matrix functions, spectral representation

Question

I've been having trouble solving this exercise: Find the solution of $$ \dot{\bf{x}}(t) = A\bf{x}(t)+\bf{c}\delta(t-1)$$ where $A = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}, \bf{c} = {(1,1,1)}, \bf{x}(t=0)= {(1,0,1)}$. Now I followed these steps: For $t > 1$ the delta term vanishes and we are left with $\dot{\bf{x}}(t) = A\bf{x}(t)$, whose general solution is given by: $$ \bf{x}(t) = e^{tA}\bf{x}(0)$$ Now I need to compute the matrix exponential. I noticed that: $$A= 2Id_3 + B, B =\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ It's easy to notice that $[1]$: $$B^k = [0], k \ge2 $$ Now we have: $$e^{tA} = \sum_{n=0}^{\infty}\frac{t^n}{n!}A^n = Id_3 + \sum_{n = 1}^{\infty} \frac{t^n}{n!}(2Id_3+B)^n$$ Now, since the identity matrix is idempotent, we get: $$ (2Id_3+B)^n = Id_3\sum_{k =0}^{n} 2^n \left(\frac{B}{2}\right)^k $$ Therefore we have: $$e^{tA} = Id_3\left(1 + \sum_{n=1}^{\infty}\frac{(2t)^n}{n!} \sum_{k =0}^{n} \left(\frac{B}{2}\right)^k \right) $$ Now I do recognize that I need to isolate the term $$\sum_{n=1}^{\infty}\frac{(2t)^n}{n!} = e^{2t}-1$$ nevertheless, I can't figure out how to compute the second sum with index $k$. The only thing I know is that for $k>2$ all the terms of the sum vanish because of condition $[1]$. But I don't know how to go on. If someone could help, It would be much appreciated.

Answer 1

First, regarding the matrix exponential: a simpler approach is to note that $2\operatorname{Id}_3$ and $B$ commute (i.e. $(2 \operatorname{Id}_3) B = B (\operatorname{Id}_3)$), from which it follows that \begin{align} \exp[t(2\operatorname{Id}_3 + B)] &= \exp[2t\operatorname{Id}_3 + tB] = \exp(2t \operatorname{Id}_3)\exp(tB) \\ & = \left(e^{2t}\operatorname{Id}_3\right) \left(I + tB\right) = e^{2t} \pmatrix{1&0&t\\0&1&0\\0&0&1}. \end{align} Second, your attempt to account for the $\delta(t-1)$ term is incorrect. Here is one approach (the use of a generalized "integrating factor") that yields the correct answer: $$ \dot{\mathbf x} = A\mathbf x(t) + \mathbf c\delta(t-1)\\ \dot{\mathbf x} - A\mathbf x(t) = \mathbf c\delta(t-1)\\ e^{-At}\dot{\mathbf x} - A e^{-At}\mathbf x = e^{-At}\mathbf c \delta(t-1)\\ \frac {d}{dt}\left[e^{-At}\mathbf x \right] = e^{-At}\mathbf c \delta(t-1)\\ e^{-At}\mathbf x = \left[e^{-At}\mathbf x\right]_{t = 0} + \int_0^t e^{-At}\mathbf c \delta(t-1)\,dt\\ e^{-At}\mathbf x = \mathbf x(0) + \int_0^t e^{-At}\mathbf c \delta(t-1)\,dt\\ \mathbf x(t) = e^{At}\mathbf x(0) + e^{At}\int_0^t e^{-At}\mathbf c \delta(t-1)\,dt. $$ Now, with the "sifting property" of $\delta$, we find that for $t > 1$ the integral term is given by $$ e^{At}\int_0^t e^{-At}\mathbf c \delta(t-1)\,dt = e^{At} e^{-A(1)}\mathbf c = e^{A(t-1)} \mathbf c. $$

Thus, the result can be expressed as follows: $$ \mathbf x(t) = \begin{cases} e^{At}\mathbf x(0) & t < 1,\\ e^{At} \mathbf x(0) + e^{A(t-1)} \mathbf c & t > 1. \end{cases} $$ How exactly you decide on the behavior at the trajectory's discontinuity (at $t = 1$) is a matter of competing conventions.