@mmcdara 

One should get the initial conditions by equating same powers of epsilons in the initial conditions I had given before:

x(0)=epsilon, x'(epsilon)=epsilon+epsilon^{-1}

you just equate the same powers of epsilon with the ansatz: x(t,epsilon)=x_{-1}(t)/epsilon+x_0(t)+x_1(t)*epsilon

To get the following initial conditions:

x_{-1}(0)=x_0(0)=0, x_1(0)=1, x_{-1}'(epsilon)=1, x_{0}'(epsilon)=0, x_{1}'(epsilon)=1.

Now, since the above initial condtions are obtained at t=epsilon, and epsilon is between 0 and 1, we might check how the solutiions behave at epsilon =0 and at epsilon=1.

I.e you can change the initial conditions instead of epsilon to be zero or one.

Thanks, I apprecaite your time.

@Carl Love thanks for the correction!

I forgot the proper word for describing what I meant.

@Thomas Richard 

OK, indeed you are correct.

I checked again and it seems I mistyped in the equation:

(a^2*sin(theta)^2 - Delta^2)*((a^2 + r^2)^2 - a^2*Delta*sin(theta)^2)*sin(theta)^2/rho^4 - 4*a^2*M^2*r^2*sin(theta)^4/rho^4;

the first appearance of Delta in this expression should be without the square... :-)

Now, it seems to be working: 1=1.

BTW, how do you suggest me to let maple showing all the steps of the this expansion?

@Thomas Richard 

You can check the post that I posted in PF here:

https://www.physicsforums.com/threads/some-algebra-in-schutzs-textbook.1006477/

There are screenshots of the textbook's pages where this expression appears in.

@Mac Dude

I know that the expression I typed should be expanded to -\Delta*\sin^2(\theta) in the end since it's written in the textbook I got this expression from. Schutz's A First Course in GR second edition on pages 309-313.

@acer I tried using the plots:display command by reading the help in maple, here's what I wrote:

HH := 5*log[10]((c/H_0)*Int(1/(A*(1+zp)^4+B*(1+z)^3+C)^(1/2)/10,                              zp=0..z,                              method=_d01ajc, epsilon=1e-5)):    F:=plot(eval(HH,[A=0, B=0.31, C=0.69, H_0=1,c=1]), z=1e-7 .. 1.0,       thickness=3, color=red, smartview=false); G:=plot(eval(HH,[A=0.7, B=0.3, C=0 H_0=1,c=1]), z=1e-7 .. 1.0,       thickness=3, color=blue, smartview=false);  H:=plot(eval(HH,[A=0.6, B=0.1, C=0.3, H_0=1,c=1]), z=1e-7 .. 1.0,       thickness=3, color=green,smartview=false);   display({F,G,H},axes=boxed,,scaling=constrained,title='3 comsological models')

But I get an error message:

Error, invalid function arguments
Typesetting:-mambiguous(HH Assign 5astlog(10)ApplyFunction((csol

  H_0)astIntApplyFunction(1sol(Aast(1 + zp)circ4 + Bast(1 + z)

  circ3 + C)circ(1sol2)sol10comma   zpequals0period;&periodzcomma

     methodequals_d01ajccomma epsilonequals1e-5))colon    FAssign

  plotApplyFunction(evalApplyFunction(HHcomma(Aequals0comma B

  equals0.31comma Cequals0.69comma H_0equals1commacequals1))comma

   zequals1e-7 period;&period 1.0comma   thicknessequals3comma

  colorequalsredcomma smartviewequalsfalse)semi GAssignplotApply\

  Function(evalApplyFunction(HHcomma(Aequals0.7comma Bequals0.3

  comma Cequals0 H_0equals1commacequals1))comma zequals1e-7

  period;&period 1.0comma  thicknessequals3comma colorequalsblue

  comma smartviewequalsfalse)semi  HAssignplotApplyFunction(eval

  ApplyFunction(HHcomma(Aequals0.6comma Bequals0.1comma Cequals

  0.3comma H_0equals1commacequals1))comma zequals1e-7

  period;&period 1.0comma  thicknessequals3comma colorequalsgreen

  commasmartviewequalsfalse)semi  displayApplyFunction

  Typesetting:-mambiguous(((FcommaGcommaH)commaaxesequalsboxed

  commacommascalingequalsconstrainedcommatitleequals(3

  comsological model)),

  Typesetting:-merror("invalid function arguments"))   )

How to fix this?

