The inner product in Sobolev space defined as
$$ \langle u,v\rangle _{H^m(\Omega)} = \int_{\Omega}\sum_{\alpha=0}^m \sum_{\beta=\alpha} D^{\beta}uD^{\beta}v d\Omega $$
where
$$D^{\alpha}u = \frac{\partial^{\alpha}u}{\partial^{\alpha_1}x_1\partial^{\alpha_2}x_2…\partial^{\alpha_n}x_n} $$
$$ \alpha_1+…\alpha_n =\alpha$$
So for example assume $ \Omega \subset R^2; m=1$ then we have
$$ \langle u,v\rangle _{H^1(\Omega)} = \int_{\Omega}uv + \frac{\partial u}{\partial x}\frac{\partial v}{\partial x} +\frac{\partial u}{\partial y}\frac{\partial v}{\partial y} d\Omega $$
if $m=2$
$$ \langle u,v\rangle _{H^2(\Omega)} = \int_{\Omega}uv + \frac{\partial u}{\partial x}\frac{\partial v}{\partial x} +\frac{\partial u}{\partial y}\frac{\partial v}{\partial y} +
\frac{\partial^2 u}{\partial x^2}\frac{\partial^2 v}{\partial x^2} +
\frac{\partial^2 u}{\partial y^2}\frac{\partial^2 v}{\partial y^2} +
\frac{\partial^2 u}{\partial x \partial y}\frac{\partial^2 v}{\partial x \partial y} d\Omega $$
My Question is why don't we need 2 before $ \frac{\partial^2 u}{\partial x \partial y}\frac{\partial^2 v}{\partial x \partial y} $ because we have another partial derivative $ \frac{\partial^2 u}{\partial y \partial x}\frac{\partial^2 v}{\partial y \partial x} $. Am I missing the point?
Also how this integral will look if $m=3$?
Best Answer
Yes, the 2 should be needed if you want to write it like that. That is because in the definition of weak derivative you "discharge" the derivatives on a function that is $\mathcal{C}^{\infty}_0$ and thus regular enough to have symmetry of partial derivatives.
If $m=3$, then you have to keep combining derivatives, i.e. $\frac{\partial^3u}{\partial x^3}\frac{\partial^3v}{\partial x^3}, \frac{\partial^3u}{\partial x^2 \partial y}\frac{\partial^3v}{\partial x^2 \partial y},$ and so on.