Failure of Clairaut's theorem where both mixed partials are defined but not equal

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Statement

For a function of two variables at a point

It is possible to have a function f of two variables x,y and a point (x_0,y_0) in the domain of f such that both the second-order mixed partial derivatives of f exist at (x_0,y_0), i.e., both the numbers f_{xy}(x_0,y_0) and f_{yx}(x_0,y_0) exist, but they are not equal.

For a function of two variables overall

It is possible to have a function f of two variables x,y such that both the second-order mixed partial derivatives f_{xy}(x,y) and f_{yx}(x,y) exist everywhere on \R^2 but they are not equal as functions, i.e., there exists a point where the values of the second-order mixed partial derivatives are not equal.

Related facts

Proof

Example

Consider the function:

f(x,y) := \left\lbrace \begin{array}{rl} \frac{xy(x^2 - y^2)}{x^2 + y^2}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0) \\\end{array}\right.

We do some computations:

Item Value Explanation
f_x(x,y) for (x,y) \ne (0,0) \frac{(x^2 + y^2)(3x^2y - y^3) - 2x^2y(x^2 - y^2)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3 - y^5}{(x^2 + y^2)^2} use the quotient rule for differentiation and simplify
f_y(x,y) for (x,y) \ne (0,0) \frac{(x^2 + y^2)(x^3 - 3xy^2) - 2xy^2(x^2 - y^2)}{(x^2 + y^2)^2} = \frac{x^5 - 4x^3y^2 - xy^4}{(x^2 + y^2)^2} use the quotient rule for differentiation and simplify
f_x(0,y) for y \ne 0 -y plug in x = 0 in the general expression for f_x(x,y) and simplify.
f_y(x,0) for x \ne 0 x plug in y = 0 in the general expression for f_y(x,y) and simplify.
f_x(0,0) 0 start with \lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x} and simplify, noting that the numerator is identically zero
f_y(0,0) 0 start with \lim_{y \to 0} \frac{f(0,y) - f(0,0)}{y} and simplify, noting that the numerator is identically zero
f_{xy}(0,0) -1 start with \lim_{y \to 0} \frac{f_x(0,y) - f_x(0,0)}{y} and simplify using the expressions obtained above for f_x(0,y) and f_x(0,0)
f_{yx}(0,0) 1 start with \lim_{x \to 0} \frac{f_y(x,0) - f_y(0,0)}{x} and simplify using the expressions obtained above for f_y(x,0) and f_y(0,0)