(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This shouldn't be so hard to do I guess, but I just cannot figure it out. The problem statement:

Prove that the special form of the discrete Laplacian operator in radial coordinates acting on a grid function [itex]u_{l,m}[/itex] at the central grid point [itex]l=0, m=0[/itex], given by:

[itex]\displaystyle\delta_{\Delta\circ}u_{0,0}=\frac{4}{N_{\theta}h_{r}^{2}} \sum_{m=0}^{N_{\theta}-1} \left(u_{1,m}-u_{0,0}\right)[/itex]

is a second-order accurate approximation to the continuous Laplacian. In order to do this, consider the values of the grid function [itex]u_{1,m}[/itex] to be values of a continuous function at the location [itex]x=h_{r}\cos mh_{\theta}, y=h_{r}\sin mh_{\theta}[/itex] and perform Taylor expansions about the point [itex]x=0, y=0[/itex]. To be clear: [itex]h_{r}[/itex] and [itex]h_{\theta}[/itex] denote the radial and angular grid spacings respectively.

2. Relevant equations

[itex]\displaystyle u_{1,m} = u\left(x,y\right) = u\left(h_{r}\cos mh_{\theta}, h_{r}\sin mh_{\theta}\right)[/itex]

[itex]\displaystyle u\left(x,y\right) = u\left(0,0\right) + x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + \frac{1}{2!} \left(x^{2}\frac{\partial^{2}u}{\partial x^{2}} + 2xy \frac{\partial^{2}u}{\partial x\partial y} + y^{2}\frac{\partial^{2}u}{\partial y^{2}}\right) + \dots[/itex]

[itex]\displaystyle \frac{\partial u}{\partial x} = \cos mh_{\theta} \frac{\partial u}{\partial h_{r}} - \frac{\sin mh_{\theta}}{h_{r}} \frac{\partial u}{\partial mh_{\theta}}[/itex]

[itex]\displaystyle \frac{\partial u}{\partial y} = \sin mh_{\theta} \frac{\partial u}{\partial h_{r}} + \frac{\cos\displaystyle mh_{\theta}}{h_{r}} \frac{\partial u}{\partial mh_{\theta}}[/itex]

[itex]\displaystyle \frac{\partial^{2}u}{\partial x^{2}} = \cos^{2}mh_{\theta} \frac{\partial^{2}u}{\partial h_{r}^{2}} - \frac{2\sin mh_{\theta}\cos mh_{\theta}}{h_{r}} \frac{\partial^{2}u}{\partial h_{r}\partial mh_{\theta}} + \frac{\sin^{2}mh_{\theta}}{h_{r}^{2}} \frac{\partial^{2}u}{\partial \left(mh_{\theta}\right)^{2}} + \frac{\sin^{2}mh_{\theta}}{h_{r}} \frac{\partial u}{\partial h_{r}} + \frac{2\sin mh_{\theta}\cos mh_{\theta}}{h_{r}^{2}} \frac{\partial u}{\partial mh_{\theta}}[/itex]

[itex]\displaystyle \frac{\partial^{2}u}{\partial y^{2}} = \sin^{2}mh_{\theta} \frac{\partial^{2}u}{\partial h_{r}^{2}} + \frac{2\sin mh_{\theta}\cos mh_{\theta}}{h_{r}} \frac{\partial^{2}u}{\partial h_{r}\partial mh_{\theta}} + \frac{\cos^{2}mh_{\theta}}{h_{r}^{2}} \frac{\partial^{2}u}{\partial \left(mh_{\theta}\right)^{2}} + \frac{\cos^{2}mh_{\theta}}{h_{r}} \frac{\partial u}{\partial h_{r}} - \frac{2\sin mh_{\theta}\cos mh_{\theta}}{h_{r}^{2}} \frac{\partial u}{\partial mh_{\theta}}[/itex]

[itex]\displaystyle \frac{\partial^{2}u}{\partial x\partial y} = \sin mh_{\theta}\cos mh_{\theta} \frac{\partial^{2}u}{\partial h_{r}^{2}} + \frac{1}{h_{r}} \left(\cos^{2}mh_{\theta} - \sin^{2}mh_{\theta}\right) \frac{\partial^{2}u}{\partial h_{r}\partial h_{\theta}} - \frac{\sin mh_{\theta}\cos mh_{\theta}}{h_{r}^{2}} \frac{\partial^{2}u}{\partial mh_{\theta}}[/itex]

[itex]\displaystyle \Delta u=\frac{\partial^{2}u}{\partial r^{2}} + \frac{1}{r^{2}} \frac{\partial^{2}u}{\partial\theta^{2}} + \frac{1}{r} \frac{\partial u}{\partial r}[/itex]

3. The attempt at a solution

From what I can make up from the problem statement, I thought the way to go to solve this problem is just to substitute for the cartesian operators and variables in terms of the radial ones up to second-order in the expansion of [itex]u_{1,m}[/itex] and then by cancellation I would be left with something which resembles the continuous Laplacian in radial coordinates. However, when I do this I get the following result:

[itex]\displaystyle u_{1,m} = u\left(h_{r}\cos mh_{\theta},h_{r}\sin mh_{\theta}\right) = u\left(0,0\right) + h_{r} \frac{\partial u}{\partial h_{r}} + \frac{1}{2!}\left[h_{r}^{2} \frac{\partial^{2}u}{\partial h_{r}^{2}} + 2h_{r}\sin^{2}mh_{\theta}\cos^{2}mh_{\theta} \frac{\partial u}{\partial h_{r}} + 2\sin mh_{\theta}\cos mh_{\theta}\left(\cos^{2}mh_{\theta} - \sin^{2}mh_{\theta}\right) \frac{\partial u}{\partial mh_{\theta}}\right][/itex]

This substitution is kind of tedious, so I have carried it out twice now in order to avoid any mistakes, but this is the result I get. The first term naturally cancels with the term [itex]u_{0,0}[/itex] from the definition and the linear term seems to be correct (shouldn't this cancel, since it shoulde be second-order accurate???). However, the second-order terms do not really fit in for some reason. The first second-order term looks sort of right (except for a factor of 2 that is), but the other two second-order terms don't make much sense. Maybe I am going about this in completely the wrong way, but I do not really no what do differently really...

If someone could help me out with this, I would be very grateful.

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# Homework Help: Taylor expansion in radial coordinates

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