# Bliss point hamiltonian function

I have the following utility function

$$u(c_t)=g(c_t-b)^2$$ for constants $$g,b>0$$

Edited: in order to be bliss utility function, It must be g is negative. But in the question g is given to be positive. I don’t know so. Please help me solving this question according to true assumption of g. (I don’t know which one true. )

I want to do the following

$$\max_{c_t} \int_0^{\infty} u(c_t)e^{-pt}$$

Subject to $$\dot{k_t} =Ak_t^a -c_t-(n+\delta)k_t$$ $$k_0>0$$ $$c_t>0$$

I set up hamiltonian function

$$H= g(c_t-b)^2 +q_t [Ak_t^a -c_t-(n+\delta)k_t]$$

where q is costate variables. $$c$$ is control variables and $$k$$ is State variables.

Focs

W.r.t $$c$$ : $$c_t[2g(c_t-b)-q_t]=0$$

W.r.t $$k$$: $$q_t[aAk_t^{a-1}-(n+\delta)]=-[\dot{q_t}-pq_t]$$

W.r.t. $$q_t$$ : $$\dot{k_t} =Ak_t^a -c_t-(n+\delta)k_t$$

Transversality condition $$\lim q_t e^{-pt}k_t=0$$ When I solve them, I found

$$\frac{\dot{q_t}}{q_t} = -[aAk_t^{a-1}-(n+p+\delta)]$$

And $$\frac{\dot{q_t}}{q_t}= \frac{\dot{c_t}}{c_t-g}$$

But I could not proceed after that. Because the denominator of the fraction is $$c_t-g$$ .

What should I get the results (equations) that I need to draw phase diagram? They are Euler equation and other differential equations.

• I assume $g$ is negative here? Otherwise that's not a bliss point utility function. – corran_horn Nov 26 '19 at 2:30