Unique solution in differential equation

Question

Given a functions g(t,T) and Q(t,T) such that

$g(t,T) = - \frac{\partial}{\partial T} \ln Q(t,T)$,

$Q(T,T) = 1 = Q(t,t)$,

T>0 and

$t \in [0,T]$

Does it follow that $Q(t,T) = exp(-\int_{t}^{T} g(t,u) du)$?

My professor gives an argument that suggests it is so, but a different way I tried suggested the instead we have $Q(t,T) = \pm exp(-\int_{t}^{T} g(t,u) du)$. Who is right? What is the flaw in the wrong one's reasoning?

My professor's:

$g(t,u) = - \frac{\partial}{\partial u} \ln Q(t,u)$

$\int_{t}^{T} g(t,u) du = \int_{t}^{T} - \frac{\partial}{\partial u} \ln Q(t,u) du$

$\int_{t}^{T} g(t,u) du = \int_{t}^{T} - \frac{\partial}{\partial u} \ln Q(t,u) du$

$- \int_{t}^{T} g(t,u) du = \ln Q(t,T) - \ln Q(t,t)$

$- \int_{t}^{T} g(t,u) du = \ln Q(t,T)$

$e^{- \int_{t}^{T} g(t,u) du} = Q(t,T)$

QED

Mine:

$g(t,u) = - \frac{\partial}{\partial u} \ln Q(t,u)$

$\int_{t}^{T} g(t,u) du = \int_{t}^{T} - \frac{\partial}{\partial u} \ln Q(t,u) du$

$- \int_{t}^{T} g(t,u) du = - \int_{t}^{T} - \frac{\partial}{\partial u} \ln Q(t,u) du$

$- \int_{t}^{T} g(t,u) du = \int_{t}^{T} \frac{\partial}{\partial u} \ln Q(t,u) du$

$- \int_{t}^{T} g(t,u) du = \int_{t}^{T} \frac{\partial}{\partial u} Q(t,u) / Q(t,u) du$

Let

$v = Q(t,u)$

$dv = \frac{\partial}{\partial u} Q(t,u)$

$- \int_{t}^{T} g(t,u) du = \ln |Q(t,T)/ Q(t,t)|$

$- \int_{t}^{T} g(t,u) du = \ln |Q(t,T)|$

$e^{- \int_{t}^{T} g(t,u) du} = |Q(t,T)|$

$\pm e^{- \int_{t}^{T} g(t,u) du} = Q(t,T)$

QED

P.S. g and Q are stochastic, and it is assumed we can swap integral and derivative (if even relevant).

Answer 1

The solution

$$Q(t,T) = - \exp \left( - \int_t^T g(t,u) \, du \right)$$

does not satisfy the initial conditions $Q(T,T) = q(t,t)=1$. Obviously,

$$Q(t,t) = Q(T,T)=-1.$$