Prove $gF$ is a conservative vector field if and only if $\nabla g$ is parallel to $F$ everywhere

Question

What I am given in the problem:

$F(x,y) = (P(x, y), Q(x, y))$ is a conservative vector field defined on $\mathbb{R}^2$

$P$ and $Q $ are smooth functions, and $F \neq(0, 0)$ for all points

Problem: if $g:\mathbb{R}^2 \to \mathbb{R}$ is a smooth function, prove that $gF$ is conservative if and only if $\nabla g$ is parallel to F everywhere.

I know that since $F$ is conservative and $P$ and $Q $ are smooth functions, then $\frac{\delta Q}{\delta x}=\frac{\delta P}{\delta y} $ and that there exists an $f$ such that $\nabla f = F$. Should I assume that the given $g$ is the potential function of F ?

Any help would be greatly appreciated!

Answer 1

If $gF$ is conservative, then

$\nabla \times (gF)= 0; \tag 1$

we have

$\nabla \times (gF) = \nabla \times (gP, gQ) = \dfrac{\partial (gQ)}{\partial x} - \dfrac{\partial (gP)}{\partial y} = (gQ)_x - (gP)_y, \tag 2$

where we have introduced subscript notation for partial derivatives in the right-most equality in (2).

We further have

$(gQ)_x - (gP)_y = g_x Q + gQ_x - g_yP - gP_y; \tag 3$

then from (1) and (2),

$(gQ)_x - (gP)_y = g_x Q + gQ_x - g_yP - gP_y = 0, \tag 4$

or

$g_xQ - g_yP = gP_y - gQ_x = g(P_y - Q_x) = -g \nabla \times (P, Q) = -g \nabla \times F = 0, \tag 5$

since $F = (P, Q)$ is itself conservative, i.e.,

$\nabla \times F = 0; \tag 6$

thus

$g_xQ - g_y P = 0; \tag 7$

now if

$\nabla g = 0, \tag 8$

then (7) trivially holds; I suppose this may be considered a special case of co-linearity since then

$\nabla g = 0 (P, Q); \tag 9$

having dealt with this eventuality, we may henceforth assume that

$\nabla g \ne 0; \tag{10}$

since

$F = (P, Q) \ne 0, \tag{11}$

at least one of $P, Q \ne 0$ at each point $(x, y)$; suppose then for the moment that

$P \ne 0; \tag{12}$

then from (7),

$g_y = \dfrac{Q}{P} g_x; \tag{13}$

thus

$\nabla g = (g_x, g_y) = (g_x, \dfrac{Q}{P} g_x) = \dfrac{g_x}{P}(P, Q) = \dfrac{g_x}{P}F; \tag{14}$

in the light of (10) we see that

$g_x \ne 0 \ne \dfrac{g_x}{P}, \tag{15}$

and thus $\nabla g$ and $F$ are co-linear in this case. A parallel argument applies in the event that

$Q \ne 0, \tag{16}$

in which case we obtain

$\nabla g = \dfrac{g_y}{Q}F, \tag{17}$

with

$g_y \ne 0, \tag{18}$

and again, $\nabla g$ and $F$ are colinear.

Answer 2

(I'm reading your $\Delta$ as $\nabla$.)

Since both ${\bf F}$ and $g$ are defined in all of ${\mathbb R}^2$ it is sufficient to look at $${\rm curl}(g\,{\bf F})=(gQ)_x-(gP)_y=g(Q_x-P_y)+(g_xQ-g_yP)\ .$$ Here the first part on the RHS vanishes by assumption on ${\bf F}$. The second part $$g_xQ-g_yP={\rm det}\left[\matrix{g_x&g_y\cr P&Q\cr}\right]$$ vanishes iff the vectors $\nabla g=(g_x,g_y)$ and $(P,Q)\ne{\bf 0}$ are linearly dependent, and this is the case iff $\nabla g(x,y)=\lambda\, {\bf F}(x,y)$. Since we want this to be true for all $(x,y)\in{\mathbb R}^2$ the claim follows.