examples of lamellar field | Physics Library

In the examples that follow, show that the given vector field U is lamellar everywhere in ℝ³ and determine its scalar potential u.

Example 1. Given

U := y i + (x + sin z) j + y cos z k.

For the rotor (curl) of the field we obtain ∇×U = | |
||⃗i ⃗j ⃗k ||
|∂- -∂ -∂ |
||∂x ∂y ∂z ||
y x+sin z ycos z = ( )
∂(ycos-z)− ∂(x+sin-z-)
∂y ∂z i + ( )
∂y-− ∂-(ycosz-)
∂z ∂x j + ( )
∂(x+sin-z-)− ∂y-
∂x ∂y k,
which is identically 0 for all x, y, z. Thus, by the definition given in the parent entry, U is lamellar.
Since ∇u = U, the scalar potential u = u(x, y, z) must satisfy the conditions

∂u ∂u ∂u ---= y, --- = x+sin z, --- = ycos z. ∂x ∂y ∂z

Thus we can write

∫ u = y dx = xy + C , 1

where C₁ may depend on y or z. Differentiating this result with respect to y and comparing to the second condition, we get

∂u- ∂C1- ∂y = x + ∂y = x + sin z.

Accordingly,

∫ C1 = sin z dy = ysin z + C2,

where C₂ may depend on z. So

u = xy + y sinz + C . 2

Differentiating this result with respect to z and comparing to the third condition yields

∂u ∂C2 ∂z-= y cosz + -∂z- = y cosz.

This means that C₂ 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 ℝ²:

U(x, y, 0) := ωy i + ωxj, ω = constant

Now, ∇×U = ||⃗ ⃗ ⃗ ||
|i∂ j∂ k∂ |
||∂x ∂y ∂z||
|ωy ωx 0 | = ( )
∂(ωx ) ∂(ωy )
------− ------
∂x ∂y k = 0, and so U is lamellar.

Therefore there exists a potential field u with U = ∇u. We deduce successively:

∂u- ∂u- ′ ′ ∂x = ωy; u (x,y,0) = ωxy + f(y); ∂y = ωx + f (y ) ≡ ωx; f(y ) = 0; f(y) = C

Thus we get the result

u(x, y, 0) = ωxy + C,

which corresponds to a particular case in ℝ².

Example 3. Given

U := axi + byj − (a + b)z)k.

The rotor is now ∇×U = | |
||⃗i ⃗j ⃗k ||
||∂∂x ∂∂y ∂∂z ||
|ax by − (a + b)z | = 0. From ∇u = U we obtain

∂u ax2 ---= ax =⇒ u = ----+ f (y, z) (1) ∂x 2

∂u by2 ∂y- = by =⇒ u = -2--+ g(z,x) (2 )

∂u z2 ---= − (a + b)z =⇒ u = − (a + b)--+ h (x, y) (3) ∂z 2

Differentiating (1) and (2) with respect to z and using (3) give

∂f (y,z) z2 ′ − (a + b)z =-------- = ⇒ f(y,z) = − (a + b)---+ F (y ) (1); ∂z 2

∂g (z,x) z2 ′ − (a + b)z =---∂z--- =⇒ g(z,x) = − (a + b) 2-+ G(x ) (2).

We substitute (1′) and (2′) again into (1) and (2) and deduce as follows:

ax2- z2- ∂u- ′ by2- by2-- z2- ′′ u = 2 − (a+b) 2 +F (y); ∂y = F (y) = by; F (y) = 2 +C1; f(y,z) = 2 − (a+b ) 2 +C1 (1 );

2 2 2 2 2 u = by--− (a+b )z-+G (x); ∂u-= G ′(x) = ax; G (x) = ax--+C2; g(z,x) = ax--− (a+b )z-+C2 (2′′); 2 2 ∂x 2 2 2

putting (1′′), (2′′) into (1), (2) then gives us

ax2 by2 z2 ax2 by2 z2 u = -2--+ -2--− (a + b)-2-+ C1, u = --2- + -2--− (a + b)2--+ C2,

whence, by comparing, C₁ = C₂ = C, so that by (3), the expression h(x,y) and u itself have been found, that is,

ax2 by2 z2 u = ---- + ----− (a + b)---+ C. 2 2 2

Unlike Example 1, the last two examples are also solenoidal, i.e. ∇⋅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

ay ax ⃗U := − -2----2⃗i + -2----2⃗j + v(z)⃗k x + y x + y

with a differentiable function v : ℝ → ℝ; if v is a constant, then U is also solenoidal.

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