0
$\begingroup$

Given two generic functions $x(t)$ and $y(t)$ I want to prove that $x(t) = y(t)$.

To do so, I take the derivative, which turn out to be: $$\dot x(t) = Ax(t) + B$$ $$\dot y(t) = Ay(t) + B$$ where $A$ and $B$ are the same in both derivatives.

Is this sufficient to say both functions are equal?

The reason I ask is that, in general, $x(t)$ and $y(t)$ could have vastly different forms (ie. $x(t)$ could be the resultant of a complicated integral $y(t)$ or something similar). Because of this, I am wondering if I have to go through the trouble of reducing $x(t)$ to $y(t)$ or vice-versa.

Thanks in advance!

Improve this question
$\endgroup$
  • $\begingroup$ Are $A,B$ constants or functions of $x,y,t$ ? $\endgroup$ – Yves Daoust Mar 4 '20 at 13:52
  • $\begingroup$ @YvesDaoust they are constants $\endgroup$ – Clark Mar 4 '20 at 15:38
1
$\begingroup$

They both satisfy the linear first order differential equation:

$\begin{equation*} f'(t) = A f(t) + B \end{equation*}$

You need at least one other condition, say prove that $x(t_0) = y(t_0)$ for some value $t_0$, or perhaps the same derivative at a point. Note that the solution is

$\begin{equation*} f(t) = c e^{A t} - B/A \end{equation*}$

here $c$ is an unknown constant, to be determined by other conditions.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ do you mean $x(t_0) = y(t_0)$? So if I can prove both functions have the same value at one point, they should have the same value at all points? (ie. from existence and uniqueness?) $\endgroup$ – Clark Mar 4 '20 at 13:13
  • 1
    $\begingroup$ @Clark, yes; and fixed. $\endgroup$ – vonbrand Mar 4 '20 at 13:15
  • $\begingroup$ Awesome, thanks! It's all coming back $\endgroup$ – Clark Mar 4 '20 at 13:18
1
$\begingroup$

If $x(t)$ and $y(t)$ are $C^1$ functions such that $x(t_0) = y(t_0)$ for some $t_0$ furthermore $x'(t) = y'(t) = f(t)$ for some function (Lipschitz-continuous) $f$, then according to the uniqueness of the solution of a differential equation you have $x(t)=y(t)$ for all $t\geq t_0$.

Improve this answer
$\endgroup$
1
$\begingroup$

By subtraction,

$$\dot x-\dot y=\dot{(x-y)}=A(x-y)$$ so that the two given functions can differ by a solution of this equation, given by

$$x-y=Ce^{At}.$$

Improve this answer
$\endgroup$
  • $\begingroup$ This doesn't contradict the other answers right? Let's say $x(t_0) = y(t_0) = 0$ where $t_0 = 0$ for simplicity. This would then give $C = 0$ and, therefore $x(t) = y(t)$ correct? $\endgroup$ – Clark Mar 4 '20 at 15:47
  • 1
    $\begingroup$ @Clark: absolutely. My goal was to show how to solve the question without integrating the original equations and focusing on the difference. (Though the benefit is tiny.) $\endgroup$ – Yves Daoust Mar 4 '20 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.