In Brunner and Strulik (2002) the authors claim, that the solution of \begin{align} \dot c &= \frac{c}{\sigma}(\alpha k^{\alpha-1} - \delta - \rho)\\ \dot k &= k^\alpha - \delta k - c \end{align}

is given by (see eq. 27) \begin{align} c(t) &= \left(1 - \frac{1}{\sigma}\right) k(t)^\alpha\\ k(t) &= \left[\frac{1}{\delta\sigma} +\left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp(-\delta(1-\alpha)t) \right]^{\frac{1}{1-\alpha}} \end{align} if $\alpha\delta\sigma = \delta + \rho$. I can verify the solution for $c(t)$ (see eg. this thread). However I'm not sure how to solve for $k(t)$. We can plug in $c(t)$ into $\dot k$, which gives \begin{align} \dot k &= \frac{1}{\sigma} k^\alpha - \delta k. \end{align}

  • How would one proceed?


With respect to Alecos answer we may solve the following ODE

\begin{align} \dot z + (1-\alpha)\delta z = \frac{1-\alpha}{\sigma}. \end{align}

The genereal solution is given by \begin{align} z(t) &= \frac{1}{\exp(\int (1-\alpha)\delta dt)}\left[\int \exp\left(\int (1-\alpha)\delta dt\right) \frac{1-\alpha}{\sigma} dt + C \right]\\ &= \frac{1}{\delta\sigma} + C\exp((\alpha-1)\delta t). \end{align}

With $z(0)$ given we pin down $C$ \begin{align} C = z(0) - \frac{1}{\delta\sigma} \end{align}

which yields the desired result \begin{align} z(t) &= \frac{1}{\delta\sigma} + \left(z(0) - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t) \\ \Longleftrightarrow \quad k(t)^{1-\alpha} &= \frac{1}{\delta\sigma} + \left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t)\\ \Longleftrightarrow \quad k(t) &= \left[\frac{1}{\delta\sigma} + \left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t)\right]^{\frac{1}{1-\alpha}}. \end{align}

  • $\begingroup$ I did not read the entire paper but it seems interesting. What I have remarked is that they make an assumption of constant saving rate (page 749). With this assumption, they can take $k$ as a "constant" term for which allows to find an analytical solution for the differential system. Otherwise, it is not possible. $\endgroup$ – optimal control Nov 23 '15 at 12:48
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    $\begingroup$ The last equation looks like a solvable differential equation. I think dividing everything by $k^{\alpha}$ and doing the following change of variable : $v=k^{1-\alpha}$, might help. This yields a first order differential equation that we know the solution of. $\endgroup$ – Louis. B Nov 23 '15 at 18:10

The differential equation

$$\dot k = \frac{1}{\sigma} k^\alpha - \delta k$$

has the structure of a Bernoulli equation. We solve it by the following transformation steps:

1) Mulitply throughout by $k^{-\alpha}$:

$$k^{-\alpha}\dot k = \frac{1}{\sigma} - \delta k^{1-\alpha} \tag{1}$$

2) Define the variable $$z \equiv k^{1-\alpha} \implies \dot z = (1-\alpha)k^{-a}\dot k \tag{2}$$

3) Combine to get

$$(1),(2) \implies \frac {1}{1-\alpha}\dot z = \frac{1}{\sigma} - \delta z $$

$$\implies \dot z + (1-\alpha)\delta z = \frac {1-\alpha}{\sigma}$$

This is a standard first-order differential equation with constant coefficients. Solve it and then reverse the change of variable to get the result.

PS: The above requires that $k \neq 0$ everywhere, which is the economically meaningful case.


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