4
$\begingroup$

Let $x(t),y\in \mathbb{R}^p,A\in \mathbb{R}^{p\times p}$, I want to solve the following system \begin{align} \frac{dx(t)}{dt}&=-A(x(t)-y) \end{align} with known $x(0)$. I have 'something' but I suspect that it is wrong. I first take \begin{align} \frac{dx(t)}{dt}+Ax(t)&=Ay\\ \exp(At)\frac{dx(t)}{dt}+\exp(At)Ax(t)&=\exp(At)Ay\\ \frac{d}{dt}\exp(At)x(t)&=\exp(At)Ay\\ \exp(At)x(t)&=x(0)+\int_0^t \exp(As) ds Ay\\ \exp(At)x(t)&=x(0)+[\exp(At)-I]Ay\\ x(t)&=\exp(-At)[x(0)-Ay]+Ay \end{align}

However I have reason to believe (from looking at some other material) that the answer 'should' be \begin{align} x(t)&=\exp(-At)[x(0)-y]+y \end{align}

Are either of these answers (either the one I derived or the one I suspect) correct? If the latter is correct, why? If neither are correct, what is the true answer?

$\endgroup$

2 Answers 2

5
$\begingroup$

Note when computing the integral that $\frac{d}{ds} \exp(As)$ is equal to $\exp(As) A$, not just $\exp(As)$, so your second-to-last line should be $\exp(At) x(t) = x(0) + [\exp(At)-I] y$ (with just $y$, not $Ay$). So the second answer is the correct one.

$\endgroup$
1
  • $\begingroup$ Thanks for pointing my mistake. My screen reader misses too often the super/under scripts $\endgroup$ Mar 29, 2022 at 7:17
3
$\begingroup$

There was some error in your calculation - I fixed it below..

I give complete calculation for everyone's understanding..

Let $x(t),y\in \mathbb{R}^p,A\in \mathbb{R}^{p\times p}$ .

To solve the differential equation: $$ {dx \over dt} = - A [x(t) - y] $$ with known $x(0)$.

We rewrite it as $$ {dx \over dt} + A x(t) = A y $$

$$ \exp(A t) \ {dx \over dt} + \exp(A t) A x(t) = \exp(A t) A y $$ i.e. $$ {d \over dt}[\exp(A t) x(t)] = \exp(A t) A y $$

Integrating both sides, we get $$ \exp(A t) x(t) = x(0) + \int\limits_{0}^t \exp(A s) A y ds $$ i.e. $$ \exp(A t) x(t) = x(0) + \left[\exp(A s) y \right]_0^t = x(0) + \exp(A t) y - y $$

This can be rewritten as $$ x(t) = \exp(-A t) [ x(0) + \exp(A t) y - y] = y + \exp(- A t) [x(0) - y] $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .