PDECS1

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Change to polar (and cylindrical) coordinates: \nabla^2 u(x,y)\,

\nabla^2 u(x,y) = u_{xx} + u_{yy}\,

The transformation is:

x = r \cos\theta, y = r \sin\theta\,

r = + \sqrt{x^2+y^2}, \theta = \tan^{-1} \frac{y}{x}\,

Remember the derivatives \frac{d}{dx}\left[\tan^{-1}\frac{y}{x}\right] = \frac{-y}{x^2+y^2}, \frac{d}{dy}\left[\tan^{-1}\frac{y}{x}\right] = \frac{x}{x^2+y^2}\,

Plug back into the original DE.

\nabla^2u(r,\theta) = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}\,

u_x = u_r r_x + u_\theta \theta_x = \frac{1}{2}(x^2+y^2)^{-1/2}\,2x\, u_r - \frac{y}{x^2+y^2}u_\theta\,

=\frac{x\sqrt{x^2+y^2}\, u_r - y u_\theta}{x^2+y^2} = \frac{x \,r\, u_r - y\, u_\theta}{r^2}\,

= \frac{r^2 \cos\theta \,u_r - r \sin \theta\, u_\theta}{r^2} = \cos \theta\, u_r - r^{-1} \sin_\theta\, u_\theta\,

Let this derivative be the differential operator D_x\,. Then,

D_x u = (\cos\theta\, D_r - r^{-1}\sin\theta\, D_\theta) u\,

Therefore

D_x D_x u = (\cos\theta\, D_r - r^{-1}\sin\theta\, D_\theta) (\cos\theta\, D_r - r^{-1}\sin\theta\, D_\theta) u\,

= \cos\theta\, D_r (\cos\theta \,u_r - r^{-1}\sin\theta \,u_\theta) - r^{-1}\sin\theta\, D_\theta (\cos\theta \,u_r - r^{-1}\sin\theta \,u_\theta) \,

= \cos^2\theta\,u_{rr} - \cos\theta\,\sin\theta\,(r^{-1}u_{\theta r} - r^{-2}u_\theta) - r^{-1}\sin\theta\,(\cos\theta\,u_{r\theta} - \sin\theta\,u_r - r^{-1}\sin\theta\,u_{\theta\theta} - r^{-1}\cos\theta\,u_\theta)\,

  • = \cos^2\theta\, u_{rr} + \cos\theta\,\sin\theta\,(-r^{-1}u_{\theta r} + r^{-2}u_\theta-r^{-1}u_{r\theta} + r^{-2}u_\theta) + \sin^2\theta\,(r^{-1}u_r + r^{-2} u_{\theta\theta} )\,

u_y = u_r \,r_y + u_\theta\theta_y = \frac{1}{2}(x^2+y^2)^{-1/2}\,2y \,u_r + \frac{x u_\theta}{x^2+y^2}\,

\frac{y\, r\, u_r + x u_\theta}{r^2} = \frac{r^2\sin\theta u_r + r\cos\theta u_\theta}{r^2} = \sin\theta\, u_r + r^{-1}\cos\theta u_\theta\,

D_y u = (\sin\theta\,D_r + r^{-1}\cos\theta\, D_\theta)u\,

D_y D_y u = (\sin\theta\,D_r + r^{-1}\cos\theta\, D_\theta)(\sin\theta\,D_r + r^{-1}\cos\theta\, D_\theta)u\,

= \sin\theta D_r(\sin\theta\,u_r + r^{-1}\cos\theta\,u_\theta) + (r^{-1}\cos\theta D_\theta (\sin\theta u_r + r^{-1}\cos\theta u_\theta)\,

= \sin^2\theta u_{rr} + \cos\theta\,\sin\theta\,(r^{-1}u_{\theta r} - r^{-2}u_\theta) + r^{-1}\cos\theta\,(\sin\theta\,u_{r\theta} + \cos\theta u_r + r^{-1}\cos\theta\,u_{\theta\theta}-r^{-1}\sin\theta\,u_\theta)\,

  • = \sin^2\theta\, u_{rr} + \cos\theta\,\sin\theta\,(r^{-1} u_{\theta r} - r^{-2}u_\theta + r^{-1}u_{r\theta}-r^{-2}u_\theta) + \cos^2\theta(r^{-1}u_r + r^{-2}u_{\theta\theta})\,

Summing the two starred lines, it is concluded that

u_{xx} + u_{yy} = u_{rr} + r^{-1}u_r + r^{-2}u_{\theta\theta}\,

To make the transformation to cylindrical coordinates, add the transformation z=z\, at the top and the term + u_{zz}\, to the last equation.

Partial Differential Equations

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