optimal control to minimise a path












1














enter image description here



I'm having issues solving this problem. Here is what I have tried so far.



$$ u=dot {x_1} + x_1 $$
$$ J= frac{1}{2} int_{0}^{t_1}((2x_1)^2+2dot {x_1} x_1 + dot {(x_1)^2})dt$$



Can I proceed and say
$$frac{d}{dt}(frac{partial L}{partial x_1})-frac{partial L}{partial dot {x_1}}=0 $$
or would this be incorrect because I don't have $t_1$?
I also think Pontryagin might help here but I'm not sure how to proceed if that's the case. Thanks in advance!






optimization control-theory hamilton-equations







share|cite|improve this question






















  • What have you tried when using Pontryagin?
    – Kwin van der Veen
    Nov 29 at 23:27
















1














enter image description here



I'm having issues solving this problem. Here is what I have tried so far.



$$ u=dot {x_1} + x_1 $$
$$ J= frac{1}{2} int_{0}^{t_1}((2x_1)^2+2dot {x_1} x_1 + dot {(x_1)^2})dt$$



Can I proceed and say
$$frac{d}{dt}(frac{partial L}{partial x_1})-frac{partial L}{partial dot {x_1}}=0 $$
or would this be incorrect because I don't have $t_1$?
I also think Pontryagin might help here but I'm not sure how to proceed if that's the case. Thanks in advance!






optimization control-theory hamilton-equations







share|cite|improve this question






















  • What have you tried when using Pontryagin?
    – Kwin van der Veen
    Nov 29 at 23:27














1












1








1


1





enter image description here



I'm having issues solving this problem. Here is what I have tried so far.



$$ u=dot {x_1} + x_1 $$
$$ J= frac{1}{2} int_{0}^{t_1}((2x_1)^2+2dot {x_1} x_1 + dot {(x_1)^2})dt$$



Can I proceed and say
$$frac{d}{dt}(frac{partial L}{partial x_1})-frac{partial L}{partial dot {x_1}}=0 $$
or would this be incorrect because I don't have $t_1$?
I also think Pontryagin might help here but I'm not sure how to proceed if that's the case. Thanks in advance!






optimization control-theory hamilton-equations







share|cite|improve this question













enter image description here



I'm having issues solving this problem. Here is what I have tried so far.



$$ u=dot {x_1} + x_1 $$
$$ J= frac{1}{2} int_{0}^{t_1}((2x_1)^2+2dot {x_1} x_1 + dot {(x_1)^2})dt$$



Can I proceed and say
$$frac{d}{dt}(frac{partial L}{partial x_1})-frac{partial L}{partial dot {x_1}}=0 $$
or would this be incorrect because I don't have $t_1$?
I also think Pontryagin might help here but I'm not sure how to proceed if that's the case. Thanks in advance!






optimization control-theory hamilton-equations




optimization control-theory hamilton-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 at 12:15









Robbie Meaney

749




749












  • What have you tried when using Pontryagin?
    – Kwin van der Veen
    Nov 29 at 23:27


















  • What have you tried when using Pontryagin?
    – Kwin van der Veen
    Nov 29 at 23:27
















What have you tried when using Pontryagin?
– Kwin van der Veen
Nov 29 at 23:27




What have you tried when using Pontryagin?
– Kwin van der Veen
Nov 29 at 23:27










1 Answer
1






active

oldest

votes


















1














Calling



$$
H(x,u,lambda) = frac{1}{2} left(u(t)^2+x(t)^2right)+lambda (t) (u(t)-x(t))
$$



we have



$$
dotlambda(t) = -frac{partial H}{partial x} = lambda(t)+x(t)\
frac{partial H}{partial u} = lambda(t) + u(t) = 0
$$



Now solving



$$
dotlambda(t) = lambda(t)+x(t)\
lambda(t) + u(t) = 0\
dot x(t)+x(t) = u(t)
$$
we obtain



$$
x(t) = frac{left(7 c_1-3 c_2right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_1-2 c_2right) cosh left(sqrt{2} tright)\
lambda(t) = frac{left(7 c_2-17 c_1right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_2-12 c_1right) cosh left(sqrt{2} tright)
$$



