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}.

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