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Let $\Omega:=\{(x_1,x_2) \subset \mathbb{R}^2 | x_2>0\}$. I want to solve the differential equation $$\begin{pmatrix} \dot{x_1} \\\dot{x_2} \end{pmatrix}=\begin{pmatrix}x_2^2-x_1^2 \\-2x_1x_2\end{pmatrix}$$I have no idea how to do that, since I have only seen linear differential equations so far. I'd appreaciate any starting point here. I have the hint to look at $z=x_1+ix_2$ but that confuses me even more. Thank you!

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    $\begingroup$ The hint is useful: can you manage to get terms like $x_2^2 - x_1^2$ and $2 x_1 x_2$ from $z$ somehow, if $z = x_1 + ix_2$? Can you rearange this equation as a single, complex differential equation: $\dot z = f(z)$? $\endgroup$ – Lurco Jun 29 '15 at 9:34
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If $z(t)=x_1 (t) + \Bbb i x_2 (t)$ then $$- z^2 = (x_2 ^2 - x_1 ^2) - 2 x_1 x_2 \Bbb i$$ Therefore, your equation can be rewritten as $\dot z = - z ^2$ or, equivalently, $$\frac{\dot z}{z^2} = -1$$ Integrate this with respect to $t$, getting $1 /z = t + C$, or $z(t) = 1/(t+C)$, with $C=a + b \Bbb i$ an integration constant. To get back to $x_1$ and $x_2$, write $$x_1 + x_2 \Bbb i = \frac1{t+a + b \Bbb i} = \frac{t+a - b \Bbb i}{(t+a)^2 + b^2}$$ Equating the real and imaginary parts, $$x_1 = \frac{t+a}{(t+a)^2 + b^2}\qquad x_2 = \frac{-b}{(t+a)^2 + b^2}$$ To have $x_2 > 0$ (as in the definition of $\Omega$), you will have to take $b<0$.

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  • $\begingroup$ @Did: Thank you, that's what happens when you copy & paste without paying attention... $\endgroup$ – Alex M. Jun 29 '15 at 10:36
  • $\begingroup$ Is it just me and my browser or does $\dot{z}$ in $\frac{\dot{z}}{z^2}$ look almost like italic $i$ to others? I got a bit confused there. $\endgroup$ – JiK Jun 29 '15 at 13:50
  • $\begingroup$ @JiK: Well, when it's written that small, $\Bbb i$ can easily be mistaken for $\dot z$. In any case, this has nothing to do with the browser, but rather with the font family and font size used for displaying (and these can be changed in the options menu of each browser). Note, also, that some nice looking fonts might not be installed on your system. $\endgroup$ – Alex M. Jun 29 '15 at 13:53
  • $\begingroup$ A good reason to use $\dot z/z^2$, and more generally $A/B$ rather than $\frac{A}B$, when not in displayed mode. $\endgroup$ – Did Jun 29 '15 at 16:49
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Observe how the right hand side looks like the real and the imaginary term of a squared complex number. In fact, if you write

$$z^2=x_1^2-x_2^2+i (2x_1x_2)$$ you can see, that if you multiply the second equation by $i$ and add it to the first one, you will get $$\dot{z}=-z^2$$ which is trivially solvable.

This is a very handy trick that can be used in physics quite frequently when there is some intrinsic rotational motion going on (or something like that).

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