May 14th update: by request, have a look at Sem#12: Trying to clarify that discounting in the "5150" problem & related just posted.

May 11th update: see paragraph on light yellow background right under the Dyn.pr. 2 headline.

Exam 2008 problem 2 and the 2015 solution note are in the https://www.uio.no/studier/emner/sv/oekonomi/ECON4140/oldexams/ folder.

But also, there are videos for 2015 problems 2 and 3 in the YouTube playlist, now made available as https://www.youtube.com/watch?v=UlNYPqS6mgg&list=PLB7Srf1SQsVi1Z71NWBBLMSMmXikOus2R&index=9 and https://www.youtube.com/watch?v=YnzzV83UnzE&list=PLB7Srf1SQsVi1Z71NWBBLMSMmXikOus2R&index=8 respectively

The "ECON5150" adaptations

I did some copying from 2020, and might not have touched it up to fit completely.

The difference equation $LaTeX: p\cdot v_{t+1}-v_t+(1-p)v_{t-1} = K$ . Recall the tasks:

Find a particular solution of the form Ht for all p, except one.
Find the general solution for all p in (0,1) except that one p.
Is it globally asymptotically stable for any p?

Solution

To find a particular solution on the linear form:
Inserting Ht we get for the left-hand side: H[p(t + 1) - t + (1-p)(t-1)] = H[t(p-1+(1-p))+p-(1-p)] = H(2p-1). So a particular solution is Kt/(2p-1) except when p=1/2. (Aside: then we need a quadratic.)
Characteristic equation pm² - m - (1-p) = 0, yields $LaTeX: 2p\,m = 1\pm \sqrt{1^2-4p(1-p)}= 1\pm \sqrt{1-4p+4p^2}=1\pm(1-2p)$ . Roots 1 and $LaTeX: \frac{1-p}p$ . Note that these are two distinct roots except when p=1/2, which we rule out. For all p that are neither 0 nor 1/2 nor 1, we therefore have $LaTeX: v_t=A+B\big(\frac{1-p}p\big)^t+\frac{K}{2p-1}t$ .
A root is 1, so not asymptotically stable (the theory is asymptotically stable if every root has complex modulus <1).
But once you have solved it, you can just notice that unless A=0 you won't have convergence to zero.
And please recall that the inhomogeneous equation inherits the stability property from the homogeneous, so the Kt/(2p-1) part is simply irrelevant to this question.

Dyn.pr. 2

UPDATE May 11th. So this solution does only mention "t0=0", i.e. it put t0=0 in $LaTeX: J_{t_0}(x)=\displaystyle\max_{u_t\geq 0}\Big\{\beta^{T-t_0}(x_T+\sqrt{x_T})+\sum_{t=t_0}^{T-1}\beta^{t-t_0}\big(u_t+\sqrt{u_t}\big)\Big\}$ . I realize this causes some confusion. The thing is, is "J_T" supposed to be current at T or discounted as if you at time zero were told "at time T you are going to inherit this problem at state x"? More generally, is $LaTeX: J_{t_0}$ intended to ... [analogous]. For a math course, I fully appreciate that you are worried to pick the right thing - I mean for econ purposes you can just be specific about what is your intended object of study. I will have a look into any confusion I may have caused and try to avoid that on the exam. ^{(Note, this is not a promise/threat that the exam will have dynamic programming at all!)}End "UPDATE MAY 11th" here, next line is as previously posted.

This is deterministic adaptation of problem 1 of that same above 5150 exam. Consider the problem $LaTeX: J_0(x)=\displaystyle\max_{u_t\geq 0}\Big\{\beta^{T}(x_T+\sqrt{x_T})+\sum_{t=0}^{T-1}\beta^{t}\big(u_t+\sqrt{u_t}\big)\Big\}$ where $LaTeX: \quad x_{t+1}=\frac{x_t-u_t}\beta,\qquad$ x₀=x>0 given.

Calculate J_T-1 and J_T-2.
By now you should be able to guess a form of the value function J_T-t(x) up to some A_T-t that does not depend on x; prove this by induction (you don't need to solve for that constant, as long as you get a difference equation for it. If you cannot guess, see cheatnote a couple of items further down, but that is the form at time=0, so you need to discount it.)
- If you chose a nice form as per the cheatnote, the difference equation for A should be "easy" to solve as well!
Let T go to infinity. State the Bellman equation for this problem.
[That problem also assumes that x stays nonnegative.]
Guess and verify a form for the value function.
- Cheatnote: this form is probably most convenient!

