Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

Hello,

How I can separate three functions (u__ru__theta and u__phi) in the equation attached below.

Please also see attached figure. I want to calculate L11L22L33.

Thanks.

SEPARATE
 

restart

B := simplify((r*(R+r*cos(theta))^2*(mu+lambda)*(diff(`u__θ`(r, theta, phi), r, theta))+2*r^2*(R+r*cos(theta))^2*(mu+(1/2)*lambda)*(diff(u__r(r, theta, phi), r, r))+r^2*(mu+lambda)*(R+r*cos(theta))*(diff(`u__φ`(r, theta, phi), phi, r))+mu*(R+r*cos(theta))^2*(diff(u__r(r, theta, phi), theta, theta))+(diff(u__r(r, theta, phi), phi, phi))*mu*r^2-3*(R+r*cos(theta))^2*(mu+(1/3)*lambda)*(diff(`u__θ`(r, theta, phi), theta))+(2*(R+2*r*cos(theta)))*r*(R+r*cos(theta))*(mu+(1/2)*lambda)*(diff(u__r(r, theta, phi), r))-r^2*sin(theta)*(mu+lambda)*(R+r*cos(theta))*(diff(`u__θ`(r, theta, phi), r))-3*r^2*(mu+(1/3)*lambda)*cos(theta)*(diff(`u__φ`(r, theta, phi), phi))-r*mu*sin(theta)*(R+r*cos(theta))*(diff(u__r(r, theta, phi), theta))-(2*(2*cos(theta)^2*r^2+2*cos(theta)*R*r+R^2))*(mu+(1/2)*lambda)*u__r(r, theta, phi)+r*`u__θ`(r, theta, phi)*sin(theta)*(3*r*(mu+(1/3)*lambda)*cos(theta)+R*mu))/(r^2*(R+r*cos(theta))^2))

``

``

 


 

Download SEPARATE


 

Hello, I bring here a problem with maple that in result the  "".

I can remove '' in final results?

Thanks


 


  ;
  ):


  X= (R+r*costheta)*sinphi):
  = (R+r*costheta)*cosphi):
  Z= (r*sintheta)):

  
  r, theta, phi,
    X, Y, Z
  ):

 


   
  
    (1/2*Curl
      
        < u__rrtheta, phi),
            u__thetartheta, phi),
            u__phirtheta, phi)
          >,
          rtheta, phi
        )
      )
    )
  );

Vector(3, {(1) = (1/2)*csgn(r)*csgn(R+r*cos(theta))*(csgn(1, R+r*cos(theta))*(R+r*cos(theta))*`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi)+((R+r*cos(theta))*(diff(`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi), theta))-r*sin(theta)*`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi))*csgn(R+r*cos(theta))-csgn(r)*r*(diff(`#msub(mi("u"),mi("&theta;",fontstyle = "normal"))`(r, theta, phi), phi)))/(r*(R+r*cos(theta))), (2) = -(csgn(1, R+r*cos(theta))*(R+r*cos(theta))*`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi)+((R+r*cos(theta))*(diff(`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi), r))+cos(theta)*`#msub(mi("u"),mi("&phi;",fontstyle = "normal"))`(r, theta, phi))*csgn(R+r*cos(theta))-(diff(`#msub(mi("u"),mi("r"))`(r, theta, phi), phi)))*csgn(R+r*cos(theta))/(2*r*cos(theta)+2*R), (3) = (1/2)*(-(diff(`#msub(mi("u"),mi("r"))`(r, theta, phi), theta))+csgn(r)*r*(diff(`#msub(mi("u"),mi("&theta;",fontstyle = "normal"))`(r, theta, phi), r))+`#msub(mi("u"),mi("&theta;",fontstyle = "normal"))`(r, theta, phi)*(r*csgn(1, r)+csgn(r)))*csgn(r)/r})

(1)

``


 

