1
$\begingroup$

Find all integral curves to the vector field $v(x,y) = x^2\, \partial_x + y\, \partial_y$. See which ones are defined for all $t \in \mathbb{R}$

So, I have to find all curves $\gamma(t)$ that satisfy $v(\gamma(t))=\gamma'(t)$. Right? Let's assume that $\gamma(t)=(\gamma_1(t),\gamma_2(t))$. By plugging it in $v$, we get the equations

$$\frac{d\gamma_1}{dt}=\gamma_1^2 \implies \gamma_1=0 \,\,\,\text{ or }\,\,\,\, \frac{d\gamma_1}{\gamma_1^2}=dt\implies\gamma_1(t)=\frac{1}{C_1-t} \,\text{ or }\, \gamma_1(t)=0$$ $$\frac{d\gamma_2}{dt}=\gamma_2 \implies \gamma_2 = C_2e^t$$

So, it seems that all integral curves for $v$, in the absence of any information about the starting point, must be of the form $\gamma(t)=(\frac{1}{C-t},Ke^t)$ or $\gamma(t)=(0,Ke^t)$.

But what about $\gamma_{\star}=(\frac{1}{t},e^t)$? This also seems to be an integral curve but I can't see how it can be found by solving the differential equations.

Where is my mistake?

Edit: I just figured the answer and realized that I had made a stupid calculational mistake. $\gamma_{\star}=(\frac{1}{t},e^t)$ doesn't satisfy $\gamma_1^2 \partial_x + \gamma_2 \partial_y = \gamma_1' \partial_x + \gamma_2' \partial_y$.

$\endgroup$
2
$\begingroup$

Your $\gamma_1$ isn't correct. $$ \frac{\mathrm{d}\gamma_1}{\gamma_1^2}=\mathrm{d}t\implies\gamma_1(t)=\frac{1}{C-t}. $$

$\endgroup$
3
  • $\begingroup$ OK. Thanks. I'll fix it. But how does this answer my question? How do you get $(\frac{1}{t},Ke^t)$ as an integral curve? That's an integral curve. Isn't it? $\endgroup$ – stressed out May 23 '19 at 20:48
  • $\begingroup$ $(t^{-1},e^t)$ is not an integral curve. At any point $(c^{-1},e^c)$, the integral curve through that point is $(1/(c-t),e^{c+t})$. $\endgroup$ – user10354138 May 23 '19 at 20:54
  • $\begingroup$ not sure what you mean. The $C$ is there for the finite-time blowup (forward or backward in time). $\endgroup$ – user10354138 May 23 '19 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.