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I am struggling to prove that my EL equation gives a local maximum. (if possible, I want to show it is a global maximum as well).

It satisfies the Legendre necessary condition for maximum, but I cannot show for sufficient conditions. I checked Gelfand and Fomin, sufficient conditions for a strong extremum. However, "no conjugate point" condition seems not holding and even my objective function has no terminal point constraint. u is a strictly concave function here and initial condition is just enough small scalar.

$$\text{fun}:=\int_0^\infty e^{-rt}\left[u(c'(t))+w-p\,c'(t)\right]e^{-b\,c(t)}\,dt. $$

$$\text{EL}:= u'(c'(t))-p=\frac{b\left[u(c'(t))+w-p\,c'(t)\right]+u''(c'(t))\,c''(t)}{r+b\,c'(t)}$$

Anonymously lost student
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1 Answers1

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This isn't a full answer, but it's too long for a comment.

In terms of the Lagrangian $L:=e^{-(r+bc)}(u(c^\prime)+w-pc^\prime)$, so$$\partial_c L=-bL,\,\partial_{c^\prime}L=e^{-(r+bc)}(\partial_{c^\prime}u-p),$$the EL equation is

$$\frac{d}{dt}\partial_{c^\prime}L=e^{-(r+bc)}(c^{\prime\prime}\partial_{c^\prime}^2u-bc^\prime(\partial_{c^\prime}u-p))$$

$$0=\partial_cL-\frac{d}{dt}\partial_{c^\prime}L=e^{-(r+bc)}\left[-b(u+w-pc^\prime)-c^{\prime\prime}\partial_{c^\prime}^2u+bc^\prime(\partial_{c^\prime}u-p)\right].$$I realise you managed that part, but don't be tempted to rearrange this into a different equation just yet. You can't classify stationary points without second derivatives, obtained from differentiating the original first derivative rather than something proportional to it. I think that was why you were struggling.

When we derive an EL equation from an action of the form $S:=\int_a^bL(c,\,c^\prime)dt$ as $0=\frac{\delta S}{\delta c}(c,\,c^\prime,\,c^{\prime\prime})=\partial_cL-\frac{d}{dt}\partial_{c^\prime}L$, we need to check whether the second functional derivative is negative for a putative maximum. In other words, we need to check$$0>\left(\frac{\partial}{\partial c}-\frac{d}{dt}\frac{\partial}{\partial c^\prime}+\frac{d^2}{dt^2}\frac{\partial}{\partial c^{\prime\prime}}\right)\times\\\left[e^{-(r+bc)}\left[-b(u+w-pc^\prime)-c^{\prime\prime}\partial_{c^\prime}^2u+bc^\prime(\partial_{c^\prime}u-p)\right]\right].$$I'll leave evaluating the right-hand side to you; you can then see whether the result follows from your parameter constraints.

J.G.
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  • Dear J.G. Thank you very much for your reply. 1) Just to be clear, there is a small mistake in your EL; there is t. I guess it was a typo. 2) on the RHS first bracket, the first two terms equal to zero by EL and the third term is zero as well, since L does not contain c''. Would you be able to check this? – Anonymously lost student Oct 28 '19 at 20:03
  • @김찬우 You'll want to use the EL equation to simplify only after you've worked out the second functional derivative. I'll make an edit soon to address the $t$. It resulted from your writing $ct$ instead of $c(t)$ before. – J.G. Oct 28 '19 at 20:08
  • Thank you. Now I get what you mean. It is similar to checking S.O.C condition, but this time it should hold for all t with the optimal functional. And this tells only about local maximum, but not global maximum. Just one more question, if I want to further check global maximum, are there typical techniques that I can apply? (perturbation, or straight numerical analysis...) – Anonymously lost student Oct 28 '19 at 20:17
  • I am sorry. Why do you multiply the second bracket? If we put the optimal functional, it should be 0. – Anonymously lost student Oct 28 '19 at 20:31
  • @김찬우 That $\times$ was just to indicate the operator acts on the contents of the next line. – J.G. Oct 28 '19 at 21:19
  • I rechecked. And the second functional derivative turns out to be 0. (just in case if you want to check: the first term is 0 by EL, and if we just take a derivate once to the third term, it is exactly same to the second term) I guess this result should be something similar to a saddle point. Is it correct? – Anonymously lost student Nov 02 '19 at 19:43