baharm31

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MaplePrimes Activity


These are replies submitted by baharm31

@baharm31 @Christian Wolinski I understand now, I have the simple relation between p and q as the third equation.

Thanks,

Baharm31

@Christian Wolinski Thanks, I was trying this but I am still struggling, could you please write more how you get p and q?

Thanks,

Baharm31

@Axel Vogt Thanks for your response, I kind of understand what you are trying to say. If we eliminate for p we get an expression for p and q but how can this give me p and q as a function of other parameters? I would like to get something like p= X and q=Y and X and Y are a function of system parameters. If I apply your elimination I get

[{p = R__n*q/(gamma__1*omega__n)}, {(16*((1/4)*(e^2*eta^2+3*e*eta*omega__n+(5/2)*omega__n^2)*(e^2*eta^2+7*e*eta*omega__n+(29/2)*omega__n^2)*L__1^2+omega__n^4*R))*q*C*gamma__1}]

I guess there is a way to use the second expression to find q. In this case, q is zero. Can you please explain more?

Regards,

Baharm31

@Kitonum Thanks for the reply

@vv Thaks. It was the correct answer. If I add c01=1, c12=1, and c02=1 then the dsolve cannot find the solution. Do I need to do something extra if I add the C matrix?

Baharm31

 

@Mariusz Iwaniuk How do you know about this?

Thanks for answering.

Bahar

 

This is how you plot it but do you need exactly 50 points? You can decide the region for Lambda values in your plot by view:

plot(P, style = point, symbol = diamond, color = black, symbolsize = 20, view = [-Pi .. Pi, -10 .. 10])


``

``

``

Simplifing the equations by assuming:

varepsilon := 1;

0

(1)

aprime_t := (1/8)*(-2*a^3*Gamma*omega__n^2*varepsilon*sin(2*varphi)-4*a*delta*omega__n^2*varepsilon*sin(2*varphi)+2*varepsilon*Y0*Omega^2*sin(phi+varphi)-8*omega__n*varepsilon*varsigma*a*Omega)/Omega^2*Omega = 0

(1/8)*(2*Y0*Omega^2*sin(varphi)-8*Omega*a*omega__n*varsigma)/Omega = 0

(2)

`ϕPrime_t` := (1/8)*(-4*a^3*Gamma*omega__n^2*varepsilon*cos(2*varphi)-6*a^3*alpha*omega__n^2*varepsilon-4*a*delta*omega__n^2*varepsilon*cos(2*varphi)-4*varepsilon*Delta*a*Omega^2+2*varepsilon*Y0*Omega^2*cos(phi+varphi))/(Omega^2*a)*Omega = 0

(1/8)*(-6*a^3*alpha*omega__n^2-4*Delta*Omega^2*a+2*Y0*Omega^2*cos(varphi))/(Omega*a) = 0

(3)

solve({`ϕPrime_t`, aprime_t}, [a, varphi])

[[a = RootOf(9*alpha^2*omega__n^4*_Z^6+12*Delta*Omega^2*alpha*omega__n^2*_Z^4+(4*Delta^2*Omega^4+16*Omega^2*omega__n^2*varsigma^2)*_Z^2-Y0^2*Omega^4), varphi = arctan(4*RootOf(9*alpha^2*omega__n^4*_Z^6+12*Delta*Omega^2*alpha*omega__n^2*_Z^4+(4*Delta^2*Omega^4+16*Omega^2*omega__n^2*varsigma^2)*_Z^2-Y0^2*Omega^4)*omega__n*varsigma/(Omega*Y0), 3*alpha*omega__n^2*RootOf(9*alpha^2*omega__n^4*_Z^6+12*Delta*Omega^2*alpha*omega__n^2*_Z^4+(4*Delta^2*Omega^4+16*Omega^2*omega__n^2*varsigma^2)*_Z^2-Y0^2*Omega^4)^3/(Omega^2*Y0)+2*Delta*RootOf(9*alpha^2*omega__n^4*_Z^6+12*Delta*Omega^2*alpha*omega__n^2*_Z^4+(4*Delta^2*Omega^4+16*Omega^2*omega__n^2*varsigma^2)*_Z^2-Y0^2*Omega^4)/Y0)]]

(4)

``


Download Example.mw

@Kitonum 

Well, I don't think your solution works in all cases. I don't want to change the font size, at least not the label font size. But I want to change the figure size, actually the horizontal axis size. It is easy to do what you did when the horizontal and vertical axis are the same size.

Thanks,

Baharm31

b.zaghari@vv You might be write that sometimes implicitplot give misleading results, however for some other parameters I checked numerically and I found what implicitplot does is correct and those extra lines do exist. Physically I have some explanation for them as well. So I am sure about it.

Based on what you said [solve is not designed for such tasks. It finds (in principle) the set {y : f(x,y)=0} and it is on your side to "assemble" the selections. Note that this could be a difficult task and the number of selections could be infinite.], is this what I have already done? and made the last plot in the attached file or you are talking about a different method?

Thanks,

Baharm31

 

 

TestBaharm31-Example.mw @Carl Love I do understand that eval, evalf, solve they might not be able to find all the solution of an equation, however it is not clear to me how implicitplot can manage to find all solutions but not eval and solve. I attached an example which shows this. So for a given value that I am looking, I can solve the equation and find some values but there are not all possible values. When I compared the graph of the solutions with implicitplot I found that implicitplot has found two more solutions which I think it is correct. I am looking for an analytical expression for all lines in the implicit plot, so it is important to me to find those solutions one by one.

Thanks

Baharm31

 

TestBaharm31.mw@Carl Love Sorry not sure why it did not work first time. I do undersatnd your point that eval might not solve a function, but is there other way to solve functions with several real and imaginary solutions? At the moment I can plot all solutions, so it means MAPLE could find them but I do not know how to access to the equation for each solution.

 

Thanks,

Baharm31

b@Carl Love 

I found your response very useful, however since my function is very complicated I cannot find solutions with eval and solve. I have several solutions as I said previously and in the main equation (Fun) there are several parameters that I try to make them zero to simply it however the main aim is finding the solutions of the main function (Fun) with all parameters.

I have uploaded my worksheet. I originally started to plot the function (Fun) (y versus x) for given parameters and I could see how many solutions I have but still I need to find the analytical expression for the figures. Probably I am just to optimistic about it.

Thanks for help,

Baharm31

 

 

I actually tried gridrefine as well but did not give me a proper result. I am doing the implicit plot since I could not find the solutions of my finction, so even using (solve), I get same memory error. Is there any other way rather than buying a memory to solve equations that have several solutions? when I plot it with implicitplot I am sure it is not giving me the plot of all possible solutions. So going back to solving it, is it correct to say:

solve(function, x) assuming A=..., B=..., C=...

So at least I can give some help to the solver. I am sure it is possible for MAPLE I just could not find how to write the code.

Regards,

Baharm31

 

Hi,

I would like to solve the above differential equations for a and phi analytically. My general question is about how with Maple I can solve these kind of equations analytically.

 

Thanks

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