Integral curves for $\vec X=\big(x\ln(x),-y\ln(y)\big)$

Question

Consider the vector field $$\vec X=\big(x\ln(x),-y\ln(y)\big)$$

Q: What are the integral curves and how can the vector field be written as a differential equation? How is it related to $\vec Y=\big(x,-y\big)?$

Attempt to write $\vec X$ as a differential equation:

$$\dot x=x\ln(x)$$

$$ \dot y=-y\ln(y)$$

I think $\vec X$ and $\vec Y$ are related by multiplying the $x-$component of $\vec Y$ by $\ln(x)$ and multiplying the $y-$component of $\vec Y$ by $\ln(y).$

Answer 1

We can find the integral curves of $\vec{X}$, as you have mentioned above

$$\begin{cases} \frac{dx}{dt} = x\ln(x) \\ \frac{dy}{dt} = -y\ln(y) \end{cases} \Rightarrow \frac{dx}{x\ln x} = \frac{dy}{-y\ln y} \Rightarrow \ln(\ln x) = -\ln (\ln y) + c \Rightarrow e^{\ln (\ln y)} = e^{-\ln(\ln x)+c} \\ \Rightarrow \ln y = \frac{A}{\ln x} \Rightarrow \ln x \ln y = A$$

The filed's tangent curves are behaving like hyperbolas.

For the $\vec{Y}$ similarly, we have

$$\begin{cases} \frac{dx}{dt} = x \\ \frac{dy}{dt} = -y \end{cases} \Rightarrow \frac{dx}{x} = \frac{dy}{-y} \Rightarrow \ln x = -\ln y + d \Rightarrow y=\frac{B}{x}$$

These two actually resemble the similar curves. Please notice that for $\vec{X}$ the doman is $\mathbb{R}^+\times \mathbb{R}^+$, in this domain we can define $x=e^t$ and $y=e^s$ where $(t,s)\in \mathbb{R}\times\mathbb{R}$ and we get $\ln x \ln y = A \Rightarrow st=A \Rightarrow s = \frac{A}{t}$, Although what you actually get in $x-y$ plane are different curves, but they are similar to hyperbolas.

That's true for integral curves, but the vector fields are different as we see in these figures

$(x\ln(x),-y\ln(y))$

$(x,-y)$

This is the MATLAB code for generating those figs.

close all force
clear 
clc
[x,y] = meshgrid(1:0.1:5,1:0.1:5);
u = x .* log(x);
v = -y .* log(y);

z = x;
w = -y;

figure
quiver(x,y,u,v)
figure
quiver(x,y,z,w)
grid on