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Let's say you have $y(v,t)$ where $v(x(s,t),t) = y(s,t)$.

I know that $\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial v}{\partial t} $

However, I see in my textbook that the second partial derivative of $y$ with respect to $t$ is $\frac{\partial^2 y}{\partial t^2} = \frac{\partial^2 v}{\partial x^2}(\frac{\partial x}{\partial t})^2 + 2(\frac{\partial^2 v}{\partial x \space \partial t}) + \frac{\partial^2 v}{\partial t^2} + \frac{\partial v}{\partial x}\frac{\partial^2 x}{\partial t^2}$

and I don't understand how to get there. At first I thought that in the first term the denominator of $\frac{\partial^2 v}{\partial x^2}$ was to the second power (yes, I am aware that that simply means the second partial derivative with rspect to x and has nothing to do with the operation of multiplying the same thing twice) because the $(\frac{\partial x}{\partial t})$ was also squared, and then I attempted to apply the normal power rule (fg)' = f'g + g'f, but I am really confused about this and any help would be really appreciated. This is from page 3 of H.F. Weinberger's Introduction to PDEs and I am just trying to relearn partial differentiation for my PDE class. Thanks

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Let $y(s,t) = v \big( x(s,t), t \big)$. For the partial derivative $\frac{\partial y}{\partial t}$, I prefer to write :

$$ \frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) + \frac{\partial v}{\partial y}\big( x(s,t),t \big) \tag{$\star$}$$

This way, we clearly see what depends on $t$. To compute $\frac{\partial^{2} y}{\partial t^{2}}$, we have to differentiate the two terms in $(\star)$ with respect to $t$.

For the first one, we have :

$$ \begin{eqnarray*} \frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) \Big] & = & \Big[\frac{\partial}{\partial t} \frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big] \frac{\partial x}{\partial t}(s,t) + \frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big[ \frac{\partial}{\partial t} \frac{\partial x}{\partial t}(s,t) \Big] \\ & = & \Big[ \frac{\partial^{2} v}{\partial^{2} x}\big( x(s,t),t) \frac{\partial x}{\partial t}(s,t) + \frac{\partial^{2} v}{\partial x \partial y}\big( x(s,t),t \big) \Big] \frac{\partial x}{\partial t}(s,t) + \\ & &\frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big[\frac{\partial^{2} x}{\partial^{2} t}(s,t) \Big] \\ & = & \frac{\partial^{2} v}{\partial x^{2}}\big( x(s,t),t \big) \left( \frac{\partial x}{\partial t}(s,t) \right)^{2} + \frac{\partial^{2} v}{\partial x \partial y}\big( x(s,t),t \big)\frac{\partial x}{\partial t}(s,t) + \\ & & \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial^{2} x}{\partial t^{2}}(s,t) \\ \end{eqnarray*} $$

Which we can summarize :

$$ \frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) \Big] = \frac{\partial^{2} v}{\partial x^{2}} \left( \frac{\partial x}{\partial t} \right)^{2} + \frac{\partial^{2} v}{\partial x \partial y} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial x} \frac{\partial^{2} x}{\partial t^{2}} \tag{1}$$

And for the second term, we have as well :

$$\frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial y}\big( x(s,t),t \big) \Big] = \frac{\partial^{2} v}{\partial y \partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) + \frac{\partial^{2} v}{\partial y^{2}}\big( x(s,t),t \big) $$

Which also writes :

$$ \frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial y}\big( x(s,t),t \big) \Big] = \frac{\partial^{2} v}{\partial y \partial x} \frac{\partial x}{\partial t} + \frac{\partial^{2} v}{\partial y^{2}} \tag{2}$$

Since $v$ is $\mathcal{C}^{2}$, from Schwartz lemma, we have :

$$ \frac{\partial^{2} v}{\partial x \partial y} = \frac{\partial^{2} v}{\partial y \partial x} $$

Hence, $(1)$ and $(2)$ give :

$$ \frac{\partial^{2} y}{\partial t^{2}} = \frac{\partial^{2} v}{\partial x^{2}} \left( \frac{\partial x}{\partial t} \right)^{2} + 2 \frac{\partial^{2} v}{\partial x \partial y} \frac{\partial x}{\partial t} + \frac{\partial^{2} v}{\partial y^{2}} + \frac{\partial v}{\partial x}\frac{\partial^{2} x}{\partial t^{2}}$$

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  • $\begingroup$ Where does dy come from under "For the first one, we have" ? randomly it appears in the denominator along with dx, but it never comes up elsewhere. Thanks for your help $\endgroup$ Oct 8 '13 at 6:34
  • $\begingroup$ Do you mean : "why is there a $\partial y$ in $\frac{\partial^{2} v}{\partial x \partial y}$" ? $\endgroup$
    – pitchounet
    Oct 8 '13 at 9:51

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