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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?

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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.

So, let's answer the question. Here you can use the Green's formula

$$ \int \int \left( u\frac{\partial^2v}{\partial x \partial y}\right)dydx=\int \int \left( u\frac{\partial}{\partial y}\left(\frac{\partial v}{\partial x}\right)\right)dydx. $$ Now, you can again apply the Green's formula on $\partial v/\partial x$. I think it's easy to understand, now.

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