Solution: the induction step.

Let me skip the first question, but assuming you have done enough of it to guess the form with discounting, as $LaTeX: \beta^{T-k}\big(x+\sqrt{A_{T-k}x}\big)$ ; it holds true in the base case k=0, so suppose for induction that it holds true at k. Then at k+1:
$LaTeX: \displaystyle J_{T-(k+1)}=\max_u\Big\{\beta^{T-(k+1)}\big(u+\sqrt u\big)+\beta^{T-k}\Big(\frac{x-u}{\beta}+\sqrt{A_{T-k}\big(\frac{x-u}{\beta}}\Big)\Big\}\\\displaystyle=\beta^{T-(k+1)}\max_u\Big\{u+\sqrt u+x-u+\sqrt{\beta A_{T-k}\big(x-u)}\Big\}\\\displaystyle=\beta^{T-(k+1)}\Big[x+\max_u\Big\{\sqrt u+\sqrt{\beta A_{T-k}\big(x-u)}\Big\}\Big]$

We will be done if the max of the curly-braced expression is a scaling of $LaTeX: \sqrt x$ . While it is possible to show this by assuming LaTeX: u=hx , factoring out $LaTeX: \sqrt x$ and observing that the maximization wrt. h is independent of x, that is a stretch to spot outright (and you cannot use this at the exam). The braced expression is concave in u with vertical tangents at u=0 and u=x, so the FOC is necessary and sufficient and yields $LaTeX: \frac1{\sqrt u}=\frac{\sqrt{\beta A_{T-k}}}{\sqrt{x-u}}$ and thus $LaTeX: (1+\beta A_{T-k}) u=x$ , which shows that indeed u is of the form hx and thus proves the form of the value function. We are done with that part of the question

To get to the difference equation for A, I guess most of you will insert: $LaTeX: \sqrt u=\frac{1}{\sqrt{1+\beta A_{T-k}}}\sqrt x$ and going back to the FOC, $LaTeX: {\sqrt{x-u}}=\sqrt{\beta A_{T-k}}\sqrt u=\frac{\sqrt{\beta A_{T-k}}}{\sqrt{1+\beta A_{T-k}}}\sqrt x$ , and so we end up at $LaTeX: \beta^{T-(k+1)}\Big[x+\sqrt{1+\beta A_{T-k}}\,\sqrt x\Big]$ and the difference equation $LaTeX: A_{T-(k+1)}=1+\beta A_{T-k}$ .
(Solution, which you weren't asked for: $LaTeX: A_{T-k}=\frac{1+\beta^k}{1+\beta}$ .)
Bellman equation for the infinite-horizon problem is the same as at zero except with the same function (rather than J₁) re-appearing on the right-hand side:
$LaTeX: J(x)=\max_u\Big\{u+\sqrt u + \beta\cdot J(\frac{x-u}{\beta})\Big\}$
The natural guess would be the limit of the finite-horizon value function. That would be the same form as above, and with $LaTeX: \lim_{T\to+\infty}A_{T-T}=\lim_{T\to+\infty}\frac{1+\beta^T}{1+\beta}=\frac1{1+\beta}$ . However, do we need to solve for $LaTeX: A_{T-k}$ in order to get to that guess?
We don't. If we make the guess $LaTeX: J(x)=x+\sqrt{Ax}$ and copy the same calculations as above, we will end with the right-hand side being $LaTeX: x+\sqrt{1+\beta A}\sqrt x$ , and the Bellman equation will hold if A = 1+beta A. Note that this does not require the above solution of the difference equation - only the constant solution of the difference equation.
(If you argue that well, solving it and taking the limit gives you confidence that this is actually the solution of the problem rather than just one function satisfying some optimality condition ... yes there is some nitpickery here.)

The "ECON5150" adaptations

The difference equation . Recall the tasks:

Solution

Dyn.pr. 2

Solution: the induction step.

The difference equation $LaTeX: p\cdot v_{t+1}-v_t+(1-p)v_{t-1} = K$ . Recall the tasks: