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Invariance of the Laplacian

Problem

Let \(n\in\mathbb{N}\), \(\Omega\subset\mathbb{R}^n\) an open subset, and \(u:\Omega\to\mathbb{R}\) a two-times continuously differentiable scalar function. Additionally, let \(A = (a_{ij}) \in\mathbb{R}^{n\times n}\) denote an orthogonal matrix and \(b\in\mathbb{R}^n\) an arbitrary shift vector. Define the affine transformation \(\varphi:\mathbb{R}^n\to\mathbb{R}^n\) with \(\varphi(x)=Ax + b\) for all \(x\in\mathbb{R}^n\). In this case the Laplacian is invariant under the transformation \(\varphi\).

\[ \Delta(u\circ\varphi) = \Delta u\circ\varphi \]

Proof

We proof the proposition by computation.

\[ \Delta(u\circ\varphi) = \nabla\cdot\nabla (u\circ\varphi) \]
\[ = \nabla\cdot\left( \mathrm{D}\varphi \cdot \nabla u\circ\varphi \right) \]
\[ = \nabla\cdot\left(A \cdot \nabla u\circ\varphi \right) \]

At this point, let us go over to a simplified Einstein notation where doubled indices demand a summation from \(1\) to \(n\).

\[ = \partial_p ( a_{pq} \cdot \partial_qu\circ\varphi ) \]
\[ = a_{pq} \cdot \partial_p (\partial_q u\circ\varphi) \]
\[ = a_{pq} \cdot \partial_r \partial_q u\circ\varphi \cdot \partial_p\varphi_r \]
\[ = a_{pq} a_{rp} \cdot \partial_r \partial_q u\circ\varphi \]

Now, we apply orthogonality of \(A\) which means \(A^\mathrm{T}A = \mathbb{I}\).

\[ = \delta_{rq} \cdot \partial_r \partial_q u\circ\varphi \]
\[ = \partial_q\partial_q u\circ\varphi = \Delta u\circ\varphi \]
Last update: November 11, 2020