examples of lamellar field

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

Example 1.  Given

$\overrightarrow{U}:=y\overrightarrow{i}+(x+\sin z)\overrightarrow{j}+y\cos z\overrightarrow{k}.$

For the rotor (http://planetmath.org/NablaNabla) (curl) of the we obtain,
which is identically $\overrightarrow{0}$ for all $x$, $y$, $z$.  Thus, by the definition given in the parent (http://planetmath.org/LaminarField) entry, $\overrightarrow{U}$ is lamellar.
Since  $\nabla u=\overrightarrow{U}$,  the scalar potential  $u=u(x,y,z)$  must satisfy the conditions

$$\frac{\partial u}{\partial x}=y,\frac{\partial u}{\partial y}=x+\sin z,\frac{\partial u}{\partial z}=y\cos z.$$

Thus we can write

$$u=\int y𝑑x=xy+C_1,$$

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

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

Accordingly,

$$C_1=\int \sin zdy=y\sin z+C_2,$$

where $C_2$ may depend on $z$.  So

$$u=xy+y\sin z+C_2.$$

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

$$\frac{\partial u}{\partial z}=y\cos z+\frac{\partial C_2}{\partial z}=y\cos z.$$

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

$$u=xy+y\sin z+C$$

expresses the required potential function.

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

$\overrightarrow{U}(x,y,\mathrm{\hspace{0.17em}0}):=\omega y\overrightarrow{i}+\omega x\overrightarrow{j},\omega =\text{constant}$

Now,  ,  and so $\overrightarrow{U}$ is lamellar.

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

$$\frac{\partial u}{\partial x}=\omega y;u(x,y,0)=\omega xy+f(y);\frac{\partial u}{\partial y}=\omega x+f^{\prime }(y)\equiv \omega x;f^{\prime }(y)=0;f(y)=C$$

Thus we get the result

$$u(x,y,\mathrm{\hspace{0.17em}0})=\omega xy+C,$$

which corresponds to a particular case in ${ℝ}^2$.

Example 3.  Given

$\overrightarrow{U}:=ax\overrightarrow{i}+by\overrightarrow{j}-(a+b)z)\overrightarrow{k}.$

The rotor is now    From  $\nabla u=\overrightarrow{U}$  we obtain

$$\frac{\partial u}{\partial x}=ax⟹u=\frac{a{x}^2}{2}+f(y,z)\mathit{\hspace{1em}}(1)$$

$$\frac{\partial u}{\partial y}=by⟹u=\frac{b{y}^2}{2}+g(z,x)\mathit{\hspace{1em}}(2)$$

$$\frac{\partial u}{\partial z}=-(a+b)z⟹u=-(a+b)\frac{z^2}{2}+h(x,y)\mathit{\hspace{1em}}(3)$$

Differentiating (1) and (2) with respect to $z$ and using (3) give

$$-(a+b)z=\frac{\partial f(y,z)}{\partial z}⟹f(y,z)=-(a+b)\frac{z^2}{2}+F(y)\mathit{\hspace{1em}}(1^{\prime });$$

$$-(a+b)z=\frac{\partial g(z,x)}{\partial z}⟹g(z,x)=-(a+b)\frac{z^2}{2}+G(x)\mathit{\hspace{1em}}(2^{\prime }).$$

We substitute $(1^{\prime })$ and $(2^{\prime })$ again into (1) and (2) and deduce as follows:

$$u=\frac{a{x}^2}{2}-(a+b)\frac{z^2}{2}+F(y);\frac{\partial u}{\partial y}=F^{\prime }(y)=by;F(y)=\frac{b{y}^2}{2}+C_1;f(y,z)=\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+C_1\mathit{\hspace{1em}}(1^{\prime \prime });$$

$$u=\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+G(x);\frac{\partial u}{\partial x}=G^{\prime }(x)=ax;G(x)=\frac{a{x}^2}{2}+C_2;g(z,x)=\frac{a{x}^2}{2}-(a+b)\frac{z^2}{2}+C_2\mathit{\hspace{1em}}(2^{\prime \prime });$$

putting $(1^{\prime \prime })$, $(2^{\prime \prime })$ into (1), (2) then gives us

$$u=\frac{a{x}^2}{2}+\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+C_1,u=\frac{a{x}^2}{2}+\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+C_2,$$

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,

$$u=\frac{a{x}^2}{2}+\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+C.$$

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

$$\overrightarrow{U}:=-\frac{ay}{x^2+y^2}\overrightarrow{i}+\frac{ax}{x^2+y^2}\overrightarrow{j}+v(z)\overrightarrow{k}$$

with a differentiable function  $v:ℝ\to ℝ$;  if $v$ is a constant, then $\overrightarrow{U}$ is also solenoidal.