What I mean by the title is often, if we have a value function like $$V(K,I) = \max_{K',I'} F(K') +\beta V(K',I')$$ the First order conditions will give us something that depends on the derivative of the value function, say $V_1(K',I')$. We deal with this by using the B-S condition to find $V_1(K,I)$, then advance it one period?
What about when $V_1(K,I)$ still depends on $V_1(K',I')$, though? (see $(*)$ below for an example, if you wish to skip the setup)
Here is an example. Note that subscripts denote derivatives, the number in the subscript being the input with respect to which we are taking the derivative. The exception to this is negative subscripts, such as $I_{-1}$, where the subscript denotes being one period in the past (in the case of $-1$). A prime denotes being one period in the future (t+1), and two primes denotes being two periods in the future.
Back to the problem: We have value function $$V(K,I_{-1}) = \max_{I,K'',C}u(C) + \beta V(K',I) \\ \text{ s.t. } C+I \leq f(K) \\ \text{ and } I =K'' - (1-\delta)K' $$ So basically we have a simple model with investment that takes two periods to build. $f(K)$ is our production function.
The FOC's give $$ u_1(c) = \lambda_1 \text{, $\lambda_1$ is the multiplier on the first constraint }\\ \lambda_2 = 0 \text{ multiplier on second constraint is zero} \\ \beta V_2(K',I) = -\lambda_1 $$ Now to get $V_2$ we use the B-S condition, which gives $$V_2(K,I_{-1}) = \beta V_1(K',I)$$ because $K' = I_{-1} + (1-\delta)K$ from the second constraint. Use the BS condition again and get $$V_1(K,I_{-1}) =f_1(K) + (1-\delta)V_1(K',I) \tag{*}$$, again, because $K' = I_{-1} + (1-\delta)K$
The previous equation is my question To reiterate, we use B-S condition to find the derivative of the value function, but in $(*)$ the B-S condition depends on what we are trying to find. How do we handle this?
I feel like $V_1(K'I)$ may just be $1$, or $0$, but then that makes $V_2$ be $\beta$ or $0$, which seems wrong...