Is There a Function G Such That G_x Equals -F_y and G_y Equals F_x in R?

[Medicol]

Suppose all second partial derivatives of $F = F (x, y)$ are continuous and $F_{xx} + F_{yy} = 0$ on an open rectangle $R$.
Show that $$F_ydx - F_xdy = 0$$ is exact on $R$, and therefore there’s a function $G$ such that
$$G_x = −F_y$$ and $$Gy = F_x$$ in $R$.
≈≈≈≈≈≈≈≈

To prove that $F_ydx + F_xdy = 0$ is exact on $R$,
I have $$F_{xx} + F_{yy} = 0$$
which is $$F_{xx}=-F_{yy}$$
Integrating both sides and cancel out the constants I obtain
$$F_x=-F_y$$
This proves the function $F_ydx - F_xdy = 0$ is exact on $R$​

Could you help me prove the existence of $G$ ? Thank you...

 

[WWGD]

Suppose all second partial derivatives of $F = F (x, y)$ are continuous and $F_{xx} + F_{yy} = 0$ on an open rectangle $R$.
Show that $$F_ydx - F_xdy = 0$$ is exact on $R$, and therefore there’s a function $G$ such that
$$G_x = −F_y$$ and $$Gy = F_x$$ in $R$.
≈≈≈≈≈≈≈≈

To prove that $F_ydx + F_xdy = 0$ is exact on $R$,
I have $$F_{xx} + F_{yy} = 0$$
which is $$F_{xx}=-F_{yy}$$
Integrating both sides and cancel out the constants I obtain
$$F_x=-F_y$$
This proves the function $F_ydx - F_xdy = 0$ is exact on $R$​

Could you help me prove the existence of $G$ ? Thank you...

If your rectangle lives in the plane, then just show the form $$F_ydx - F_xdy = 0$$ is closed, since in a rectangle, every closed form is exact. If you haven't seen this, what results can you use?

 

[Borek]

$$F_{xx}=-F_{yy}$$
Integrating both sides and cancel out the constants

How do you know constants are identical?

 

[Medicol]

I answered so, I meant to choose 2 exact constants to give both a go.
I'm thinking because I already answered the first which is also the main part of the problem, I may continue to put "the given function has become exact on R, so there must be a function G that satisfies both of the given conditions in R"

Is this a correct solution ?