# PlanetPhysics/Examples of Lamellar Field

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

Example 1. \, Given

${\displaystyle {\begin{matrix}{\vec {U}}\,:=\,y\,{\vec {i}}+(x+\sin {z})\,{\vec {j}}+y\cos {z}\,{\vec {k}}.\end{matrix}}}$

For the rotor (curl) of the field we obtain [/itex]\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}$\displaystyle ,\\ which is identically $\vec{0}$ for all ${\displaystyle x}$, ${\displaystyle y}$, ${\displaystyle z}$.\, Thus, by the definition given in the parent entry, ${\displaystyle {\vec {U}}}$ is lamellar.\\ Since \,${\displaystyle \nabla {u}={\vec {U}}}$,\, the scalar potential \,${\displaystyle u=u(x,\,y,\,z)}$\, must satisfy the conditions ${\displaystyle {\frac {\partial {u}}{\partial {x}}}=y,\quad {\frac {\partial {u}}{\partial {y}}}=x\!+\!\sin {z},\quad {\frac {\partial {u}}{\partial {z}}}=y\cos {z}.}$ Thus we can write ${\displaystyle u=\int y\,dx=xy+C_{1},}$ where ${\displaystyle C_{1}}$ may depend on ${\displaystyle y}$ or ${\displaystyle z}$. Differentiating this result with respect to ${\displaystyle y}$ and comparing to the second condition, we get ${\displaystyle {\frac {\partial {u}}{\partial {y}}}=x+{\frac {\partial {C_{1}}}{\partial {y}}}=x+\sin {z}.}$ Accordingly, ${\displaystyle C_{1}=\int \sin {z}\,dy=y\sin {z}+C_{2},}$ where ${\displaystyle C_{2}}$ may depend on ${\displaystyle z}$.\, So ${\displaystyle u=xy+y\sin {z}+C_{2}.}$ Differentiating this result with respect to ${\displaystyle z}$ and comparing to the third condition yields ${\displaystyle {\frac {\partial {u}}{\partial {z}}}\,=\,y\cos {z}+{\frac {\partial {C_{2}}}{\partial {z}}}\,=\,y\cos {z}.}$ This means that ${\displaystyle C_{2}}$ is an arbitrary constant. Thus the form ${\displaystyle u=xy+y\sin {z}+C}$ expresses the required potential function.\\ Example 2. \, This is a particular case in ${\displaystyle \mathbb {R} ^{2}}$: ${\displaystyle {\begin{matrix}{\vec {U}}(x,\,y,\,0)\,:=\,\omega y\,{\vec {i}}+\omega x\,{\vec {j}},\quad \omega ={\mbox{constant}}\end{matrix}}}$ Now,\;$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ \omega y & \omega x & 0 \end{matrix}\right| = \left(\frac{\partial(\omega x)}{\partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}${\displaystyle ,\,andso}$\vec{U}${\displaystyle islamellar.Thereforethereexistsapotentialfield$u}$ with\, ${\displaystyle {\vec {U}}=\nabla {u}}$.\, We deduce successively: ${\displaystyle {\frac {\partial {u}}{\partial {x}}}=\omega y;\;\;u(x,y,0)=\omega xy+f(y);\;\;{\frac {\partial {u}}{\partial {y}}}=\omega x+f'(y)\equiv \omega x;\;\;f'(y)=0;\;\;f(y)=C}$ Thus we get the result ${\displaystyle u(x,\,y,\,0)=\omega xy+C,}$ which corresponds to a particular case in ${\displaystyle \mathbb {R} ^{2}}$.\\ Example 3. \, Given ${\displaystyle {\begin{matrix}{\vec {U}}\,:=\,ax{\vec {i}}+by{\vec {j}}-(a+b)z){\vec {k}}.\end{matrix}}}$ The rotor is now\,$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ ax & by & -(a+b)z \end{matrix}\right|= \vec{0}.${\displaystyle \;From\,}$\nabla u=\vec{U}${\displaystyle \,weobtain[itex]{\frac {\partial u}{\partial x}}=ax\;\implies \;u={\frac {ax^{2}}{2}}+f(y,z)\quad (1)}$ ${\displaystyle {\frac {\partial u}{\partial y}}=by\;\implies \;u={\frac {by^{2}}{2}}+g(z,x)\quad (2)}$ ${\displaystyle {\frac {\partial u}{\partial z}}=-(a+b)z\;\implies \;u=-(a+b){\frac {z^{2}}{2}}+h(x,y)\quad (3)}$ Differentiating (1) and (2) with respect to ${\displaystyle z}$ and using (3) give ${\displaystyle -(a+b)z={\frac {\partial f(y,z)}{\partial z}}\;\implies \;f(y,z)=-(a+b){\frac {z^{2}}{2}}+F(y)\quad (1');}$ ${\displaystyle -(a+b)z={\frac {\partial g(z,x)}{\partial z}}\;\implies \;g(z,x)=-(a+b){\frac {z^{2}}{2}}+G(x)\quad (2').}$ We substitute ${\displaystyle (1')}$ and ${\displaystyle (2')}$ again into (1) and (2) and deduce as follows: ${\displaystyle u={\frac {ax^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+F(y);\;\;{\frac {\partial u}{\partial y}}=F'(y)=by;\;\;F(y)={\frac {by^{2}}{2}}+C_{1};\;\;f(y,z)={\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{1}\quad (1'');}$ ${\displaystyle u={\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+G(x);\;\;{\frac {\partial u}{\partial x}}=G'(x)=ax;\;\;G(x)={\frac {ax^{2}}{2}}+C_{2};\;\;g(z,x)={\frac {ax^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{2}\quad (2'');}$ putting ${\displaystyle (1'')}$, ${\displaystyle (2'')}$ into (1), (2) then gives us ${\displaystyle u={\frac {ax^{2}}{2}}+{\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{1},\quad u={\frac {ax^{2}}{2}}+{\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{2},}$ whence, by comparing,\, ${\displaystyle C_{1}=C_{2}=C}$,\, so that by (3), the expression ${\displaystyle h(x,y)}$ and ${\displaystyle u}$ itself have been found, that is, ${\displaystyle u={\frac {ax^{2}}{2}}+{\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C.}$

Unlike Example 1, the last two examples are also solenoidal, i.e.\, ${\displaystyle \nabla \cdot {\vec {U}}=0}$,\, 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 ${\displaystyle {\vec {U}}\,:=\,-{\frac {ay}{x^{2}+y^{2}}}{\vec {i}}+{\frac {ax}{x^{2}+y^{2}}}{\vec {j}}+v(z){\vec {k}}}$ with a differentiable function \,${\displaystyle v:\mathbb {R} \to \mathbb {R} }$;\, if ${\displaystyle v}$ is a constant, then ${\displaystyle {\vec {U}}}$ is also solenoidal.