Which means adjoint problem of a differential equation? I wanted to know if anyone can help me with the following problem:
Get the adjoint problem (differential equation and boundary conditions) for the problem given by:
$$\frac{d^2 u}{dx^2}=f(x)$$ $$0<x<1$$ $$u(0)=\frac{du}{dx}(0)=0$$
Actually I do not know to be the "adjoint problem", any help, example or reference would help me too.

You need to find an operator adjoint to the given one
$$
L = \frac {d^2}{dx^2}
$$
Condition on real adjoint operator is 
$$
\left \langle Lu, v\right \rangle = \left \langle u, L^*v\right \rangle \Longleftrightarrow \int_0^1 \left(Lu\right ) v\ dx = \int_0^1 u \left ( L^* v\right) dx
$$
Now, just do the integration by parts, twice.
$$
\int_0^1 u'' v dx = \left . u' v \right |_0^1 - \int_0^1 u'v'dx = \left . u' v \right |_0^1 - \left . u v' \right |_0^1 + \int_0^1 u v'' dx
$$
Expand boundary conditions
$$
\left . u' v \right |_0^1 - \left . u v' \right |_0^1 = \left ( \left . u' v \right |_1 - \left . u v' \right |_1 \right )- \left( \left . u' v \right |_0 - \left . u v' \right |_0 \right ) = \left . u' v \right |_1 - \left . u v' \right |_1
$$
Analyzing it you can deduce that if $\left . v\right |_1 = \left . v' \right |_1 = 0$, then
$$
\left \langle Lu, v\right \rangle = \int_0^1 u'' v dx = \int_0^1 uv'' dx = \left \langle u, L^*v\right \rangle
$$
so, final answer is
$$
L^* = \frac {d^2}{dx^2}
$$
with boundary conditions
$$
v(1) = \frac {dv}{dx}(1) = 0
$$
or in terms of ODE
$$
\frac {d^2v}{dx^2} = f \\
v(1) = \frac {dv}{dx}(1) = 0,\qquad x \in [0,1]
$$
