# Differential 1-form and system of PDEs

I am working with the following difference of 1-forms $$d\omega = \alpha - \beta$$ where $d$ is the exterior derivative and $\alpha = ydz$ and $\beta = (y-2y\sin^2(x))dx + (\sin(x)\cos(x))dy + ydz$.

So \begin{align*} d\omega &=ydz - [(y-2y\sin^2(x))dx + (\sin(x)\cos(x))dy + ydz] \\ &= (2y\sin^2(x)-y)dx + (-\sin(x)\cos(x))dy +0dz. \end{align*}

My goal is to determine $\omega$. I have tried to anti-differentiate each term separately so that when the exterior derivative is taken, the result is $d\omega$. However, I was told that I will need to generate three partial differential equations for $\omega$ and then possibly use Maple or MatLab to solve the system of PDEs.

I am not sure how I am to generate these three PDEs from $$d\omega = (2y\sin^2(x)-y)dx + (-\sin(x)\cos(x))dy$$ and was hoping someone knows how to do this.

Since $d\omega$ is a 1-form, $\omega$ must be a 0-form, or a function. The total derivative is then given by \begin{equation} d\omega = \frac{\partial \omega}{\partial x}dx + \frac{\partial \omega}{\partial y}dy + \frac{\partial \omega}{\partial z}dz, \end{equation} so you need to set \begin{equation} \frac{\partial \omega}{\partial x} = 2y\sin^2(x)-y \end{equation} and \begin{equation} \frac{\partial \omega}{\partial y} = -\sin(x)\cos(x), \end{equation} and also require that $\partial\omega/\partial z=0$, and then solve for $\omega$.
Well, if I understand properly, $\omega$ is a function and thus $d\omega$ is an exact one-form. Since $d\omega = \alpha - \beta$, $$d(d\omega) = 0 = d\alpha - d\beta, \,\, \text{ that is } \,\, d\alpha = d\beta.$$ This is a necessary condition (and in the case of simply connected spaces, as is the case here, it is a sufficient one) that guarantees that if $d\alpha - d\beta = 0$ then there exists a function $\omega$ such that $d\omega = \alpha - \beta$. Thus, before you start, you need to check that $d\alpha = d\beta$. If that's not true, then you cannot find $\omega$. But if it is true, and your forms look like $$\alpha = \alpha_1 dx + \alpha_2 dy + \alpha_3dz$$ $$\beta = \beta_1 dx + \beta_2 dy + \beta_3dz$$ then $$\alpha - \beta = (\alpha_1 - \beta_1) dx + (\alpha_2 - \beta_2) dy + (\alpha_3-\beta_3) dz$$ $$d\omega = \frac{\partial \omega}{\partial x} dx + \frac{\partial \omega}{\partial y} dy + \frac{\partial \omega}{\partial z} dz$$ and then $d\omega= \alpha-\beta$ can be written component-wise \begin{align} \frac{\partial \omega}{\partial x} &= \alpha_1 - \beta_1\\ \frac{\partial \omega}{\partial y} &= \alpha_2 - \beta_2\\ \frac{\partial \omega}{\partial z} &= \alpha_3 - \beta_3\\ \end{align} and you need to integrate them. From the first equation $$\omega(x,y,z) = \int_{x_0}^{x}\Big(\alpha_1(\tilde{x},y,z) - \beta_1(\tilde{x},y,z)\Big)d\tilde{x} + g_1(y,z) = f_1(x,y,z) + g_1(y,z),$$ where $f_1(x,y,z) = \int_{x_0}^{x}\Big(\alpha_1(\tilde{x},y,z) - \beta_1(\tilde{x},y,z)\Big)d\tilde{x}$ and $g_1(y,z)$ is an unknown function only of $y$ and $z$. To find it, plug in the second equation $$\frac{\partial \omega}{\partial y} (x,y,z) = \frac{\partial f_1}{\partial y}(x,y,z) + \frac{\partial g_1}{\partial y}(y,z) = \alpha_2(x,y,z) - \beta_2(x,y,z),$$ express the derivative for the unknown function $g_1$: $$\frac{\partial g_1}{\partial y}(y,z) = \alpha_2(x,y,z) - \beta_2(x,y,z) - \frac{\partial f_1}{\partial y}(x,y,z),$$ the right hand side is guaranteed to be independent of $x$ by the closeness condition $d\alpha-d\beta = 0$. Then integrate with respect to $y$ to obtain and expression for $g_1(y,z)$ of the form $g_1(y,z) = g_2(y,z) + h_1(z)$, where $g_2(y,z)$ is a known function obtain by integration of the right hand side of the equation above and thus $\omega(x,y,z) = f_1(x,y,z) + g_2(y,z) + h_1(z)$ is plugged in the last equation of $$\frac{\partial \omega}{\partial z}(x,y,z) = \frac{\partial f_1}{\partial z}(x,y,z) + \frac{\partial g_2}{\partial z}(y,z) + \frac{\partial h_1}{\partial z}(z)= \alpha_3(x,y,z) - \beta_3(x,y,z)$$ and obtain an expression for the derivative of $h_1$ $$\frac{\partial h_1}{\partial z}(z) = \alpha_3(x,y,z) - \frac{\partial f_1}{\partial z}(x,y,z) - \frac{\partial g_2}{\partial z}(y,z) - \beta_3(x,y,z)$$ which is again guaranteed to depend only on $z$, so integrate it and obtain the last function $h_1(z)$. Thus $\omega(x,y,z) = f_1(x,y,z) + g_2(y,z) + h_1(z)$