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Consider the following function:

e[w_] := 1/(2 w (w + a i W) (1 + a i w W)) (2 w^2 + 2 a i w^3 W + a^2 i^2 (-1 + 2 w^2) W^2 - \[Sqrt](a i W (4 (i + lambda) w^3 + 4 a i w^2 (1 + 2 i + 2 lambda + (i + lambda) w^2) W + 4 a^2 i^2 w (1 + i + lambda + 2 (i + lambda) w^2) W^2 + a^3 i^3 (1 + 4 (i + lambda) w^2) W^3)))

I'm trying to do two things. First, I would like to verify whether $e$ is an increasing and concave function of $w$ under the assumption of $a \in [0,1]$, $i \in [0,1]$, $\lambda \in [0,1]$, $w>0$, $W>0$. That is, $\frac{\partial e}{\partial w}>0$, $\frac{\partial^2 e}{\partial w^2}<0$.

Here is my code for this:

Clear["Global`*"]; e[w_]:= 1/(2 w (w + a i W) (1 + a i w W)) (2 w^2 + 2 a i w^3 W + a^2 i^2 (-1 + 2 w^2) W^2 - \[Sqrt](a i W (4 (i + lambda) w^3 + 4 a i w^2 (1 + 2 i + 2 lambda + (i + lambda) w^2) W + 4 a^2 i^2 w (1 + i + lambda + 2 (i + lambda) w^2) W^2 + a^3 i^3 (1 + 4 (i + lambda) w^2) W^3))); Assuming[a > 0 && a < 1 && i > 0 && i < 1 && lambda > 0 && lambda < 1 && w > 0 && W > 0, FullSimplify@Reduce[e'[w] > 0]] Assuming[a > 0 && a < 1 && i > 0 && i < 1 && lambda > 0 && lambda < 1 && w > 0 && W > 0, FullSimplify@Reduce[e''[w] < 0]]

Second, I would like to find $w$ that solves $$\frac{\partial e}{\partial w}=\frac{e}{w}$$ Here is my code for this:

Assuming[a > 0 && a < 1 && i > 0 && i < 1 && lambda > 0 && lambda < 1 && w > 0 && W > 0, FullSimplify@Solve[e'[w] == e[w]/w, w]]

These codes are running forever. Any help please?

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  • $\begingroup$ Which is running forever, the solver or the simplifier? $\endgroup$ Commented Jul 3, 2024 at 4:59
  • $\begingroup$ @MichaelE2, I guess it is the solver. $\endgroup$ Commented Jul 3, 2024 at 5:15
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    $\begingroup$ @ppp Function e[w] only depends on two (not four!) parameters a i W and i+lambda $\endgroup$ Commented Jul 3, 2024 at 8:53
  • $\begingroup$ @Ulrich Neumann, thanks, your finding that it only depends on two parameters is very useful! $\endgroup$ Commented Jul 4, 2024 at 11:36

1 Answer 1

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Function e[w]only depends on two parameters a i W and i+lambda

e[w_] := (2 w^2 + 2 a i w^3 W + a^2 i^2 (-1 + 2 w^2) W^2 - \[Sqrt](a i W (4 (i + lambda) w^3 + 4 a i w^2 (1 + 2 i + 2 lambda + (i + lambda) w^2) W + 4 a^2 i^2 w (1 + i + lambda + 2 (i + lambda) w^2) W^2 + a^3 i^3 (1 + 4 (i + lambda) w^2) W^3))) /. {W -> aiw/(a i), lambda -> ip\[Lambda] - i} // Simplify

enter image description here

Its possible now to plot the regions

RegionPlot3D[e'[w] > 0 , {aiw, 0.1, 10}, {ip\[Lambda], 0.1, 10}, {w, 0.1, 20},Evaluated -> True, PlotLabel -> "e'[w]>0",AxesLabel -> {"a i W", "i+lambda", "w"}]

enter image description here

RegionPlot3D[e''[w] < 0 , {aiw, 0.1, 10}, {ip\[Lambda], 0.1, 10}, {w, 0.1, 20}, Evaluated -> True, PlotLabel -> "e''[w]<0",AxesLabel -> {"a i W", "i+lambda", "w"}]

enter image description here

addendum

second question

cond = Evaluate[e'[w] - e[w]/w]; ContourPlot3D[ cond , {aiw, 0.1, 10}, {ip\[Lambda], 0.1, 10}, {w,0.1, 10}, AxesLabel -> {"a i W", "i+lambda", "w"},Contours -> { 0} ]

enter image description here

It looks like cond==0can't be fullfilled in this parameterrange!

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  • $\begingroup$ Ulrich Neumann, thanks! According to the plots, it seems e'>0 for the entire region examined, and e''<0 for only a partial region, is this correct? One problem with this simulation approach is that we cannot claim it as generally applicable since the parameter regions are not exhaustive. Maybe an analytical approach is not possible? $\endgroup$ Commented Jul 3, 2024 at 10:54
  • $\begingroup$ Can you please help with my second question as well? Thank you so much. $\endgroup$ Commented Jul 3, 2024 at 10:57
  • $\begingroup$ @ppp I tried to answer your second question in my modifoed answer $\endgroup$ Commented Jul 3, 2024 at 13:29
  • $\begingroup$ Ulrich Neumann, thanks for the answer to the second question. But I am wondering if there is a way to find the analytical solution for w explicitly? $\endgroup$ Commented Jul 3, 2024 at 21:54
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    $\begingroup$ @ppp The condition w e'[w]==e[w] is an expression with a Sqrt term. Eliminating this Sqrt by squaring you get an expression which can be solved using Solve, but solution contains several Root objects. $\endgroup$ Commented Jul 4, 2024 at 11:42

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