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Suppose we have a $1$-dimensional differential inequality

$$\frac{dx}{dt} \leq x - x^3 $$

We can apply the Comparison principle to claim that if $y(t)$ is the solution to $\frac{dy}{dt} = y - y^3$, then $x(t) \leq y(t)$ (assuming $x(0) \leq y(0)$. Can we extend similar argument to a $2$-dimensional system? For example, let us consider the following system of equations

$$\frac{dx_1}{dt}=x_1-x_2, \quad \quad \quad \frac{dx_2}{dt} \leq x_1 + x_2 - \frac{x_2^4}{x_1^2}$$

Is the solution to following

$$\frac{dy_1}{dt}=y_1-y_2, \quad \quad \quad \frac{dy_2}{dt} = y_1 + y_2 - \frac{y_2^4}{y_1^2}$$

related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that $x_2(t) \leq y_2 (t)$ and then subsequently use it to claim $x_1(t) \geq y_1(t)$?

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I don't know about the specific example you give. However, the answer is "no" for general problems, even if they are linear. Consider the structure:
\begin{align} x' &= f(x,y)\\ y' &=g(x,y) \leq h(x,y) \end{align} Specifically, compare the following two systems:

System 1

\begin{align} x' &= x-y\\ y' &= \underbrace{2x-y-1}_{g(x,y)} \end{align}

System 2

\begin{align} w' &= w-z\\ z' &= \underbrace{2w-z}_{h(w,z)} \end{align} Suppose initial conditions are $(x(0),y(0))=(w(0),z(0))=(0,0)$. Then $(w(t),z(t))=(0,0)$ for all $t \geq 0$. However, $(x(t),y(t))$ has solution: \begin{align} x(t) &= 1-\cos(t)\\ y(t) &= 1-\sin(t) - \cos(t) \end{align} and indeed $y(t)$ takes both positive and negative values over $t \in [0, \infty)$. So we cannot say that $y(t) \leq \underbrace{z(t)}_{0}$ for all $t$.

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  • $\begingroup$ Thank you for your answer. I am still trying to figure out why the comparison lemma based argument doesn't work. To solve the coupled ODE, we can plug in the solution of $x(t)$ in terms of $y(t)$ as $x(t)=e^{t}\int_{\tau=0}^{t}e^{-\tau}y(\tau)d\tau$ ($x(0)=0$ assumed). Same could be done for $w(t)$. Now we have two equations as: $y'=2e^{t}\int_{\tau=0}^{t}e^{-\tau}y(\tau)d\tau - y(t)-1, z'=2e^{t}\int_{\tau=0}^{t}e^{-\tau}z(\tau)d\tau - z(t)$. Based on the above counter example, we cannot say that $y(t) \leq z(t)$ even though we have $y(0)=z(0)$ and $y'(t) < z'(t)$. $\endgroup$ – K. Ghusinga Jul 14 '16 at 16:10
  • $\begingroup$ Is it because $y'$ and $z'$ involve integrals of $y$ and $z$ over past values, and in some sense are not functions of only current values of $y(t)$ and $z(t)$? $\endgroup$ – K. Ghusinga Jul 14 '16 at 16:18
  • $\begingroup$ Yes, the two integrals $\int_0^t e^{-\tau} y(\tau)d\tau$ and $\int_{0}^te^{-\tau}z(\tau)d\tau$ are comparing state values over the interval $[0, t]$, rather than instantaneous state values. In the counter-example above, $z(\tau)=0$ for all $\tau$, while the integral of $y(\tau)$ starts out negative (corresponding to your intuition) but, because $y()$ changes, it affects changes in $x()$, which then affects changes in $y()$ again, so that $y()$ eventually goes positive. I chose eigenvalues of $i$ and $-i$ so that we would get oscillatory behavior. $\endgroup$ – Michael Jul 14 '16 at 17:33

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