Suppose I have two ODEs, $$\frac{dy}{dt}=f(y), \ \ \ \frac{dx}{dt}= Tf(x),$$ where $t$ is the time and $T$ is the terminal time. Also $x(0)=y(0)=c$.

How to show that $y(T)=x(1)$? This means the value of $y(T)$ for the LHS ODE is equal to $x(1)$ of the RHS ODE.

My effort: I only know that for any $t_p\in [0,T]$, LHS ODE $$\frac{dy}{dt}|_{t=t_p}= f(y(t_p))$$however, RHS ODE $$\frac{dx}{dt}|_{t=t_p}= Tf(x(t_p)),$$ i.e., at each point, the slope is $T$-times the previous one.

But I still have no idea about how to show $x(1) = y(T)$?

Consider an example:

Suppose $x(0)=1$, and consider the following MATLAB code

tf = 4;
x0 = 1;
tspan = [0 tf];
sspan = [0 1];
x0 = 1;
[t,x] = ode45(@(t,x) x, tspan, x0);
[s,y] = ode45(@(s,y) tf*y, sspan, x0);

enter image description here

In this case, $x(4)=x(1)$ (blue line is the second system.)

I just edited my question so I am sorry for the first answer from Kavi Rama Murthy.

  • $\begingroup$ This is awesomely unclear. It would be good to use different letters for the two equations: $x'=f(x)$, $y'=Tf(y)$... $\endgroup$ Commented Jun 28, 2019 at 22:45
  • $\begingroup$ Btw you can't possibly show $y(1) = x(T)$, because you didn't specify initial conditions - both equations have infinitely many solutions. $\endgroup$ Commented Jun 28, 2019 at 22:47
  • $\begingroup$ @DavidC.Ullrich Sorry for that, I modify it a bit. $\endgroup$
    – Denny
    Commented Jun 29, 2019 at 2:01
  • 1
    $\begingroup$ @Denny You changed the question completely and then commented that my answer is wrong. You should mention in your question that you have edited the question. $\endgroup$ Commented Jun 29, 2019 at 4:45
  • $\begingroup$ @KaviRamaMurthy I added that in my question. Sincerely sorry for my confusing statement of the previous edition of my questions! $\endgroup$
    – Denny
    Commented Jun 29, 2019 at 4:52

1 Answer 1


I start with the claim that $y(T \times t) = x(t)$. Here, we can check that $ t = 0$ we obtain $y(0) = x(0) = c$ for some constant $c$. Furthermore, at $t = 1$, we have $y(T) = x(1)$ and we are done.

To verify that, we just have to show that under this claim, the two differential equations are equivalent (ie it satisfy both equations).

From $\frac{dy(\xi)}{d\xi} = f(y(\xi))$, we pick $\xi = Tt$. Hence,

$\frac{dy(\xi)}{d\xi} = f(y(\xi)) \iff \frac{dy(Tt)}{d(Tt)} = f(y(Tt)) \iff \frac{d(x(t))}{d(Tt)} = f(x(t)) \iff \frac{d(x(t))}{d(t)} = Tf(x(t))$

Thus, $y(T \times t) = x(t)$ indeed satisfy both differential equations.

  • $\begingroup$ You want to verify the claim that $y(Tt) = x(t)$. Why could you say $T(f(x(t))) = Tf(y(Tt))$? You have not done this, right? $\endgroup$
    – Denny
    Commented Jun 29, 2019 at 2:40
  • 1
    $\begingroup$ I feel that I just have to verify the claim, ie I want to say that all my solutions of $y$ and $x$ satisfy this relation of $y(Tt) = x(t)$. I can indeed check that this holds for both equations. Simply said, the two differential equations are related under a linear transformation. So if I want to probe the behavior in one of the differential equation, I can use the transformation to probe the behavior in the other equation. $\endgroup$
    – HK Tan
    Commented Jun 29, 2019 at 2:55

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