Systems of Differential Equations — Question 7

Question 7 – Solution

Strategy. Express the terminal state as three moments, then use orthogonality to prove optimality over the full control class.

Step 1: Derive the terminal moments. For $X=(y,y',y'')^T$, $X'=NX+e_3u$, where $N$ has ones just above the diagonal and zeros elsewhere. Since $N^3=0$, $$e^{Nr}=\begin{pmatrix}1&r&r^2/2\\0&1&r\\0&0&1\end{pmatrix},\qquad X(1)=\int_0^1\begin{pmatrix}(1-t)^2/2\\1-t\\1\end{pmatrix}u(t)\,dt.$$ The required three moments are $(1,0,0)^T$.

Step 2: Solve within the quadratic space. For $u=a+bt+ct^2$, the constraints are $$a/6+b/24+c/60=1,\quad a/2+b/6+c/12=0,\quad a+b/2+c/3=0.$$ Their solution is $$\boxed{u_*=60-360t+360t^2,\qquad y_*=10t^3-15t^4+6t^5.}$$ The moment kernels span all quadratics. A quadratic in the homogeneous null space is orthogonal to itself and hence zero, proving uniqueness here.

Step 3: Prove global minimum effort. For any admissible $u=u_*+v$, all three terminal moments of $v$ vanish. Because $u_*$ is in their span, $\int_0^1u_*v=0$. Therefore $$\boxed{J(u)=J(u_*)+\int_0^1v^2\,dt\geq J(u_*)=720.}$$ Equality holds exactly when $v=0$ almost everywhere. This argument only uses square integrability and moment constraints, not polynomial form for $v$.

Step 4: Verify the endpoint motion. We have $y_*'=30t^2(1-t)^2$ and $y_*''=60t(1-t)(1-2t)$, so the two endpoint states are exactly $(0,0,0)$ and $(1,0,0)$; differentiating again gives $u_*$. The graph uses $u_*/60$ so the control and displacement are both readable.