# Scalar product of two gradients?

If I have $$\nabla u(x,y) \cdot \nabla u(x,y),$$ can I distribute nabla as $$\nabla (u(x,y) \cdot u(x,y))= \nabla \rvert u(x,y)\lvert ^2 \quad \text{?}$$ or $$(\nabla \cdot \nabla)u(x,y)=\nabla^2u(x,y) \quad \text{?}$$ or $$\lvert \nabla u(x,y) \rvert ^2= \lvert \nabla\rvert^2\lvert u(x,y)\rvert^2 \quad \text{?}$$

• Just try your formulas with some simple functions, like $u(x,y)=x-y$ or $u(x,y)=xy$. – celtschk Feb 21 '17 at 9:42

## 1 Answer

Of course $$\nabla(u(x,y)\cdot u(x,y))=\nabla|u(x,y)|^2,$$ because you just rewrote the term the operator is acting on. The second one is more or less a definition. As $\nabla$ is an operator, scalar products are not defined a priori. But you define it to be the same as applying the operator twice because it works well with the way of writing $\nabla$as $(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$.

Assuming that

$$|\nabla|^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$

I would disagree one the last one. The left side contains a first derivative squared, the right side is a second derivative of a square. In general $$(f^2)''=(2ff')'=2ff''+2(f')^2\not=(f')^2.$$

Maybe the author just defined $|\nabla|^2$ in such a way, so that it agrees. But in my understanding $\nabla\cdot\nabla=|\nabla|^2$.

• Thanks! I updated my post. In a book I saw $\lvert \nabla u(x,y) \rvert ^2= \lvert \nabla\rvert^2\lvert u(x,y)\rvert^2$. Is i wrong? – JDoeDoe Feb 21 '17 at 10:13