# Integration by parts with cross derivatives

I wish to solve the following simplified problem in the context of Weak Formulations

$$\large \iint(u \frac{\partial ^2 v}{\partial x ^2})dxdy + \iint(u \frac{\partial ^2 v}{\partial x \partial y})dxdy = 0$$

I know from Green's first identity we can write

$$\large \iint(u \frac{\partial ^2 v}{\partial x ^2})dxdy = \int (uv\hat{n}_x)ds - \iint (\frac{\partial v}{\partial x} \frac{\partial u}{\partial x})dxdy$$

But what about the other term? I suspect I could say $$\large \frac{dxdy}{\partial x \partial y} = 1$$ and thus

$$\large \iint(u \frac{\partial ^2 v}{\partial x \partial y})dxdy = u v$$

But I have never seen any Weak Form like $$\int_{\Omega} f(u,v) d \Omega + uv = \int_{\Gamma} g(v) d \Gamma$$, so I'm not sure. Can someone please explain it?

The one you have suspected is wrong, as take a simple example $$v=x+y$$ and $$u=1$$. Then, $$\int \int \left( u\frac{\partial^2v}{\partial x \partial y}\right)dxdy=0$$ but your right-hand side is non-zero.
$$\int \int \left( u\frac{\partial^2v}{\partial x \partial y}\right)dydx=\int \int \left( u\frac{\partial}{\partial y}\left(\frac{\partial v}{\partial x}\right)\right)dydx.$$ Now, you can again apply the Green's formula on $$\partial v/\partial x$$. I think it's easy to understand, now.