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If $u$ = $f(x,y)$, where $x = e^s\cos t$ and $y = e^s\sin t$, show that $$\frac{\partial^2u}{\partial x^2}+ \frac{\partial^2u}{\partial y^2} = e^{-2s}\left[\frac{\partial^2u}{\partial s^2}+\frac{\partial^2u}{\partial t^2}\right] $$

I'm not sure where to start. Any ideas?

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1 Answer 1

up vote 2 down vote accepted

Do you know the chain rule?

$$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}\\ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y} $$

or the inverse

$$ \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\\ \frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t} $$

You can use the second pair of rules, then solve for the desired quantities, or use the first pair of rules after obtaining $s(x,y), t(x,y)$ from the definitions of $x,y$.

Then you should proceed in a similar way for the second derivatives.

Edit

Going into the calculations:

\begin{align} \frac{\partial u}{\partial s}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\\ &=\frac{\partial u}{\partial x}e^s\cos t+\frac{\partial u}{\partial y}e^s\sin t=\\ &=x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\\ \frac{\partial u}{\partial t}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=\\ &=-\frac{\partial u}{\partial x}e^s\sin t+\frac{\partial u}{\partial y}e^s\cos t=\\ &=-y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y} \end{align}

This is true for every function $u$, so

\begin{align} \frac{\partial}{\partial s}&=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\\ \frac{\partial}{\partial t}&=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y} \end{align}

so we have

\begin{align} \frac{\partial^2u}{\partial s^2}&=\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)\\ \frac{\partial^2u}{\partial t^2}&=\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\left(-y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y}\right) \end{align}

I suppose that from here you can go on without help.

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mmm.. i don't fully understand how to use the chain rule formula. And i have trouble coming up with an equation for $f(x,y)$. Can you elaborate more on this? –  uohzxela Oct 28 '12 at 10:36
    
@uohzxela: I edited the answer. –  enzotib Oct 28 '12 at 11:38
    
thank you very much! –  uohzxela Oct 28 '12 at 13:09

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