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I need to show that next equation stands:

$$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = \frac{1}{r}\frac{\partial}{\partial r} (r\frac{\partial u}{\partial r} ) +\frac{1}{r^2}(\frac{\partial^2 u}{\partial \theta^2} )$$ where $u=f(x,y)$ , $x = r\cos\theta$ , $y = r\sin\theta$ , $r = \sqrt{x^2+y^2}$ , $\theta = \arctan(\frac{y}{x})$

I got lost where I came around $$du=dx(\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2})+dy(\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2})\\\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2},\ \ \ \ \ \ \ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2}$$

I just can't figure out, what is the next step and how do I go on with this. Could you, please, help me?

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up vote 1 down vote accepted

Some general points:

  • I prefer subscript notation for such computations, it simplifies both writing and typing.
  • It is important to choose the direction of transformation: left to right or right to left.

For example, it is easy to differentiate $x,y$ in terms of $r,\theta$. It's not as easy the other way around (and besides, the formula $\theta=\arctan\frac{y}{x}$ is not 100% correct). So it makes sense to use the chain rule in the form $u_r=u_x x_r+u_y y_r$ and $u_\theta=u_x x_\theta+u_y y_\theta$, which is the opposite direction to what you tried.

For example, we see that $u_r=u_x \cos\theta+u_y \sin\theta$. This formula holds for any smooth function, so we can use it again with $ru_r$ in place of $u$. Although not strictly necessary, it helps to notice that $ru_r=xu_x+yu_y$.

$(ru_r)_r=(ru_r)_x \cos\theta+(ru_r)_y \sin\theta = (xu_x+yu_y)_x \cos\theta+(xu_x+yu_y)_y \sin\theta$, and so on...

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I'm sorry for not being clear enough. The direction is left-to-right and the goal is to convert the left side to polar coordinates. – Michael Sazonov Jul 4 '12 at 22:36

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