3
$\begingroup$

From Sydsaeter / Hammond (Further Mathematics for Economic Analysis, 2008, 2nd ed., p. 293):

$$ \max \int\limits_{0}^T [N(\dot{x}(t)) + \dot{x}(t)f(x(t))] e^{-rt} dt $$ where N and f are $C^1$ functions, r and T positive constants, $x(0) = x_0$, and $x(T)=x_T$. Deduce the Euler Equation: $\frac{d}{dt} N'(\dot{x}) = r [N'(\dot{x} + f(x)] $

Now, first of all, I'm not really sure how to understand 'deduce' here. Does it mean that I should bring the equation into the general Euler-equation form

$$ \frac{\delta F}{\delta x} = \frac{d}{dt} \frac{\delta F}{\delta \dot{x}}$$

?

If so, I end up with

$$ \dot{x} e^{-rt} f'(x) = \frac{d}{dt} N'(\dot{x})$$

which doesn't look too promising.

Am I totally misunderstanding the task here, or is my derivative wrong? Thanks!

$\endgroup$
1
$\begingroup$

I think you did it wrong. The left side is correct. The right side is $\displaystyle\frac{d}{dt}[N'(\dot{x})+f(x)]e^{-rt}$ where $\displaystyle\frac{d}{dt}$ is being applied to the whole expression. This gives $\displaystyle e^{-rt}\frac{d}{dt}[N'(\dot{x})+f(x)]-re^{-rt}[N'(\dot{x})+f(x)]$, hence $\displaystyle \frac{d}{dt}[N'(\dot{x})+f(x)]-r[N'(\dot{x})+f(x)]=\dot{x}f'(x)$, and using chain rule on the left side, this is $\displaystyle \frac{d}{dt}N'(\dot{x})+\dot{x}f'(x)-r[N'(\dot{x})+f(x)]=\dot{x}f'(x)$, hence $\displaystyle \frac{d}{dt}N'(\dot{x})=r[N'(\dot{x})+f(x)]$

$\endgroup$
1
$\begingroup$

The estremals for the functional $F$ are the solution of the Euler--Lagrange equation which you have correctly written. It seems to me that you have erroneously computed $\dfrac{\delta F}{\delta\dot x}$. I have reported below my computations: $$\frac{\delta F}{\delta x}=\dot x f'(x)e^{-rt},$$
$$\frac{\delta F}{\delta\dot x}=(N'(\dot x)+f(x))e^{-rt},$$
$$\frac{d}{dt}\frac{\delta F}{\delta\dot x}=e^{-rt}\left(\frac{d}{dt} N'(\dot x)+f'(x)\dot x-rN'(\dot x)-rf(x)\right).$$

Therefore $$0=\frac{d}{dt}\frac{\delta F}{\delta\dot x}-\frac{\delta F}{\delta x}=e^{-rt}\left(\frac{d}{dt} N'(\dot x)-rN'(\dot x)-rf(x)\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.