I've been looking at the cake eating problem over a finite horizion and have been trying to figure out if we can derive a policy function for such a problem. My work is written below.
sequence form of such a problem is:
$$\max _{c_t,x_{t+1}}U(c_t)=\sum_{i=0}^T\beta^tu(c_t),\ \ \ \ \ \ 1>\beta>0$$ $$s.t. x_{t+1}=x_{t}-c_t$$
supposing that the instantaneous utility function for such preferences are $u(c_t)=\ln(c_t)$ it follows that the function equation for such a problem is:
$$v(x_{t+1})=\max_{x_{t+1}}\{\ln(x_t-x_{t+1})+\beta v(x_{t+1})\}$$
In this case im supposing $T=2$ (we have three periods starting from period zero) and suppose $x_0>0$. Solving the problem from backwards induction we know. $$v(x_2)=\ln(x_2)$$ it follows from this result that for the previous periods value function we have: $$v(x_1)=\max_{x_2}\{\ln(x_1-x_2)+\beta v(x_2)\}$$ $$v(x_1)=\max_{x_2}\{\ln(x_1-x_2)+\beta \ln(x_2)\}$$
after taking the first order condition for this result and solving for $x_2$ we get: $$x^* _2=\frac{\beta x_1}{1+\beta}$$
we know now our value function is properly defined as:
$$v(x_1)=\ln\left(x_1-\frac{\beta x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta} \right)$$
Simplifying:
$$v(x_1)=\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) $$
For our $v(x_0)$ we follow the same procedure before
$$v(x_0)=\max_{x_1}\{\ln(x_0-x_1)+\beta v(x_1)\}$$ $$v(x_0)=\max_{x_1} \left \{\ln(x_0-x_1)+\beta \left[\ln\left(\frac{ x_1}{1+\beta}\right)+\beta \ln\left(\frac{\beta x_1}{1+\beta}\right) \right]\right\}$$
upon solving such an optimization problem we find:
$$x_1^*=\frac{(\beta+\beta^2)x_0}{1+\beta+\beta^2}$$
From this work that I've done above we can see that the policy functions change each period.
Given the solving above do time invariant policy functions exist?
