Doubts about the solution of the ODE $\dot{x}=Ax+Bu$

Question

I am studying control theory and I have a doubt about the linear ODE $$\dot{x}(t)=Ax(t)+Bu(t) \quad x(t_0)=x_0, \qquad (a)$$ where $x\in\mathbb{R}^n$, $u\in\mathbb{R}^m$,$A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times m}$. I know that the solution of $(a)$ is: $$x(t)=e^{At}x_0 + \int_{t_0}^t{e^{A(t-\tau)}Bu(\tau)}d\tau. \qquad (b)$$

Instead, if I take an ODE of the form: $$\dot{z}(t)=(A+B)z(t) \quad z(t_0)=z_0, \qquad (c)$$ where $x,A,B$ as before ($n=m$), then I know that this time the solution is: $$z(t)=e^{(A+B)t}z_0.$$

The question is that if I treat $(c)$ as $(a)$ with $Bz(t)\doteq Bu(t)$, I expect $(b)$ to still hold, in particular I expect to write the solution of (c) as: $$z(t)=e^{At}z_0 + \int_{t_0}^t{e^{A(t-\tau)}Bz(\tau)}d\tau,$$ which instead is not true.

Why this doesn't hold? Is it because the term $e^{At}z_0$ should be instead $e^{(A+B)t}z_0$?

Answer 1

Consider the algebraic equation $x^2 = 2ax + b$. The solutions to this one are $a \pm \sqrt{a^2+b}$. Now let $b = x$. The solutions are $x = 0,2a+1$, while if you plug in $x$ in the previous solution, you get $x = a\pm\sqrt{a^2+x}$. So they're different?

But wait, $x = a\pm\sqrt{a^2+x}$ isn't a solution at all--there's $x$ on both sides. And we notice that if we take the $+$ sign, that equation is solved by $x = 2a+1$, and if we take the $-$ sign, it's solved by $x = 0$. So in fact it does have the correct solutions, but we had to solve an algebraic equation equivalent to the original quadratic to get there.

The same issue occurs with $$ z(t)=e^{At}z_0 + \int_{t_0}^t{e^{A(t-\tau)}Bz(\tau)}d\tau. $$ There's a $z$ on both the left- and right-hand sides, so this is an integral equation, not an explicit solution. It turns out the solution to this integral equation is, in fact, $z_0e^{(A+B)t}$. But as before, this wasn't all that helpful, as we still had to solve an integral equation to get there, and this integral equation is equivalent to the original differential equation.

Answer 2

If you look at (a) with $z(t) = \begin{bmatrix} x(t)\\u(t)\end{bmatrix}$, you see that it is actually $$ \dot{x}(t) = (A+B)z(t), $$ not $$ \dot{z}(t) = (A+B)z(t), $$ since obviously $\dot{x}\neq \dot{z}$. This should be obvious, since not even the dimensions of $x$ and $z$ are the same.

Not sure what you mean by $Bz(t) = Bu(t)$. Since (c) is a homogeneous ODE, the solution is as you stated $$z(t) = e^{(A+B)t} z_0.$$ However, renaming $z$ does not suddenly change the fact that (c) is homogeneous.

Answer 3

We can check that $$z(t)=e^{(A+B)t}z_0,\tag{*}$$ and $$z(t)=e^{At}z_0 + \int_{t_0}^t{e^{A(t-\tau)}Bz(\tau)}d\tau\tag{**}$$ are equivalent indeed. Let $t=t_0$. Then we have $z(t_0)=z_0$ in both cases. Now differentiate the right-hand sides w.r.t. $t$. In the first case we have $$\frac{d}{dt}e^{(A+B)t}z_0=(A+B)e^{(A+B)t}z_0=(A+B)z(t). $$ In the second case, \begin{multline*}\frac{d}{dt}\left[e^{At}z_0 + \int_{t_0}^t{e^{A(t-\tau)}Bz(\tau)}d\tau\right]=\frac{d}{dt}e^{At}\left[z_0 + \int_{t_0}^t{e^{-A\tau}Bz(\tau)}d\tau\right]\\=Ae^{At}\left[z_0 + \int_{t_0}^t{e^{-A\tau}Bz(\tau)}d\tau\right]+e^{At}{e^{-At}Bz(t)}=Az(t)+Bz(t)=(A+B)z(t).\end{multline*} So we see that both expressions give the solution to the same DE, but written in a different way.