A market for a commodity is modelled by the demand and supply

Question

So, our solution attempt:

We found the particular solution, which is $\frac{3k}{(2k-r)}$ Then we used formula: $p_t=p^* + (p_0-p^*)\cdot\left(-\frac{b}a\right)^T$ And found that $p_t = 1$

However, we can’t understand how to prove that $k<1+\frac{r}{2}$

Can anyone help us?

Answer 1

Once you substitute the demand and supply functions into the difference equation, you get: $$p_t=(1+r-2k)p_{t-1}+3k.$$ Hence the condition for the coefficient of $p_{t-1}$ must be: $$|1+r-2k|<1 \Rightarrow \frac{r}{2}<k<1+\frac{r}{2}.$$

The solution of the difference equation is: $$p_t=\left(1-\frac{3k}{2k-r}\right)(1+r-2k)^t+\frac{3k}{2k-r}$$ When $t\to \infty$, the first term tends to $0$. Hence: $$\frac32<\frac{3k}{2k-r}<3 \iff 6k-3r<6k<12k-6r \iff r>0 \ \text{and} \ \ k>r.$$

Answer 2

\begin{align}p_t&=(1+r)p_{t-1}+k(3-p_{t-1}-p_{t-1})\\ &=(1+r-2k)p_{t-1}+3k \\ &=(1+r-2k)[(1+r-2k)p_{t-2}+3k)+3k \\ &= (1+r-2k)^2p_{t-2}+3k\sum_{i=0}^1(1+r-2k)^i \\ &=(1+r-2k)^tp_0 +3k\sum_{i=0}^{t-1}(1+r-2k)^i\end{align}

Now use geometric series to conclude when does it converge.