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.

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    $\begingroup$ Can you provide a citation for the first equation? Because I've never seen the diffusion equation applied to a stress field before - it doesn't make sense to me. $\endgroup$ – lemon Aug 21 '17 at 11:13
  • $\begingroup$ @lemon It is derived by my own work and will be soon published. I'm afraid that I cannot share with you more details -- I hate copyrights and do not believe in intellectual property but have to stick with stupid university rules :( Viva free/open source and anarchism. $\endgroup$ – alifornia Aug 21 '17 at 23:29
  • $\begingroup$ @lemon yet basically, the first equation is based on a (time-dependent process of) stress evolution in metals. My confusion , however, arises from the fact that the Laplacian of stress field is well knownly related to the Laplacian f temperature field. Yet, it is merely in space while the first stress evolution relationship is in space-time. Mathematically, it works perfectly fine when I plug the solution of the second equation in the first one, but I do not get it fully. Any idea or input? Much appreciated. $\endgroup$ – alifornia Aug 21 '17 at 23:34

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.

  • $\begingroup$ Thanks for your input. "This means that the Laplacian operators are unnecessary"? I have a crazy temperature distribution with non zero first and second gradients. How should I relate stress filed with temperature field then? $\endgroup$ – alifornia Aug 22 '17 at 1:00
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    $\begingroup$ Why does the "craziness" of the temperature distribution matter? In the absence of mechanical strain, the stress at each point should just be proportional to the relative temperature there. Normally one writes $$ \sigma = -\frac{E\alpha}{1-2\nu} \Delta T$$ so I'm not sure about your proportionality constant, but the relationship stands - you shouldn't need to differentiate both sides. $\endgroup$ – J. Murray Aug 22 '17 at 1:16
  • $\begingroup$ It's indeed necessary to take Laplacian if I need a pure relationship between only stress and temperature fields. Interestingly, I do not have even considered thermal expansion and have no equation for that in my analysis. In other words, stress is highly related to temperature but through a completely different process than thermal expansion. This process creates non zero first gradient of stress (that sort of can be seen as BCs to gradient of thermal stress equation). Yet still the equation I provided is always true! I'm very confused though but does that now make sense why I did that? $\endgroup$ – alifornia Aug 22 '17 at 3:23
  • $\begingroup$ Not to me. By what mechanism does the temperature influence the internal stress of a material if not through thermal expansion? $\endgroup$ – J. Murray Aug 22 '17 at 3:39
  • $\begingroup$ Sure thing. Temperature gradients create a driving force in microstructure metals called thermomigration which drives atomic flux and therefore cause stress. In nano dimension it is a huge stress (up to 500MPa). It is described through $J=\frac{DC}{kT}\frac{Q}{T}\nabla T$ where $D$, $k$ and $Q$ are the diffusion coefficient, Boltzmann's constant and the specific heat of transfer. This can be seen more less as a body force (maybe not). $\endgroup$ – alifornia Aug 22 '17 at 5:21

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 $$ \nabla^2 \left( \frac{2E} {3(v-1)} \alpha \nabla^2 T\right) = 0 $$


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