PlanetPhysics/Examples of Lamellar Field

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

Example 1. \, Given

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

For the rotor ([[../Curl/|curl]]) of the field we obtain </math>\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}Failed to parse (syntax error): {\displaystyle ,\\ which is identically <math>\vec{0}} for all $x$, $y$, $z$.\, Thus, by the definition given in the parent entry, $\overrightarrow{U}$ is lamellar.\\ Since \,$\mathrm{\nabla }u=\overrightarrow{U}$,\, 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 second condition, 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 constant. Thus the form $u=xy+y\sin z+C$ expresses the required potential [[../Bijective/|function]].\\

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

$\begin{array}{c}\overrightarrow{U}(x,y,0):=\omega y\overrightarrow{i}+\omega x\overrightarrow{j},\omega =\text{constant}\end{array}$

Now,\; </math>\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}$,andso$\vec{U}$islamellar.Thereforethereexistsapotentialfield<math>u$ with\, $\overrightarrow{U}=\mathrm{\nabla }u$.\, 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 ${ℝ}^2$.\\

Example 3. \, Given

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

The rotor is now\, </math>\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}$weobtain<math>\frac{\partial u}{\partial x}=ax⟹u=\frac{a{x}^2}{2}+f(y,z)(1)$ $\frac{\partial u}{\partial y}=by⟹u=\frac{b{y}^2}{2}+g(z,x)(2)$ $\frac{\partial u}{\partial z}=-(a+b)z⟹u=-(a+b)\frac{z^2}{2}+h(x,y)(3)$ Differentiating (1) and (2) with respect to $z$ and using (3) give $-(a+b)z=\frac{\partial f(y,z)}{\partial z}⟹f(y,z)=-(a+b)\frac{z^2}{2}+F(y)(1^{\prime });$ $-(a+b)z=\frac{\partial g(z,x)}{\partial z}⟹g(z,x)=-(a+b)\frac{z^2}{2}+G(x)(2^{\prime }).$ We substitute $(1^{\prime })$ and $(2^{\prime })$ again into (1) and (2) and deduce as follows: $u=\frac{a{x}^2}{2}-(a+b)\frac{z^2}{2}+F(y);\frac{\partial u}{\partial y}=F^{\prime }(y)=by;F(y)=\frac{b{y}^2}{2}+C_1;f(y,z)=\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+C_1(1^{″});$ $u=\frac{b{y}^2}{2}-(a+b)\frac{z^2}{2}+G(x);\frac{\partial u}{\partial x}=G^{\prime }(x)=ax;G(x)=\frac{a{x}^2}{2}+C_2;g(z,x)=\frac{a{x}^2}{2}-(a+b)\frac{z^2}{2}+C_2(2^{″});$ putting $(1^{″})$, $(2^{″})$ 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 [[../SolenoidalVectorField/|solenoidal]], i.e.\, $\mathrm{\nabla }\cdot \overrightarrow{U}=0$,\, which physically may be interpreted as the [[../ContinuityEquation/|continuity equation]] of an incompressible fluid flow.\\

Example 4. \, An additional example of a [[../LamellarField/|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.