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I'm reading a book of differential geometry that states that it's easy to get the integral curves of this vector field:

$$X(x,y)=(x^2-y^2,2xy)$$

But proceeding the way the book says, first I have to take that equation as

$$\frac{dx}{dt}=x^2-y^2$$ $$\frac{dy}{dt}=2xy$$

But from that I don't know what more to do to proceed. I know that the integral curves must be circles of center $(0,t)$, and the vector field seems as this

enter image description here

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    Well, because in this case it's easy to solve the differential equation $\frac{dy}{dx} = \frac{\dots}{\dots}$ – 2017-01-03
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    So I don't need to consider the differential equations written there? – 2017-01-03
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    Well, you have to. The problem asks you about integral curves of this vector field. Integral curves can be found from solution of system of ODEs. The system of ODEs could be reduced to the first order ODE which can be solved. – 2017-01-03
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    You may use the chain rule to find $\frac{dy}{dx}$ and then solve it as a homogeneous ODE. By the way, which software do you use to create such image? – 2017-01-03
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    @Sameh I did it with Mathematica, but you can do something very similar in python using matplotlib and the ODE solver of sympy. – 2017-01-03
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    @MonsieurGalois thank you. – 2017-01-03
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    You're welcome. The utility with Mathematica is that it has a one-line command to do this stuff. – 2017-01-03

2 Answers 2

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Note that the system implies $$y'(x)=\frac{2xy(x)}{x^2-y(x)^2}.$$ (This is the approach mentioned by @Sameh Shenawy in the comments.) This can be solved by standard methods. For example, setting $y(x)=xf(x)$ we arrive at the separable differential equation $$f'(x) = \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2}.$$ The relevant integral may be done by a partial fraction expansion.

In my opinion the method put forward by @WimC is much preferable.

Addendum

The gory detail ... \begin{align*} f'(x) &= \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2} \\ \frac{1-f^2}{f(1+f^2)}df &= \frac{dx}{x} \\ \frac{(1+f^2)-2f^2}{f(1+f^2)} df &= \frac{dx}{x} \\ \int\left(\frac{1}{f} - \frac{2f}{1+f^2}\right)df &= \int\frac{dx}{x} \\ \ln f - \ln(1+f^2) &= \ln\frac{x}{2r} \\ \ln\frac{f}{1+f^2} &= \ln\frac{x}{2r} \\ \frac{f}{1+f^2} &= \frac{x}{2r} \\ \frac{y/x}{1+y^2/x^2} &= \frac{x}{2r} \\ \frac{y}{x^2+y^2} &= \frac{1}{2r} \\ x^2+y^2-2ry &= 0 \\ x^2 + y^2 - 2ry + r^2 &= r^2 \\ x^2 + (y-r)^2 &= r^2. \end{align*} (The integration constant was chosen to have the form $-\ln 2r$ for convenience.)

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    Could you explain the standard methods you're referring? – 2017-01-04
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    @MonsieurGalois: Added. – 2017-01-04
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Indentifying $(x,y)$ with $z=x + y \, \mathrm{i}$ the differential equation reads $$\frac{\partial z}{\partial t}=z^2.$$ Rewriting this we find $$\frac{-\partial z^{-1}}{\partial t} = \frac{\partial z}{\partial t} z^{-2} = 1$$ and so $$-z^{-1}= t+z_0 \textrm{ or } z = -\frac1{t+z_0}$$ for some fixed initial value $z_0 \in \mathbb{C}$. For $t \in \mathbb{R}$ these describe circular trajectories as expected.

Alternative 1: transform the vector field under the inversion

$$\sigma(x, y) = \left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$

This map is an involution ($\sigma^2 = \mathrm{id}$) and its Jacobian is $$D\sigma_{(x,y)}=\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix}.$$

Therefore it maps the element $$\begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix}$$ at point $(x, y)$ to the element

$$\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix} \begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$$ at point $\sigma(x,y)$. The trajectories of the transformed vector field are easy to find. Then apply the inversion $\sigma$ to find the trajectories of the original vector field.

Alternative 2: Since you know that the trajectories are circular start with the ansatz $$\begin{cases} x = -R \sin \alpha \\ y = R + R \cos \alpha \end{cases}$$ for some function $\alpha$ of $t$. Then the given differential equations are consistent and both lead to $$\frac{\partial \alpha}{\partial t} = 2R \, (1 + \cos\alpha).$$ One solution to this equation is $\alpha(t) = 2 \arctan(2R \, t)$. Substituting this in the expressions for $x$ and $y$ gives the trajectory

$$\left(\frac{-t}{t^2 + (2R)^{-2}}, \frac{(2R)^{-1}}{t^2+(2R)^{-2}}\right).$$

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    I see that this solution is correct, but I don't think this is the author's "easy" way to get the integral curves. – 2017-01-03
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    @MonsieurGalois: "Easy" is in the eye of the beholder; the components of your vector field are obviously the real and imaginary parts of the complex squaring map. ;) – 2017-01-05
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    @AndrewD.Hwang That's obviously true, but the section where this problem is added is the first of the book and it only contains an explanation about calculus (that could be the explained in Spivak) and topology (four theorems from the Munkres book), not more. That's the reason of why I don't think that this is the "easy" way. – 2017-01-05