I have the function fun[x_,y_, t_] = Cos[x] + Exp[I y] Sin[x] Sin[t]; such that $0\le {\rm x}\le \pi$, $0\le {\rm y}\le 2\pi$, and $0\le {\rm t} \le 10$. I compute the derivative of the real part of this function as derivativefun[x_,y_, t_] = D[Re[fun[x,y, t]], t];. Now I want to write the following algorithm to be executed on Mathematica:

Step1: Select a value of x and y say x=0 and y=0, find derivativefun[0,0, t]; Integrate over those intervals of t for which derivativefun[0,0, t] is positive (The integral will be a function of t).

Step2: Select another value of x and y say x=0.5 and y=0.5 find derivativefun[0.5,0.5, t];Integrate over those intervals of t for which derivativefun[0.5,0.5, t] is positive (again, the integral will be a function of t).

Step3: Do the same for the entire range of $0\le {\rm x} \le\pi$ and $0\le {\rm y} \le 2\pi$.

Step4: Finally, choose that integral (as a function of t) which is the maximum of all.

To sum up, I need to compute the integral of the derivative of the real part of fun[x_,y_, t_]over the range of ${\rm t}$ where the derivative is an increasing function. This is to be done for all values of parameters ${\rm x}$ and ${\rm y}$ within their given ranges. Finally that integral is to be chosen which is the maximum of all.

Edit: The derivative is to be taken of the absolute value of the function fun[x_,y_, t_].

Edit2: Since there are issues with the derivative of the Abs[]. It turns out that real part of the function fun[x_,y_, t_] would also work for me. Therefore, I have edited the question, changing Abs[] to Re[].

  • $\begingroup$ Your function is complex, so what does it mean to integrate over the intervals where the function is positive? Also, what have you tried? $\endgroup$ – Carl Woll May 16 '19 at 17:05
  • $\begingroup$ Thanks, @CarlWoll. It should be the absolute value of the derivative of function. Edited my question. $\endgroup$ – H. Kenan May 17 '19 at 4:44

The definition of derivativefun[x_,y_, t_] = D[Abs[fun[x,y, t]], t]; seems to be wrong, because t is used as function argument and as variable of differentiation.

Try (avoid use of Abs[] because Abs'[] isn't defined)

derivativefun[x_, y_ ] :=Evaluate[ComplexExpand[D[Sqrt[# Conjugate[#]] &[ fun[x, y, t]], t]]];

which gives what you want

Plot[derivativefun[1, 0]  , {t, 0, 10}]

enter image description here

The integration (for examplary x,y ) over positive derivative follows to

NIntegrate[ Max[0, derivativefun[1, 0] ], {t, 0, 10}]


The question changed from Abs to Re, which simplifies the problem considerably.

real part of fun[x_, y_, t_]

D[ComplexExpand[Re[Cos[x] + Exp[I y] Sin[x] Sin[t]], TargetFunctions -> {Re, Im}], t]
(*Cos[t] Cos[y] Sin[x]*)

f[x_, y_] := NIntegrate[Max[0, Cos[t] Cos[y] Sin[x]], {t, 0,10}]
Plot3D[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi}]

enter image description here

The maximum of this surface is waht you are looking for!

  • $\begingroup$ Thanks, @Ulrich Neumann. It turns out that Re[] would equally work for my problem. Therefore, I have edited the question, replacing Abs[] by Re[]. $\endgroup$ – H. Kenan May 17 '19 at 7:20
  • $\begingroup$ That simplifies the problem considerably! $\endgroup$ – Ulrich Neumann May 17 '19 at 7:37
  • $\begingroup$ Thanks, @Ulrich Neumann. That looks good. However, I want just a 2-D plot, which shows the Integral (for a particular x=x_0 and y=y_0) as a function of t, such that this integral is the maximum from all other integrals (wich are for other values of x and y). $\endgroup$ – H. Kenan May 17 '19 at 9:34
  • $\begingroup$ After integrating over t there is no dependancy on t anymore! The value of this integral is plotted in my answer. $\endgroup$ – Ulrich Neumann May 17 '19 at 9:37
  • $\begingroup$ Can't one talk about the indefinite integral here? I mean without giving limits? $\endgroup$ – H. Kenan May 17 '19 at 10:02

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