I have a question concerning the Biot-Savart law: For the 3D Euler equations, that is,
$\partial_t u + (\nabla \cdot u) u = -\nabla p $
$\nabla \cdot u = 0$,
the Biot-Savart law declares a relation between the velocity and its curl, =curl() via the Biot-Savart operator, that is
$u(x) = \int \omega(y) \times \frac{(x-y)}{|x-y|^3} dy$.
Now, my question is, when one applies $\nabla_x$ on both sides of the equation, that is, $\nabla_x u(x)= \nabla_x(\int \omega(y) \times \frac{(x-y)}{|x-y|^3} dy)$, this should be apparently equal to the following:
$\nabla_x(\int \omega(y) \times \frac{(x-y)}{|x-y|^3} dy) = \int \omega(y) \times (\frac{(x)}{|y|^3} - 3 \frac{<y,x>}{|y|^5}dy$.
I have tried many different derivations, but I can't really seem to find the final result.
I would really appreciate if someone could help, thank you in advance!
Let $\omega(y) = [\omega_1,\omega_2,\omega_3]$.
$$V = \omega(y) \times \frac{x-y}{|x-y|^3} = $$ $$ \left[\omega_2 \frac{(x_3-y_3)}{|x-y|^3}- \omega_3 \frac{(x_2-y_2)}{|x-y|^3}, \omega_1 \frac{(x_3-y_3)}{|x-y|^3}- \omega_3 \frac{(x_1-y_1)}{|x-y|^3}, \omega_1 \frac{(x_2-y_2)}{|x-y|^3}- \omega_2 \frac{(x_1-y_1)}{|x-y|^3} \right]$$
$$\nabla_x V = -3\left(\omega_2 \frac{x_1(x_1-y_1)(x_3-y_3)}{|x-y|^5}- \omega_3 \frac{x_1(x_1-y_1)(x_2-y_2)}{|x-y|^5} \right)- 3\left(\omega_1 \frac{x_2(x_2-y_2)(x_3-y_3)}{|x-y|^5}- \omega_3 \frac{x_2(x_2-y_2)(x_1-y_1)}{|x-y|^5}\right) -3\left( \omega_1 \frac{x_3(x_3-y_3)(x_2-y_2)}{|x-y|^5}- \omega_2 \frac{x_3(x_3-y_3)(x_1-y_1)}{|x-y|^5} \right)$$
Now i think we need to combine the coefficient of $\omega_i$ and match the coefficient of $\omega_i$ between LHS and RHS. Can u calculate the cross product of RHS similarly and post in the question. I will then try to see if both matches. I can do it now itself but i feel there are some typos in your right hand side for example denominator is based on $|y|$. Is it $|x-y|$ ? Can u please fix typos and post the cross product of right hand side in a similar fashion ?