When is partial differentiation commutitive

Question

Consider a function: $$f:\mathbb R ^2\to\mathbb R $$ when does $$\dfrac {\partial f(x,t)}{\partial t\partial x}=\dfrac {\partial f(x,t)}{\partial x\partial t}$$

Thinking about it in terms of the limit definition of derivative, and thinking about it as taking slices of surfaces and measuring the slope on the edge, seems to be giving me the feeling that answer is always.

However I have some memories of there being requirements like piecewise continuous, and smooth.

What is the full set of conditions?

It's true where the partial derivatives are all continuous. (But not an iff statement as far as I'm aware). — ah11950, Mar 25 '14 at 01:44
ah11950: That sounds like an answer (not a comment). Why don't you post it as an answer? — Frames Catherine White, Mar 25 '14 at 01:48
The point is that the discrete version of your formula, that is $$\Delta_t\Delta_x f = \Delta_x \Delta_t f, $$ where $\Delta_t f(x, t)=f(x, t +h)$ for a fixed increment $h$, is always true. However, to infer from this that partial derivatives commute you need a limiting process and so you need at least some continuity. — Giuseppe Negro, Mar 25 '14 at 02:10

score 4 · Accepted Answer · answered Mar 25 '14 at 01:58

4

It's true where the mixed partial derivatives are all continuous. (But not an iff statement as far as I'm aware).

An example of a function that fails to satisfy equality of mixed partial derivatives is

$$f(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} \; &\text{if}\; (x,y) \neq 0\\ 0 &\text{if}\; (x,y) = 0\end{cases}$$

At the origin we have $f_{xy}(0,0) = 1 \neq -1 = f_{yx}(0,0)$

answered Mar 25 '14 at 01:58

ah11950

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How did you calculate $f_{xy}, f_{yx}$ so easily? Is there some special technique to this? – PythonSage Mar 31 '20 at 23:56

When is partial differentiation commutitive

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