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Suppose we have a system of linear ODEs represented in state-space form as follows: $$ \dot{x}(t) = Ax(t) + Bu(t), \; x(t_0)=x_0. $$ The well-known solution, valid for $t\in[t_0,\infty)$, is $$ x(t) = e^{A(t-t_0)}x_0 + \int_{t_0}^t e^{A(t-\sigma)}Bu(\sigma)d\sigma. $$ What is the general form of the solution when $x(t_f)=x_f$ is specified instead, and we want to know the solution for $t\in(-\infty,t_f]$?

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In order to travel back in time you first need to get a Flux capacitor. –  Asaf Karagila Jan 17 '13 at 17:10
    
Been trying to get my hands on one of those.. Hard to come by. –  CCL Jan 17 '13 at 17:34

1 Answer 1

up vote 3 down vote accepted

If we suppose that $u$ is defined in an interval $(a,b)$( can be $\mathbb R$ ) and $t_0\in (a,b)$, then the solution of

$$ \dot{x}(t) = Ax(t) + Bu(t), \; x(t_0)=x_0. $$

is

$$ x(t) = e^{A(t-t_0)}x_0 + \int_{t_0}^t e^{A(t_0-s)}Bu(s)ds. $$

FOR ALL $t\in (a,b)$, not only $t\geq t_0$.

To see this

write the equation like this

$$x'-Ax=Bu(t)$$

Now we apply $e^{(t_0-t)A}$ in the equation

$$e^{(t_0-t)A}(x'-Ax)=e^{(t_0-t)A}Bu(t)\implies \dfrac{d}{dt}e^{(t_0-t)A}x=e^{(t_0-t)A}Bu(t)$$

Therefore

$$\int_{t_0}^{t}\dfrac{d}{ds}e^{(t_0-s)A}xds=\int_{t_0}^te^{(t_0-s)A}Bu(s)ds$$

and we get

$$ x(t) = e^{A(t-t_0)}x_0 + \int_{t_0}^t e^{A(t_0-s)}Bu(s)ds. $$

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To prove this, do you just differentiate $x(t)$ using the Leibniz rule? I have a book that says (for $x(0)=0$) that the solution is $$x(t) = -\int_0^t e^{A(t-\sigma)}Bu(\sigma)d\sigma, \; t\in(-\infty,0].$$ This must be a typo, since if you differentiate, you get $$\dot{x}(t) = Ax(t) - Bu(t)$$ –  CCL Jan 17 '13 at 16:41
    
this helped you? –  user52188 Jan 17 '13 at 17:07
    
Working through the last step... On the left you have $$e^{(t_0-t)A}x(t) - x(t_0) = \int_{t_0}^t B u(s) ds.$$ Which gives $$x(t) = e^{A(t-t_0)}x(t_0) + e^{A(t-t_0)}\int_{t_0}^t B u(s) ds$$ How does the solution at the top follow from this? –  CCL Jan 17 '13 at 17:16
    
I wrong see again –  user52188 Jan 17 '13 at 17:21
    
Ah yes I see. But you mean $(t-s)$ instead of $(t_0 - s)$ in the final solution. –  CCL Jan 17 '13 at 17:29

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