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I learned the proof of the fact that 3-D laplacian is invariant under all rigid motions in space in Strauss's Partial Differential Equations:

Any rotation in three dimensions is given by $${\bf x'}=B{\bf x},$$ where $B$ is an orthogonal matrix. The laplacian is $$\triangle u=\sum_{i=1}^3 u_{ii}=\sum_{i,j=1}^3\delta_{ij}u_{ij}$$ where the subscripts on $u$ denote partial derivatives. Therefore, $$\triangle u=\sum_{k,l}\Big(\sum_{i,j}b_{ki}\delta_{ij}b_{lj}\Big)u_{k'l'}=\sum_{k,l}\delta_{kl}u_{k'l'} =\sum_{k}u_{k'k'}.$$

Added: I believe the chain rule is needs here, but I don't see how it is applied here.

Here is my question:

How can I get the first equality, i.e. $$\triangle u=\sum_{k,l}\Big(\sum_{i,j}b_{ki}\delta_{ij}b_{lj}\Big)u_{k'l'}?$$

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The factor $\sum_{i,j} b_{ki} \delta_{ij} b_{lj}$ is just the (k,l)th entry of the matrix $BIB^{-1}$ (because $B^{-1} = B^T$ when B is orthogonal), i.e. $I$, and so is $\delta_{kl}$, though I don't quite understand what k' and l' mean that's different to k and l, so maybe that doesn't answer your question... –  Billy Aug 1 '11 at 17:14
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The first equality is the chain rule. (If this is not obvious, perhaps the notation needs to be improved.) –  Zhen Lin Aug 1 '11 at 17:15
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I would think the whole thing could be written in matrix notation, without these sums and indices. –  Michael Hardy Aug 1 '11 at 17:33
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It looks to me that you are trying to show something like $B^{-1} \triangle ~ B = \triangle $. –  Alice Aug 1 '11 at 17:35

1 Answer 1

up vote 3 down vote accepted

I strongly dislike the use of undecorated subscripts for partial derivatives, so I will write $\partial_j$ for the partial differentiation operator $\partial / \partial x_j$. For convenience I will use the summation convention for repeated indices.

Firstly, we have the change-of-coordinates equation $\tilde{x}_{i'} = B_{i' j} x_j$. The chain rule states $$\frac{\partial}{\partial x_j} = \frac{\partial \tilde{x}_{i'}}{\partial x_j} \frac{\partial}{\partial \tilde{x}_{i'}}$$ so, writing $\tilde{\partial}_{i'}$ for the operator $\partial / \partial \tilde{x}_{i'}$, putting these together, we have $\partial_j = B_{i' j} \tilde{\partial}_{i'}$. The Laplacian is defined by $\nabla^2 = \delta_{i j} \partial_i \partial_j$, so substituting the previous equation in we obtain $\nabla^2 = \delta_{i j} B_{i k'} B_{j l'} \tilde{\partial}_{k'} \tilde{\partial}_{l'}$. By orthogonality, $\delta_{i j} B_{i k'} B_{j l'} = B_{j k'} B_{j l'} = \delta_{k' l'}$, so we are done.

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+1. Thanks for your proof. It's much clear than the one in the book for me. I still curious about my original question:how can one get the first equality? –  Jack Aug 2 '11 at 4:50
    
@Jack: It's the bit starting at ‘substituting the previous equation in’. It is literally just the chain rule. –  Zhen Lin Aug 2 '11 at 6:11

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