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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
For the rotor (curl) of the field 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 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 $\mathbb{R}^2$ :
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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 field $u$ 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
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 $$-(a+b)z = \frac{\partial f(y,z)}{\partial z} \; \implies \; f(y,z) =
-(a+b)\frac{z^2}{2}+F(y) \quad(1');$$ $$-(a+b)z=\frac{\partial g(z,x)}{\partial z} \; \implies \; g(z,x) = -(a+b)\frac{z^2}{2}+G(x) \quad (2').$$ 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.
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