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I should maximize the following integral (by determine the optimal time $t_o$)

$C(u)=\int_0^T e^{-rt}(1-u(t))x(t)dt$

with $u(t)=1$ if $t<t_{o}$ and $u(t)=0$ if $t_{o}<t$. under $x´(t)=u(t)x(t)$, $x(0)=x_0>0$. $0<r<1$

This means that the integral from $0$ to $t_o$ is $0$. This problem reduces to

$C(u)=\int_{t_0}^T e^{-rt}x(t)dt$

and if I solve the differential equation then I get $\frac{dx}{x}=u dt$ and with that $x(t)=e^{\int_0^t u dt+c}$ and for $c=log(x_0)$ and for $t>t_o$ is $\int_0^t udt=t_o$. So $x(t)=e^{t_o+log(x_0)}$

Is that correct? How can I determine the optimal time $t_0$?

Brian
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1 Answers1

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First, you solved the differential equation incorrectly, or else haven't simplified it enough to really see what's going on. The equation for $x$ says that $x$ grows exponentially with rate constant $1$ until $t_0$, when it stops evolving entirely. Hence

$$x(t) = \begin{cases} x_0 e^t & 0 \leq t \leq t_0 \\ x_0 e^{t_0} & t > t_0\end{cases}.$$

That can simply be plugged into the integral. Given the simplification you performed, you get (by some abuse of notation):

$$C(t_0)=\int_{t_0}^T e^{-rt} x_0 e^{t_0} dt = x_0 e^{t_0} \int_{t_0}^T e^{-rt} dt.$$

The integral can be done explicitly:

$$C(t_0) = \frac{x_0}{r}e^{t_0} \left ( e^{-rt_0}-e^{-rT} \right ).$$

Now the problem has been reduced to an ordinary one variable optimization problem (at least for each fixed $T$ and $r$).

Ian
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