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I know that in physics the divergence of a vector over a volume is equal to the dot product of the vector by the normal to the surface, over the surface, ie:

$$\int_{\Omega} \nabla \cdotp \vec{u}\ d\Omega = \int_{\Gamma} u\ \cdotp n\ d\Gamma$$ In my case my vector is the gradient of a scalar function $f(\alpha)$ depending of the scalar $\alpha$ and i'm in the following case: $$\int_{\Omega} \alpha\nabla \cdotp \nabla f(\alpha)\ d\Omega$$ I've a scalar inside on my integrale, infront of my divergence. Can i write: $$\int_{\Omega} \alpha\nabla \cdotp \nabla f(\alpha)\ d\Omega = \int_{\Gamma} \alpha \nabla f(\alpha)\cdotp n\ d\Omega $$ or is this false ? Note that $\alpha$ is variable through $\Omega$. Thanks.

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  • $\begingroup$ If $\alpha$ is a scalar, then how is $\nabla f(\alpha)$ defined? Do you mean $f$ is a scalar function of the vector $\alpha$, where $\alpha$ is in $\Omega$? $\endgroup$ – runway44 Jun 13 at 5:36
  • $\begingroup$ $\alpha$ is a scalar, varying through my pysicial space $\Omega$. $f$ is a scalar function. $\nabla f(\alpha)$ is the vector ${\frac{\partial f(\alpha)}{\partial x}\vec{e_x}\ \ \frac{\partial f(\alpha)}{\partial y}\vec{e_y}\ \ \frac{\partial f(\alpha)}{\partial z}\vec{e_z}}$ $\endgroup$ – pomme Jun 13 at 5:46
  • $\begingroup$ If your physical space $\Omega$ is comprised of scalars, then is $\Omega$ just a subset of the real number line $\Bbb R$? I.e. it is one-dimensional? Or do you mean $\alpha$ is itself is a scalar function of the vector in $\Omega$, as in $\alpha(x,y,z)$? $\endgroup$ – runway44 Jun 13 at 5:50
  • $\begingroup$ Yes, $\alpha$ is a scalar function of the vector in Ω, as in α(x,y,z) $\endgroup$ – pomme Jun 13 at 5:54
  • $\begingroup$ That's something you should mention in your question. And then the answer is that your theorem does not imply your desired equality, since the theorem does not have any scalar function in front of the divergence. $\endgroup$ – runway44 Jun 13 at 6:07

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