} \,:=\, y\,\vec{i}+(x+\sin{z})\,\vec{j}+y\cos{z}\,\vec{k}.$    
For the rotor (curl) of the field we obtain $\displaystyle\nabla\!\times\!\vec{U} = \left\vert\begin{matrix} \vec{i} & \vec{... ...tial(x\!+\!\sin{z})}{\partial{x}}-\frac{\partial{y}}{\partial{y}}\right)\vec{k}$,
which is identically $\vec{0}$ for all $x$, $y$, $z$.  Thus, by the definition given in the parent entry, $\vec{U}$ is lamellar.
Since   $\nabla{u} = \vec{U}$,  the scalar potential   $u = u(x,\,y,\,z)$  must satisfy the conditions

$\displaystyle \frac{\partial{u}}{\partial{x}} = y,\quad \frac{\partial{u}}{\partial{y}} = x\!+\!\sin{z},\quad \frac{\partial {u}}{\partial{z}} = y\cos{z}.$
Thus we can write

$\displaystyle u = \int y\,dx = xy+C_1,$
where $C_1$ may depend on $y$ or $z$. Differentiating this result with respect to $y$ and comparing to the second condition, we get

$\displaystyle \frac{\partial{u}}{\partial{y}} = x+\frac{\partial{C_1}}{\partial{y}} = x+\sin{z}.$
Accordingly,

$\displaystyle C_1 = \int\sin{z}\,dy = y\sin{z}+C_2,$
where $C_2$ may depend on $z$.  So

$\displaystyle u = xy+y\sin{z}+C_2.$
Differentiating this result with respect to $z$ and comparing to the third condition yields

$\displaystyle \frac{\partial{u}}{\partial{z}} \,=\, y\cos{z}+\frac{\partial{C_2}}{\partial{z}} \,=\, y\cos{z}.$
This means that $C_2$ is an arbitrary constant. Thus the form

$\displaystyle u = xy+y\sin{z}+C$
expresses the required potential function.

Example 2.  This is a particular case in $\mathbb{R}^2$:

$\displaystyle \vec{U}(x,\,y,\,0) \,:=\, \omega y \,\vec{i}+ \omega x \,\vec{j}, \quad \omega =$   constant    
Now, $\displaystyle\nabla\!\times\!\vec{U} = \left\vert\begin{matrix} \vec{i} & \vec{... ...a x)}{\partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}$,  and so $\vec{U}$ is lamellar.

Therefore there exists a potential field $u$ with  $\vec{U}=\nabla{u}$.  We deduce successively:

$\displaystyle \frac{\partial{u}}{\partial{x}} = \omega y; \;\; u(x,y,0) = \omeg... ...tial{u}}{\partial{y}}= \omega x+f'(y)\equiv \omega x; \;\; f'(y)=0; \;\; f(y)=C$
Thus we get the result

$\displaystyle u(x,\,y,\,0) = \omega xy+C,$
which corresponds to a particular case in $\mathbb{R}^2$.

Example 3.  Given

$\displaystyle \vec{U} \,:=\, ax\vec{i}+by\vec{j}-(a+b)z)\vec{k}.$    
The rotor is now  $\displaystyle\nabla\!\times\!\vec{U} = \left\vert\begin{matrix} \vec{i} & \vec{... ...c{\partial}{\partial{z}}\ ax & by & -(a+b)z \end{matrix}\right\vert= \vec{0}.$ From  $\nabla u=\vec{U}$  we obtain

$\displaystyle \frac{\partial u}{\partial x} = ax \; \implies \; u = \frac{ax^2}{2}+f(y,z) \quad(1)$

$\displaystyle \frac{\partial u}{\partial y} = by \; \implies \; u = \frac{by^2}{2}+g(z,x) \quad(2)$

$\displaystyle \frac{\partial u}{\partial z} = -(a+b)z \; \implies \;u = -(a+b)\frac{z^2}{2}+h(x,y) \quad(3)$
Differentiating (1) and (2) with respect to $z$ and using (3) give

$\displaystyle -(a+b)z = \frac{\partial f(y,z)}{\partial z} \; \implies \; f(y,z) = -(a+b)\frac{z^2}{2}+F(y) \quad(1');$

$\displaystyle -(a+b)z=\frac{\partial g(z,x)}{\partial z} \; \implies \; g(z,x) = -(a+b)\frac{z^2}{2}+G(x) \quad (2').$
We substitute $(1')$ and $(2')$ again into (1) and (2) and deduce as follows:

$\displaystyle u = \frac{ax^2}{2}-(a+b)\frac{z^2}{2}+F(y); \;\; \frac{\partial u... ...{by^2}{2}+C_1; \;\; f(y,z) = \frac{by^2}{2}-(a+b)\frac{z^2}{2}+C_1 \quad (1'');$

$\displaystyle u = \frac{by^2}{2}-(a+b)\frac{z^2}{2}+G(x); \;\; \frac{\partial u... ...c{ax^2}{2}+C_2; \;\; g(z,x) = \frac{ax^2}{2}-(a+b)\frac{z^2}{2}+C_2\quad (2'');$
putting $(1'')$, $(2'')$ into (1), (2) then gives us

$\displaystyle u = \frac{ax^2}{2}+\frac{by^2}{2}-(a+b)\frac{z^2}{2}+C_1, \quad u = \frac{ax^2}{2}+\frac{by^2}{2}-(a+b)\frac{z^2}{2}+C_2,$
whence, by comparing,  $C_1 = C_2 = C$,  so that by (3), the expression $h(x,y)$ and $u$ itself have been found, that is,

$\displaystyle u = \frac{ax^2}{2}+\frac{by^2}{2}-(a+b)\frac{z^2}{2}+C.$

Unlike Example 1, the last two examples are also solenoidal, i.e.  $\nabla\cdot\vec{U}=0$,  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

$\displaystyle \vec{U} \,:=\, -\frac{ay}{x^2+y^2}\vec{i}+\frac{ax}{x^2+y^2}\vec{j}+v(z)\vec{k}$
with a differentiable function   $v:\mathbb{R}\to\mathbb{R}$;  if $v$ is a constant, then $\vec{U}$ is also solenoidal.



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Cross-references: lamellar field, continuity equation, solenoidal, function, curl, scalar, vector field

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