# Bi Laplacian operator explicitly form

let $u \in H^2(\mathbb{R}^n)$. My question is: please how we can write explicitly the operator $$u- \Delta u + \Delta^2 u?$$ We can write it into the form $\sum_{|\alpha|\leq 2} (-1)^{|\alpha|} D^{2 \alpha }u?$

## 1 Answer

What you are writing is not the operator. The operator is $$\mathcal{L} = 1-\Delta +\Delta^2$$ and it acts on functions $u\in H^2(\mathbb{R}^n)$. You are also missing coefficients in your sum. It should be $$\mathcal{L}=\sum_{\alpha\leq2} a_{\alpha}(-1)^{|\alpha|} D^{2\alpha}.$$ From there, assuming the notation $D_j = \partial_{x_j}$, you have $\mathcal{L}$ as written above with $a_{I_0}=1$ where $I_0 = 0\in\mathbb{R}^n$ (or all multi-indices of length 0), $a_{I_1}=1,$ where $I_1$ is all multi-indices of length 1, and $a_{I_4}=1, a_{I_2}=2$, where $I_2$ is of the form $I_2 = (0,\cdots,0,1,0,\cdots,0,1,0)$ (i.e. two of the indices are 1) and $I_4$ is of the form $I_4=(0,\cdots,0,2,0,\cdots,0)$ (i.e. only one of the indices in the multi-index is 2 and the rest are 0).

• please how we write $\Delta^2 u$ explicitly in dimension $n=2$? Commented Mar 22, 2018 at 19:28
• $\Delta^2 = \partial_{x_1}^4 + \partial_{x_2}^4 + 2\partial_{x_1}^2\partial_{x_2}^2$. Commented Mar 22, 2018 at 19:37
• Thabk you so much Commented Mar 22, 2018 at 20:44