According to an economic model, the budget $b(t)$ at time $t\geq 0$ in a household is chosen to maximise the lifetime utility $U[b]=\int_{0}^{\infty}e^{\beta t}u(c(t))dt$, where $u(c)\geq 0$ is the household utility function, $\beta>0$ is the discount rate, a constant, and $c(t)$ is the household consumption satisfying the budgetary equation $b'(t)=\gamma b(t)+\omega-c(t), b(0)=b_{0}$. In this equation, $\gamma>0$ is the bank interest rate, $\omega>0$ is the household wage (both assumed constant in this model), and $b_{0}>0$ is the initial household budget $b(0)$. Note that $b(t)$ may be negative if the household is in debt, but $c(t)>0$. The initial household consumption is $c(0)=c_{0}>0$.
The budget $b(t)$ is subject to a No-Ponzi condition $\lim_{t \to \infty} e^{-\gamma t}b(t)\geq 0$ (which prevents the household financing current consumption through indefinitely borrowing and rolling over debt). Throughout this question you may assume that the usual theory of the calculus of variations is valid for this model on the infinite time interval $t\in [0, \infty)$.
Use the budgetary equation to express the functional $U[b]$ in terms of $b$ and $b'$, then calculate the Euler-Lagrange equation, and show that it may be written as a differential equation in $c(t)$:
$u''(c(t))c'(t)=(\beta-\gamma)u'(c(t))$.
Here's my work:
By using the budgetary equation $b'(t)=\gamma b(t)+\omega-c(t), b(0)=b_{0}$, we have that $c(t)=b'(t)-\gamma b(t)-\omega$, so the functional $U[b]$ in terms of $b$ and $b'$ is $U[b]=\int_{0}^{\infty}e^{-\beta t}u(b'(t)-\gamma b(t)-\omega)dt$.
To calculate the Euler-Lagrange equation, we have $F(t, b, b')=e^{-\beta t}u(b'(t)-\gamma b(t)-\omega)$.
Note that $\frac{\partial F}{\partial b'}=e^{-\beta t}u'(b'(t)-\gamma b(t)-\omega)\cdot b''(t)$ and $\frac{\partial F}{\partial b}=e^{-\beta t}u'(b'(t)-\gamma b(t)-\omega)\cdot (-\gamma b'(t))$.
From the work above so far, I need to find $\frac{d}{dt}(\frac{\partial F}{\partial b'})$. How should I find this? Is my work above correct up to here?
Best Answer
So we have the functional $U$ given by
$$U[b]=\int_0^\infty e^{-\beta t}u(b'(t)-\gamma b(t)-\omega)\,\mathrm{d}t.$$
The corresponding Lagrangian $\mathcal{L}$ of this functional is given by
$$\mathcal{L}(t,b,b')=e^{-\beta t}u(b'-\gamma b-\omega).$$
The relevant partial derivatives of $\mathcal{L}$ are given by
$$\partial_2\mathcal{L}(t,b,b')=-\gamma e^{-\beta t}u'(b'-\gamma b-\omega)$$
and
$$\partial_3\mathcal{L}(t,b,b')=e^{-\beta t}u'(b'-\gamma b-\omega).$$
Furthermore then
$$\frac{\mathrm{d}}{\mathrm{d}t}\partial_3\mathcal{L}(t,b(t),b'(t))=-\beta e^{-\beta t}u'(b'(t)-\gamma b(t)-\omega)+(b''(t)-\gamma b'(t))e^{-\beta t}u''(b'(t)-\gamma b(t)-\omega).$$
With $c(t)=b'(t)-\gamma b(t)-\omega$, this means that
$$\partial_2\mathcal{L}(t,b(t),b'(t))=-\gamma e^{-\beta t}u'(c(t))$$
and
$$\frac{\mathrm{d}}{\mathrm{d}t}\partial_3\mathcal{L}(t,b(t),b'(t))=-\beta e^{-\beta t}u'(c(t))+c'(t)e^{-\beta t}u''(c(t)).$$
Consequently, by subtracting the second from the first, the Euler-Lagrange equation becomes
\begin{align*} 0 &=-\gamma e^{-\beta t}u'(c(t))+\beta e^{-\beta t}u'(c(t))-c'(t)e^{-\beta t}u''(c(t)) \\ &=(\beta-\gamma)e^{-\beta t}u'(c(t))-c'(t)e^{-\beta t}u''(c(t)). \end{align*}
Dividing everything by $e^{-\beta t}$ and rearranging this becomes
$$u''(c(t))c'(t)=(\beta-\gamma)u'(c(t)),$$
which is what we wanted to show.