# Euler-Lagrange equation for a constant generalized coordinate

I have an action integral that I want to optimize of the form $$S = \int_0^T \left( a(t)q^2 -2b(t)q + \gamma \dot{q}^2 \right) dt \,.$$ I know that in fact $$\dot{q} = 0$$, in which case the critical points can easily be found by plugging this in, setting $$\frac{d S}{dq}=0$$, and solving for $$q$$. This gives the result, $$q = \left( \int_0^T a(t)dt \right)^{-1} \int_0^T b(t) dt.$$ Suppose instead that I want to solve this with the Euler-Lagrange equation. The fact that $$q$$ is constant can be imposed by the constraint $$f(q)=q-c=0$$. Applying the Euler-Lagrange equations gives $$2a(t)q - 2b(t) - 2 \gamma \ddot{q} = \lambda \,.$$ Differentiating the constraint equation twice shows that $$\ddot{q}=0$$. Plugging this in gives $$2a(t)q - 2 b(t) = \lambda \,,$$ from which it is already apparent that there is a problem since the left-hand side is a function of $$t$$ and the right hand side is not. This can be solved by taking the derivative of both sides and solving for $$q$$, which results in $$q = \frac{\dot{b}}{\dot{a}} \,,$$ which is incorrect. What have I done incorrectly here?

Edit: The reason I am asking is that I have a similar problem where $$q$$ and $$b$$ are vectors, $$a$$ is a matrix, and I know that some of the entries of $$q$$ are constant while others are not. So I need to make sure that I am handling the constant entries correctly when using the Euler-Lagrange equations.

Note on boundary conditions: I don't know if this matters, but there are no natural constraints to impose on $$q$$ at the boundaries, so the boundaries are free, which leads to the requirement of the Transversality conditions $$\frac{\partial L}{\partial \dot{q}}\big|_{t=0} = 0$$ and $$\frac{\partial L}{\partial \dot{q}}\big|_{t=T} = 0$$, where the Lagrangian $$L$$ is given by $$2b(t)q^2 -2a(t) + \gamma \dot{q}^2$$. This imposes the boundary conditions $$\dot{q}(T)=\dot{q}(0)=0$$.

The Lagrange multiplier $$\lambda$$ is time-dependent, so when you take the derivative of the equation $$2a(t) q - 2 b(t) = \lambda$$, $$\lambda$$ does not drop out of the equation. I don't see any way of solving the resulting equations for $$q$$, since you have 2 variables $$\lambda$$ and $$q$$ and one equation. All the constraint equation tells you is that $$q$$ is constant but it will not help you determine what this constant is.