Notation laplace operator squared $\Delta^2$

Question

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?

Answer 1

$\nabla^4$ 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.

Answer 2

$$\Delta f =\nabla \cdot (\nabla f)$$ Set $g=\Delta f=u_{xx}+u_{yy}$, then $$\Delta ^2 f=\Delta g=\nabla\cdot (\nabla g)=\nabla \cdot (\nabla (u_{xx}+u_{yy}))= \nabla \cdot (\nabla u_{xx}))+\nabla \cdot (\nabla u_{yy})=\underbrace{\Delta u_{xx}}_{u_{xxxx}+u_{xxyy}}+\underbrace{\Delta u_{yy}}_{=u_{yyxx}+u_{yyyy}}=u_{xxxx}+2u_{xxyy}+u_{yyyy}$$