2
$\begingroup$

Mathematica has found that the maximum of the following function is negative using NMaximize, but I am trying to prove that the maximum is negative (i.e. that the function is negative for all parameter values) - since it is an important step in a proof that I am writing:

Maximize[{(-1 + E^(δ ρ)) λ + (-1 +  E^(δ λ) + E^(δ ρ) - E^(δ (2 λ + ρ))) ρ, ρ > 0 && δ > 0 && λ > 0}, {ρ, λ, δ}]

However, Mathematica will not run this maximization, and it quickly returns the input as its output.

As an alternative method, I have tried to get Mathematica to simultaneously solve the first-order conditions of the maximization problem:

Solve[{-1 + E^(δ λ) - E^(δ (2 λ + ρ)) (1 + δ ρ) +  E^(δ ρ) (1 + δ (λ + ρ)) == 0, ρ (E^(δ λ) λ + E^(δ ρ) (λ + ρ) - E^(δ (2 λ + ρ)) (2 λ + ρ)) == 0, -1 + E^(δ ρ) + E^(δ λ) δ ρ - 2 E^(δ (2 λ + ρ)) δ ρ == 0}, {ρ, δ, λ}]

The Solve operation has been running on a HPC for a few hours now - does anyone have any advice as to how to make this run faster? I have been given that the advice that "approximate" solutions found through NMaximize would not suffice for the proof.

Thank you

-- I have also tried putting this through Matlab, but I am getting a lot of error messages.

$\endgroup$
3
$\begingroup$

Not a complete answer but an observation that may help to to solve your actual problem.

The expression

a = (-1 + E^(δ ρ)) λ + (-1 + E^(δ λ) + E^(δ ρ) - E^(δ (2 λ + ρ))) ρ

is nonpositive for all nonnegative δ, ρ, λ if and only if the following expression is nonnegative for all nonnegative x, y:

b[x_,y_] = δ a /. {ρ -> x/δ, λ -> y/δ} // Simplify

-(-1 + E^y) (-1 + E^x + E^(x + y)) x + (-1 + E^x) y

That reduces the number of variables by one. Might be helpful.

Edit:

The following should prove your claim:

The above substitution reduces the problem to showing that $b(x,y) \leq 0$ for all $x \geq 0$ and all $y\geq 0$, where

$$b(x,y) := \left(e^x-1\right) y-x \left(e^y-1\right) \left(e^{x+y}+e^x-1\right).$$

We have

$$b(x,0) = 0$$

and

$$\frac{\partial b}{\partial y}(x,0) = e^x (1-2 x)+x-1 \leq 0$$

for all $x \geq 0$. Moreover,

$$\frac{\partial^2 b}{\partial y^2}(x,y) = x \left(-e^y\right) \left(4 e^{x+y}-1\right) \leq 0$$ for all $x \geq 0$ and $y \geq 0$.

Hence we obtain $$\frac{\partial b}{\partial y}(x,y) = \frac{\partial b}{\partial y}(x,0) + \int_0^y \frac{\partial^2 b}{\partial y^2}(x,t) \, \mathrm{d} t \leq \frac{\partial b}{\partial y}(x,0) \leq 0$$

for all $x \geq 0$, $y\geq 0$ and

$$b(x,y) = b(x,0) + \int_0^y \frac{\partial b}{\partial y}(x,t) \, \mathrm{d} t \leq b(x,0) = 0.$$

$\endgroup$
  • $\begingroup$ Thank you - this looks rather promising, I think it will take me a bit to process the argument. I'm not quite sure that I understand the first step though - can you please explain how showing that the second expression is nonnegative shows that the first expression is nonnegative? $\endgroup$ – Elke Oct 17 '18 at 13:54
  • $\begingroup$ That's quickly done: We multiply the first expression by $\delta \geq 0$, so this does not change signs. Moreover we only rescale $\lambda$ and $\varrho$ by $\delta \geq 0$, so their signs will also not be flipped. If you like, you can treat the case $\delta = 0$ first. Then expression a equals 0. $\endgroup$ – Henrik Schumacher Oct 17 '18 at 13:58

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.