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

Currently, what I've done:

$$\begin{align} f_x &= (f_s, f_t), \\f_s &= e^s \cos t, \\f_t &= -e^s \sin t, \\(f_x)_x &= ((f_s)_s, (f_t)_t), \\(f_s)_s &= e^s \cos t, \\(f_s)_t &= -e^s \sin t \\(f_t)_s &= -e^s \sin t \\ (f_t)_t &= -e^s \cos t \\ \ \\f_y &=(f_s, f_t), \\f_s &= e^s \sin t, \\f_t &= e^s \cos t, \\(f_y)_y &= ((f_s)_s, (f_t)_t), \\(f_s)_s &= e^s \sin t, \\(f_s)_t &= e^s \cos t \\(f_t)_s &= e^s \cos t \\ (f_t)_t &= -e^s \sin t \end{align}$$

Could tutors over here advise me whether I am on the right track, as I have no idea how to proceed from here. Thanks :)

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I think the formulas you want to use are $f_s=f_xx_s+f_yy_s$, $f_t=f_xx_t+f_yy_t$. I don't think $f_x=(f_s,f_t)$ and the like will get you anywhere. –  Gerry Myerson Oct 29 '12 at 6:17
    
Thanks Gerry for point out that to me. I have an idea of how to proceed from here :D –  melyong Oct 29 '12 at 9:41

2 Answers 2

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It helps when we get some grip on the things going on here. We are given a real-valued function $$(x,y)\mapsto u(x,y)$$ defined in some part of the $(x,y)$-plane. Now the variables $x$ and $y$ are replaced by new variables $s$ and $t$ according to $$x:=e^s\cos t\ ,\quad y=e^s\sin t\ .$$ (One may view this as a map of the $(s,t)$-plane into the $(x,y)$-plane.) In this way the function $u$ becomes a function of the new variables, called the pullback of $u$: $$\tilde u(s,t):=u\bigl(x(s,t),y(s,t)\bigr)\ .$$ The partial derivatives of $\tilde u$ with respect to $s$ and $t$ are obtained by means of the chain rule: $$\tilde u_s= u_x\,x_s+u_y\,y_s\ ,\quad \tilde u_t=u_x\,x_t+u_y\,y_t\ .$$ Now we are told to look at $\tilde D:=\tilde u_s^2 +\tilde u_t^2$. We get $$\eqalign{\tilde D&=u_x^2(x_s^2+x_t^2)+2u_xu_y(x_s y_s+x_ty_t)+u_y^2(y_s^2+y_t^2)=\ldots\cr &=e^{2s}(u_x^2+u_y^2)\cr &=e^{2s}\, D\ .\cr}$$

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You are mistaken at differentiating $f$, because \begin{gather} f_s =\frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial s}, \\ f_t =\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial t} \end{gather}

Write $u=f(e^s \cos{t}, \, e^s \sin{t})$ and apply the chain rule calculating $\dfrac{\partial f }{\partial s},\, \dfrac{\partial f }{\partial s},\, \dfrac{\partial ^2 f}{\partial s^2}, \, \dfrac{\partial ^2 f}{\partial t^2}.$ For example, $$\frac{\partial f }{\partial s}=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial s}=e^s \cos{t} \frac{\partial u}{\partial x} +e^s\sin{t}\frac{\partial u}{\partial y},\\ \frac{\partial f }{\partial t}=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial s}=-e^s \sin{t} \frac{\partial u}{\partial x} +e^s\cos{t}\frac{\partial u}{\partial y}.$$

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Thank you so much M.Strochyk! You point out where i went wrong. :D –  melyong Oct 29 '12 at 9:40

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