Unrest said:
Suppose it's:
k d2T/dx2 + q. = 0
If we fix q.=0 then we can apply two T=constant BCs and determine the T field.
I don't think we have any disagreements about that.
But what if we fix T over part of the domain and leave q. unknown, while we fix q.=0 over the rest of the domain.
The way I would think about that is to say that you have two different equations on the two domains. In one equation the unknown variable is q, on the other one it is T.
So you need a complete set of boundary conditions for each equations. Of course on the interface between the domains, the boundary conditions presumably say that q and T (and possibly some of their derivatines as well) are consistent at the common boundary. That really means you have a
coupled system of two equations, though it's a rather unusual type of coupled system.
I have a (free) book which says that BCs apply at t>0. And the wording implies that ICs are not a special case of BCs. Maybe this is just a personal choice?
It's hard to make any comment on that without knowing the context. FWIW I would have said the description "initial conditions" is just a subset of the more general term "boundary conditions".
If you are talking about the heat flow or diffusion equation, there is no reason why you
must have explicit initial conditions of the form "the temperature everywhere in the region at time 0 is given by the function T(x,y,z)." You could have conditions like "the temperature distribution at time 0 is the same everywhere as the temperature distribution at time 1, and the mean temperature averaged over the whole region and between times 0 and 1 is zero". In other situations you may have boundary conditions like "the temperature is always finite", to exclude solutions where the temperature is proportional to 1/t when the time t is small, or temperatures that grow exposnentially as the time goes to infinity.
Or depends if you're a mathematician or an engineer?
Hm... I have a math degree, but I spent all my working life in engineering companies. So I don't know how to answer that!