Question

Show that
(vr .∇) vr = 1/2(∇v^2)− vr × (∇×vr)
Note: here vr= vector r

Answer 1

Answer on Question #39182 – Math – Other

1. Show that

$$(\vec{v} \vec{\nabla}) \vec{v} = \frac{1}{2} \vec{\nabla} v^2 - \vec{v} \times [\vec{\nabla} \times \vec{v}]$$

Proof.

Let simplify the expressions in Cartesian coordinates.

If $\vec{v} = v_1 \vec{i} + v_2 \vec{j} + v_3 \vec{k}$, than $X$-component of the left-side of the equation:

$$\left(v_1 \frac{\partial}{\partial x} + v_2 \frac{\partial}{\partial y} + v_3 \frac{\partial}{\partial z}\right) v_1 = v_1 \frac{\partial v_1}{\partial x} + v_2 \frac{\partial v_1}{\partial y} + v_3 \frac{\partial v_1}{\partial z}.$$

The vector product:

$$[\vec{\nabla} \times \vec{v}] = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_1 & v_2 & v_3 \end{vmatrix} = \vec{i} \left(\frac{\partial v_3}{\partial y} - \frac{\partial v_2}{\partial z}\right) - \vec{j} \left(\frac{\partial v_3}{\partial x} - \frac{\partial v_1}{\partial z}\right) + \vec{k} \left(\frac{\partial v_2}{\partial x} - \frac{\partial v_1}{\partial y}\right).$$

$X$-component of the right-side of the equation:

$$\begin{aligned} & \frac{1}{2} \frac{\partial}{\partial x} \left(v_1^2 + v_2^2 + v_3^2\right) - \left[ v_2 \cdot \left(\frac{\partial v_2}{\partial x} - \frac{\partial v_1}{\partial y}\right) + v_3 \cdot \left(\frac{\partial v_3}{\partial x} - \frac{\partial v_1}{\partial z}\right) \right] = \\ & = v_1 \frac{\partial v_1}{\partial x} + v_2 \frac{\partial v_2}{\partial x} + v_3 \frac{\partial v_3}{\partial x} - v_2 \frac{\partial v_2}{\partial x} + v_2 \frac{\partial v_1}{\partial y} - v_3 \frac{\partial v_3}{\partial x} + v_3 \frac{\partial v_1}{\partial z} = v_1 \frac{\partial v_1}{\partial x} + v_2 \frac{\partial v_1}{\partial y} + v_3 \frac{\partial v_1}{\partial z}. \end{aligned}$$

The $X$- and $Y$-components can be checked in the same way.

Answer: The statement is correct.