I'm trying to derive an HJB from a discrete time setting. At some point, I am left with

$$ \lim_{\Delta\to 0} \frac{v(c_{t+\Delta}, u_{t+\Delta}, t+\Delta) - v(c_{t}, u_{t}, t)}{\Delta}$$

and am not sure what to do. If $\Delta$ was only in one argument, this would be a partial differential. My hunch is that this is the total derivative in $t$, but I dont know how to show that.

How do I proceed with the expression above?

  • $\begingroup$ If I understood well your question, you try to show where the expression comes from. It is the basic difference quotient. The logic of total derivative can be understood by this one. Is it what you are looking for ? $\endgroup$ Sep 21, 2015 at 14:17
  • $\begingroup$ @optimalcontrol "The logic of total derivative can be understood by this one.' which is what I am looking for. I'd like to see the derivation of the total derivative, starting from the above expression. $\endgroup$
    – FooBar
    Sep 21, 2015 at 14:40
  • $\begingroup$ I have showed the most part on my papers but I will try to write down on LateX tomorrow when I have time. Also, altough the content is about economics, I think your question fits better for mathematics.stackexchange.com because what you are looking for as an explanation is purely mathematics. $\endgroup$ Sep 21, 2015 at 22:12

1 Answer 1


You can separate your function in three terms by writing \begin{align} & v(c_{t+\Delta},u_{t+\Delta},t+\Delta)-v(c_t,u_t,t) = \\ & v(c_{t+\Delta},u_{t+\Delta},t+\Delta)-v(c_t,u_{t+\Delta},t+\Delta) \\ + & v(c_{t},u_{t+\Delta},t+\Delta)-v(c_t,u_t,t+\Delta) \\ + & v(c_t,u_t,t+\Delta)-v(c_t,u_t,t) \end{align} When you divide by $\Delta$ and take the limit $\Delta \rightarrow 0$, the first expression converges to $\dfrac{\partial v}{\partial c} \dfrac{dc}{dt}$, the second expression to $\dfrac{\partial v}{\partial u} \dfrac{du}{dt}$, and the third expression to $\dfrac{\partial v}{\partial t}$. Therefore your total derivative equals \begin{equation} \dfrac{\partial v}{\partial c} \dfrac{dc}{dt} + \dfrac{\partial v}{\partial u} \dfrac{du}{dt} + \dfrac{\partial v}{\partial t} \end{equation}


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