User:Egm6321.f12.team4.harris/Homework7 R*7.6

Problem R*7.6 – Determining Laplacian of u in Spherical Coordinates Using Math/Physics Convention
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Statement
The Laplacian of u, as shown in Problem R*7.3, is written in spherical coordinates as


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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(cos\theta)^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 cos\theta} \frac{\partial }{\partial \theta}(cos\theta \frac{\partial u}{\partial \theta}) $$     (7.6.1)
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In the above equation, $$ \theta $$ is defined from the $$ r\theta $$ -plane up to a point, P, in the $$ r, \theta, \phi $$ coordinate system. In other words, it is the latitudinal definition, and it ranges from $$ \theta \in [\frac{-\pi }{2}, \frac{\pi }{2}] $$. For this problem, determine the Laplacian of u in spherical coordinates according to the mathematical and physics convention for $$ \theta $$.

Solution
The mathematical and physics convention for $$ \theta $$ defines $$ \bar{\theta} $$ from the positive $$ \phi $$ axis to the point, P, in the $$ r, \theta, \phi $$ coordinate system. In other words, it is the angle measured from the "North Pole," and it ranges from $$ \bar{\theta} \in [0, \pi ] $$. Thus one can see that the relation for transforming $$ \theta $$ to $$ \bar{\theta } $$ is


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$$ \bar{\theta } := \frac{\pi }{2} - \theta $$     (7.6.2)
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Plugging Equation 7.6.2 into Equation 7.6.1 for $$ \theta $$ yields


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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(cos(\frac{\pi }{2} - \theta))^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 cos(\frac{\pi }{2} - \theta)} \frac{\partial }{\partial (\frac{\pi }{2} - \theta)}(cos(\frac{\pi }{2} - \theta) \frac{\partial u}{\partial (\frac{\pi }{2} - \theta)}) $$     (7.6.3)
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Using the trigonometric identity


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$$ cos(\alpha + \beta ) = cos\alpha cos\beta - sin\alpha sin\beta $$     (7.6.4)
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one can see that


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$$ cos(\frac{\pi }{2} + (-\theta ) ) = \underbrace{cos(\frac{\pi }{2})}_{0} cos(-\theta )- \underbrace{sin(\frac{\pi }{2})}_{1} \underbrace{sin(-\theta )}_{-sin\theta} = sin \theta $$     (7.6.5)
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Using Equation 7.6.5, Equation 7.6.2 becomes
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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(sin\theta )^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 sin\theta } \frac{\partial }{\partial \theta}(sin\theta \frac{\partial u}{\partial \theta}) $$     (7.6.6)
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Thus Equation 7.6.6 is the Laplacian of u in spherical coordinates in the mathematical and physics convention.

Author and References

 * Solved and typed by -- Kaitlin Harris and Rui Che
 * Reviewed by -- Rui Che