How to prove that a null controllable region is convex?

Question

I have to do the following exercise: Now, I tried to prove it by using Lagrange's Formula for continuous-time systems: $$x(t) = e^{At} x(0) + \int^{t}_{0}e^{A(t-\tau)}Bu(\tau)d\tau$$ Since $x(T)=0$, we have that $$e^{AT} x(0) = - \int^{T}_{0}e^{A(t-\tau)}Bu(\tau)d\tau$$ Now, in order to prove convexity, we choose two initial states $x_1,x_2$ and we take their convex combination $(1-\lambda)x_1+\lambda x_2$ (of course with $\lambda \in [0;1]$). Then we consider $\bar{T} = max(T_1,T_2)$, where $T_1$ and $T_2$ are associated to $x_1$ and $x_2$. We have: $$x(\bar{T})=e^{A\bar{T}}[(1-\lambda)x_1+\lambda x_2]+\int^{\bar{T}}_{0}e^{A(\bar{T}-\tau)}Bu(\tau)d\tau=\\-(1-\lambda)\int^{T_1}_{0}e^{A(T_1-\tau)}Bu_1(\tau)d\tau-\lambda\int^{T_2}_{0}e^{A(T_2-\tau)}Bu_2(\tau)d\tau+\int^{\bar{T}}_{0}e^{A(\bar{T}-\tau)}Bu(\tau)d\tau$$

where $u_1$ and $u-2$ are associated to $x_1$ and $x_2$. The above formula should we equal to 0 in order for the equality to hold, but I don't know how to proceed from this point on

Answer 1

In order to prove convexity of $\mathcal{C}$ we need to prove that

$$\lambda x_{0,1}+(1-\lambda)x_{0,2}\in\mathcal{C}$$

for all $x_{0,1},x_{0,2}\in\mathcal{C}$.

So, assume that $x_{0,1},x_{0,2}\in\mathcal{C}$. Then there exist times $T_1,T_2$ and $u_1,u_2\in\mathcal{U}$ such that

$$0=\exp(AT_i)x_{0,i}+\int_0^{T_i}\exp(A(T_i-s))Bu_i(s)ds\qquad \qquad (i=1,2)$$

Assume also without loss of generality that $T_1\geq T_2$. If we now select the control input

$$u(t)=\left\{\array{\lambda u_{1}(t)+(1-\lambda)u_{2}(t)\:, \quad t\in[0,T_2]\\ \lambda u_1(t)\:,\quad \qquad \qquad \qquad t\in(T_2,T_1]}\right.$$

then the state vector starting from $\lambda x_{0,1}+(1-\lambda)x_{0,2}$ will have the following value at $t=T_1$ $$x(T_1)=\exp(AT_1)(\lambda x_{0,1}+(1-\lambda)x_{0,2})+\int_0^{T_1}{\exp(A(T_1-s))Bu(s)ds}\\ =\lambda \exp(AT_1)x_{0,1}+(1-\lambda)\exp(AT_1)x_{0,2}+\lambda\int_0^{T_1}{\exp(A(T_1-s))Bu_1(s)ds} \\+(1-\lambda)\int_0^{T_2}\exp(A(T_1-s))Bu_2(s)ds$$

Using now the transitive property of the transition matrix we have $$\exp(AT_1)=\exp(A(T_1-T_2))\exp(AT_2)\\ \exp(A(T_1-s))=\exp(A(T_1-T_2))\exp(A(T_2-s))$$ and therefore $$x(T_1)=\lambda \exp(AT_1)x_{0,1}+\lambda\int_0^{T_1}{\exp(A(T_1-s))Bu_1(s)ds} \\+(1-\lambda)\exp(A(T_1-T_2))\exp(AT_2)x_{0,2}\\+(1-\lambda)\exp(A(T_1-T_2))\int_0^{T_2}\exp(A(T_2-s))Bu_2(s)ds\\=\lambda \left[\exp(AT_1)x_{0,1}+\lambda\int_0^{T_1}{\exp(A(T_1-s))Bu_1(s)ds}\right]\\ +(1-\lambda)\exp(A(T_1-T_2))\left[\exp(AT_2)x_{0,2}+\int_0^{T_2}\exp(A(T_2-s))Bu_2(s)ds\right]=0.$$ Since also $u\in\mathcal{U}$ as $|u(t)|\leq \lambda |u_1(t)|+(1-\lambda)|u_2(t)|\leq 1$ the proof is completed.