Lagrange formula for solving linear differential equations

Question

I am studying control systems, and my textbook uses "Lagrange's formula" for solving time-continuous linear systems in "state-space". Below are the equations presented:

$$\dot{x}(t) = Ax(t) + Bu(t)$$

$$y(t) = Cx(t) + Du(t)$$

where A,B,C and D are matrices of coefficients, but let's assume they are all singular to make things simpler.

Assuming that $t > t_0$ and $x(t_0) = x_{t_0}$,the formula used to calculate $x(t)$ is the following:

$$x(t) = e^{A(t-t_0)}x_{t_0} + \int_{t_0}^t e^{A(t-\tau)}Bu(\tau) d\tau $$

This formula is very similar to another formula I learned in calc 2 for an identical purpose, albeit this next formula is defined for all values of t and contains an indefinite integral in place of the definite integral in the formula above

$$ x(t) = e^{-At} \int Bu(t)e^{At} dt $$

It is obvious to me that these formulas are strongly connected, but whereas I understand how the second one is derived, I cannot say the same about the first one. And what is up with using $\tau$ as the variable for intergration? I've been told it is a "dummy variable", but it was presented as a fact, rather than a proven result of calculus.

I've also tried to google this so-called "Lagrange formula", but unfortunately I haven't had any success with it. Can anybody please help me understand this?

Answer 1

It is simply notation. The indefinite integral is computed using the anti-derivative: $$ \int f(t) \, dt = F(t) + C $$ where $F'(t) = f(t)$, and $C$ is a constant. Then the definite integral is $$ \int_a^b f(t) \, dt = F(b) - F(a) .$$ In this second equation, the $t$ is a 'dummy variable.' I could just as well write $$ \int_a^b f(\tau) \, d\tau = F(b) - F(a) .$$ In particular, to avoid confusion, we make sure that the dummy variable is not the same as the variables we use in the limits (or anywhere else). So if I want to write $$ \int_{t_0}^t f(\tau) \, d\tau = F(t) - F(t_0) $$ it would be confusing if I used the letter $t$ as the dummy variable.

And look $$ \int f(t) \, dt = \int_{t_0}^t f(\tau) \, d\tau + C_1 $$ where $C_1 = C + F(t_0)$.

Finally, we have this formula: $$ g(t) \int f(t) \, dt = g(t) \left(\int_{t_0}^t f(\tau) \, d\tau + C_1\right) = \int_{t_0}^t g(t) f(\tau) \, d\tau + C_1 g(t) .$$ This last formula would be nonsense if we didn't change the dummy variable to something other than $t$.

Answer 2

The first formula is obtained by applying the Lagrange method of constant variation:

First, the ODE is linear so it can be solved as

$$ \cases{ \dot x_h = A x_h\\ \dot x_p = A x_p + B u\\ x = x_h + x_p } $$

The homogeneous has as solution $x_h(t) = e^{A t}C_0$ (here $C_0$ is a constant array) now making $x_p = e^{A t}C_0(t)$ (note that in the method "variation of constants" due to Lagrange, we assume now that $C_0=C_0(t)$ is an array of functions $t$ dependent) and substituting we have

$$ Ae^{At}C_0(t)+e^{At}C_0'(t) = Ae^{At}C_0(t) + B u(t)\Rightarrow C_0'(t) = e^{-A t}B u(t) $$

and then

$$ C_0(t) = \int_0^t e^{-A\tau}B u(\tau) d\tau $$

and finally

$$ x = x_h+x_p = e^{A t}C_0 + e^{A t}\int_0^t e^{-A\tau}B u(\tau) d\tau = e^{A t}C_0 + \int_0^t e^{A(t-\tau)}B u(\tau) d\tau = e^{At}\int e^{-A \tau}B u(\tau) d\tau $$

note that here $C_0$ is a constant array