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Please help me with how to start with this question: A particle of unit mass moves along the x-axis subject to a force $u(t)$. We want to transfer the particle from rest at the origin, to rest as x=1 in unit time, so that the effort involved , measured by $J=\int_0^1 (u(t))^2 dt $ is minimised. Denote $x_1=x$ (position) and $x_2=x'$ (velocity), we get the state equations: $$ \begin{pmatrix} x_1' \\ x_2'\end{pmatrix} = \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0\\1 \end{pmatrix}u(t). $$ (a) investigate which controls of the type $u(t)=a + bt +ct^2$, $a,b,c$ are real, can be used in this positioning problem. i.e. for which $a,b,c$ is it possible to reach the target in time?

(b) prove that for all of the controls of this type, the control $u(t)=6(1-2t)$ is optimal.

Natalie
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1 Answers1

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So, $u(t)=a+bt+ct^2$ is the acceleration. Given the initial state $x(0)=0$, $x'(0)=0$, the velocity function is $x'(t)=at+bt^2/2+ct^3/3$ and the position function is $x(t)=at^2/2+bt^3/6+ct^4/12$. The constraints $x'(1)=0$ and $x(1)=1$ amount to $$a+b/2+c/3 =0 \tag1$$ $$a/2+b/6+c/12=1\tag2$$ The solution space is $(a,b,c)=(1,18,-30)+k(1,-6,6)$, where $k\in\mathbb R$. One way to proceed from here is to write $u(t)=u_0(t)+kv(t)$ where $u_0(t)=1+18 t-30t^2$ and $ v(t)= k(1-6t+6t^2)$. The function $ \int_0^1 (u_0(t)+kv(t))^2\,dt$ is a convex quadratic polynomial in $k$. Its only critical point is the minimum. Since $$ \frac{d}{dk} \int_0^1 (u_0(t)+kv(t))^2\,dt = 2\int_0^1 (u_0(t)v(t)+ kv(t)^2)\,dt $$ the minimum is attained at $$ k = -\frac{\int_0^1 u_0(t)v(t)\,dt }{ \int_0^1 v(t)^2 \,dt} $$

ˈjuː.zɚ79365
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