3
$\begingroup$

I am trying to understand the way Smith demonstrates that the general solution to his equation (12) is (13) (see page 6).

(12) \begin{eqnarray} \dot{z} &=& (1- \alpha)\left[1-\left(\delta + \frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z\right] \end{eqnarray}

In the Appendix I demonstrate that the general solution to Equation (12) is:

(13)\begin{eqnarray} z &=& \frac{1}{\bar{x}+\delta} 2F1(1-\alpha,1,d;\omega)+B\bar{x}^{\alpha-1}e^{-(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{1-\alpha} \end{eqnarray}

The Appendix: A.1 \begin{eqnarray} \dot{z} &=& -(1- \alpha)\left(\delta + \frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z \end{eqnarray} This can be integrated to find the complementary solution:(A.2) \begin{eqnarray} z_c &=& \bar{x}^{\alpha-1}e^{-(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{1-\alpha} \end{eqnarray}

To find the particular solution to equation (12), I will use the method of variation of parameters. Conjecture that the particular solution is $z_p$ $=$ $z_c$$\Psi$, where $\Psi$ is an unknown function of time. Substituting this conjecture into equation (12), it follows that: (A.3) \begin{eqnarray} \dot{\Psi} &=& \frac{1-\alpha}{z_c} &=& (1-\alpha)\bar{x}^{1-\alpha}e^{(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{\alpha-1} \end{eqnarray}

First, is (A.2) then just an integrated version of (A.1) or are there any other steps involved?

Second, I really do not see how he substituted the conjecture into 12 and how can he obtain (A.3) from this substitution? Especially here I am really lost and am lacking the imagination how he came up with $\dot{\Psi}$. Did he substitute $z_p$ for $\dot{z}$ or $z$ in (12)?

Citation: [Smith, William. (2006). A Closed Form Solution to the Ramsey Model. Contributions to Macroeconomics. 6.]

$\endgroup$

1 Answer 1

1
$\begingroup$

Let's solve the differential equation (12).

As a first step, we look for a "simpler" differential equation, namely (A.1). (A.1) can be written $$\frac{\dot{z}}{z}=-(1-\alpha)\left(\delta+\frac{\bar{x}e^{-\bar{x}t}}{e^{-\bar{x}t}+\bar{x}A}\right)$$ On the left-hand side, you have the derivative of $\ln(z)$. You can integrate to obtain the form of any solution $z_c$: $$\ln(z_c)=-(1-\alpha)\left(\delta t-\ln(e^{-\bar{x}t}+\bar{x}A)\right)+ constant$$ You obtain $z_c=\kappa e^{-(1-\alpha)\delta t}\left(e^{-\bar{x}t}+\bar{x}A\right)^{1-\alpha}$, where $\kappa$ is the exponential of the constant in the previous equation. This equation is equivalent to (A.2) for a particular constant, such that $\kappa=\bar{x}^{\alpha-1}$. This choice of the constant comes from some boundary conditions on $z$ (which should be stated in the paper).

As a second step, we use the method of variation of parameters, meaning we are looking for a solution of (12), $z_p$, that has a particular form, $z_p=z_c.\Psi$. If we find an expression for $\Psi$, then we have found a solution of (12). We substitute $z_p$ in (12): $$\dot{z_c}.\Psi+z_c.\dot{\Psi}=(1-\alpha)\left[1-\left(\delta+\frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z_c.\Psi\right]$$ This expression simplifies since $z_c$ satisfies (A.1): $$z_c.\dot{\Psi}=(1-\alpha)$$ This is (A.3). Then, I guess you we have to find a solution $\Psi$ of this equation, and we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.