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baharm31

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12 years, 12 days
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font...

baharm31 60
June 24 2016
0 0

@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

Numerical calculations an comparison bet...

baharm31 60
January 18 2016
0 0

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

 

 

implicitplot and solve...

baharm31 60
January 18 2016
0 7

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

 

Woksheet...

baharm31 60
January 16 2016
0 0

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

eval for functions with several solution...

baharm31 60
January 15 2016
0 0

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

 

 

finding the solution of a system...

baharm31 60
January 14 2016
0 0

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

 

Explanation...

baharm31 60
December 09 2014
0 0

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

Explanation about the above problem...

baharm31 60
December 09 2014
0 0

Hi,

I am trying to solve these two equations. My main goal is finding a way to solve two differential equation together analytically and find a and phi. I do not want the numerical method. Thanks.

 

Save values into a vector...

baharm31 60
August 22 2014
0 0

@Preben Alsholm Thanks, I am working on that. I want to save w(x,t) values into a vector and then find the FFT (Fast Fourier Transform), I am not sure how that works. Previously you showed me how to look at each values but how can I save them for all t and x values in a vector form. It seems for FFT I need that vector. Sorry if it is not relevent to PDE problem.

Thanks,

 

 

Solved with that standard form...

baharm31 60
August 19 2014
0 0

@Preben Alsholm That is fine now and it works. I just put the integral and Dirac function back. For Dirac function it seems it works as I want, as I see in the graphs but as you explained before it is just zero. Also it seems integral is fine as well. However I am guessing there should be a way I see that they are generating values. For example in Matlab I can see if any function generating something or not. Just by looking at the numerical output of that specific part. Is that possible in Maple? I just want to check if the integral and Dirac function are working properly.

Many thanks.Euler-Bernoulli_Beam-solved.mw

 

v(x,0)=f(x) in bcs1 and bcs2...

baharm31 60
August 18 2014
0 0

@Preben Alsholm I think I did not quite get the v(x,t) bcs, as you have written it, bcs union {v(x,0)=f(x)}. Can you please look at my code now. I did what you said. Where did I do wronf for BCS?

Define PDE Euler-Bernoulli Beam
 

``

> 

restart:

Parametrs of piezoelectric and cantilever beam
 

NULL

> 

Ys := 70*10^9: # Young's Modulus structure

> 

Yp := 11.1*10^10: # Young's Modulus pieazo
> 

ha := -0.00125: # Position
> 

hb := 0.001: # Position
> 

hc := 0.0015: # Position
> 

d31 := -180*10^(-12): # Piezoelectric constant
> 

b := 0.01: #Width of the beam
> 

tb := 0.002:
> 

epsilon33 := 15.92*10^(-9):
> 

hp :=0.00025: # Position
> 

hpc := 0.00125: # Position
> 

YI := b*(Ys*(hb^3- ha^3)+Yp*(hc^3-hb^3))/3: # Bending stiffness of the composit cross section
> 

cs := 0.564: # The equivqlent coefficient of strain rate damping

> 

ca := 0: # Viscous air damping coefficient
> 

Ibeam := (b * tb^3 )/12: # The equivalent moment of inertia
> 

m := 0.101: # Mass of the structure
> 

upsilon := - Yp*d31*b*(hc^2-hb^2)/(2*hp): # Coupling term
> 

lb := 0.57:# Length of the structure (Cantilever Beam)
> 

lp := 0.05:# Length of the Piezoelectric
> 

R:= 10000: # Shunted resistor

Electrical circuit equation
 
> 

PDE1:=(epsilon33 * b*lp / hp) * diff(v(t), t) + (v(t)/R)+ (d31*Yp*hpc*b* diff(w(x, t),$(x, 2))*diff(w(x, t), t))=0:
> 

NULL

PDE Equation
 
> 

fn := 3.8:# Direct Excitation frequency;
> 

wb(x,t) := 0.01*sin(fn*2*Pi*t):#Direct Excitation;
> 

plot(wb(x,t),t = 0 .. 0.25*Pi,labels = [t,wb], labeldirections = ["horizontal", "vertical"], labelfont = ["HELVETICA", 15], linestyle = [longdash], axesfont = ["HELVETICA", "ROMAN", 10], legendstyle = [font = ["HELVETICA", 10], location = right],color = black):
> 

