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- Thread starter Poirot
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- Jan 29, 2012

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First, "u''(x)" and "u''(t)" is bad notation. What you have written implies that u is a function of a single variable and that you have chosen to call that variable "x" in one case and "t" in the other. What you mean should be written as [tex]u_{xx}+ u_{tt}= 0[/tex].

You convert from rectangular coordinates to polar coordinates using the chain rule.

$\dfrac{\partial u}{\partial x}= \dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+ \dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}$

Of course, $r= (x^2+ t^2)^{1/2}$ so that

$\dfrac{\partial r}{\partial x}= (1/2)(x^2+ t^2)^{-1/2}(2x)= \dfrac{x}{\sqrt{x^2+ t^2}}= \dfrac{r \cos(\theta)}{r}= \cos(\theta)$

and $\theta= \arctan(t/x)$ so that

$\dfrac{\partial \theta}{\partial x}= \dfrac{1}{1+ t^2/x^2}(-t/x^2)= \dfrac{-t}{x^2+ t^2}= \dfrac{-r \sin(\theta)}{r^2}= -\dfrac{1}{r}\sin(\theta)$

Now repeat that to get $\partial^2 u/\partial x^2$ in terms of r and $\theta$ and do the same thing for the derivatives with respect to t.

(Your choice of "t", which more often represents "time", as the second variable is a little odd. Normally, if you want to change to polar coordinates you are thinking of a geometric situation with x and y coordinates. You are, of course, welcome to use whatever letters you want but I have used $t= r \sin(\theta)$.)

You convert from rectangular coordinates to polar coordinates using the chain rule.

$\dfrac{\partial u}{\partial x}= \dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+ \dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}$

Of course, $r= (x^2+ t^2)^{1/2}$ so that

$\dfrac{\partial r}{\partial x}= (1/2)(x^2+ t^2)^{-1/2}(2x)= \dfrac{x}{\sqrt{x^2+ t^2}}= \dfrac{r \cos(\theta)}{r}= \cos(\theta)$

and $\theta= \arctan(t/x)$ so that

$\dfrac{\partial \theta}{\partial x}= \dfrac{1}{1+ t^2/x^2}(-t/x^2)= \dfrac{-t}{x^2+ t^2}= \dfrac{-r \sin(\theta)}{r^2}= -\dfrac{1}{r}\sin(\theta)$

Now repeat that to get $\partial^2 u/\partial x^2$ in terms of r and $\theta$ and do the same thing for the derivatives with respect to t.

(Your choice of "t", which more often represents "time", as the second variable is a little odd. Normally, if you want to change to polar coordinates you are thinking of a geometric situation with x and y coordinates. You are, of course, welcome to use whatever letters you want but I have used $t= r \sin(\theta)$.)

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- Jan 26, 2012

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You could start by sorting out your notation. Laplace's equation is usually writtenLet u=u(x,t). Then laplace's equation is u''(x) + u''(t)=0 I want to switch to polar co-0rdinates so u=u(r,theta) and deduce laplace's equation from the cartesian form.

I tried using the chain rule but couldn't quite get there.

\[u_{xx}(x,y)+u_{yy}(x,y)=0\]

or:

\[\nabla^2 u=0 \]

\(t\) usually represents time and does not appear in Laplace's equation.

You seem to be conflating Laplace's equation with the wave equation. Could you please post the original question exactly as asked?

CB

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