Consider the growth equation: $ \dot{x} = tu $, with $x(0)=0$ and $x(1)=1$, and with the cost function: $ J= \int_0^1 u^2 dt $.
Show that $u^*=3t$ is a successful control, with $x^*=t^3$ and $J^*=3$ the corresponding trajectory and cost. If $u=u^* + v $ is another successful control, show that $\int_0^1 vt dt = 0 $ and by finding the cost for this control, show that $u^*$ is the optimal control.
To show that $u^* = 3t$ is a successful control (in the sense that it satisfies the boundary conditions), we can plug it into the state equation, giving $\dot{x} = 3t^2$. Integrating $\dot{x}$ to get $x^*(t)$ gives $x^*(t) = \int_{0}^{t} 3\tau^2 d\tau = t^3$. Then we see that $x^* = t^3$ (as expected from the problem statement). Plugging $t = 0$ and $t = 1$ into $x^*$, we have $x^*(0) = 0$ and $x^*(1) = 1$ which together mean that $u^*$ is a successful control.
To find the value of $J^*$, we examine $J^* = \int_{0}^{1} u^2dt = \int_{0}^{1} 9t^2dt = 3t^3\bigg|_{0}^{1} = 3$.
Now take $u = u^* + v$. If this is a successful control, then it must also generate a state trajectory which satisfies the boundary conditions. Then \begin{equation} x(t) = \int_{0}^{t} t(u^* + v)dt = \int_{0}^{t} tu^* dt + \int_{0}^{t} tvdt = t^3 + \int_{0}^{t}tvdt \end{equation} where we have used the fact that $\int_{0}^{t} tu^*dt = t^3$. This clearly satisfies $x(0) = 0$. To check $x(1) = 1$, we have \begin{equation} x(1) = 1 + \int_{0}^{1} tvdt = 1. \end{equation} Then it must be the case that $\int_{0}^{1}tvdt = 0$.
Looking at $J = \int_{0}^{1} (u^* + v)^2dt$, to minimize $J$, we'd like to minimize the term $(u^* + v)^2$ at each point in time while still satisfying the boundary conditions and the condition that $\int_{0}^{1}tvdt = 0$. We can achieve this by setting $v = 0$ for all time. Then $J = 3$ in this case as well. Then no other successful control has a lower cost than $u^*$ and $u^*$ is indeed the optimal control.