$P_y = Q_x$ but $P_x, Q_y$ not continuous?

Question

Stewart - Calculus

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Suppose all the assumptions hold but $P_x$ or $Q_y$ is not continuous, is the conclusion necessarily false?

If so, why? Also, please give an example.

If not, please give an example where the conclusion still holds.

Answer 1

We don't need $P_x$ or $Q_y$ continuous, but at least one of $P_y$ or $Q_x$ continuous. For an example consider the function $$f(x,y):=\left\{\eqalign{&{x^2y-xy^3\over x^2+y^2}\qquad(x^2+y^2\ne0) \cr &0\qquad\qquad\ \qquad(x=y=0)\cr}\right.$$ Then ${\bf F}:=\nabla f$ is a conservative vector field with $$P(x,y)={x^4y+4x^2y^3-y^5\over(x^2+y^2)^2},\quad Q(x,y)={x^5-4x^3y^2-xy^4\over(x^2+y^2)^2}\qquad(x^2+y^2\ne0)$$ and $P(0,0)=Q(0,0)=0$ (check this!). One now computes $$P_y(0,0)=\lim_{h\to0}{P(0,h)-P(0,0)\over h}=\lim_{h\to0}{-h^5/h^4\over h}=-1\ ,$$ and similarly $Q_x(0,0)=1$.