Optimal control
Ifp( 1 )>0 then aspis decreasingp(t)>0 for every 0<t< 1 ,
which implies thatu1. In this casex=u= 1 .Using thatx( (^0) )= 0 x(t)=t,which is not feasible as the boundary condition
x( 1 )=0 is not valid. Hencep( 1 )< 0.
Ifp( (^0) )<0 then aspis decreasingp(t)<0 for everyt.Hence one
has thatu 1 and therefore asx=u= 1 .Using that
x( 0 )=0 one has thatx(t)= tand hence the boundary condition
x( 1 )=0 is not valid again. Hencep( 0 )> 0.