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maple does not work at all

it displays this error

Error, (in StringTools:-FormatMessage) unknown option MAPLE
 

car_2som_opp := proc (U::list, V::list)  #construction d'un carré connaissant 2 sommets opposés 
local dist, eqCerU, eqCerV, r, sol, X, Y; 
dist := proc (M, N) sqrt(add((M[i]-N[i])^2, i = 1 .. 2)) end proc;
r := dist(U, V)/sqrt(2); 
eqCerU := (x-U[1])^2+(y-U[2])^2 = r^2; 
eqCerV := (x-V[1])^2+(y-V[2])^2 = r^2;
sol := solve([eqCerU, eqCerV], [x, y],allsolutions,explicit);  
map(allvalues,sol): 
X := [subs(op(sol[1]), x), subs(op(sol[1]), y)]; 
Y := [subs(op(sol[2]), x), subs(op(sol[2]), y)]; 
display(plot([U, X, V, Y, U],scaling = constrained, axes = none)) 
end proc:

car_2som_opp([-5,6],[7,-3]);"error
 

Need help solving this problem with a maple proc using the Crank–Nicolson method for the differential part and any other quadrature  for the integral part and thank you so much in advance any ideas or thoughts would be helpful

How I can convert attached maple code into series form to the latex format.

I want to convert to this format series form) not for certain M and N.

 

 

Hi. what is reason maple unable to integral in answer of dsolve? when I try to use dsolve for solve my equation in answer of maple there are expressions of integral that isnot calculated

&int;(-625 R^2 ro (-2 cp3 x1 lambdaopt+sin((pi t)/10) cp2+(16 cp2)/5) (-cos((pi t)/10)+cos(2 pi)+((-t/5+4) x1+(8 t)/25-32/5) pi) lambdaopt pi b11 (e)^(-((16+5 sin((pi t)/10)) cp1)/(10 lambdaopt x1))+625 (-R^2 k11 (cos((pi t)/10))^3+k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) (cos((pi t)/10))^2+((32 sin((pi t)/10) R^2 k11)/5+(281 R^2 k11)/25+4 lambdaopt^2 (a11 x1+a13 x3)) cos((pi t)/10)-(32 k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) sin((pi t)/10))/5+(281 (x1-8/5) (R^2 k11+(100 lambdaopt^2 (a11 x1+a13 x3))/281) pi t)/125) x1^2)&DifferentialD;t

Hi,

I'm trying to solve an expression with only one unknown variable, but for some reason solve and fsolve are unable to return a solution.

The expression I'm trying to solve is:

p := k*T*m__hhw*(ln(1+exp(E__fv-E__hh0)/(k*T))+ln(1+exp(E__fv-E__hh1)/(k*T)))/(Pi*h__bar^2*L__z) = 0.3e25

where all variables are predefined and am trying to solve for E__fv. However, when I use solve, I get the error: "Warning, solutions may have been lost", and when I try to use fsolve, it simply returns the expression as an answer and I am unable to find the numerical value for E__fv. Any helps or tips are appreciated.

If it helps, the defined variable values are:

k = 1.3806E-23;

T = 300;

h__bar = 1.05456E-34

L__z = 6E-9

m__hhw = 3.862216E-31;

E__hh0 = 3.012136E-21

E__hh1 = 1.185628E-20

 

A similar expression in the previous line was able to solve correctly and return a numerical value, so I'm not sure why solve/fsolve can't solve this one.

Similar solved expression:

solve(0.3e25 = m__cw*k*T*ln(1+exp(E__fc-E__c0)/(k*T))/(Pi*h__bar^2*L__z), E__fc);
                          -42.88488490
 

I have this polynomial equation: (x-2)^2*(x-3)+epsilon =0, I want to draw a bifurcation diagram in the (epsilon , x) plane.

 

How to implement this in maple 2018?

 

Thanks!

