Computation of adjoint operator I am struggeling with the computation of the adjoint operator $T^*$ to the following given operator $T$:
Let $T: \mathbb{R}^n \rightarrow L^2(0,1), u_0 \mapsto u(t)$, where $u(t)$ is given via
\begin{align*}
u'(t) &= Au, t>0\\
u(0) &= u_0
\end{align*}
I know that the adjoint operator is defined through the equation
\begin{align*}
 \langle T^*v, u_0 \rangle_2 = \langle v, Tu_0\rangle_{L^2(0,1)}
\end{align*}
Since $T$ is defined in an implicit way through the given differential equation, I assumed that $T^*$ has to be definite in a similar way through another ODE. I tried the following:
\begin{align*}
\langle T^*v, u_0 \rangle_2 = \langle v, Tu_0\rangle_{L^2(0,1)} = \int_0^1 v(t) \cdot Tu_0(t) dt = \int_0^1 v(t) \cdot u(t) dt 
\end{align*}
Then I tried to use integration by parts but that didn't seem to help since there are no derivatives in the integral. Can someone give me a hint?
Edit: I am looking for a representation of $T^*$ that derives from the solution of the corresponding adjoint problem. So I tried to derive equations that have to be satisfied by $v$. Unfortunately, I wasn't very successful so far.
Thanks in advance:)
 A: In this specific case you can find the explicit formula for $T$: $u(t)=e^{At}u_{0}$. Here $e^{At}$ is defined in terms of usual Taylor expansion of the exponential function, evaluated at the matrix $At$. By Picard-Lindelof theorem the solution $u(t)$ must be uniquely given, and the particular choice $u(t)=e^{At}u_{0}$ satisfies the equation so it must be the unique solution. (Or, you can directly show it must be the solution by verifying that $e^{-At}u(t)$ must have zero derivative, so must be a constant.)
Therefore, in your equation
$$
  \left\langle T^{*}v,u_{0} \right\rangle_{2}
  = \int_{0}^{1}v(t)\cdot u(t)\,dt,
$$
the right-hand side is equal to
$$
  \int_{0}^{1} v(t)\cdot e^{At}u_{0}\,dt = \left(\int_{0}^{1}e^{A^{*}t}v(t)\,dt\right)\cdot u_{0}
$$
(here we used the identity $(e^{At})^{*}=e^{A^{*}t}$), thus you conclude
$$
  T^{*}v = \int_{0}^{1}e^{A^{*}t}v(t)\,dt.
$$
Probably it is also possible to obtain a similar formula when the ODE is of more general form. Not so sure what precisely such a statement would be.
Edit
Regarding your comment. Consider the ODE
\begin{align*}
  w'(t) &= -A^{*}w(t) - v(t), \\
  w(1) &= 0.
\end{align*}
Then with some similar computations, you can see that the solution is given as
$$
  w(t) = e^{-A^{*}t}\int_{t}^{1}e^{A^{*}\tau}v(\tau)\,d\tau.
$$
Then $T^{*}v$ is precisely $w(0)$. In other words, you are considering the linear ODE with the coefficient matrix $A^{*}$ instead of $A$, while the input $v$ acts as the "forcing term", and you are solving the ODE "backward", so you are fixing the "final value" instead of the "initial value". The minus signs in the ODE also accounts for the direction of time being backward. Then the output of $T^{*}$ is the initial value of the ODE.
(The "general case" I mentioned in the end of the answer before the edit probably would be something similar.)
Now $T^{*}$ being given in terms of the solution $w$ of the ODE above can be derived directly as follows, which is probably what you want. Since
$$
  v(t) = -A^{*}w(t) - w'(t),
$$
we can write
\begin{align*}
  \left\langle T^{*}v,u_{0} \right\rangle_{2}
  &= -\int_{0}^{1}A^{*}w(t)\cdot u(t) + w'(t)\cdot u(t)\,dt \\
  &= -\int_{0}^{1}w(t)\cdot Au(t) + w'(t)\cdot u(t)\,dt \\
  &= -\int_{0}^{1}(w(t)\cdot u(t))'dt \\
  &= -(w(1)\cdot u(1) - w(0)\cdot u(0)) \\
  &= w(0)\cdot u_{0},
\end{align*}
thus we get $T^{*}v = w(0)$.
