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Need to prove $\operatorname{div}(\nabla u)=\nabla ^2 u$ where $u=g(x,y,z)$

The RHS is the Lapacian which we were told is a vector. But $\nabla u=(g_x,g_y,g_z)$ and the divergence of that is $g_{xx}+g_{yy}+g_{zz}$ which is not a vector. I don't get how it can equate then...

Siminore
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snowman
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  • The Laplacian takes a scalar valued function and gives back a scalar valued function. If the function is vector valued, then its Laplacian is vector valued. I abhor the del squared notation that you've used for this reason. It's completely incorrect notation and it can be confusing. – Cameron Williams Apr 28 '15 at 14:02
  • Sorry I don't understand what you are saying. We were told that if we have a differentiable scalar field $u$, the Laplacian of $u$ is $$\nabla ^2 u = \bigg( \frac{\partial ^2 u}{\partial x^2},\frac{\partial ^2 u}{\partial y^2},\frac{\partial ^2 u}{\partial z^2} \bigg)$$ – snowman Apr 28 '15 at 14:04
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    That is not correct at all. What you wrote as the Laplacian in your post is correct. – Cameron Williams Apr 28 '15 at 14:05
  • Clearly my lecturer wrote the completely wrong thing then. Thanks. – snowman Apr 28 '15 at 14:42
  • @CameronWilliams Would it interest you to know that $\nabla^2 = \nabla \cdot \nabla + \nabla \wedge \nabla$ (which is completely analogous to the vector square $\vec a^2 = \vec a\cdot \vec a + \vec a \wedge \vec a$) in geometric algebra/ calculus? So at least in that system, it makes a lot of sense to me. (Note that if you have equality of mixed partials, then $\nabla \wedge \nabla F = 0$ for any multivector field $F$ and thus this reduces to the regular expression for $\nabla^2$) –  Apr 28 '15 at 14:53
  • @Bye_World I'm aware of this but when $\nabla^2$ is written, it is usually meant as $\Delta$, i.e. $\nabla\cdot\nabla$. – Cameron Williams Apr 28 '15 at 16:00

2 Answers2

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The "Laplacian" is an operator that can operate on both scalar fields and vector fields. The operator on a scalar can be written,

$$\nabla^2 \{\} = \nabla \cdot (\nabla \{\})$$

which will produce another scalar field.

The operator on a vector can be expressed as

$$\nabla^2 \{\} = \nabla (\nabla \cdot \{\})\,\,-\nabla \times (\nabla \times \{\})$$

which will produce another vector field.

In Cartesian coordinates, both operators can be written

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

where it is evident that operation on a scalar (vector) field transforms into a scalar (vector) field.

Mark Viola
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  • I saw once that the laplacian was written like $\Delta { }$, is this true? – snowman Apr 28 '15 at 14:44
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    @snowman Yes, some use the notation "$\Delta$" to represent the Laplacian. – Mark Viola Apr 28 '15 at 14:45
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    What about vector fields in dimension more than 3? – Hesam May 07 '17 at 13:15
  • @user5y292 The OP's question is about the operator on $\mathbb{R}^3$. – Mark Viola May 07 '17 at 17:29
  • I'm struggling with the same question. In the last equation $\nabla^2{}$ acts on a 3-dimensional vector and returns a sum which is a scalar. So it seems that the operation on a vector returns a scalar, which doesn't agree with your last sentence. What am I not understanding? – jeffery_the_wind Dec 03 '20 at 14:18
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Lapacian of an Nth Rank Tensor is another Nth Rank Tensor. This means Lapacian of a Scalar Field is another Scalar Field. Laplacian of Vector Field is another Vector Field and so on.

Arup Hore
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