Which Formula for the Internal Elastic Energy is Correct? I am confused by the internal energy of a crystal. Lets say I have a isotropically stressed crystal. The total differential of the internal energy per unit volume V is defined as
$$du = \sigma d\epsilon$$
$\epsilon$ being the strain and $\sigma$ the stress. Because the internal energy is a homogeneous function, this implies
$$u=\sigma \epsilon$$
But if I integrate $du = \sigma d\epsilon$ from $0$ to some strain $\epsilon$ using the constitutive relation $\sigma = C \epsilon$, $C$ being some constant, I arrive at
$$ u = \frac{1}{2} \sigma \epsilon $$
Now I am very confused. Which equation is the correct one, $u=\frac{1}{2}\sigma \epsilon$ or $u=\sigma \epsilon$. Or which equation applies to which situation? Probably I didnt get some really important point here :) 
Thanks for the help.

UPDATE
Ok, I think I now have understood my problem and solved it. In case anyone else ever wonders about this I will post my answer her. But first I want to reframe the question and clarify some assumptions. I will then post my answer beneath.
So, let there be a system given with some time dependent volume $V$ that we want to compress or expand isothermically and reversibly. That is $dT = 0$ and $dS=0$. Furthermore, for simplicity of the argument, I will assume a homogeneous crystal and homogeneous deformation, so the work of the deformation is $-pdV$. That said, we have for the change of internal energy
$$
dU = -p dV \tag{1} \label{eq1}
$$
After division by the reference volume $V_0$ with the definition $J=V/V_0$ and $u=U/V_0$, this reads
$$
du=-pdJ' \tag{2} \label{eq2}
$$
We will assume a constitutive relation for the pressure as being
$$
p(J) = - K ln(J) \tag{3} \label{eq3}
$$
Where $K$ is the bulk modulus. Using the small strain approximation, that is $J\approx 1$, this reads
$$
p(J) = - K (J-1) \tag{4} \label{eq4}
$$
We now integrate eq. \eqref{eq2} from $J'=1$ to $J'=J$ and get 
$$
\Delta u = K/2(J-1)^2 \tag{5} \label{eq5}
$$
On the other hand, $U$ is a homogeneous function of order $1$, that is $U(\lambda V) = \lambda U(V)$ and therefor, according to Eulers Theorem, we got 
$$
U=-pV \tag{6} \label{eq6}
$$
This holds independently of the function $p(J)$ and after devision by $V_0$ and usage of \eqref{eq4} we end up with
$$
\Delta u = K(J-1)^2 \tag{7} \label{eq7}
$$
Which, obviously, is not the same as \eqref{eq5}. Why is that so? Which formula is correct?
 A: Now, I think I understood my missunderstandings and want other people to participate from that :)
First of all, the equation $6$ in the question is not the whole truth. Because $U$ is also a homogeneous function of $S$ and $N$, $N$ being the total number of particles in the system, the function $U$ can be written as
$$
U = TS-pV+N\mu \tag{8} \label{eq8}
$$
$\mu$ is the chemical potential of the system. Keeping in mind, that $dS=0$ because of reversibility and $dT=0$ and $dN = 0$ we calculate for the change of internal energy between the states $0$ and $1$, $1$ being the finale state of volume $V$ and $0$ the initial state of voume $V_0$
$$
\Delta u = -p_1 J + p_0 + N \Delta \mu = -p(J) J + p(1) + N \Delta \mu = -p(J)J + N \Delta \mu \tag{9} \label{eq9}
$$
So we have to calculate $\Delta \mu$ to get the full picture. This can be done by considering the Gibbs-Duhem relation. Since $du = TdS-pdJ+\mu d\rho$, $\rho := N/V_0$, the Gibbs-Duhem relation reads
$$
sdT-Jdp+\rho d \mu = 0 \tag{10} \label{eq10}
$$
Since $dT=0$ and using equation $4$ from the question, this then leads to
$$
d \mu = \frac{J}{\rho} dp = \frac{J}{\rho} p_J dJ' = -K \frac{J}{\rho} dJ' \tag{11} \label{eq11}
$$
After integration from $J'=1$ to $J'=J$ this gives
$$
\Delta \mu = - \frac{K}{2 \rho} (J^2-1) \tag{12} \label{eq12}
$$
Inserting \eqref{eq12} and the equation $4$ from the question into \eqref{eq9} we end up with
$$
\Delta u = K/2(J-1)^2 \tag{13} \label{eq13}
$$
So, in the end the equations are, of course, consistent. 
