Htykuut

I am trying to solve the following problem (I choose a numerical example). The state $x$ is governed by a differential equation linear in $xin [0,1]$. Two variables, $theta in[0,1]$ and $rhoin[0,1)$, can change the state using a control $mu in [0,1]$. I want to choose $mu$ such that the time it takes to reach $x=1$ is minimized.

Let me state the problem:
$$ min_{mu, T} int_0^T mathrm dt$$
subject to
$$ dot{x}=tfrac{1}{2}left(xleft(tfrac{4}{1-rho}-5right)+5theta -1right)$$
$$ dot{theta} = (1-theta)mu$$
$$ dot{rho} = (1-rho)(1-mu)$$
$$ x(0)=x_0, x(T)=1, theta(0)=rho(0)=0$$
Then I set up the Lagrangian as:
$$L=1+lambda_1 tfrac{1}{2}left(xleft(tfrac{4}{1-rho}-5right)+5theta -1right)+ lambda_2(1-theta)mu+lambda_3(1-rho)(1-mu)-w_1 mu -w_2(1-mu)$$
This is how I proceed. I write the conditions for an optimum as:
$$ frac{partial L}{partial mu}=lambda_2(1-theta)-lambda_3(1-rho)-w_1+w_2=0 quad (1) \ frac{partial L}{partial x}=frac{lambda_1}{2}left(frac{4}{1-rho}-5right)=-dot{lambda_1} quad (2) \ frac{partial L}{partial theta}=frac{5lambda_1}{2} -lambda_2 mu=-dot{lambda_2} quad (3) \ frac{partial L}{partial rho}=frac{lambda_1}{2}frac{4 x}{(1-rho)^2} -lambda_3 (1-mu)=-dot{lambda_3} quad (4) \ w_1geq 0, w_1 mu=0 \ w_2geq 0, w_2(1-mu)=0$$

The complementary slackness conditions suggest that $lambda_2(1-theta)>(<)lambda_3(1-rho)$ implies $mu=0 (1)$.
At this point I feel stuck. I did differentiate (1) one more time w.r.t. the time to get:
$$
dot{lambda}_2(1-theta)-lambda_2(1-theta)mu-dot{lambda}_3(1-rho)+lambda_3(1-rho)-dot{w}_1+dot{w}_2=0
$$
Since (I think) $-dot{w}_1+dot{w}_2=0$, I can substitute the remaining expression into equations (3) and (4), but then $lambda_1 = dot{lambda}_1=0$.

What is wrong here?

Any help / hints / suggestions / admonishments would be great!

Since your cost function is not a function of $mu$ and the differential equations are only linear in $mu$ then the partial derivative of the Lagragian with respect to $mu$ is not going to be a function of $mu$. So when you set the partial derivative equal to zero in equation $(1)$ you can not find an expression for an optimal $mu$ as a function of the (co-)states.

What equation $(1)$ actually tries to do is minimize the Lagragian with respect to $mu$. Now when there are constraints on the control variable you can also minimize the Lagragian directly, while also taking the constraints on $mu$ into consideration. Also I would think that the co-states $w_1$ and $w_2$ are not needed once you solve for $mu$ as mentioned above.

Optimal time problem includes terminal time condtion
$$[L]_{t=T^{*}} = 0,$$
where $T^*$ is the optimal time. You should also use this equation.