 

> 

 

> 

 

> 

PDE2 := YI*diff(w(x, t),$(x, 4))+ cs*Ibeam*diff(w(x, t),$(x, 4))*diff(w(x, t), t)+ ca* diff(w(x, t), t) + m * diff(w(x, t),$(t, 2))+ upsilon*v(t) =-m*diff(wb(x, t),$(t, 2))-ca*diff(wb(x, t), t):#PDE
> 

tmax := 0.3:
> 

xmin := 0:
> 

xmax := lb:
> 

N := 1:#NUMBER OF NODE POINT
> 

bc1 := dw(xmin, t) = 0:
> 

bc2 := dw(xmax, t) = 0:
> 

bc3 := w(xmin, t) = 0:
> 

ic1 := wl(x, 0) = 0:

Maple's pdsolve command
 

 

> 

PDE1a:=subs(v(t)=v(x,t),PDE1);

0.3184000000e-7*(diff(v(x, t), t))+(1/10000)*v(x, t)-0.2497500000e-3*(diff(diff(w(x, t), x), x))*(diff(w(x, t), t)) = 0

 (2.1)
> 

PDE2a:=subs(v(t)=v(x,t),PDE2);

1.567812500*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.3760000000e-11*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(diff(w(x, t), t))+.101*(diff(diff(w(x, t), t), t))+0.4995000000e-3*v(x, t) = 0.583376e-1*sin(7.6*Pi*t)*Pi^2

 (2.2)
> 

bcs2 := { w(x,0)=0 , D[2](w)(x,0)=0 , w(0, t) = rhs(bc1), D[1](w)(0, t)= rhs(bc1), D[1,1](w)(lb,t) = rhs(bc2), D[1,1,1](w)(lb,t) = rhs(bc2),v(x,0) = 0}: # Boundary conditions for PDE2.
> 

bcs1 := { w(x,0)=0 , w(0, t) = rhs(bc1), D[1](w)(0, t)= rhs(bc1),v(x,0) = 0}: # Boundary conditions for PDE1.
> 

PDE:=[PDE1,PDE2]:
> 

BCS:=[bcs1,bcs2]:
> 

PDES := pdsolve({PDE1a,PDE2a}, BCS,numeric);

> 

 

Error, (in pdsolve/numeric/process_IBCs) invalid initial/boundary condition: {v(x, 0) = 0, w(0, t) = 0, w(x, 0) = 0, (D[1](w))(0, t) = 0}

  
> 

PDES:-plot3d(t = 0 .. tmax, x = 0 .. xmax, axes = boxed, orientation = [-120, 40], shading = zhue, transparency = 0.3);

Error, `PDES` does not evaluate to a module

  

``


Download Euler-Bernoulli_Beam-PROBPLEM.mw

PDE1 and PDE2...

baharm31 60
August 18 2014
0 0

I have two PDEs with two unknowns w(x,t) and v(t), so I want to splve them together. Is that possible, if we ignore Dirac, or other things which cause problems. just focus on solving two PDEs with two unknowns. 

Thanks,

PDE2...

baharm31 60
August 18 2014
0 0

@Preben Alsholm It is:

PDE2.mw

"PDE2 := YI*diff(w(x, t),$(x, 4))+ cs*Ibeam*diff(w(x, t),$(x, 4))*diff(w(x, t), t)+ ca* diff(w(x, t), t) + m * diff(w(x, t),$(t, 2))+ upsilon*v(t)*(1) =-m*diff(wb(x, t),$(t, 2))-ca*diff(wb(x, t), t);#PDE"
 

``

``



Download PDE2.mw

 

PDE2...

baharm31 60
August 18 2014
0 0

@Preben Alsholm But even without Dirac, PDE2 has v(t) inside it wich is realated to PDE1. So I cannot solve them seperatly. Can I?

fdiff...

baharm31 60
August 18 2014
0 0

@Preben Alsholm I could not really get why you are explaining this? You mean I should change the way that I plotted my graph or it is to do with my main problem, solving these two PDE's together? I am a bit lost now what you want to achive by that?

Thanks,

 

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