# PlanetPhysics/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

${\begin{matrix}{\vec {U}}\,:=\,y\,{\vec {i}}+(x+\sin {z})\,{\vec {j}}+y\cos {z}\,{\vec {k}}.\end{matrix}}$ 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 $x$ , $y$ , $z$ .\, Thus, by the definition given in the parent 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 constant. Thus the form $u=xy+y\sin {z}+C$ expresses the required potential function.\\ Example 2. \, This is a particular case in $\mathbb {R} ^{2}$ : ${\begin{matrix}{\vec {U}}(x,\,y,\,0)\,:=\,\omega y\,{\vec {i}}+\omega x\,{\vec {j}},\quad \omega ={\mbox{constant}}\end{matrix}}$ Now,\;$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ \omega y & \omega x & 0 \end{matrix}\right| = \left(\frac{\partial(\omega x)}{\partial{x}}-\frac{\partial(\omega y)}{\partial{y}}\right)\vec{k}=\vec{0}$,\,andso$ \vec{U}$islamellar.Thereforethereexistsapotentialfield$u$ with\, ${\vec {U}}=\nabla {u}$ .\, We deduce successively: ${\frac {\partial {u}}{\partial {x}}}=\omega y;\;\;u(x,y,0)=\omega xy+f(y);\;\;{\frac {\partial {u}}{\partial {y}}}=\omega x+f'(y)\equiv \omega x;\;\;f'(y)=0;\;\;f(y)=C$ 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 ${\begin{matrix}{\vec {U}}\,:=\,ax{\vec {i}}+by{\vec {j}}-(a+b)z){\vec {k}}.\end{matrix}}$ The rotor is now\,$\nabla\!\times\!\vec{U} = \left|\begin{matrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\ ax & by & -(a+b)z \end{matrix}\right|= \vec{0}.$\;From\,$ \nabla u=\vec{U}$\,weobtain[itex]{\frac {\partial u}{\partial x}}=ax\;\implies \;u={\frac {ax^{2}}{2}}+f(y,z)\quad (1)$ ${\frac {\partial u}{\partial y}}=by\;\implies \;u={\frac {by^{2}}{2}}+g(z,x)\quad (2)$ ${\frac {\partial u}{\partial z}}=-(a+b)z\;\implies \;u=-(a+b){\frac {z^{2}}{2}}+h(x,y)\quad (3)$ Differentiating (1) and (2) with respect to $z$ and using (3) give $-(a+b)z={\frac {\partial f(y,z)}{\partial z}}\;\implies \;f(y,z)=-(a+b){\frac {z^{2}}{2}}+F(y)\quad (1');$ $-(a+b)z={\frac {\partial g(z,x)}{\partial z}}\;\implies \;g(z,x)=-(a+b){\frac {z^{2}}{2}}+G(x)\quad (2').$ We substitute $(1')$ and $(2')$ again into (1) and (2) and deduce as follows: $u={\frac {ax^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+F(y);\;\;{\frac {\partial u}{\partial y}}=F'(y)=by;\;\;F(y)={\frac {by^{2}}{2}}+C_{1};\;\;f(y,z)={\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{1}\quad (1'');$ $u={\frac {by^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+G(x);\;\;{\frac {\partial u}{\partial x}}=G'(x)=ax;\;\;G(x)={\frac {ax^{2}}{2}}+C_{2};\;\;g(z,x)={\frac {ax^{2}}{2}}-(a+b){\frac {z^{2}}{2}}+C_{2}\quad (2'');$ putting $(1'')$ , $(2'')$ 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.