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I'm trying to wrap my head around this problem - the interplay between $\nabla$ and $\Delta$ is doing my head in. It says to use the divergence theorem.

Prove that

$$\int_\Omega u \cdot \Delta v\, dx + \int_\Omega \nabla u \cdot \nabla v \,dx = \int_{\delta \Omega}u\nabla v \cdot n\, d\sigma$$

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It seems reasonable to compute $\nabla\cdot(u\nabla v)$ (where $u\nabla v$ is found in the boundary integral) and compare it with what is integrated on the left hand side. –  Harald Hanche-Olsen Oct 6 '12 at 7:37
    
That works, provided that I can then use the product rule for the divergence of scalar valued functions. That is, provided that u is a scalar and v is a vector field. This isn't specified, but it's also not stated, so I'm happy with this line of logic. Thanks greatly for your hint –  Ronan Oct 6 '12 at 9:40
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