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If I have function $u: \mathbb{R}^n \longrightarrow \mathbb{R}$ smooth, does it always hold that: $$\nabla^2(\nabla u)= \nabla(\nabla^2 u)$$

this is true in general?

Porufes
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  • Gradient of $u$ is a vector field in $\mathbb{R}^n$. How do you define Laplacian of a vector field $X=(X_1,...,X_n)$? Do you define it as $\Delta X=(\Delta X_1,...,\Delta X_n)$? – Paul Feb 16 '14 at 23:41
  • yep well at least in cartesian coordinates, http://en.wikipedia.org/wiki/Vector_Laplacian – Porufes Feb 17 '14 at 02:31

1 Answers1

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This is true for a $C^3$ function defined on $\mathbb{R}^n$. To see this, note that $$\tag{1}\Delta (\nabla u)=\Delta\Big(\frac{\partial u}{\partial x_1},\frac{\partial u}{\partial x_2},...,\frac{\partial u}{\partial x_n}\Big) =\Big(\Delta(\frac{\partial u}{\partial x_1}),\Delta(\frac{\partial u}{\partial x_2}),...,\Delta(\frac{\partial u}{\partial x_n})\Big).$$ On the other hand, we have $$\tag{2}\nabla(\Delta u)=\Big(\frac{\partial }{\partial x_1}(\Delta u),\frac{\partial}{\partial x_2}(\Delta u),...,\frac{\partial}{\partial x_n}(\Delta u)\Big).$$

Since $u$ is $C^3$, we have for $1\leq i\leq n$ $$\Delta(\frac{\partial u}{\partial x_i})= \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2}(\frac{\partial u}{\partial x_i})=\sum_{j=1}^n\frac{\partial^3u}{\partial x_j^2\partial x_i}\\ =\sum_{j=1}^n\frac{\partial^3u}{\partial x_i\partial x_j^2} =\frac{\partial}{\partial x_i}\Big(\sum_{j=1}^n\frac{\partial^2u}{\partial x_j^2}\Big) =\frac{\partial}{\partial x_i}(\Delta u), $$ which implies that $$\Delta (\nabla u)=\nabla(\Delta u)$$ by $(1)$ and $(2)$.

Paul
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