1
$\begingroup$

Please correct me if I am wrong in the following. Let us say I have an unconstrained system with a Lagrangian $\bar L$ and a corresponding Hamiltonian $\bar H$.

$${{\dot x}_i} = {u_i}\qquad \quad i = 1,...,n$$

Let the cost functional be $ J\left( x \right) = \int\limits_{{t_0}}^{{t_f}} {\bar L\left( {t,x,\dot x} \right)dt} $. It can be shown that the the above system satisfies Euler-Lagrange equation under optimality. From the canonical equations we have $$\dot p^* = - {{\bar H}_x^*} = {{\bar L}_x^*}$$ Then we use the fact that ${\bar H}$ has a stationary point with respect to $u^*$ and from there we get $ p^* = {\bar L_{\dot x}^*} $ and so Euler-Lagrange is satisfied.

Now we introduce a set of constraints: $$ \begin{align*} &{{\dot x}_i} = {f_i}\left( {t,{x_1},...,{x_n},{u_1},...,{u_{n - k}}} \right), & i = 1,...,k & \\ &{{\dot x}_{k + i}} = {u_i},& i=1,...,n-k& \end{align*} $$

I am trying to show that Euler-Lagrange still holds with the augmented Lagrangian.

$$ \bar L\left( {t,x,\dot x} \right) + \sum\limits_{t = 1}^k {\lambda _i^*\left( t \right)\left( {{{\dot x}_i} - {f_i}\left( {t,{x_1},...,{x_b},{{\dot x}_{k + 1}},...,{{\dot x}_n}} \right)} \right)} $$

and new Hamiltonian. I have got the Hamiltonian in the form, $$ H = \left[ {\begin{array}{*{20}{c}} {{{\left( {p + \lambda } \right)}^T}}&p^T \end{array}} \right]\left[ {\begin{array}{*{20}{c}} f \\ u \end{array}} \right] - \bar L + {\lambda ^T}\dot x$$

I can see that canonical equation ${{\dot x}^*} = {H_p}^*$ still holds and have derived the second one as, $${{\dot p}^*} = {{\bar L}_x}^* - {\left( {{p^*} + {\lambda ^*}} \right)^T}{\left. {\frac{{\partial f}}{{\partial x}}} \right|^*}$$

but I am not quite sure how to proceed from here. The lone $\dot x$ term in the Hamiltonian feels very weird to me.

$\endgroup$
0
$\begingroup$

I think I have solved it. It was like a strenuous exercise in book-keeping. Folks are welcome to point out any errors in here.

