I have a PDE $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + u(x,t)$ for $\forall x \in [a,b], \forall t \in [0,T]$
[Edited for comment from @user10354138] If I substitute $x=a$ to the PDE, we have $\frac{\partial u}{\partial t}(a,t) = \frac{\partial^2 u}{\partial x^2}(a,t) + u(a,t)$. Is it still a boundary condition at $x=a$?
The reason I ask this question is that I thought boundary conditions give extra information and we will use these extra information to determine any coefficients/functions generated from integration. However, in this case, this "boundary condition" does not bring us any extra information because the solution of the PDE must satisfy the PDE itself.
I get this question from reading this finance note. In page 4, definition 2.2 and Remark, it says "The boundary condition at $y = 0$ is obtained by formally plugging in $y = 0$ into (4)".
Usually, the function needs to satisfy the PDE (without IC+BC) on an open set. The "evaluation" at the boundary is a restriction but not so much in terms of a usual BC, but more in terms of restrictions on the functional space and the existence of the limit.
Generally speaking such a condition will make existence results and uniqueness results harder.
Regarding the paper you linked: the formal evaluation is actually an implicit condition regarding the growth of the second order partial derivatives when $y\to0$.