Calculating second partial derivate of a multivariable function.

Question

I have a doubt resolving Apostol Vol.2, sec 22, prob 4:

The equations $u =f(x, y), x = X(s, t), y = Y(s, t)$ define $u$ as a function of $s$ and $t$, say $u = F(s, t).$ Find formulas for the partial derivatives $\frac{\partial^2F}{\partial s\partial t}$ and $\frac{\partial^2F}{\partial t^2}$

I was able to get the formula for $\frac{\partial^2F}{\partial t^2}$, but for $\frac{\partial^2F}{\partial s\partial t}$ I seem to have troubles, here's my solution:

$\frac{\partial^2F}{\partial s\partial t}=\frac{\partial f}{\partial x}\frac{\partial^2X}{\partial s\partial t}+\frac{\partial^2f}{\partial x^2}\frac{\partial X}{\partial t}\frac{\partial X}{\partial s}+\frac{\partial^2f}{\partial y\partial x}\frac{\partial Y}{\partial s}\frac{\partial X}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial^2Y}{\partial s\partial t}+\frac{\partial^2f}{\partial x\partial y}\frac{\partial X}{\partial s}\frac{\partial Y}{\partial t}+\frac{\partial^2f}{\partial y^2}\frac{\partial Y}{\partial s}\frac{\partial Y}{\partial t}$

But I think this is wrong given a solution I found on the internet, is my solution right or where could I have gone wrong?

Thanks.

Answer 1

The derivation of $\frac{\partial^2 F}{\partial s\partial t}$ is sound.

We obtain \begin{align*} \color{blue}{\frac{\partial^2 F}{\partial s\partial t}} &=\frac{\partial}{\partial s}\left(\frac{\partial F}{\partial t}\right)\\ &=\frac{\partial}{\partial s}\left(\frac{\partial f}{\partial x}\,\frac{\partial X}{\partial t} +\frac{\partial f}{\partial y}\,\frac{\partial Y}{\partial t}\right)\\ &=\left(\frac{\partial}{\partial s}\left(\frac{\partial f}{\partial x}\right)\right)\,\frac{\partial X}{\partial t} +\left(\frac{\partial f}{\partial x}\right)\,\frac{\partial ^2 X}{\partial s\partial t}\\ &\qquad+\left(\frac{\partial}{\partial s}\left(\frac{\partial f}{\partial y}\right)\right)\,\frac{\partial Y}{\partial t} +\left(\frac{\partial f}{\partial y}\right)\,\frac{\partial ^2 Y}{\partial s\partial t}\\ &=\left(\frac{\partial^2 f}{\partial x^2}\,\frac{\partial X}{\partial s} +\frac{\partial^2 f}{\partial y\partial x}\,\frac{\partial Y}{\partial s}\right)\,\frac{\partial x}{\partial t} +\left(\frac{\partial f}{\partial x}\right)\,\frac{\partial^2 X}{\partial s\partial t}\\ &\qquad+\left(\frac{\partial^2 f}{\partial x\partial y}\,\frac{\partial X}{\partial s} +\frac{\partial^2 f}{\partial y^2}\,\frac{\partial Y}{\partial s}\right)\,\frac{\partial Y}{\partial t} +\left(\frac{\partial f}{\partial y}\right)\,\frac{\partial^2 Y}{\partial s\partial t}\\ &\,\,\color{blue}{=\frac{\partial^2 f}{\partial x^2}\,\frac{\partial X}{\partial s}\,\frac{\partial X}{\partial t} +\frac{\partial ^2f}{\partial y\partial x}\,\frac{\partial Y}{\partial s}\,\frac{\partial X}{\partial t}+\frac{\partial^2 f}{\partial x\partial y}\,\frac{\partial X}{\partial s}\,\frac{\partial Y}{\partial t} +\frac{\partial^2 f}{\partial y^2}\,\frac{\partial Y}{\partial s}\,\frac{\partial Y}{\partial t}}\\ &\qquad \,\,\color{blue}{+\frac{\partial f}{\partial x}\,\frac{\partial^2 X}{\partial s\partial t} +\frac{\partial f}{\partial y}\,\frac{\partial^2 Y}{\partial s\partial t}}\tag{1} \end{align*} in accordance with OPs derivation.

It is also convenient to make a plausibility check by calculating a specific example in two ways. We consider \begin{align*} u=f(x,y)=x^2+2y&\qquad x=X(s,t)=4s+3t\\ &\qquad y=Y(s,t)=2st \end{align*} We have according to (1) \begin{align*} &\left(\frac{\partial^2 }{\partial X^2}(x^2+2y)\right)\left(\frac{\partial }{\partial s}(4s+3t)\right)\left(\frac{\partial }{\partial t}(4s+3t)\right)\\ &\qquad+\left(\frac{\partial ^2}{\partial Y\partial X}(x^2+2y)\right)\left(\frac{\partial }{\partial s}(2st)\right)\left(\frac{\partial }{\partial t}(4s+3t)\right)\\ &\qquad+\left(\frac{\partial^2 }{\partial X\partial Y}(x^2+2y)\right)\left(\frac{\partial }{\partial s}(4s+3t)\right)\left(\frac{\partial }{\partial t}(2st)\right)\\ &\qquad+\left(\frac{\partial^2 }{\partial Y^2}(x^2+2y)\right)\left(\frac{\partial }{\partial s}(2st)\right)\left(\frac{\partial }{\partial t}(2st)\right)\\ &\qquad+\left(\frac{\partial }{\partial X}(x^2+2y)\right)\left(\frac{\partial^2 }{\partial s\partial t}(4s+3t)\right)\\ &\qquad+\left(\frac{\partial }{\partial Y}(x^2+2y)\right)\left(\frac{\partial^2 }{\partial s\partial t}(2st)\right)\\ &=(2)(4)(3)+(0)(2t)(3)+(0)(4)(2s)+(0)(2t)(2s)+(2x)(0)+(2)(2)\\ &=24+4\\ &\,\,\color{blue}{=28}\tag{2} \end{align*}

On the other hand we obtain \begin{align*} u=F(s,t)&=(4s+3t)^2+2(2st)\\ &=16s^2+28st+9t^2\\ \\ \frac{\partial^2 F}{\partial s\partial t}&=\frac{\partial ^2 }{\partial s\partial t}\left(16s^2+28st+9t^2\right)\\ &=\frac{\partial }{\partial s}\left(\frac{\partial }{\partial t}\left(16s^2+28st+9t^2\right)\right)\\ &=\frac{\partial }{\partial s}\left(28s+18t\right)\\ &\,\,\color{blue}{=28} \end{align*} in accordance with (2).