In books about Pseudo-differential operators, they use many times
$\triangle^k u$
but i have a question, what means this really
option A
$\displaystyle{\sum_{|\alpha|=k}}{\, \frac{k!}{\alpha!} \left({\frac{\partial^2 }{\partial x_{1}^2}}\right)^{\alpha_{1}} \! \! \! \! ...\left({\frac{\partial^2 }{\partial x_{n}^2}}\right)^{\alpha_{n}} \! \! \! u(x)} $
option B
$\displaystyle{\sum_{|\alpha|=k}}{\, \frac{k!}{\alpha!} \left({\frac{\partial^{2\alpha_{1}} }{\partial x_{1}^{2\alpha_{1}}}}\right)...\left({\frac{\partial^{2\alpha_{n}} }{\partial x_{n}^{2\alpha_{n}}}}\right) u(x)} $
Option A means: for each $i$, take $\left({\dfrac{\partial^2 }{\partial x_{i}^2}}\right)$ and make $\left({\dfrac{\partial^2 }{\partial x_{i}^2}}\right)...\left({\dfrac{\partial^2 }{\partial x_{i}^2}}\right)$, $\alpha_{i}$ times, (just multiply)
Option B means: for each $x_{i}$ take derivate $2\alpha_{i}$-th
thanks so much