Solution to a linear system

Solution to a linear system#

(Note: useful properties of the matrix exponential can be found here)

Let \(A\in\mathbb{R}^{n\times n}\), \(B\in\mathbb{R}^{n\times m}\), and let \(u:[0,\infty)\to\mathbb{R}^m\) be integrable. Consider the linear system

\[ \dot x(t)=Ax(t)+Bu(t), \qquad x(0)=x_0. \]

Claim: The function

\[ x(t)=e^{At}x_0+\int_0^t e^{A(t-\tau)}Bu(\tau)\,d\tau \]

is a solution to the above ordinary differential equation.

Proof:

Step 1: Verify the initial condition.

At \(t=0\),

\[ x(0)=e^{A\cdot 0}x_0+\int_0^0 e^{A(0-\tau)}Bu(\tau)\,d\tau =Ix_0+0=x_0, \]

where we used the fact that \(e^{0}=I\) (see link above).

Step 2: Differentiate and verify the ODE.

Differentiate term by term. Using

\[ \frac{d}{dt}e^{At}=Ae^{At}, \]

(see link above) we obtain

\[ \frac{d}{dt}\big(e^{At}x_0\big)=Ae^{At}x_0. \]

For the integral term, apply the Leibniz rule to

\[ \int_0^t e^{A(t-\tau)}Bu(\tau)\,d\tau. \]

Let

\[ g(t,\tau)=e^{A(t-\tau)}Bu(\tau). \]

Then

\[ \frac{d}{dt}\int_0^t g(t,\tau)\,d\tau = g(t,t)+\int_0^t \frac{\partial}{\partial t}g(t,\tau)\,d\tau. \]

Compute each term:

\[ g(t,t)=e^{A(t-t)}Bu(t)=Bu(t), \]

and

\[ \frac{\partial}{\partial t}g(t,\tau) = A e^{A(t-\tau)}Bu(\tau). \]

Therefore,

\[ \frac{d}{dt}\int_0^t e^{A(t-\tau)}Bu(\tau)\,d\tau = Bu(t)+\int_0^t A e^{A(t-\tau)}Bu(\tau)\,d\tau. \]

Combining the derivatives,

\[ \dot x(t) = Ae^{At}x_0 + Bu(t) + \int_0^t A e^{A(t-\tau)}Bu(\tau)\,d\tau. \]

Factor out \(A\):

\[ \dot x(t) = A\left(e^{At}x_0+\int_0^t e^{A(t-\tau)}Bu(\tau)\,d\tau\right) + Bu(t). \]

Since the expression in parentheses is exactly \(x(t)\), we conclude

\[ \dot x(t)=Ax(t)+Bu(t). \]

Because both the differential equation and the initial condition are satisfied, the given expression is indeed a solution.