differential-geometrydifferential-topologyordinary differential equationssmooth-manifoldsVector Fields
Integral curve of vector field $W = x^2 \frac{\partial}{\partial x}$.
My textbook is saying the unique integral curve of the vector field given by $W = x^2 \frac{\partial}{\partial x}$ starting at $(1,0)$ is:
$\gamma(t) = (\frac{1}{1-t},0)$.
This confuses me a bit, clearly $W_{(1,0)}=(2,0)$ and yet $\gamma(0)=(1,0)$. What am I not understanding here?
How about a vector field in the plane with a stable limit cycle?
$$ \frac{dr}{dt} = r(1-r) \\ \frac{d \theta}{dt} = r $$ If I'm not mistaken, this is a continuous vector field with an unstable singularity at the origin such that any initial condition in the punctured disk $\{1 > r > 0\}$ tends toward the periodic orbit $\{r = 1\}$.
The vector field $X = w_{1}\left(q_{1}\frac{\partial}{\partial p_{1}} - p_{1}\frac{\partial}{\partial q_{1}}\right) + w_{2}\left(q_{2}\frac{\partial}{\partial p_{2}} - p_{2}\frac{\partial}{\partial q_{2}}\right)$ is essentially comprised of the rotational fields flowing clockwise about the origin in the $p_1q_{1}$-plane and the $p_{2}q_{2}$-plane, respectively. For example, the plot of the vector field $ Y = q\frac{\partial}{\partial p} - p\frac{\partial}{\partial q}$ in the $pq$-plane is given below.
The rotational symmetry of the field is evident and the factors of $w_{1}$ an $w_{2}$ merely scale the speed with which one flows around the integral curves.
Recognizing this, you should be able to find the integral curves geometrically. The only trick will be to get your curve passing through your desired initial point.
Taking the initial point to be $P(1, 0, 1, 0)$, you should find the integral curve to be $$ \gamma(t) = \left(p_{1}(t), q_{1}(t), p_{2}(t), q_{2}(t)\right) $$ where \begin{align*} p_{1}(t) = \cos w_{1}t\\ q_{1}(t) = -\sin w_{1}t\\ p_{2}(t) = \cos w_{2}t\\ q_{2}(t) = \sin w_{2}t.\\ \end{align*}
Now take the initial point to be $P(p_{1}, 0, p_{2}, 0)$, where $p_{1}$ and $p_{2}$ are both greater than zero. Then the integral curve should be built out of the circle of radius $p_{1}$ centered at the origin in the $p_{1}q_{1}$-plane and the circle of radius $p_{2}$ centered at the origin in the $p_2q_{2}$-plane. Again, both circles should be traced out clockwise.
That is, you should have $$ \gamma(t) = \left(p_{1}(t), q_{1}(t), p_{2}(t), q_{2}(t)\right) $$ where \begin{align*} p_{1}(t) &= p_{1}\cos w_{1}t\\ q_{1}(t) &= -p_{1}\sin w_{1}t\\ p_{2}(t) &= p_{2}\cos w_{2}t\\ q_{2}(t) &= -p_{2}\sin w_{2}t.\\ \end{align*}
You should be able to pivot directly from here to the desired curves. You merely need to amp up your rotational speed in both planes.
In particular, let $r_{i}$ denote the distance from a point in the $p_1q_{i}$-plane to the origin in the $p_{i}q_{i}$-plane (i.e., $r_{i} = \sqrt{p_{i}^2 + q_{i}^2}$) and $\alpha_{i}$ denote the oriented angle measured clockwise from the $p_{i}$ axis in the $p_{i}q_{i}$-plane.
Then the integral curve through a beginning at a point $P(p_{1}, q_{1}, p_{2}, q_{2})$ should take the form $$ \gamma(t) = \left(p_{1}(t), q_{1}(t), p_{2}(t), q_{2}(t)\right), $$ where \begin{align*} p_{1}(t) &= r_{1}(0)\cos\left(w_{1}\left(t + \frac{\alpha_{1}(0)}{w_{1}}\right)\right)\\ q_{1}(t) &= -r_{1}(0)\sin\left(w_{1}\left(t + \frac{\alpha_{1}(0)}{w_{1}}\right)\right) \\ p_{2}(t) &= r_{2}(0)\cos\left(w_{2}\left(t + \frac{\alpha_{2}(0)}{w_{2}}\right)\right)\\ q_{1}(t) &= -r_{2}(0)\sin\left(w_{2}\left(t + \frac{\alpha_{2}(0)}{w_{2}}\right)\right), \\ \end{align*} and the angles $\alpha_{i}(0)$ are angle measures of the projection $(p_{i}, q_{i})$ in the indicated polar coordinates of the $p_{i}q_{i}$-plane.
Further, note that the quantities $r_{i}(t)= \sqrt{\left(p_{i}(t)\right)^2 + \left(q_{i}(t)\right)^2}$, $i= 1, 2$, are conserved quantities for the flow.
Best Answer
If $\gamma$ is to be an integral curve, then at any point $p$ on $\gamma$, the value of $W$ at that point corresponds to the derivative of $\gamma$ at that point.
Specifically, for $t = 0$, when we have the point $p = \gamma(0) = (1,0)$, the derivative $\gamma'(0)$ is indeed equal to $W_p = (1^2, 0)$. The same will be true for any other value of $t$, which makes $\gamma$ the integral curve you're after.