Given the system, we define the Hamiltonian as, $$H = {p_1}{f_1}\left( {{x_1},{x_2},u} \right) + {p_2}u - L\left( {{x_1},{x_2},{f_1}\left( {{x_1},{x_2},u} \right),u} \right)$$ and so the canonical equations give us, $$\frac{{\partial H}}{{\partial u}} = {p_1}\frac{{\partial {f_1}}}{{\partial u}} + {p_2} - \frac{{\partial L}}{{\partial {f_1}}}\frac{{\partial {f_1}}}{{\partial u}} - \frac{{\partial L}}{{\partial u}} \equiv 0$$ along with $${{\dot p}_1} = - {p_1}\frac{{\partial {f_1}}}{{\partial {x_1}}} + \frac{{\partial L}}{{\partial {x_1}}} + \frac{{\partial L}}{{\partial {f_1}}}\frac{{\partial {f_1}}}{{\partial {x_1}}}\quad {\text{and }}\quad {{\dot p}_2} = - {p_1}\frac{{\partial {f_1}}}{{\partial {x_2}}} + \frac{{\partial L}}{{\partial {x_2}}} + \frac{{\partial L}}{{\partial {f_1}}}\frac{{\partial {f_1}}}{{\partial {x_2}}}$$ Note that from the augmented Langrangian we have, $${{\bar L}_{{x_1}}} = {L_{{x_1}}} - \lambda \frac{{\partial {f_1}}}{{\partial {x_1}}}\qquad {\text{and }}\qquad {{\bar L}_{{{\dot x}_1}}} = {L_{{{\dot x}_1}}} - \lambda $$ To satisfy E-L we need, $$\frac{d}{{dt}}\left( {{{\bar L}_{{{\dot x}_1}}}} \right) = {{\bar L}_{{x_1}}}$$ and so we set $\lambda = - {p_1} + {L_{{{\dot x}_1}}}$. Then ${{\bar L}_{{{\dot x}_1}}} = {L_{{{\dot x}_1}}} - \lambda = {L_{{{\dot x}_1}}} - \left( { - {p_1} + {L_{{{\dot x}_1}}}} \right) = {p_1}$. So, $$\frac{d}{{dt}}\left( {{{\bar L}_{{{\dot x}_1}}}} \right) = {{\dot p}_1} = - {p_1}\frac{{\partial {f_1}}}{{\partial {x_1}}} + \frac{{\partial L}}{{\partial {x_1}}} + \frac{{\partial L}}{{\partial {f_1}}}\frac{{\partial {f_1}}}{{\partial {x_1}}} = - {p_1}\frac{{\partial {f_1}}}{{\partial {x_1}}} + {L_{{x_1}}} + {L_{{{\dot x}_1}}}\frac{{\partial {f_1}}}{{\partial {x_1}}}$$ and we also have, $${{\bar L}_{{x_1}}} = {L_{{x_1}}} - \lambda \frac{{\partial {f_1}}}{{\partial {x_1}}} = {L_{{x_1}}} - \left( {{p_1} + {L_{{{\dot x}_1}}}} \right)\frac{{\partial {f_1}}}{{\partial {x_1}}} = {L_{{x_1}}} - {p_1}\frac{{\partial {f_1}}}{{\partial {x_1}}} + {L_{{{\dot x}_1}}}\frac{{\partial {f_1}}}{{\partial {x_1}}}$$

So E-L is satisfied in the first variable.

To see that the second E-L equation is also satisfied we note from the canonical equation that it can be written as, $$\frac{{\partial H}}{{\partial u}} = {p_1}\frac{{\partial {f_1}}}{{\partial u}} + {p_2} - \frac{{\partial L}}{{\partial {f_1}}}\frac{{\partial {f_1}}}{{\partial u}} - \frac{{\partial L}}{{\partial u}} = {p_1}\frac{{\partial {f_1}}}{{\partial {{\dot x}_2}}} + {p_2} - {L_{{{\dot x}_1}}}\frac{{\partial {f_1}}}{{\partial {{\dot x}_2}}} - {L_{{{\dot x}_2}}} \equiv 0$$

Then we note by comparison that $${{\bar L}_{{{\dot x}_2}}} = {L_{{{\dot x}_2}}} + \left( { - {p_1} + {L_{{{\dot x}_1}}}} \right)\frac{{\partial {f_1}}}{{\partial {{\dot x}_2}}} = {L_{{{\dot x}_2}}} - {p_1}\frac{{\partial {f_1}}}{{\partial {{\dot x}_2}}} + {L_{{{\dot x}_1}}}\frac{{\partial {f_1}}}{{\partial {{\dot x}_2}}} = {p_2}$$ and so we need $\frac{d}{{dt}}\left( {{{\bar L}_{{{\dot x}_2}}}} \right) = {{\dot p}_2} = {{\bar L}_{{x_2}}}$ And that this is indeed the case is established because, $${{\bar L}_{{x_2}}} = {L_{{x_2}}} - \lambda \left( { - \frac{{\partial {f_1}}}{{\partial {x_2}}}} \right) = {L_{{x_2}}} + \lambda \frac{{\partial {f_1}}}{{\partial {x_2}}} = {L_{{x_2}}} + \left( { - {p_1} + {L_{{{\dot x}_1}}}} \right)\frac{{\partial {f_1}}}{{\partial {x_2}}} = {L_{{x_2}}} - {p_1}\frac{{\partial {f_1}}}{{\partial {x_2}}} + {L_{{{\dot x}_1}}}\frac{{\partial {f_1}}}{{\partial {x_2}}}$$ which is $\dot p_2$ as we can see by comparing with previous canonical equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.