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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.]

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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.

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