# 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 (http://planetmath.org/NablaNabla) (curl) of the we obtain $\displaystyle\nabla\!\times\!\vec{U}=\left|\begin{matrix}\vec{i}&\vec{j}&\vec{% k}\\ \frac{\partial}{\partial{x}}&\frac{\partial}{\partial{y}}&\frac{\partial}{% \partial{z}}\\ y&x\!+\!\sin{z}&y\cos{z}\end{matrix}\right|\\ =\left(\frac{\partial(y\cos{z})}{\partial{y}}-\frac{\partial(x\!+\!\sin{z})}{% \partial{z}}\right)\vec{i}+\left(\frac{\partial{y}}{\partial{z}}-\frac{% \partial(y\cos{z})}{\partial{x}}\right)\vec{j}+\left(\frac{\partial(x\!+\!\sin% {z})}{\partial{x}}-\frac{\partial{y}}{\partial{y}}\right)\vec{k}$,
which is identically $\vec{0}$ for all $x$, $y$, $z$.  Thus, by the definition given in the parent (http://planetmath.org/LaminarField) entry, $\vec{U}$ is lamellar.
Since  $\nabla{u}=\vec{U}$,  the scalar potential  $u=u(x,\,y,\,z)$  must satisfy the conditions

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

Thus we can write

 $u=\int y\,dx=xy+C_{1},$

where $C_{1}$ may depend on $y$ or $z$. Differentiating this result with respect to $y$ and comparing to the second condition, we get

 $\frac{\partial{u}}{\partial{y}}=x+\frac{\partial{C_{1}}}{\partial{y}}=x+\sin{z}.$

Accordingly,

 $C_{1}=\int\sin{z}\,dy=y\sin{z}+C_{2},$

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 $\mathbb{R}^{2}$:

 $\displaystyle\vec{U}(x,\,y,\,0)\,:=\,\omega y\,\vec{i}+\omega x\,\vec{j},\quad% \omega=\mbox{constant}$

Now,  $\displaystyle\nabla\!\times\!\vec{U}=\left|\begin{matrix}\vec{i}&\vec{j}&\vec{% k}\\ \frac{\partial}{\partial{x}}&\frac{\partial}{\partial{y}}&\frac{\partial}{% \partial{z}}\\ \omega y&\omega x&0\end{matrix}\right|=\left(\frac{\partial(\omega x)}{% \partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}$,  and so $\vec{U}$ is lamellar.

Therefore there exists a potential $u$ with  $\vec{U}=\nabla{u}$.  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

 $u(x,\,y,\,0)=\omega xy+C,$

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  $\displaystyle\nabla\!\times\!\vec{U}=\left|\begin{matrix}\vec{i}&\vec{j}&\vec{% k}\\ \frac{\partial}{\partial{x}}&\frac{\partial}{\partial{y}}&\frac{\partial}{% \partial{z}}\\ ax&by&-(a+b)z\end{matrix}\right|=\vec{0}.$  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

 $-(a+b)z=\frac{\partial f(y,z)}{\partial z}\;\implies\;f(y,z)=-(a+b)\frac{z^{2}% }{2}+F(y)\quad(1^{\prime});$
 $-(a+b)z=\frac{\partial g(z,x)}{\partial z}\;\implies\;g(z,x)=-(a+b)\frac{z^{2}% }{2}+G(x)\quad(2^{\prime}).$

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

 $u=\frac{ax^{2}}{2}-(a+b)\frac{z^{2}}{2}+F(y);\;\;\frac{\partial u}{\partial y}% =F^{\prime}(y)=by;\;\;F(y)=\frac{by^{2}}{2}+C_{1};\;\;f(y,z)=\frac{by^{2}}{2}-% (a+b)\frac{z^{2}}{2}+C_{1}\quad(1^{\prime\prime});$
 $u=\frac{by^{2}}{2}-(a+b)\frac{z^{2}}{2}+G(x);\;\;\frac{\partial u}{\partial x}% =G^{\prime}(x)=ax;\;\;G(x)=\frac{ax^{2}}{2}+C_{2};\;\;g(z,x)=\frac{ax^{2}}{2}-% (a+b)\frac{z^{2}}{2}+C_{2}\quad(2^{\prime\prime});$

putting $(1^{\prime\prime})$, $(2^{\prime\prime})$ 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.

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