 

Hello,
Anyone has an idea what is wrong with the following code procedure?
the result of u(x) should be a continuous function of x (and it is continuous when solved numerically)

THANKS IN ADVANCE,
Gil Soffer
 

restart;
Heaviside(0):=1:Heaviside(0.):=1:
dats:={s(x)=-Dt*(x+h/2)+st};
eqs:={diff(u(x),x)=Heaviside(sY-s(x))*s(x)/E+Heaviside(s(x)-sY)*(sY/E+(s(x)-sY)/Esec),
      u(-h/2)=y0};
sol0:=simplify(dsolve(subs(dats,eqs),u(x)));
sol1:=int(lhs(eqs[1]),x)=eval(student:-simpson(subs(x=_x,subs(dats,rhs(eqs[1]))),_x=-h/2..x,40))+y0:

dat1:={Dt=-0.1,Esec=1,E=3,sY=0.8,st=0.05,y0=5,h=10};
plot([subs(evalf(subs(dat1,sol0)),u(x)),subs(evalf(subs(dat1,sol1)),u(x))],x=-5..5,title=u(x),legend=[symbolic,numeric]);
plot(subs(subs(dat1,subs(dats,eqs[1])),diff(u(x),x)),x=-5..5,title=diff(u(x),x));


Hello guys, 

I have a probelm with computing an integral by maple. I dont know why maple cannot compute.

 

integral.mw

Thank you for your attention

Best

With this application developed entirely in Maple using native syntax and embedded components for science and engineering students. Just replace your data and you're done.

Pearson_Coeficient.mw

Lenin Araujo Castillo

Ambassador of Maple

 

Hello,

do not work well and U functions are not replaced with series form.

Please see equation 5.

Also, How me can differential with respect to the constant Amnr], Bmnr], Cmnr] as shown in   attached figure?

For Differentiation I need a

Diff.pdf

I can use ApproximateInt for the integral?

approximate_int
 

restart

``

 

"f[1,1](r,theta):=(sin(-4.700000000 10^(-6)+4.700000000 r)-0.1369508410 sinh(-4.700000000 10^(-6)+4.700000000 r)) cos(6 theta):"

"L[1, 1](r, theta):=-2* (((&PartialD;)^2)/(&PartialD;r^2) f[1,1](r,theta))+7* f[1,1](r,theta)+5 *f[1,1](r,theta)-(2 *6 (((&PartialD;)^2)/(&PartialD;theta^2) f[1,1](r,theta)))/r+(0.6 (((&PartialD;)^4)/(&PartialD;r^2&PartialD;theta^2) f[1,1](r,theta)))/4+(.5 (((&PartialD;)^4)/(&PartialD;theta^4) f[1,1](r,theta)))/4"

proc (r, theta) options operator, arrow, function_assign; -2*(diff(f[1, 1](r, theta), r, r))+12*f[1, 1](r, theta)-12*(diff(f[1, 1](r, theta), theta, theta))/r+.6*(diff(f[1, 1](r, theta), r, r, theta, theta))/4+.5*(diff(f[1, 1](r, theta), theta, theta, theta, theta))/4 end proc

(1)

``

``

 

for w to 1 do for s to 1 do k[w, s] := (int(int(L[w, s](r, theta)*f[w, 1](r, theta), theta = 0 .. 2*Pi), r = 0 .. 1))/(int(int(f[w, 1](r, theta)^2, theta = 0 .. 2*Pi), r = 0 .. 1)); print([w, s] = %) end do end do

[1, 1] = 0.3929199233e-1*(int(0.1005309649e-16*(2329569981.*r*cos(4.700000000*r)^2-0.9913063750e15*r*cos(4.700000000*r)*sin(4.700000000*r)+0.1054581250e21*r*sin(4.700000000*r)^2-328995293.4*r*cos(4.700000000*r)*cosh(4.700000000*r)+0.6999899860e14*r*cos(4.700000000*r)*sinh(4.700000000*r)+0.6999899860e14*r*sin(4.700000000*r)*cosh(4.700000000*r)-0.1489340396e20*r*sin(4.700000000*r)*sinh(4.700000000*r)+1363855.810*r*cosh(4.700000000*r)^2-0.5803641743e12*r*cosh(4.700000000*r)*sinh(4.700000000*r)+0.6174086961e17*r*sinh(4.700000000*r)^2+2982150000.*cos(4.700000000*r)^2-0.1269000000e16*cos(4.700000000*r)*sin(4.700000000*r)+0.1350000000e21*sin(4.700000000*r)^2-816815901.0*cos(4.700000000*r)*cosh(4.700000000*r)+0.1737906172e15*cos(4.700000000*r)*sinh(4.700000000*r)+0.1737906172e15*sin(4.700000000*r)*cosh(4.700000000*r)-0.3697672707e20*sin(4.700000000*r)*sinh(4.700000000*r)+55931812.29*cosh(4.700000000*r)^2-0.2380077119e14*cosh(4.700000000*r)*sinh(4.700000000*r)+0.2531996935e19*sinh(4.700000000*r)^2)/r, r = 0 .. 1))

