How to find the exact controllability of the beam equation, using some Ingham result?

Question

Set $(T,L) \in (0,\infty)^{2}$. Consider the beam equation $$\left \{ \begin{array}{ll} { z }_{ tt }\quad +{ z }_{ xxxx }=0,\quad \left( t,x \right) \in \left( 0,T \right) \times \left( 0,L, \right) \quad\\ z\left( t,0 \right) =0,\quad z\left( t,L \right) =0,\quad t\in \left( 0,T \right)\\ { z }_{ x }\left( t,0 \right) =0,\quad { z }_{ x }\left( t,L \right) =u\left( t \right) ,\quad t\in \left( 0,T \right) \quad\\ z\left( 0,x \right) ={ z }_{ 0 }\left( x \right) ,\quad { z }_{ t }\left( 0,x \right) ={ z }_{ 1 }\left( x \right) ,\quad x \in \left( 0,L \right) \quad \end{array}\right.$$

Let $(z_{0},z_{1}) \in L^{2}(0,L) \times H^{-2}(0,L)$ and $(z_{0T},z_{1T}) \in L^{2}(0,L)\times H^{-2}(0,L)$. Then, there is $u \in L^{2}(0,T)$ such that the solution by transposition $z \in C([0,T];L^{2}(0,L))\cap C^{1}([0,T];H^{-2}(0,L))$ of the beam equation satisfies $z(T,\cdot)=z_{0T}(\cdot)$ and $z_{t}(T,\cdot)=z_{1T}(\cdot)$.

1. The control $u$ has to be found, but not by the multipliers method, but using an Ingham result.

2. I have investigated and use the vectors and eigenvalues by defining an operator, but I don't know how to find these vectors and eigenvalues.

3. Once with these eigenvalues and eigenvectors, use an Ingham result to find the observability inequality.

4. Once with this establish the form of control $u$.