Now including the contour conditions $x(0) = 1, x(t_f) = 2$ we obtain



$$
c_1= 2 sqrt{2} coth left(sqrt{2} t_fright)-4 sqrt{2} text{csch}left(sqrt{2} t_fright)+3\
c_2= 5sqrt{2} coth left(sqrt{2} t_fright)-10 sqrt{2} text{csch}left(sqrt{2} t_fright)+7
$$



Now from the conditions



$$
H_{t_f} = 0\
u(t_f)+lambda(t_f) = 0
$$



we obtain



$$
lambda(t_f) = 2left(-1pmsqrt 2right)
$$



and then follows



$$
t_f = frac{ln 2}{sqrt 2}
$$



etc.






share|cite|improve this answer





















  • Have you used Pontryagin for this after maximising the Hamiltonian?
    – Robbie Meaney
    Dec 10 at 9:53






  • 1




    @RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
    – Cesareo
    Dec 10 at 10:34











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Calling



$$
H(x,u,lambda) = frac{1}{2} left(u(t)^2+x(t)^2right)+lambda (t) (u(t)-x(t))
$$



we have



$$
dotlambda(t) = -frac{partial H}{partial x} = lambda(t)+x(t)\
frac{partial H}{partial u} = lambda(t) + u(t) = 0
$$



Now solving



$$
dotlambda(t) = lambda(t)+x(t)\
lambda(t) + u(t) = 0\
dot x(t)+x(t) = u(t)
$$
we obtain



$$
x(t) = frac{left(7 c_1-3 c_2right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_1-2 c_2right) cosh left(sqrt{2} tright)\
lambda(t) = frac{left(7 c_2-17 c_1right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_2-12 c_1right) cosh left(sqrt{2} tright)
$$



Now including the contour conditions $x(0) = 1, x(t_f) = 2$ we obtain



$$
c_1= 2 sqrt{2} coth left(sqrt{2} t_fright)-4 sqrt{2} text{csch}left(sqrt{2} t_fright)+3\
c_2= 5sqrt{2} coth left(sqrt{2} t_fright)-10 sqrt{2} text{csch}left(sqrt{2} t_fright)+7
$$



Now from the conditions



$$
H_{t_f} = 0\
u(t_f)+lambda(t_f) = 0
$$



we obtain



$$
lambda(t_f) = 2left(-1pmsqrt 2right)
$$



and then follows



$$
t_f = frac{ln 2}{sqrt 2}
$$



etc.






share|cite|improve this answer





















  • Have you used Pontryagin for this after maximising the Hamiltonian?
    – Robbie Meaney
    Dec 10 at 9:53






  • 1




    @RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
    – Cesareo
    Dec 10 at 10:34
















1














Calling



$$
H(x,u,lambda) = frac{1}{2} left(u(t)^2+x(t)^2right)+lambda (t) (u(t)-x(t))
$$



we have



$$
dotlambda(t) = -frac{partial H}{partial x} = lambda(t)+x(t)\
frac{partial H}{partial u} = lambda(t) + u(t) = 0
$$



Now solving



$$
dotlambda(t) = lambda(t)+x(t)\
lambda(t) + u(t) = 0\
dot x(t)+x(t) = u(t)
$$
we obtain



$$
x(t) = frac{left(7 c_1-3 c_2right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_1-2 c_2right) cosh left(sqrt{2} tright)\
lambda(t) = frac{left(7 c_2-17 c_1right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_2-12 c_1right) cosh left(sqrt{2} tright)
$$



Now including the contour conditions $x(0) = 1, x(t_f) = 2$ we obtain



$$
c_1= 2 sqrt{2} coth left(sqrt{2} t_fright)-4 sqrt{2} text{csch}left(sqrt{2} t_fright)+3\
c_2= 5sqrt{2} coth left(sqrt{2} t_fright)-10 sqrt{2} text{csch}left(sqrt{2} t_fright)+7
$$



Now from the conditions



$$
H_{t_f} = 0\
u(t_f)+lambda(t_f) = 0
$$



we obtain



$$
lambda(t_f) = 2left(-1pmsqrt 2right)
$$



and then follows



$$
t_f = frac{ln 2}{sqrt 2}
$$



etc.






