School ofAutomation,XJTU1(xT[ATP + PA+P-PBR-1BTP+ Q]xJz2+ [u + R-1BT Px]TR[u + R-1BT Pxl)dt +x(T)Stx(T)如果取ATP+PA + P=PBR-1BTP+O = 0u* = -R-1BTPx有17CAIYUANLI
School of Automation, XJTU © CAI YUANLI 17 𝐽 ≥ 1 2 ∫ { 𝑇 𝑡0 𝑥 𝑇 [𝐴 𝑇𝑃 + 𝑃𝐴 + 𝑃̇ − 𝑃𝐵𝑅 −1𝐵 𝑇𝑃 + 𝑄]𝑥 + [𝑢 + 𝑅 −1𝐵 𝑇𝑃𝑥] 𝑇𝑅[𝑢 + 𝑅 −1𝐵 𝑇𝑃𝑥]}d𝑡 + 1 2 𝑥 𝑇 (𝑇)𝑆𝑇𝑥(𝑇) 如果取 𝐴 𝑇𝑃 + 𝑃𝐴 + 𝑃̇ − 𝑃𝐵𝑅 −1𝐵 𝑇𝑃 + 𝑄 = 0 𝑢 ∗ = −𝑅 −1𝐵 𝑇𝑃𝑥 有
School ofAutomation,XJTU=xT(T)STx(T)711V(T) - V(to) =[uT Ru + xT Qx]dt = -J* +(T)STX(Tx22Jto112*(T)P(T)x(T) -专xT(to)P(to)x(to) =2xT(T)STx(T) -J取P(T) = ST， 可导出 J* ==xT(to)P(to)x(to)。18CAIYUANLI
School of Automation, XJTU © CAI YUANLI 18 𝐽 ∗ ≥ 1 2 𝑥 𝑇 (𝑇)𝑆𝑇𝑥(𝑇) 𝑉(𝑇) − 𝑉(𝑡0 ) = − 1 2 ∫ [ 𝑇 𝑡0 𝑢 𝑇𝑅𝑢 + 𝑥 𝑇𝑄𝑥]d𝑡 = −𝐽 ∗ + 1 2 𝑥 𝑇 (𝑇)𝑆𝑇𝑥(𝑇) 1 2 𝑥 𝑇(𝑇)𝑃(𝑇)𝑥(𝑇) − 1 2 𝑥 𝑇(𝑡0 )𝑃(𝑡0 )𝑥(𝑡0 ) = 1 2 𝑥 𝑇(𝑇)𝑆𝑇𝑥(𝑇) − 𝐽 ∗ 取𝑃(𝑇) = 𝑆𝑇，可导出 𝐽 ∗ = 1 2 𝑥 𝑇(𝑡0 )𝑃(𝑡0 )𝑥(𝑡0 )
School ofAutomation,XJTU一个简单的拦截（或交会）问题。给定例 2 i=a(t),j=u,adt2tC1、C2、t,为常数。试求使J为最小时的a（u，y，t）。19CCAIYUANLI
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SchoolofAutomation,XJTU解a(t)=-A(t)u(t)-A(t)y(t))(1/c2)+(1/c,)(t,-t)2+(1/3)(t,-t)3Ae=D(t,-t)A,=(1/c)(tr-)+(1/2)(t-t)D(t-t)式中D(t,-1) -=[1-+→ (t,-t) +-+t,-]-+(t,-t)。20CCAIYUANLI
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School of Automation,XJTU评论。如c→0，则(t,)不受控制，我们有1As=(1/c.)+t,-t A,= 0;u(t)→a=【(1/c）+t,-订）这就是“应增速度”控制。如c→，则u（t)不受控制，而(t,-t)2t,-tA.=A,=1/cα+号(t,-1)§1/c2+1(t,-t)3221CCAIYUANLI
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