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I have the following expression (in a numerical context)

$$\Delta_h u(x) = \Delta u(x) + \frac{h^2}{12} \Delta^2 u(x) + O(h^4)$$

The $\Delta$ is the Laplace operator so $\Delta u = u_{xx}+u_{yy}$.

But what is $\Delta^2$?

In the context it would make sense (but it is not really a strong indication) for it to be $$\Delta^2 u = u_{xxxx}+u_{yyyy}$$ but when I first saw it I just thougth it was

$$\Delta^2 u = \Delta (\Delta u) = \Delta (u_{xx}+u_{yy}) = u_{xxxx}+2u_{xxyy}+u_{yyyy}.$$

Which one is correct? Can you provide references where one or the other is used?

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$\nabla^4$ is called the biharmonic operator, (http://mathworld.wolfram.com/BiharmonicEquation.html, http://mathworld.wolfram.com/BiharmonicOperator.html) and is used in the theory of elasticity and in approximation theory.

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  • $\begingroup$ What is the biharmonic operator? The question mentions two operators... (I know the link spells out which one, but it is generally best not to write answers which do not mean anything without a link, for the link might well cease to work in the future) $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '14 at 5:59
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$$\Delta f =\nabla \cdot (\nabla f)$$ Set $g=\Delta f=u_{xx}+u_{yy}$, then $$\Delta ^2 f=\Delta g=\nabla\cdot (\nabla g)=\nabla \cdot (\nabla (u_{xx}+u_{yy}))= \nabla \cdot (\nabla u_{xx}))+\nabla \cdot (\nabla u_{yy})=\underbrace{\Delta u_{xx}}_{u_{xxxx}+u_{xxyy}}+\underbrace{\Delta u_{yy}}_{=u_{yyxx}+u_{yyyy}}=u_{xxxx}+2u_{xxyy}+u_{yyyy}$$

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  • $\begingroup$ Thats exacly what I've written in my question, but can you provide a reference where it is used that way? $\endgroup$ – flawr Nov 23 '14 at 14:51
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    $\begingroup$ @flawr, if $L$ is an operator, $L^2$ denotes rather universally the composition of $L$ with itself. $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '14 at 6:00

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