If $\vec{\nabla} \times \langle P,Q,Q \rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is exact differential form.

Question

If $\vec{\nabla} \times \langle P,Q,R\rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is an exact differential form.

My attempt:-

If $Pdx+Qdy+Rdz$ is an exact differential form. Then there exists $U(x,y,z): dU=Pdx+Qdy+Rdz$. From the mixed partial theorem, It is easy to prove that $curl \langle P,Q, R\rangle=\vec{0}$.

If $\vec{\nabla} \times \langle P,Q,R \rangle=\vec{0}$

$\implies R_y=Q_z,R_x=P_z $ and $Q_x=P_y$. How do I prove the existence of scalar function $U$:$dU=Pdx+Qdy+Rdz$?

Answer 1

This is quite standard. However, you need an additional assumption to prove your result. The easiest is to assume that the domain on which $dU$ is defined is a star domain.

You can then find $U$ by simply integrating $dU$ from $\vec{r}_0= (x_0,y_0,z_0)$ to $\vec{r}=(x,y,z)$ along the path $$\vec{\gamma(}t) = t\vec{r}_0+ (1-t)\vec{r},\qquad t\in[0,1].$$ In particular, we set $$U(\vec{r}) = \int_{\gamma}(Pdx+Qdy+Rdz)\,. \tag{1} $$

Now, what is left to show is that $dU$ is indeed $(P,Q,R)$. The relevant property thereby is that in (1) the value of $U$ is independent of he path choses (and only depends on the endpoints). This fact is due to Stokes theorem and the vanishing of the curl. The result then follows from the fundamental theorem of calculus.

Answer 2

In general we know that $$dU = U_x dx + U_y dy + U_z dz $$ then you want to find $U$ such that $$P = U_x, Q = U_y, Z = U_z.$$ Note that from the curl equation you get that $U_{xy} = U_{yx}$ and $U_{xz}=U_{zx}$ and $U_{yz}=U_{zy}.$ From the first equation you get that: $$U = \int P(x,y,z)dx + \phi(y,z)$$ From the second equation you get that: $$Q = U_y = \frac{\partial }{\partial y}\int Pdx + \phi_y \implies \phi_y =Q - \frac{\partial }{\partial y}\int Pdx.$$ Since $\phi_y$ is a function of $y$ and $z$ we have that $\phi_{yx} = 0$ and thus $$\frac{\partial }{\partial x}\left(Q -\frac{\partial }{\partial y}\int Pdx\right)=0.$$ Thus $$\phi (y,z) = \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$ Thus our current expression for $U$ is
$$U = \int P(x,y,z)dx + \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$ Now recall that $$Z = U_z = \frac{\partial }{\partial z}\int P(x,y,z)dx + \frac{\partial }{\partial z}\int Q(x,y,z)dy-\frac{\partial }{\partial z}\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h_z(z).$$ Thus $$\frac{dh}{dz} = Z - \frac{\partial }{\partial z}\int P(x,y,z)dx - \frac{\partial }{\partial z}\int Q(x,y,z)dy+\frac{\partial }{\partial z}\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt.$$ And so $$h(z) = \int Z(x,y,z) dz - \int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right) dt - \int \frac{\partial }{\partial t}\left(\int Q(x,y,z)dy\right)dt+ \int \frac{\partial }{\partial m}\left[ \int\frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt \right]dm.$$ Thus we have $$U = \int P(x,y,z)dx + \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + \int Z(x,y,z) dz - \int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right) dt - \int \frac{\partial }{\partial t}\left(\int Q(x,y,z)dy\right)dt+ \int \frac{\partial }{\partial m}\left[ \int\frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt \right]dm$$