Ngô Quốc Anh

March 7, 2011

An identity involving inner product of gradients

Filed under: Uncategorized — Ngô Quốc Anh @ 22:41

In this short note, we prove the following identity

For any functions $\eta, u$, it holds

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = |\nabla (\eta u){|^2} - {u^2}|\nabla \eta {|^2}.$

The proof is elementary. By the product rule for the gradient, we know that

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \nabla u \cdot \big((\eta u)\nabla \eta + \eta \nabla (\eta u)\big).$

Thus

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \eta u\nabla \eta \cdot \nabla u + \eta \nabla u \cdot \nabla (\eta u).$

The term $\eta \nabla u \cdot \nabla (\eta u)$ can be rewritten as follows

$\displaystyle\eta \nabla u \cdot \nabla (\eta u) = \nabla (\eta u) \cdot \nabla (\eta u) - u\nabla \eta \cdot \nabla (\eta u) = |\nabla (\eta u){|^2} - u\nabla \eta \cdot \nabla (\eta u).$

We then have

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = \eta u\nabla \eta \cdot \nabla u + |\nabla (\eta u){|^2} - u\nabla \eta \cdot \nabla (\eta u).$

Keep in mind that

$\displaystyle\eta u\nabla \eta \cdot \nabla u - u\nabla \eta \cdot \nabla (\eta u) = - u\nabla \eta \cdot (\nabla (\eta u) - \eta \nabla u) = - u\nabla \eta \cdot (u\nabla \eta ).$

Therefore

$\displaystyle\nabla u \cdot \nabla ({\eta ^2}u) = |\nabla (\eta u){|^2} - {u^2}|\nabla \eta {|^2}.$