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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?

mythealias
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2 Answers2

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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.

Pragabhava
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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
  • the chain rule equation seems to be correct. i got the same thing in my textbook. – question Nov 11 '12 at 21:25
  • they got $\partial^2 u / \partial x^2$ instead of $\partial^2 u / \partial s^2$ – question Nov 11 '12 at 21:36
  • @Thomas As the answer says, the only way the first calculation is correct is to assume that all variables remain constant. If not, there is no way that is correct. In thermodynamics, the notation should be $\left(\frac{\partial f}{\partial x}\right)v = 2x$, so it's clear that the quantity $v$ is _held constant (i.e. the pressure). – Pragabhava Nov 11 '12 at 21:53
  • @Matthew I've completed the calculation. You should deduce it yourself, and as an excercise, calculate $\frac{\partial^2 u}{\partial s \partial t}$ and $\frac{\partial^2 u}{\partial t^2}$. – Pragabhava Nov 11 '12 at 22:30
  • thank you so much for your help, but my feeble mind still has difficulty grasping the concepts. – question Nov 11 '12 at 22:50
  • @Matthew No problem. I took me a long time to understand it too. If you do lots of excercises (as I did), and read the theorems carefully, eventually you'll get it. – Pragabhava Nov 11 '12 at 22:53
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It is chain rule. Just write functions in a vector valued form nad use the chain rule.

dmm
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