Conservative Vector Fields — Question 2

Question 2

Consider $$\mathbf F(x,y)=\langle y^2,\,2xy+x\rangle$$ on $\mathbb R^2$.

Tasks

1. Test whether $\mathbf F$ is conservative.

2. Identify the exact obstruction.

3. Explain why no potential function can exist even locally near any point.

Question 2 – Solution

Strategy. A continuously differentiable gradient field must satisfy equality of the cross partials.

Step 1: Compute the cross partials Let $P=y^2$ and $Q=2xy+x$. Then $$P_y=2y, \qquad Q_x=2y+1.$$ Since $P_y\ne Q_x$ at every point, the necessary condition fails. $$\boxed{\mathbf F\text{ is not conservative on }\mathbb R^2.}$$

Step 2: Identify the contradiction If $\mathbf F=\nabla f$ for a twice continuously differentiable $f$, then $$f_x=P, \qquad f_y=Q,$$ and hence $f_{xy}=P_y$ and $f_{yx}=Q_x$. Equality of mixed partials would require $2y=2y+1$, which is impossible.

Step 3: Local conclusion The mismatch is the constant $1$ everywhere, not merely at an isolated point. Every open neighborhood contains points where the necessary equality fails, so no local potential exists on any such neighborhood.

Verification Integrating $P$ gives $f=xy^2+g(y)$, whose $y$-derivative is $2xy+g'(y)$. Matching $Q$ would require $g'(y)=x$, impossible because $g'$ cannot depend on $x$.