Easy question on math of diffusion equation I have the following well-known diffusion equation:
$$\frac{\partial{\sigma}}{\partial t}=D\nabla^2\sigma$$
where $\sigma$ is the hydrostatic stress. I also know the relationship between stress and temperature as follows:
$$\nabla^2\sigma=\frac{2E}{3(v-1)} \alpha \nabla^2 T$$
My confusion though arises from the fact that my function $T(x,y)$ for temperature distribution does not have a notation of time (i.e. $t$), and therefore the second equation will results in $\sigma(x,y)$. Yet, the first equation says that $\sigma$ is a time dependent quantity (i.e. $\sigma(x,y,t)$. I have not problem with solving it but it does not fully make sense to me. If $\sigma$ is indeed space and time dependent, how $t$-dependency will not appear the second equation yet in the first one. Sorry if it's too trivial, fundamental and mathematical confusion.
 A: From a purely mathematical viewpoint, those two equations are only trivially coupled - you can solve the diffusion equation for $\sigma$ without any reference whatsoever to $T$.  Once you have $\sigma$, you can use it as a source term for the Poisson equation to find $T$ - though in general, $T$ will be time-dependent.
From a physical viewpoint, the thermal stress in the material is proportional to the temperature (or really, to $\Delta T \equiv T-T_{ref}$, where $T_{ref}$ is some reference temperature).  This means that the Laplacian operators are unnecessary, and it also means that the temperature field must be time dependent (unless the stress and the temperature are already in a steady-state configuration).
I don't understand why you seem to insist that the temperature field is not time-dependent.  The equations you wrote don't necessarily require $T$ to change with time from a purely mathematical perspective, but the physics of the situation imply that it must.
A: Let us first consider a given time dependent temperature $T(x, y, t)$. Your second equation relates two divergences
$$
\nabla^2 \sigma = \frac{2E} {3(v-1)} \alpha \nabla^2 T
$$
This equation is to be solved for each instant of time. This is because there is no explicit time derivative, so that different instants of time are independent. However, as time changes $\nabla^2 T$ may change (and the boundary conditions applied to the PDE may change too) and as such different instants of time may have different solutuons $\sigma(x, y, t) $. Thus $\sigma$ is, in principle, time dependent. 
If $T(x, y) $ as stated in your question (I'm not sure if that's an assertion or erroneous conclusion) then, assuming $E$ and $v$ are time independent too, and all of these are known, then we can rewrite the above equation as
$$
\nabla^2 \sigma = f(x, y) 
$$
for some known $f$. Then we can solve this by considering
$$
\sigma(x, y, t) = \sigma_0(x,y) + g(x, y, t)
$$
where 
$$
\nabla^2 \sigma_0 = f(x, y) 
$$
is a solution to the equation, and 
$$
\nabla^2 g(x, y, t) =0
$$
is some solution to the Laplace equation which  adjusts the solution for any time dependence in the boundary conditions or time evolution equation for $\sigma$.
Inserting the form for $\sigma$ into your first equation yields
$$
\frac{\partial g} {\partial t} = D f(x, y)
$$
thus the distribution of temperature affects the time evolving part of $\sigma$. However, taking the Laplacian of this equation yields
$$
\frac{\partial \nabla^2 g} {\partial t} =  D \nabla^2 f(x, y)
$$
thus
$$
0=\nabla^2 f(x, y)
$$
and thus if $T$ really is time independent, then we require
$$
\nabla^2 \left( \frac{2E} {3(v-1)} \alpha \nabla^2 T\right)  = 0
$$
