What is the result when Laplace operator is applied on Laplace operator, is it

$$\nabla^2(\nabla^2 r)= \nabla^4r = \frac{\partial^4 r}{\partial x^4} + \frac{\partial^4 r}{\partial y^4} + \frac{\partial^4 r}{\partial z^4} $$

or something else with more terms ?


  • $\begingroup$ I think we need to define what $\cdot$ means in this context. $\endgroup$ – manooooh Dec 12 '18 at 9:16

The way you put it is $$ \nabla^2\left( \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}\right)= \frac{\partial^4 r}{\partial x^4} + \frac{\partial^4 r}{\partial y^4} + \frac{\partial^4 r}{\partial z^4}+2\frac{\partial^4r}{\partial x^2\partial y^2} +2\frac{\partial^4r}{\partial x^2\partial z^2} +2\frac{\partial^4r}{\partial y^2\partial z^2}. $$


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