Attached is a copy of p181 of Strauss Partial Differential Equation. This is part of proof of Mean Value Property using Green's 1st identity:
[tex]\int_Dv\nabla u+\nabla v\cdot\nabla u \;dV=\oint_A v\frac {\partial u}{\partial r}dA[/tex]
Let ##v=1## and let ##u(r,\theta,\phi)## be a harmonic function ie ##\nabla^2 u=0## and ##u## has continuous 1st and 2nd partial derivatives.
[tex]\Rightarrow\;\oint_A \frac {\partial u}{\partial r}dA=0[/tex]
This is (5) on the top of the copy of the book.
If you follow the steps, the next step shows integration of
[tex]\int_0^{2\pi}\int_0^{\pi}\frac {\partial u(a,\theta,\phi)}{\partial r}\;a^2\;\sin\theta\; d\theta \;d\phi\;=0,\hbox{ Where the integration is on surface at }\;r=a[/tex]
My question is in the very next step when the book claimed since it is equal to zero, then ##u(a,\theta,\phi)## is not a function of ##r##, therefore you can move the ##\frac{\partial}{\partial r}## outside the integral. That's all reasonable. But then the next line you see the equation using ##u(r,\theta,\phi)##, which is function of ##r##!! How can you do that?
Please explain to me why, Thanks
Alan
[tex]\int_Dv\nabla u+\nabla v\cdot\nabla u \;dV=\oint_A v\frac {\partial u}{\partial r}dA[/tex]
Let ##v=1## and let ##u(r,\theta,\phi)## be a harmonic function ie ##\nabla^2 u=0## and ##u## has continuous 1st and 2nd partial derivatives.
[tex]\Rightarrow\;\oint_A \frac {\partial u}{\partial r}dA=0[/tex]
This is (5) on the top of the copy of the book.
If you follow the steps, the next step shows integration of
[tex]\int_0^{2\pi}\int_0^{\pi}\frac {\partial u(a,\theta,\phi)}{\partial r}\;a^2\;\sin\theta\; d\theta \;d\phi\;=0,\hbox{ Where the integration is on surface at }\;r=a[/tex]
My question is in the very next step when the book claimed since it is equal to zero, then ##u(a,\theta,\phi)## is not a function of ##r##, therefore you can move the ##\frac{\partial}{\partial r}## outside the integral. That's all reasonable. But then the next line you see the equation using ##u(r,\theta,\phi)##, which is function of ##r##!! How can you do that?
Please explain to me why, Thanks
Alan
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