Solow growth model in economics

The Solow Model - Introduction

Problem 1.

Production function: $$Y=F(K,L)$$ $\textbf{Y}$: output, $\textbf{K}$: capital stock, $\textbf{L}$: labour force $$k=K/L$$ $\textbf{k}$: capital/labour ratio $$y=Y/L$$ $\textbf{y}$: output/labour ratio $$\frac{Y}{L}=\frac{F(K,L)}{L}=F\left(\frac{K}{L},1\right)=F(k,1)=f(k)$$ $$y=f(k)$$ $$f(0)=0,\;f'(k)>0,\;f''(k)<0,\;k>0$$ Assumptions: $$\dot{L}=nL,\;L(0)=L_{0}$$ $\textbf{n}$: constant growth rate of labour force $$S=sY$$ $\textbf{S}$: savings as a constant fraction of output $$S=I$$ $\textbf{I}$: investment, which is the change in the capital stock plus replacement investment: $$I=\dot{K}+\delta K$$ $$K(0)=K_{0}$$

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Differentiating the variable $k$ with respect to time $$\frac{dk}{dt}=\dot{k}=\dot{\left(\frac{K}{L}\right)}=\frac{L\dot{K}-K\dot{L}}{L^{2}}=\frac{\dot{K}}{L}-\frac{K}{L}\cdot \frac{\dot{L}}{L}=\frac{K}{L}\cdot \frac{\dot{K}}{K}-\frac{K}{L}\cdot \frac{\dot{L}}{L}=k\;\left(\frac{\dot{K}}{K}-\frac{\dot{L}}{L}\right)$$ $$\dot{K}=I-\delta K=S-\delta K=sY-\delta K$$ $$\frac{\dot{K}}{K}=\frac{sY-\delta K}{K}=\frac{sY}{L}\left(\frac{L}{K}\right)-\delta =\frac{sf(k)}{k}-\delta$$ $$\frac{\dot{L}}{L}=\frac{nL}{L}=n$$ $$\dot{k}=sf(k)-\delta k-nk=sf(k)-\left(n+\delta \right)k$$ Initial conditions: $$k(0)=\frac{K_{0}}{L_{0}}=k_{0}$$ We use a Cobb-Douglas production function: $$Y=aK^{\alpha }L^{1-\alpha },\;0<\alpha <1$$ $$\frac{Y}{L}=a\left(\frac{K}{L}\right)^{\alpha }$$ $$y=f(k)=ak^{\alpha }$$ $$\dot{k}=sak^{\alpha }-\left(n+\delta \right)k$$ $$\dot{k}+\left(n+\delta \right)k=sak^{\alpha }$$ This is a Bernoulli equation, which takes the general form $$\frac{dk}{dt}+kP(t)=k^{\alpha }Q(t)$$ $$k^{-\alpha }\dot{k}+k^{1-\alpha }P(t)=Q(t)$$ $$k^{-\alpha }\dot{k}+\left(n+\delta \right)k^{1-\alpha }=sa$$ Defining the following transformation $$v=k^{1-\alpha }$$ we obtain $$\dot{v}=\left(1-\alpha \right)k^{-\alpha }\dot{k}$$ or $$\dot{k}=\frac{k^{\alpha }}{1-\alpha }\dot{v}$$ $$k^{-\alpha }\frac{k^{\alpha }}{1-\alpha }\dot{v}+\left(n+\delta \right)v=sa$$ $$\frac{\dot{v}}{1-\alpha }+\left(n+\delta \right)v=sa$$ $$\dot{v}+\left({1-\alpha }\right)\left(n+\delta \right)v=\left({1-\alpha }\right)sa$$ which is a linear differential equation in $v$.\;Solution: $$v(t)=\frac{sa}{n+\delta }+\left(v_{0}-\frac{sa}{n+\delta }\right)e^{-\left({1-\alpha }\right)\left(n+\delta \right)t}$$ which satisfies the initial condition $$v_{0}=k_{0}^{1-\alpha }$$ $$k^{1-\alpha }=\frac{sa}{n+\delta }+\left(k_{0}^{1-\alpha }-\frac{sa}{n+\delta }\right)e^{-\left({1-\alpha }\right)\left(n+\delta \right)t}$$ Hence $$k(t)=\left[\frac{sa}{n+\delta }+e^{-\left({1-\alpha }\right)\left(n+\delta \right)t}\left(k_{0}^{1-\alpha }-\frac{sa}{n+\delta }\right)\right]^{\frac{1}{1-\alpha }}$$ $$y (t) =ak^{\alpha }=a\left[\frac{sa}{n+\delta }+e^{-\left({1-\alpha }\right)\left(n+\delta \right)t}\left(k_{0}^{1-\alpha }-\frac{sa}{n+\delta }\right)\right]^{\frac{\alpha }{1-\alpha }}$$