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Consider the scalar fields $f(\theta, \phi)$ and $g(\theta, \phi)$ defined on the unit surface $S^2$, where $\theta$ is the co-latitude ($0$ at the north pole and $\pi$ at the south) and $\phi$ is the length. Show or a found counterexample such that:

$$\int_{S^2} (\nabla_s g \circ \nabla_s g)f(\theta, \phi)\sin\theta d\theta d\phi=-∫_{S^2}(fg∇^2_sg+g∇_sf \circ \nabla_sg)\sin\theta d\theta d\phi$$

where $\nabla_s$ is the operator gradient on $S^2$ in spherical coordinates $(\theta, \phi)$ $\circ$ denotes the vector or dot product and $\nabla^2_s$ is the Laplacian on $S^2$ (Hint: use Cartesian coordinates).

Usig gradiennt in spherical coordinate and doing dot product I have this: $$\int_{S^2} (\nabla_s g \circ \nabla_s g)f(\theta, \phi)\sin\theta d\theta d\phi=\int_{S^2}(\dfrac{\partial g}{\partial \phi})^2 f(\theta, \phi)sin\theta d\theta +\int_{S^2}(\dfrac{\partial g}{\partial \theta})^2\dfrac{1}{(\sin\phi)^2}f(\theta, \phi)sin\theta d\theta d\phi$$ please help I dont know wheter I'm right or wrong and how to continue, Do I use dot point or vector product?, when and how use the cartesian coordinates.

Hugus
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What I would do is to write $$ \int_{S^2} f(\theta,\phi) \sin(\theta) \, d\theta \, d\phi = \int_{\mathbb R^3} f(x) \rho(x) \, dx ,$$ where $\rho(x)$ is an integrable radially symmetric function for which $\int_{\mathbb R^n} \rho(x) \, dx = 4\pi$ and $\rho(x) = 0$ is a neighbourhood of the origin, and $f(x)$ is extended to $\mathbb R^3$ by the formula $f(x) = f(x/|x|)$. Then apply the standard integration by parts formulas.

We did this in the appendix of this paper: http://www.math.missouri.edu/~stephen/preprints/pde-sphere.html

Stephen Montgomery-Smith
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  • I know that I can use Green's Identities pero I don't how to... please help me... – Hugus Nov 09 '13 at 01:58
  • Just do integration by parts. If you have a functions $f(x)$ and $g(x)$ defined on $\mathbb R^n$, and $f(x)$ and $g(x)$ are zero for $|x|>R$, then $\int_{\mathbb R^n} (\nabla f) g , dx = - \int_{\mathbb R^n} f (\nabla g) , dx$. – Stephen Montgomery-Smith Nov 09 '13 at 15:20
  • And Green's identities, Stokes Theorem, etc - when you get down to it, all they are is integration by parts, or the fundamental theorem of calculus. – Stephen Montgomery-Smith Nov 09 '13 at 15:22