I am reading an article and suddenly it talks about one-form annihilated by a vector field. I just want to understand the concept of annihilator method so I am trying to make an easy example to verify if I understood the concept. So my question is: are my steps right?
I want to verify that $u(x)=ce^x$ is a solution of the differential equation $u'=u$. So the one-form annihilated by the corresponding vector field of differential equation is (a multiple of) $\theta=du-udx$.
Now i have to verify that $\theta$ vanishes on the curve $(x,ce^x)$ and so I calculated the associated vector field $X=(1,ce^x)$(or in local coordinate $X=\partial_x+ce^x\partial_u$) to calculate $(\theta,X)$ where $(.,.)$ is the interior product. So
$(\theta,X)=(du-udx, \partial_x+ce^x\partial_u)=(du,\partial_x+ce^x\partial_u)-u(dx,\partial_x+ce^x\partial_u)=ce^x-u$.
where the second equality holds by linearity and third equality for $(dx,du)$ is a dual basis.
So replacing $u(x)=ce^x$ in the last term, I have $(\theta,X)=0$ and then $u(x)$ is an integral curve for the equation $u'=u$.
Are these steps right? Thxxx
It's pretty much correct, except that the way you have written it makes it a little ambiguous where you are evaluating the contraction $(\theta, X)$. This manifests in the unmotivated equating of $u=c e^x$ at the end. Adding in the point of evaluation explicitly $$ (\theta_{(x,ce^x)},X_{(x,ce^x)}) = (du-ce^x dx, \partial_x+ce^x\partial_u) = -ce^x+ce^x = 0. $$