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Let $U \subset \Bbb R^2$ be an open set, $f \in C^4(U), g \in C_c^2(U)$ non-negative functions and

$$\Delta^2 : C^4(U) \rightarrow C(U),$$ $$\Delta^2 := {d^4 \over dx^4} + 2 {d^4 \over dx^2dy^2} + {d^4 \over dy^4},$$

be the biharmonic operator. Show that

$$\int_U (\Delta^2 f) g \ d\lambda^2 = \int_U (\Delta f) (\Delta g) d\lambda^2.$$

We can figure out that

$$\int_U (\Delta f) (\Delta g) d\lambda^2 = \int_U {d^2f \over dx^2} {d^2g \over dx^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dy^2} {d^2g \over dx^2} + {d^2f \over dy^2} {d^2 g \over dy^2}.$$

If it is necessary, I'll add more steps, but for the sake of clarity, just let me say that for the left side, I used the linearity of the integral and then pushed a part of the differential operators that stuck to $f$ at first on $g$, and by using the linearity again, I receive:

$$\int_U {d^2f \over dx^2} {d^2g \over dx^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dy^2} {d^2 g \over dy^2}.$$

Everything seems correct except the third term. I received ${d^2f \over dx^2} {d^2g \over dy^2}$, but I need ${d^2f \over dy^2} {d^2g \over dx^2}$.

Does anyone see "directly" where I am wrong here?

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    You must use integration by parts or, even better, the divergence theorem. Then you will use the assumption that $g$ has compact support. – 2017-02-02
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    Which one would be easier? It's an old exam question, and integration by parts sounds a little bit hard in this case (considering that this excercise yielded only a few points). – 2017-02-02

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As your function g is compactly supported, we can find a ball $B=B_R(0)$ such that $supp(g) \subset B$. As a consequence, we have $g|_{\partial B}=0$ including its derivatives. Also, note that $supp(fg) \subset supp(g)$. We know want to consider the divergence theorem on $B$, which has a sufficiently smooth and compact boundary. Note that the right and the left hand side of the statement you are trying to proof both become $0$ outside of $B$.

It is a consequence of the divergence theorem, which has Greens identity as a consquence. To see that: $$ div(g \nabla \Delta f)=<\nabla g, \nabla \Delta f>+g\Delta^2f $$ Now that g is $C^2_C$, we can use the divergence theorem and the the usual greens identity (note that $g=0$ on the boundary as well as close to it): $$ 0=\int_{\partial B} fg d\vec{S}=\int_B div(g \nabla \Delta f)= \\\int_B<\nabla g, \nabla \Delta f>+g\Delta^2f= \\ -\int_B \Delta f \Delta g + \int_{\partial B} (\partial_v g) \Delta f +\int_B g \Delta^2 f $$ Then, we see that for any function $\phi$ s.t. the follwing expresion is integrable: $$ \int_U \phi g=\int_{supp(g)} \phi g=\int_B \phi g $$ Rearranging and using that the support of $g$ is within B and U gives you:
$$ \int_U \Delta g \Delta f=\int_U g \Delta^2f $$ Please doublecheck if I did make a mistake as well. Edit: I just noted that you should set $f,g=0$ outside of $U$, so everything you do is well defined

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    Our professor told us that it wouldn't be possible to solve this with the divergence theorem for some reason. – 2017-02-09
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    Ah okay, I didnt double check the requirements. Our set $U \subset \mathbb{R}^2$ might not have a boundary/be not smooth enough there. Ill leave this answer up for discussion purposes and put a disclaimer on it. – 2017-02-09
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    I edited something and a possible solution, I will return to it later. – 2017-02-09
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    @F.Conrad: Indeed, just enclose $\text{supp } g$ by some hypersurface $S \subset U$ and let $V$ be the set of points inside (but not on) $S$. Apply the corollary to the divergence theorem, using the fact that $g=0$ outside of $V$ (and, in particular, $g \big| _S = 0$). – 2017-02-09
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    I salvaged the proof for now. See if it fits your needs @Julian – 2017-02-09
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    Thanks for our effort! I can't point out a mistake for now, but as I said, my professor clearly stated that this couldn't be solved by applying the divergence theorem. Was he wrong there? – 2017-02-10
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    Maybe he meant that you cant do it straight-forward, but you need some form of integration by parts/fundamental theorem of calculus. If you find the time, Id be interested to know why it you cant do it with the divergence theorem. – 2017-02-10
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    Hm, that's strange then. Do you have any other solution in mind that doesn't use the divergence theorem? I have tried to look it up on the internet, but didn't find anything. – 2017-02-10
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    I cant think of any "simple" solution that doesnt use any form of integration by parts, which requires some form of boundaries or that $fg$ is compactly supported. Problem is that you dont really know anything about $U$ except that the set is open. – 2017-02-10