Shentao YANG

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Question about the deviation of laplacian
« on: November 05, 2016, 06:51:32 PM »
Can any one explain in detail to me how we get these two formula in section 6.3:
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.3.html

$$\int\!\!\!\int \Delta u \cdot v\,dxdy = - \int\!\!\!\int \nabla u \cdot \nabla v\,dxdy$$

$$\int\!\!\!\int\!\!\!\int \Delta u \cdot v{\rho ^2}\sin (\phi )\,d\rho d\phi d\theta = - \int\!\!\!\int\!\!\!\int ( {u_\rho }{v_\rho } + {1 \over {{\rho ^2}}}{u_\phi }{v_\phi } + {1 \over {{\rho ^2}\sin (\phi )}}{u_\theta }{v_\theta }){\rho ^2}\sin (\phi )\,d\rho d\phi d\theta = \int\!\!\!\int\!\!\!\int ( {({\rho ^2}\sin (\phi ){u_\rho })_\rho } + {(\sin (\phi ){u_\phi })_\phi } + {({1 \over {\sin (\phi )}}{u_\theta })_\theta })v\,d\rho d\phi d\theta .$$

By the way, I think the equation $(6)'$ in
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.3.html#mjx-eqn-eq-6.3.6
is wrong, I guess the denominator of the last term should be ${\rho ^2}{\sin ^2}(\varphi )$ instead of ${\rho ^2}{\sin}(\varphi )$

Victor Ivrii

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Re: Question about the deviation of laplacian
« Reply #1 on: November 05, 2016, 09:21:20 PM »
The first equality is due to integration by parts and Gauss formula. The second  equation is the first one rewritten in spherical coordinates. Please check again if there is any misprint

Shentao YANG

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Re: Question about the deviation of laplacian
« Reply #2 on: November 05, 2016, 11:10:46 PM »
Misprint at where? I guess the link is pointing to a wrong equation, equation $(6)$ is correct, but $(6)'$, I guess, leave out a square.