How to prove in a rigorous way that: $$|u|=1 \implies \nabla|u|^2 = 0 \implies (\nabla u)^Tu=0$$ and then $\forall v$ $$\nabla u : \nabla((u.v)\cdot u)= |\nabla u|^2 (u \cdot v)$$
Asked
Active
Viewed 55 times
1
-
What is $u$? I'm guessing it's a vector field here, but is it smooth/differentiable/continuous? When you write $|u| = 1$, do you mean "$|u(x)| = 1$ for all $x$"? – BaronVT Dec 19 '13 at 20:25
-
u is a sufficiently smoth vector field and |u|=1 means |u(x)|=1 for all x – aflous Dec 19 '13 at 21:10
-
I can see that if |u| is constant then the same is true for |u|², hence its gradient is zero. and I think we can write $ \nabla|u|^2 = 2(\nabla u)^Tu$? – aflous Dec 19 '13 at 22:12
-
Yes, that is correct. – BaronVT Dec 20 '13 at 00:41
1 Answers
1
Using the Einstein summation convention:
$$\nabla u : \nabla((u.v)\cdot u) = \frac{\partial u_i}{\partial x_j} \frac{\partial}{\partial x_j}((u_k v_k) u_i) = \frac{\partial u_i}{\partial x_j}\frac{\partial u_i}{\partial x_j} (u_k v_k) + \frac{\partial u_i}{\partial x_j} u_i \frac{\partial}{\partial x_j}(u_k v_k) = |\nabla u|^2 (u\cdot v) $$ because $$ \frac{\partial u_i}{\partial x_j} u_i = \frac12 \frac{\partial}{\partial x_j}|u_i|^2 = 0 .$$
Stephen Montgomery-Smith
- 26,430
- 2
- 35
- 64