Evaporating tungsten wire 
Consider an incandescent bulb having a thin filament of tungsten that is heated to high temperature by 
  passing an electric current. The hot filament emits black-body radiation. The filament is 
  observed to break up at random locations after a sufficiently long time of operation due to 
  non-uniform evaporation of tungsten from the filament. The bulb is powered at constant voltage.

Can we solve this situation quanititatively for the temperature gradient or the resistance varitaion with temperature? Before the wire breaks, will it emit light of shorter wavelength? Is there any equation or concept regarding rate of evaporation of solids?
Is it possible to predict these variations qualitatively atleast?
Also, can tungsten evaporate? It has the highest melting point among all elements after all. And usually, isn't that  vapour thing coming off of heated tungsten due to oxidation of tungsten?
 A: The non-uniform evaporation probably originates from a small variation in thickness of the filament, or possibly local variations in the degree of perfection of the crystal structure. Now thinner parts of the filament will have a greater than average resistance per unit length, and so will dissipate more power ($P=I^2R$). So they will get hotter, as they will tend to the steady state when power in = Power out, that is $$I^2R=\sigma A_{\text{surf}} T^4.$$
[Here we're assuming that the 'hot spot' in the filament radiates as a black body.] Because it's hotter the peak wavelength of emission is lower (Wien's law: $\lambda_{\text{peak}} \propto T^{-1}$), which I hope answers one of your questions.
Now the key point to note is this: we have an unstable situation... Because the thin parts of the filament are hotter, tungsten evaporates (sublimes) faster from them than from the rest of the filament. So they become thinner still, get hotter still – and so on. 
Using a simple approximate formula linking sublimation rate to temperature, we could no doubt develop these ideas quantitatively.