(2)

``


 

Download approximate_int.mw

 

How I can replace  u__0r, theta, t) with f1, 1(r, theta) in attached file.

I want in I have only f1,1] function.

Thanks 


 

````

"f[1, 1](r, theta):=`u__0`(r, theta,t)  "

proc (r, theta) options operator, arrow, function_assign; u__0(r, theta, t) end proc

(1)
``````````

"L[1, 1](r, theta):=-`A__0`*(&PartialD;)/(&PartialD;r) (F*(&PartialD;)/(&PartialD;r)`u__0`(r,theta))-1/(2)*`A__0`*(&PartialD;)/(&PartialD;r) (`K__1`*`u__0`(r,theta))+1/(2)*`A__0`*`K__1`*(&PartialD;)/(&PartialD;r)`u__0`(r,theta)-1/(2)*`A__0`*(&PartialD;)/(&PartialD; r) (`H__1`*`u__0`(r,theta))+1/(2)*`A__0`*`H__1`*(&PartialD;)/(&PartialD;r)`u__0`(r,theta)+`K__3`*`A__0`*`u__0`(r,theta)-1/(2)*`A__0`*(&PartialD;)/(&PartialD; r) (`K__4`*`u__0`(r,theta))+1/(2)*`A__0`*`K__4`*(&PartialD;)/(&PartialD;r)`u__0`(r,theta)+`A__0`*`K__5`*`u__0`(r,theta)-2*`A__0`*(&PartialD;)/(&PartialD; theta) ((`H__2`)/(r)*(&PartialD;)/(&PartialD;theta)`u__0`(r,theta))+(1)/(4)*`A__0`*l^(2)*((&PartialD;)^(2))/(&PartialD; r &PartialD; theta)(mu*((&PartialD;)^(2))/(&PartialD;r &PartialD;theta)`u__0`(r,theta))+(1)/(4)*`A__0`*l^(2)*((&PartialD;)^(2))/(&PartialD;theta^(2))(mu*((&PartialD;)^(2))/(&PartialD; theta^(2))`u__0`(r,theta))+rho*`A__0`*`K__16`*((&PartialD;)^(2))/(&PartialD;t^(2))`u__0`(r,theta);"

proc (r, theta) options operator, arrow, function_assign; -A__0*(diff(F*(diff(u__0(r, theta), r)), r))-(1/2)*A__0*(diff(K__1*u__0(r, theta), r))+(1/2)*A__0*K__1*(diff(u__0(r, theta), r))-(1/2)*A__0*(diff(H__1*u__0(r, theta), r))+(1/2)*A__0*H__1*(diff(u__0(r, theta), r))+K__3*A__0*u__0(r, theta)-(1/2)*A__0*(diff(K__4*u__0(r, theta), r))+(1/2)*A__0*K__4*(diff(u__0(r, theta), r))+A__0*K__5*u__0(r, theta)-2*A__0*(diff(H__2*(diff(u__0(r, theta), theta))/r, theta))+(1/4)*A__0*l^2*(diff(mu*(diff(u__0(r, theta), r, theta)), r, theta))+(1/4)*A__0*l^2*(diff(mu*(diff(u__0(r, theta), theta, theta)), theta, theta))+rho*A__0*K__16*(diff(u__0(r, theta), t, t)) end proc

(2)

``


 

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