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When Evans goes through an analysis of the "heat ball" he evaluates the following integral:

$$ B = r^{-n-1}\int\int_{E(r)}4u_s\sum_{i=1}^{n}y_i\psi_{y_i} dyds$$ $$ = - r^{-n-1}\int\int_{E(r)}4nu_s\psi +4\sum_{i=1}^{n}u_{sy_i}y_i\psi dyds$$

Where $E(r)$ defines a "ball-like" region in $\mathbf{R}^{n+1}$ space and $\psi$ is zero on the boundary of this region. It's some integral by parts formula - but I'm not seeing it. Any thoughts?

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    Evaluate the integral term by term. Integrate by parts - the boundary term vanishes by the boundary condition and you'll get a sum by using the product rule on $u_sy_i$. I bet when you pull the integral back out of the sum and regroup you'll get the Rhs of your equality. – 2017-02-22
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    @Neal I After IBP I get: $-4r^{-n-1}\int\int_{E(r)}\sum^{n}_{i=1}u_{sy_{i}}\psi=-4r^{-n-1}\int\int_{E(r)}\psi\sum^{n}_{i=1}[- u_s + (u_sy_i)_{y_i}]$. The first term matches up to sign. The second term still doesnt match. – 2017-02-23
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    You want to do integration by parts on the triple product $u_s y_i \psi$. – 2017-05-03

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