Thnaks!

@acer I still don't understand how to plot three plots in one graph, can you please write an example to me?

Thanks!

@acer How can I plot all the three plots with 3 different A,B,C parameters in each graph?

All three different plots in the same graph.

Thnanks!

@Mariusz Iwaniuk I am a member of so many forums, that if I do'nt use something frequently I am bound to forget.

@Kitonum I meant that I want to plot two functions:

1.

f(x):=a*x*(1-x)+(x/4)*ln(x)+(1-x)*ln(1-x)

which satisfies f'(x1)=f'(x2)=b=(f(x1)-f(x2))/(x1-x2)

in the interval [x1,x2].

2. f(x):=a*x*(1-x)+(x/4)*ln(x)+(1-x)*ln(1-x)

where in order to find the parameter a one needs to solve f'(x0)=0 find x0 with respect to a and then find a with the condition f''(x0)=0.

Is this better understood?

Can you implement this for me?

Thanks, appreciate your help.

@Kitonum I have the following follow-up question to the original opening question.

I have the function f(x):=a*x*(1-x)+(x/4)*ln(x)+(1-x)*ln(1-x)

which satisfies f'(x1)=f'(x2)=b=(f(x1)-f(x2))/(x1-x2)

I want to find the graph of f above in the interval [x1,x2] or [x2,x1] alongside the same function f but which also satisfies at the point of intersection with the above graph f''(x)=0.

How to implement this in maple code?

Thanks in advance!

@Preben Alsholm 

Well I am trying to solve the following problem from Murdock's Perturbations: Theory and Methods.

"Exercise 2.6.1 (a)

Letting f(y)=y^3, investigate the correct and incorrect perturbation problems as follows.

Attempt to solve (2.6.3)-(2.6.4) for y1. Obtain the differential equation for y1, find its general solution, and show that the constants of integration cannot be chosen to satisfy the boundary conditions."

where (2.6.3) y''+n^2 \pi^2 y = epsilon*f(y) , y(0)=0 , y(1)=0;

(2.6.4)y ~ A sin(n\pi *x) + epsilon*y1(x)+epsilon^2*y2(x)+...

So I plugged the first-order in epsilon ansatz of A sin(n\pi*x)+epsilon*y1(x); so I get the following ode:

y1''+n^2 pi^2 * y1 = A^3 sin^3(n*pi*x), y1(0)=y1(1)=0.

So it seems both of you solved my problem.

I need to remember this option of assuming n::integer

Thanks.

@Kitonum how to find those values algebraically?

I mean I have: epsilon = sqrt((x-2)^2(3-x))

and then I plug the values x=0, x=2 or 3 and get:

+-2*sqrt(3), 0.

Are those the values I am looking for?

I am not sure understand why?

@Carl Love thanks.

Perhaps you know how to answer my following question from Murdock's text which I asked on stackexchange and didn't get an answer.

"Using the graph of y=(x-2)^2(x-3), sketch the bifurcation diagram of (x-2)^2(x-3)+epsilon^2=0 for all real epsilon.

Notice that this diagram is symmetrical about the x axis. Find the values of epsilon at the two pair pair bifurcation points".

My question is about the last sentence how to find these values?

@Carl Love Well the equation is epsilon = -(x-2)^2(x-3), in that case how to use the Bifurcation command here, I've only seen an example for the logisitc equation.