PlanetPhysics/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 $$\begin{matrix} \vec{U} \,:=\, y\,\vec{i}+(x+\sin{z})\,\vec{j}+y\cos{z}\,\vec{k}. \end{matrix}$$ For the rotor (curl) of the field we obtain \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}$$,\\ which is identically \vec{0} for all x, y, z.\, Thus, by the definition given in the parent entry, \vec{U} is lamellar.\\ Since \,,\, the scalar potential \,\, must satisfy the conditions

Thus we can write

where C_1 may depend on y or z. Differentiating this result with respect to y and comparing to the second condition, we get

Accordingly,

where C_2 may depend on z.\, So

Differentiating this result with respect to z and comparing to the third condition yields

This means that C_2 is an arbitrary constant. Thus the form

expresses the required potential function.\\

Example 2. \, This is a particular case in : Now,\; $$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ \omega y & \omega x & 0 \end{matrix}\right| = \left(\frac{\partial(\omega x)}{\partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}$$,\, and so $$\vec{U}$$ is lamellar.

Therefore there exists a potential field u with\, .\, We deduce successively:

Thus we get the result

which corresponds to a particular case in .\\

Example 3. \, Given The rotor is now\, $$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ ax & by & -(a+b)z \end{matrix}\right|= \vec{0}.$$\; From\, $$\nabla u=\vec{U}\, we obtain $$\frac{\partial u}{\partial x} = ax \; \implies \; u = \frac{ax^2}{2}+f(y,z) \quad(1)$$ $$\frac{\partial u}{\partial y} = by \; \implies \; u = \frac{by^2}{2}+g(z,x) \quad(2)$$ $$\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$$ and using (3) give $$-(a+b)z = \frac{\partial f(y,z)}{\partial z} \; \implies \; f(y,z) = -(a+b)\frac{z^2}{2}+F(y) \quad(1');$$ $$-(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')$$ again into (1) and (2) and deduce as follows: $$u = \frac{ax^2}{2}-(a+b)\frac{z^2}{2}+F(y); \;\; \frac{\partial u}{\partial y} = F'(y) = by; \;\; F(y) = \frac{by^2}{2}+C_1; \;\; f(y,z) = \frac{by^2}{2}-(a+b)\frac{z^2}{2}+C_1 \quad (1'');$$ $$u = \frac{by^2}{2}-(a+b)\frac{z^2}{2}+G(x); \;\; \frac{\partial u}{\partial x} = G'(x) = ax; \;\; G(x) = \frac{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)$$ into (1), (2) then gives us $$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$$ itself have been found, that is, $$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 $$\vec{U} \,:=\, -\frac{ay}{x^2+y^2}\vec{i}+\frac{ax}{x^2+y^2}\vec{j}+v(z)\vec{k}$$ with a differentiable function \,$$v:\mathbb{R}\to\mathbb{R}$$;\, if $$v$$ is a constant, then $$\vec{U}$$ is also solenoidal.