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I was reading a book on mathematical methods for physics by riley hobson and bence and I encountered the following paragraph - Paragraph

The proof I attempted of that is this - Suppose we want to find the value of $ f(x,y) $ when we increment both x and y by small amounts dx and dy. We can either first increase x, then y or the other way around. The two methods should give the same result : $ \Delta f = (\partial f/\partial x)_{x,y}dx + (\partial f/\partial y)_{x+dx,y}dy = (\partial f/\partial y)_{x,y}dy + (\partial f/\partial x)_{x,y+dy}dx $ Which implies - $ ( (\partial f/\partial x)_{x,y+dy} - (\partial f/\partial x)_{x,y} )dx = ( (\partial f/\partial y)_{x+dx,y} - (\partial f/\partial y)_{x,y} )dy. $ All I glean from this is that the terms next to dx and dy should be zero, and all that says is that the first derivatives are continuous.

Neeladri Reddy
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The problem with your argument is that you don't take limits anywhere: it isn't true that you can just treat $dx$ and $dy$ as 'small' (it is 'usually' - i.e. when the function is 'nice', but not always).

$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)_{x,y}=\lim_{dx \to 0}\frac{\frac{\partial f}{\partial y}_{x+dx,y}-\frac{\partial f}{\partial y}_{x,y}}{dx}=\lim_{dx \to 0}\frac{\displaystyle\lim_{dy \to 0}\frac{f_{x+dx,y+dy}-f_{x+dx,y}}{dy}-\displaystyle\lim_{dy \to 0}\frac{f_{x,y+dy}-f_{x,y}}{dy}}{dx}$$

To find the other mixed partial derivative, you need to exchange the order of the limits and this is 'usually' fine, but not in every case: that's why you need the second derivatives continuous.

If you want to follow this through with an example, use $$f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$$ for $x$, $y$ not both equal to $0$ and $f(0,0)=0$. If you evaluate them at $(0,0)$, you should get $1$ for the mixed partial shown above, but $-1$ for the other.

mcd
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  • I am not clear at all what's going on. I haven't done any real analysis. My argument is essentially this(which wasn't explained properly in my question) - I wrote the increment in f in two different ways using differentials. My first question is is that allowed and rigorous? If that's allowed then the equation resulting from the equality of the two ways of writing $\Deltaf$ would mean that I can convert that into second derivatives and I am done. Am I correct? Please go slow haha. – Neeladri Reddy Feb 03 '23 at 07:49
  • "I wrote the increment in f in two different ways using differentials. My first question is is that allowed and rigorous?" I'm afraid not. Working out increments in a function value using only its first derivatives is (almost) always an approximation, and sometimes, as here, not a useful one. I'm not sure that you can make much progress with justifying these results without learning some real analysis, and this is probably not the topic to tackle first. – mcd Feb 03 '23 at 17:30
  • I would love to learn real analysis, but I don't have much time. Is it necessary for a physics undergrad to know real analysis? – Neeladri Reddy Feb 04 '23 at 08:07