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Is there a easy composition Rule for laplacian?

$u : R^2 \rightarrow R$

$I : R^2 \rightarrow R^2$

so? $Δu(I(x))=...$

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1 Answer 1

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We just compute, using the chain rule. We write $I^j$ for the $j$-th component $I^j \colon \mathbb R^d \to \mathbb R$ of $I$, that is $I = (I^1, \ldots, I^d)$. \begin{align*} \Delta(u \circ I)(x) &= \sum_{i=1}^d \partial_i^2(u \circ I)(x)\\ &= \sum_{i=1}^d \partial_i \left(\sum_{j=1}^d \partial_j u \circ I \cdot \partial_i I^j\right)(x)\\ &= \sum_{i=1}^d\left(\sum_{j,k=1}^d (\partial_k\partial_j u \circ I)(x) (\partial_k u\circ I)(x)\cdot \partial_i I^j(x)\cdot \partial_i I^k(x) + \sum_{j=1}^d (\partial_j u \circ I)(x) \cdot \partial_i^2 I^j\right) \end{align*} So, if we write $Du$ for the gradient and $D^2u$ for the Hessian of $u$, $\partial_i I \colon \mathbb R^d \to \mathbb R$ and $\Delta I \colon \mathbb R^d \to \mathbb R$ for the componentwise partial derivative and Laplacian of $I$, we have (${}^t$ denoting transpose) $$ \Delta(u \circ I)(x) = \sum_{i=1}^d \partial_i I(x)^t D^2 u\bigl(I(x)\bigr)\partial_i I(x) + Du\bigl(I(x)\bigr)^t \Delta I(x) $$

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Please, break long expressions into more lines. Thanks. MathJax or browser truncates it. –  vesszabo Nov 30 '12 at 13:01

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