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. $\endgroup$ Nov 23, 2015 at 12:48
  • 1
    $\begingroup$ 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, 2015 at 18:10

1 Answer 1


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) Multiply 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.


Your Answer

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

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