When I learnt the continuous time and discrete time neoclassical model, it’s feels so tempting to try to put them together in a uniform view. It’s not trivially easy though. And here’s one way to do it.
Continuous-Time Neoclassical Growth Model
Given $k_0$, $f(\cdot), U(\cdot)$: $$ \begin{align*} & \max_{\lbrace c_t\rbrace_{t\ge 0}} \int_0^\infty e^{-\rho t} U(c_t),\text{d}t\cr & \text{s.t. }\dot k_t = f(k_t) - \delta k_t - c_t \end{align*}\tag{1} $$ Taking (appropriate) derivative of the Hamiltonian function solves this problem — in general, for any continuous-time problem in the following form: $$ \begin{align*} V^\star (x_0) :=& \max_{\lbrace u_t\rbrace_{t\ge 0}} \int_0^\infty e^{-\rho t} h(x_t, u_t),\text{d}t\cr & \text{s.t. }\dot x_t = g(x_t, u_t). \end{align*} $$ then, the following first-order optimality conditions are necessary (under regularity assumptions) and sufficient (under regularity and convexity assumptions) for the path $\lbrace x_t, u_t\rbrace_{t\ge 0}$ to be optimal: $$ \begin{align*} & \text{let } H(x, u, \lambda) = h(x, u) + \lambda^Tg(x, u)\cr & \text{FOCs}:\begin{cases}H_u(x_t, u_t, \lambda_t) = 0\cr \dot \lambda_t = \rho \lambda_t - H_x(x_t, u_t, \lambda_t)\cr \dot x_t = H_\lambda (x_t, u_t, \lambda_t) = g(x_t, u_t). \end{cases} \end{align*} $$ For $(1)$, the first order optimality conditions are $$ (1)’s\text{ FOCs}:\begin{cases} U’(c_t) = \lambda_t\cr \dot \lambda(t) = \rho\lambda_t - \lambda_t(f’(k_t) - \delta)\cr \dot k_t = f(k_t) - \delta k_t - c_t. \end{cases} $$
Discrete-Time Neoclassical Growth Model
Fix $k_0$, $U(\cdot), f(\cdot)$: $$ \begin{align*} & \max_{\lbrace c_t\rbrace_{t\in\N}} \sum_{t = 1}^\infty \beta^tU(c_t)\cr & \text{s.t. }k_{t + 1} = f(k_t) - c_t+ (1 - \delta)k_t \end{align*}\tag{2} $$ Now, for general discrete time model, fix (per-unit-time) objective $U(x, u)$ and the dynamic: $x_{t + 1} = \Gamma(x_t, u_t)$: $$ \begin{align*} V^{\star}(x_0) := & \max_{\lbrace x_t\rbrace_{t\in\N}} \sum_{t = 0}^\infty \beta^t F(x_{t}, x_{t + 1})\cr \text{where }& F(x, y) := \max_{y = \Gamma(x, u)} U(x, u) \end{align*} $$ The Euler equation for optimality is (note: technically you’d still need tranversality) $$ \text{FOCs}:\begin{cases} F_y(x_t, x_{t + 1}) + \beta F_x(x_t, x_{t + 2}) = 0, \forall t & \text{(Euler Equations)}\cr \lim_{t \to \infty} \beta^t F_x(x_t, x_{t + 1})x_t = 0 & \text{(Transversality)} \end{cases} $$ So, (part of) the discrete version of specific neoclassical growth model’s optimality condition looks like $$ (2)’s\text{ EE}:\ U’(c_t) = \beta U’(c_{t + 1})(1 - \delta + f’(k_{t + 1})) $$
Connecting the two
Starting from the discrete time model’s EE: $$ \begin{align*} U’(c_t) & = \beta U’(c_{t + 1})(1 - \delta + f’(k_{t + 1}))\cr & \text{(let $\Delta = 1$)}\cr \Leftrightarrow U’(c_t) & = \frac1{1 + \rho \Delta} U’(c_{t + \Delta})(1 - \delta \Delta + \Delta f’(k_{t + \Delta}))\tag{3} \end{align*} $$ (Note: Fix $\Delta = 1$, $\beta =e^{-\rho \Delta}$. Then for small $\Delta$, $e^{-\rho \Delta} \approx \frac 1{1 + \rho \Delta}$.)
Now, notice that if we take $\Delta \to 0$, $(3)$’s RHS simply reduce to $U’(c_t)$ and it’s trivial. Therefore instead we need to do a little manipulation: $$ \begin{align*} (3) & \Leftrightarrow \rho\Delta U’(c_t) = U’(c_{t + \Delta}) - U’(c_t) + U’(c_{t + \Delta})\Delta(- \delta + f’(k_{t + \Delta}))\cr & \Leftrightarrow \rho U’(c_t) = \frac{U’(c_{t + \Delta}) - U’(c_t)}{\Delta} + U’(c_{t + \Delta})(- \delta + f’(k_{t + \Delta}))\cr & \quad \text{push }\Delta \to 0 \cr & \Leftrightarrow \rho U’(c_t) = U’’(c_t)\dot c_t + U’(c_{t})(- \delta + f’(k_{t})) \tag{4} \end{align*} $$ So we’ve established that $(2)$’s EE (part of its FOCs) is somehow equivalent to $(4)$, while $(4)$ is already somewhat continuous — the last step to connect $(1)’s$ FOCs: (copying it one more time here) $$ (1)’s\text{ FOCs}:\begin{cases} U’(c_t) = \lambda_t & (i) \cr \dot \lambda(t) = \rho\lambda_t - \lambda_t(f’(k_t) - \delta) & (ii) \cr \dot k_t = f(k_t) - \delta k_t - c_t & (iii) \text{ (ignore this one)} \end{cases} $$ is just plug in $(i)$ into $(ii)$: $$ \begin{align*} (ii) \Leftrightarrow &\ \lambda(t) = \rho\lambda_t - \lambda_t(f’(k_t) - \delta)\cr \Leftrightarrow &\ \rho\lambda_t = \lambda(t) + \lambda_t(f’(k_t) - \delta)\cr & \text{plug in }(i)\cr \Leftrightarrow &\ \rho U’(c_t) = U’’(c_t) \dot c_t + U’(c_t)(f’(k_t)-\delta)\cr \Leftrightarrow & \ (4)\approx (2)’s \ EE. \end{align*} $$ Viola.