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I want to find all the primitives of this differential form:

$\omega\left(x,y,z\right)=\left(y^{3}-x^{3}\right)dx\land dy+\left(x-2z\right)dy\land dz+\left(2z-y\right)dz\land dx$

I know that:

$d\left(Fdx+Gdy+Hdz\right)=\left(G_{x}-F_{y}\right)dx\land dy+\left(H_{y}-G_{z}\right)dy\land dz+\left(F_{z}-H_{x}\right)dz\land dx$

So, in this case:

$G_{x}-F_{y}=y^{3}-x^{3}$

$H_{y}-G_{z}=x-2z$

$F_{z}-H_{x}=2z-y$

After that I don't know how to continue. Any pdf related to that question would be appreciated :)

Jack Talion
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1 Answers1

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Instead of trying to solve the system of partial differential equations that you’ve developed, you can find an antiderivative of $\omega$ using the algorithm described here.

Starting from $(y^3-x^3)\,dx\wedge dy + (x-2z)\,dy\wedge dz + (2z-y)\,dz\wedge dx$, step one of the algorithm produces $$t^3(y^3-x^3)\,(t\,dx+x\,dt)\wedge(t\,dy+y\,dt) + t(x-2z)\,(t\,dy+y\,dt)\wedge(t\,dz+z\,dt) + t(2z-y)\,(t\,dz+z\,dt)\wedge(t\,dx+x\,dt).$$ Expanding this and applying step two yields $$t^3(y^3-x^3)(tx\,dt\wedge dy-ty\,dt\wedge dx) + t(x-2z)(ty\,dt\wedge dz-tz\,dt\wedge dy) + t(2z-y)(tz\,dt\wedge dx-tx\,dt\wedge dz).$$ Collecting terms and integrating, we then have $$F = \int_0^1 z(2z-y)t^2-y(y^3-x^3)t^4\,dt \\ G = \int_0^1 x(y^3-x^3)t^4-z(x-2z)t^2\,dt \\ H = \int_0^1 y(x-2z)t^2-x(2z-y)t^2\,dt.$$ I’ll leave evaluating these integrals to you. This gives you one antiderivative of $\omega$, but just as indefinite integrals of scalar-valued functions have an arbitrary constant of integration, we can add any form $\tau$ with $d\tau=0$ to this to get another antiderivative. In particular, since we ended up with a 1-form, we can add $df$ to it, where $f$ is an arbitrary scalar-valued function defined on the same domain.

amd
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