share|cite|improve this answer





















  • Have you used Pontryagin for this after maximising the Hamiltonian?
    – Robbie Meaney
    Dec 10 at 9:53






  • 1




    @RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
    – Cesareo
    Dec 10 at 10:34














1












1








1






Calling



$$
H(x,u,lambda) = frac{1}{2} left(u(t)^2+x(t)^2right)+lambda (t) (u(t)-x(t))
$$



we have



$$
dotlambda(t) = -frac{partial H}{partial x} = lambda(t)+x(t)\
frac{partial H}{partial u} = lambda(t) + u(t) = 0
$$



Now solving



$$
dotlambda(t) = lambda(t)+x(t)\
lambda(t) + u(t) = 0\
dot x(t)+x(t) = u(t)
$$
we obtain



$$
x(t) = frac{left(7 c_1-3 c_2right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_1-2 c_2right) cosh left(sqrt{2} tright)\
lambda(t) = frac{left(7 c_2-17 c_1right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_2-12 c_1right) cosh left(sqrt{2} tright)
$$



Now including the contour conditions $x(0) = 1, x(t_f) = 2$ we obtain



$$
c_1= 2 sqrt{2} coth left(sqrt{2} t_fright)-4 sqrt{2} text{csch}left(sqrt{2} t_fright)+3\
c_2= 5sqrt{2} coth left(sqrt{2} t_fright)-10 sqrt{2} text{csch}left(sqrt{2} t_fright)+7
$$



Now from the conditions



$$
H_{t_f} = 0\
u(t_f)+lambda(t_f) = 0
$$



we obtain



$$
lambda(t_f) = 2left(-1pmsqrt 2right)
$$



and then follows



$$
t_f = frac{ln 2}{sqrt 2}
$$



etc.






share|cite|improve this answer












Calling



$$
H(x,u,lambda) = frac{1}{2} left(u(t)^2+x(t)^2right)+lambda (t) (u(t)-x(t))
$$



we have



$$
dotlambda(t) = -frac{partial H}{partial x} = lambda(t)+x(t)\
frac{partial H}{partial u} = lambda(t) + u(t) = 0
$$



Now solving



$$
dotlambda(t) = lambda(t)+x(t)\
lambda(t) + u(t) = 0\
dot x(t)+x(t) = u(t)
$$
we obtain



$$
x(t) = frac{left(7 c_1-3 c_2right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_1-2 c_2right) cosh left(sqrt{2} tright)\
lambda(t) = frac{left(7 c_2-17 c_1right) sinh left(sqrt{2} tright)}{sqrt{2}}+left(5 c_2-12 c_1right) cosh left(sqrt{2} tright)
$$



Now including the contour conditions $x(0) = 1, x(t_f) = 2$ we obtain



$$
c_1= 2 sqrt{2} coth left(sqrt{2} t_fright)-4 sqrt{2} text{csch}left(sqrt{2} t_fright)+3\
c_2= 5sqrt{2} coth left(sqrt{2} t_fright)-10 sqrt{2} text{csch}left(sqrt{2} t_fright)+7
$$



Now from the conditions



$$
H_{t_f} = 0\
u(t_f)+lambda(t_f) = 0
$$



we obtain



$$
lambda(t_f) = 2left(-1pmsqrt 2right)
$$



and then follows



$$
t_f = frac{ln 2}{sqrt 2}
$$



etc.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 at 14:06









Cesareo

8,0483516




8,0483516












  • Have you used Pontryagin for this after maximising the Hamiltonian?
    – Robbie Meaney
    Dec 10 at 9:53






  • 1




    @RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
    – Cesareo
    Dec 10 at 10:34


















  • Have you used Pontryagin for this after maximising the Hamiltonian?
    – Robbie Meaney
    Dec 10 at 9:53






  • 1




    @RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
    – Cesareo
    Dec 10 at 10:34
















Have you used Pontryagin for this after maximising the Hamiltonian?
– Robbie Meaney
Dec 10 at 9:53




Have you used Pontryagin for this after maximising the Hamiltonian?
– Robbie Meaney
Dec 10 at 9:53




1




1




@RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
– Cesareo
Dec 10 at 10:34




@RobbieMeaney Yes The hamiltonian is $H(x,u,lambda)$
– Cesareo
Dec 10 at 10:34


















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