2

I am trying to prove the following

Given a function $f:\Bbb{R}^2\to \Bbb{R}^2$ whose second derivatives are continuous, $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}$$

I am aware that proofs for this are found in most books on multivariable calculus. However, I'm trying to come up with a proof on my own.

My strategy is to basically prove that $$\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial x\partial y}$$ should be equal to $0$. What I did was that I performed a change of variables- let $m=x+y$ and $n=x-y$. Then clearly ${x=\frac{m+n}{2}}$ and $y=\frac{m-n}{2}$. I now found the difference $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}$$ As $$\frac{\partial f}{\partial m}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)$$ and $$\frac{\partial f}{\partial n}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)$$ we see that $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\frac{1}{2}\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$

My aim would then be this: To prove that $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$ Clearly this is only possible if $\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=0$, as $\frac{1}{2}\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$ if and only if $\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=0$.

Is this a viable strategy? Does anyone know why $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$ without assuming the result we are trying to prove?

  • In response to your last question: because they are both zero? – Angina Seng Mar 11 '18 at 01:56
  • @LordSharktheUnknown- I meant without assuming the result we're trying to prove –  Mar 11 '18 at 01:59
  • And I meant that I didn't think this was a viable approach :-) – Angina Seng Mar 11 '18 at 02:00
  • @AyushKhaitan I see what you were trying to do : probably change the variables and hope that such a change would lead to something useful. Unfortunately, your choice $m,n$ is not useful, because it is merely a "rephrasing" of the original problem in new coordinates, whence it is not helpful. What you must find, is some interesting consequence of the partial derivatives being continuous, which is done by the original proof. – Sarvesh Ravichandran Iyer Mar 11 '18 at 03:34

0 Answers0