examples of lamellar field

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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