Properties of a heat reservoir S. Salinas, Introduction to Statistical Physics (Springer, 2001) while discussing variational principles of thermodynamics considers a system with internal energy U, entropy S in equilibrium with a heat reservoir with internal energy $U_R$, entropy $S_R$ and fixed temperature $T$. The composite system is closed. To quote (p. 57): "The heat reservoir is described by the equations $dU_R=TdS_R$; $d^2U_R=0$, and $d^2 S_R=0$."
Why does the second derivative of the internal energy (and entropy) of the reservoir equal zero ?
 A: The second differential referenced by Salinas is defined as follows.  If $U=U(S,V)$, then
$$\mathrm dU = \left(\frac{\partial U}{\partial S}\right)_V \mathrm dS + \left(\frac{\partial U}{\partial V}\right)_S \mathrm dV \equiv T \mathrm dS - p\mathrm dV$$
$$\mathrm d^2 U = \left(\frac{\partial^2 U}{\partial S^2}\right)_V (\mathrm dS)^2 + 2\left(\frac{\partial^2 U}{\partial S \partial V}\right) \mathrm dV \mathrm dS + \left(\frac{\partial^2 U}{\partial V^2}\right)_S (\mathrm dV)^2$$
$$\equiv \left(\frac{\partial T}{\partial S}\right)_V (\mathrm dS)^2 + 2\left(\frac{\partial T}{\partial V}\right)_S \mathrm dV \mathrm dS -\left(\frac{\partial p}{\partial V}\right)_S (\mathrm dV)^2$$
where we note that $\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial p}{\partial S}\right)_V$ via the equality of mixed partial derivatives.

Why does the second derivative of the internal energy (and entropy) of the reservoir equal zero ?

Salinas is referring to a reservoir whose volume is fixed, which means that $dV_R = 0$.  This leads us immediately to $\mathrm dU_R = T_R \mathrm dS_R$ and $\mathrm d^2 U_R = \left(\frac{\partial T_R}{\partial S_R}\right)_{V_R} (\mathrm dS_R)^2$.  However, the defining characteristic of a reservoir is that it is so large relative to the system that its temperature can be considered constant.  Heat flow from the system to the reservoir produces no change in temperature, and so $\left(\frac{\partial T_R}{\partial S_R}\right)_{V_R} = 0$.  As a result, $\mathrm d^2 U_R = 0$. Performing an identical expansion for $\mathrm d^2 S_R$ yields that $\mathrm d^2 S_R=0$ as well.
