Let $\omega = x dx - ydy$ be a $1$-form on $\mathbb{R}^2$. Show that it's exact and find $f$ such that $\omega = df$.

Question

Let $\omega = x dx - ydy$ be a $1$-form on $\mathbb{R}^2$. Show that it's exact and find $f$ such that $\omega = df$.

To find such $f$ we need to show that $$xdx-ydy = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$ and this is equivalent to finding $f$ such that $$\frac{\partial f}{\partial x} = x$$ and $$\frac{\partial f}{\partial y} = -y.$$

Now I don't know much about differential equations so how do we solve this kind of system to find $f$? If $f$ were a single variable function we could just integrate both sides of the equations, but $f$ is now a map $\mathbb{R}^2 \to \mathbb{R}$ so I guess we need to use some other methods here.

Answer 1

You can see the problem as an ODE when you fix $y\in\mathbb{R}$. Let $y$ be fixed. Let us call $f_y:\mathbb{R}\to\mathbb{R}$, $x\mapsto f(x,y)$. This is a one-variable function which derivative is $f_y'(x)=\frac{\partial f}{\partial x}(x,y) = x$, so we must have that $f_y(x) = f(x,y)= \frac{x^2}{2} + C(y)$, where $C(y)$ is a “constant of integration” which can depend on $y$.

Now, returning to the general setting, since $f$ is differentiable w.r.t. $y$, we have that $C(y) = f(x,y) - \frac{x^2}{2}$ is also differentiable. Hence: $$\frac{\partial f}{\partial y}(x,y) = C’(y) = -y$$ So we conclude that $C(y) = \frac{-y^2}{2} + k$, where $k\in \mathbb R$ is a constant of integration. Finally, $f(x,y) = \frac{x^2-y^2}{2} + k$.

You can verify that this $f$ indeed satisfies $\mathrm{d}f = \omega$.

Answer 2

You should check that the 1-form is, at least, closed: $d\omega=0$.

Then, you can integrate your first equation with respect to $x$ to obtain:

$$f(x,y) = \frac{1}{2}x^2 + g(y),$$

where $g(y)$ is an arbitrary function of $y$. Taking the partial derivative of $f$ with respect to $y$, we have:

$$\frac{\partial f}{\partial y} = g'(y),$$ Equating this to $-y$, we get:

$$g'(y) = -y.$$

and integrating this equation with respect to $y$, we obtain:

$$g(y) = -\frac{1}{2}y^2 + C,$$

where $C$ is a constant of integration.

Thus, the function $f(x,y)$ is given by:

$$f(x,y) = \frac{1}{2}x^2 - \frac{1}{2}y^2 + C.$$