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I don't understand the second step at all. Where did the $\partial^2 u/ \partial x^2$ come from and why do we have six terms?

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Second step of what? I don't see any numbered steps. –  Imray Nov 11 '12 at 19:44
    
visit this search result and see if any of the questions is of help. –  user31280 Nov 11 '12 at 19:52
    
the step where the 2 terms turn into 6 terms –  question Nov 11 '12 at 19:55
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I would be helpful if we had more context, e.g. what these variables $s,t,u$ are and their "dependencies". It doesn't help that this is using $\partial y / \partial x$ notation. –  wj32 Nov 11 '12 at 20:21
    
If u=f(x,y) where x=escost and y=essint show that d2u/dx2+d2u/dy2 = e-2s[d2u/ds2+d2u/dt2 –  question Nov 11 '12 at 20:47
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The multivariable chain rule states that, if $x = x(s,t)$, $y = y(s,t)$, and $u = u(x,y)$, then \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 t} &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} \end{align} To calculate the second derivatives \begin{align} \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial s}\right) &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s}\right) \\ &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial s}\right) + \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \frac{\partial y}{\partial s}\right)\\ &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x}\right) \frac{\partial x}{\partial s} + \frac{\partial u}{\partial x}\frac{\partial}{\partial s}\left(\frac{\partial x}{\partial s}\right) + \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y}\right) \frac{\partial y}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial}{\partial s}\left(\frac{\partial y}{\partial s}\right) \end{align} where the first step is the distribution of the derivative, the second is the product rule for differentiation.

Now, $$ \frac{\partial}{\partial s}\left(\frac{\partial x}{\partial s}\right) = \frac{\partial^2 x}{\partial s^2}, \qquad \frac{\partial}{\partial s}\left(\frac{\partial y}{\partial s}\right) = \frac{\partial^2 y}{\partial s^2} $$ and, using the multivaraible chain rule again \begin{align} \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x}\right) &= \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial s} = \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial y} \frac{\partial y}{\partial s} = \frac{\partial^2 u}{\partial x^2} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y \partial x} \frac{\partial y}{\partial s} \\ \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y}\right) &= \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial s} = \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial y} \frac{\partial y}{\partial s} = \frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y^2} \frac{\partial y}{\partial s} \end{align}

Supposing $\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}$ and substituting into $\frac{\partial^2 u}{\partial s^2}$, we have $$ \frac{\partial^2 u}{\partial s^2} = \frac{\partial^2 u}{\partial x^2} \left(\frac{\partial x}{\partial s}\right)^2 + 2\frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} \frac{\partial y}{\partial s} + \frac{\partial^2 u}{\partial y^2} \left(\frac{\partial y}{\partial s}\right)^2 + \frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial s^2} + \frac{\partial u}{\partial y}\frac{\partial^2 y}{\partial s^2} $$

Why don't you calculate $\frac{\partial^2 u}{\partial s \partial t}$, $\frac{\partial^2 u}{\partial t^2}$ and see if you've improved your understanding?


The whole answer:

Let $v(s,t) = \frac{\partial u}{\partial x} e^s \cos t + \frac{\partial u}{\partial y} e^s \sin t$, where $x = x(s,t)$ and $y = y(s,t)$. Then \begin{align} \frac{\partial v}{\partial s} &= \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x} e^s \cos t + \frac{\partial u}{\partial y} e^s \sin t\right)\\ &= \frac{\partial^2 u}{\partial s \partial x} e^s \cos t + \frac{\partial u}{\partial x} \frac{\partial}{\partial s}\big(e^s \cos t\big) + \frac{\partial^2 u}{\partial s \partial y} e^s \sin t + \frac{\partial u}{\partial x} \frac{\partial}{\partial s}\big(e^s \sin t\big)\\ &= \left(\frac{\partial^2 u}{\partial x^2} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y \partial x} \frac{\partial y}{\partial s}\right) e^s \cos t + \frac{\partial u}{\partial x}e^s \cos t \\ &\hskip2in + \left(\frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y^2} \frac{\partial y}{\partial s}\right) e^s \sin t + \frac{\partial u}{\partial y}e^s \sin t \end{align} From the form of $v$ and your image, I'm assuming $x(s,t) = e^s \cos t$ and $y(s,t) = e^s \sin t$. In such case, \begin{align} \frac{\partial v}{\partial s} &= \frac{\partial^2 u}{\partial x^2} e^{2 s}\cos^2 t + 2 \frac{\partial^2 u}{\partial x \partial y} e^{2s} \cos t \sin t + \frac{\partial^2 u}{\partial y^2} e^{2 s}\sin^2 t + \frac{\partial u}{\partial x}e^s \cos t + \frac{\partial u}{\partial y}e^s \sin t \end{align} Finally, from the form of $v$, I think $v(s,t) = \frac{\partial u}{\partial s}$, but that information wasn't provided by the OP.

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ah i didn't know you could apply the differentiation rule to a partial derivative –  question Nov 11 '12 at 20:28
    
@Thomas I was missing a factor of two in the crossed term. Assuming the cross derivatives are the same, it is correct. If not, the term $2 \frac{\partial^2 u}{\partial x \partial y}$ needs to be substituted by $\left(\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y \partial x}\right)$ –  Pragabhava Nov 11 '12 at 20:28
    
@Pragabhava: My point is that the so-called multivariable chain rule doesn't tell you how to find the partial derivative bu the regular (or total) derivative. The way you have written it, you have only partial derivatives. –  Thomas Nov 11 '12 at 20:30
    
@Thomas Nope. It is a partial derivative allright. See Courant's Differential and Itegral Calculus (vol II. p. 73). If $u = u\big(s,t,x(s,t),y(s,t)\big)$, then the total derivative $$\frac{d u}{d s} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s} \color{red}{+ \frac{\partial u}{\partial s}}$$ –  Pragabhava Nov 11 '12 at 20:48
    
i plugged in the data, and i don't get the same thing, i get a similar expression, but i don't get the same thing –  question Nov 11 '12 at 21:24
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It is chain rule. Just write functions in a vector valued form nad use the chain rule.

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what do you mean? –  question Nov 11 '12 at 19:48
    
which chain rule are you referring to? –  question Nov 11 '12 at 19:49
    
en.wikipedia.org/wiki/… –  dmm Nov 11 '12 at 20:00
    
i wasn't referring to the first step, my question has nothing to do with the chain rule. –  question Nov 11 '12 at 20:04
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@Matthew It has everything to do with the chain rule. See my answer for clarification. –  Pragabhava Nov 11 '12 at 20:24
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