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.

Title	examples of lamellar field
Canonical name	ExamplesOfLamellarField
Date of creation	2013-03-22 17:39:25
Last modified on	2013-03-22 17:39:25
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Example
Classification	msc 26B12
Synonym	example of scalar potential
Synonym	determining the scalar potential
Related topic	Curl