Is there a formula for $ (I - \Delta)^k (fg) $ where $ \Delta $ is the Laplacian and $k \in Z $ similar to the Newton-Leibnitz formula for the derivatives of a product?
Action of $ (I - \Delta)^k$ on a product of functions
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1$I$ and $-\Delta$ commute so you can expand $(I - \Delta)^k$ using the binomial theorem. There is also product formula for the Laplacian $\Delta(fg) = f\Delta g + 2 \nabla f \cdot \nabla g + g \Delta f$. I think you should be able to derive a formula using these two facts (though it might be messy). – 2012-10-23
1 Answers
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Let's assume we are in $\mathbb{R}^d$ and $\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_d)\in\mathbb{N}^d$ is a multi-index. We define the differential operator \begin{equation} D^\alpha=(\frac{\partial}{\partial x_1})^{\alpha_1}(\frac{\partial}{\partial x_2})^{\alpha_2}\cdots(\frac{\partial}{\partial x_d})^{\alpha_d}. \end{equation}
Then we have the Leibniz formular \begin{equation} D^{\alpha}(fg)=\sum_{\beta \le \alpha} \frac{\alpha !}{\beta !(\alpha-\beta)!}D^{\beta}fD^{\alpha-\beta}g, \end{equation} where $\beta \le \alpha$ if and only if $\beta_j\le\alpha_j$ for each $j$ and $\alpha !=(\alpha_1!)(\alpha_2!)\cdots(\alpha_d!)$.
With this, you can just expand and apply the formula, (as mentioned in the comment, $I$ commutes with everything).