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Minton and Smith, in "Calculus" define the chain rule for full derivatives $\frac {dz} {dt}$ as it follows:

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Vretblad, however, in "Fourier Analysis and its Applications", mentions an "easy exercise in applying the chain rule" in an expansion of a partial derivativeenter image description here:

The question is: can the chain rule, originally defined only on $\frac {dz} {dt}$, be extended to $\frac {\partial z} {\partial t}$, or is Vretblad applying the chain rule on a full derivative somehow?

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    $\begingroup$ Remark: $\frac{\partial z}{\partial t}$ is defined as an ordinary one-dimensional derivative of $\phi(t)=z(t,...)$, so all rules from one-dimensional calculus are applied to it. $\endgroup$ – A.Γ. Oct 15 '15 at 9:57
  • $\begingroup$ I think your question is not much about "chain rule for partial derivatives" or "extension of chain rule to $\partial z/\partial t$", but more about "what kind of chain rule is Vretblad using when suggesting an easy exercise". Is my feeling right? $\endgroup$ – A.Γ. Oct 15 '15 at 10:07
  • $\begingroup$ I could ask it this way, as it would probably solve my doubt, but, then it wouldn't be relevant to other SE users. The way Vretblad speaks about the Chain Rule has led me to believe the Chain Rule he is using is the same as the one presented in Minton and Smith. $\endgroup$ – Bruno Schiavo Oct 15 '15 at 10:41
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    $\begingroup$ It is the same. $\endgroup$ – A.Γ. Oct 15 '15 at 10:49
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Let me put it in this way: $u$ is a function of $x,t$, but can be thought of as a function of the new variables $\xi$ and $\eta$ after the change. Vretblad is using the standard physical formalism and keeps the same name for the function $u(x,t)$ and $u(\xi,\eta)$, so we get the (terrible from the mathematical point of view) identity $$ u(x,t)=u(\xi(x,t),\eta(x,t)). $$ Derivating both sides wrt $x$ (using the chain rule in the RHS) we get $$ u_x=\frac{\partial u}{\partial \xi}\underbrace{\frac{\partial\xi}{\partial x}}_{=1}+\frac{\partial u}{\partial \eta}\underbrace{\frac{\partial\eta}{\partial x}}_{=1}= \frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta}. $$ Doing it once again and applying the chain rule to both terms in the RHS gives you $$ u_{xx}=\color{red}{\frac{\partial}{\partial x}\Bigl(\frac{\partial u}{\partial \xi}\Bigr)}+\color{blue}{\frac{\partial}{\partial x}\Bigl(\frac{\partial u}{\partial \eta}\Bigr)}=\color{red}{\frac{\partial^2 u}{\partial \xi^2}\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\eta\partial\xi}\frac{\partial\eta}{\partial x}}+\color{blue}{\frac{\partial^2 u}{\partial \xi\partial\eta}\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\eta^2}\frac{\partial\eta}{\partial x}}=... $$ I hope you can continue after that.

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  • $\begingroup$ Why is $\frac{\partial \eta}{\partial x }$ and $\frac{\partial \xi}{\partial x } =1$? $x = \frac {\eta + \xi}{2}$, they should be $2$ $\endgroup$ – Bruno Schiavo Oct 17 '15 at 5:24
  • $\begingroup$ @BrunoSchiavo Just derivate $\xi(x,t)=x-ct$ and $\eta(x,t)=x+ct$ wrt $x$. $\endgroup$ – A.Γ. Oct 17 '15 at 7:51
  • $\begingroup$ Why should we derivate $\xi(x,t)=x-ct$ and not $x = \frac {\eta + \xi}{2}$ (which comes from the sum of the two equations you wrote)? $\endgroup$ – Bruno Schiavo Oct 17 '15 at 10:40
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    $\begingroup$ @BrunoSchiavo Because the chain rule says $\frac{\partial\xi}{\partial x}$ and not anything else. I think you should work out better understanding of the chain rule first. $\endgroup$ – A.Γ. Oct 17 '15 at 12:17

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