Analysis of the semiclassical (SC) momentum rate equations Plotting the ICs and BCs and examining sensitivity to the Re and Im forces MRB: 24/2/2020, 27/2/2020, 2/3/2020. We examine solution of the SC version of the momentum rate equations, in which O terms for  are removed. A high level of sensitivity to ICs and BCs makes solution finding difficult. >  restart; >  with(PDETools): with(CodeTools):with(plots): We set up the initial conditions: >  ICu := {u(x, 0) = .1*sin(2*Pi*x)}; ICv := {v(x, 0) = .2*sin(Pi*x)}; (1) >  plot([0.1*sin(2*Pi*x),0.2*sin(Pi*x)],x = 0..2, title="ICs:\n u(x,0) (red), v(x,0) (blue)",color=[red,blue],gridlines=true);   The above initial conditions represent a positive velocity field  (blue) and a colliding momentum field (red).   Here are the BCs >  BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t))}; (2) >  BCv := {v(0,t) = 0.5*sin(2*Pi*t),v(2,t)=-0.5*sin(2*Pi*t)};   (3) >  plot([0.5*(1-cos(2*Pi*t)),0.5*sin(2*Pi*t),-0.5*sin(2*Pi*t)],t=0..1,color=[red,blue,blue],linestyle=[dash,dash,dot],title="BCs:\n u(0,t) (red-dash),\n v(0,t) (blue-dash), v(1,t) (blue-dot)",gridlines=true); >    We can now set up the PDEs for the semiclassical case. >  hBar:= 1:m:= 1:Fu:= 0.2:Fv:= 0.1:#1.0,0.2 >  pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu; (4) >  pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;   (5) >  ICu:={u(x,0) = 0.1*sin(2*Pi*x)};   (6) >  ICv:={v(x,0) = 0.2*sin(Pi*x/2)};   (7) >  IC := ICu union ICv;   (8) >  BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t)), D[1](u)(2,t) = 0.1*cos(2*Pi*t)}; (9) >  BCv := {v(0,t) = 0.2*(1-cos(2*Pi*t))};   (10) >  BC := BCu union BCv;   (11) We now set up the PDE solver: >  pds := pdsolve({pdeu,pdev},{BC[],IC[]},time = t,range = 0..2,numeric);#'numeric' solution (12) >  Cp:=pds:-animate({[u, color = red, linestyle = dash],[v,color = blue,linestyle = dash]},t = 30,frames = 400,numpoints = 400,title="Semiclassical momentum equations solution for Re and Im momenta u(x,t) (red) and v(x,t) (blue) \n under respective constant positive forces [0.2, 0.1] \n with sinusoidal boundary conditions at x = 0, 1 and sinusoidal initial conditions: \n time = %f ", gridlines = true,linestyle=solid):Cp; Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0): Newton iteration is not converging   (13) Observations on the quantum case: The classical equation for  is independent of the equation for .   (red) is a solution of the classical Burgers equation subject to a force 0.2, but  is NOT influenced by .  On the otherhand,  (blue) is a solution of the quantum dynamics equation subject to force 0.1 and is influenced by .   This one way causality (u )  is a feature of the semiclassical case, and it emphasises the controlling influence of the classical , which modulates the quantum solution for .  Causally, we have u.   The initial conditions are of low momentum amplitude:0.1 for the classical  (red) field and.2 for  (blue)  but their influence is soon washed out by the boundary conditions  and  that drive the momentum dynamics.   The temporal frequency of the boundary condition on the -field is twice that of the classical -field. This is evident in the above blue transient plot. Moreover, the boundary condition on the classical -momentum (red), drives that field in the positive direction, initially overtaking the quantum  field, as consistent with the applied forces [0.2, 0.1]. Although initially of greater amplitude than the classical field, the  momentum field is asymptotically of the same amplitude as the  field, but has greater spatial and temporal frequency, owing to the boundary conditions.   Referring to the semiclassical momentum rate equations, we note that the classical field  (red) modulates the quantum momentum rate equation for . >

 

Download SC-plots.mw

I am having difficulty getting solutions to a pair of PDEs.  Would anyone like to cast an eye over the attached file, incase I am missing something.

Thanks

Melvin

f := proc (a, b) options operator, arrow; (1/2)*b+(1/2)*arccos(sin(2*a-b)/tan(b)) end proc; dis := proc (A, B) options operator, arrow; sqrt(inner(A-B, A-B)) end proc; bisA := proc (A, B, C) local b, c, M; b, c := dis(A, C), dis(A, B); M := (b*B+C*c)/(b+c); x*(A[2]-M[2])+y*(M[1]-A[1])+A[1]*M[2]-A[2]*M[1] end proc; P := proc (a0, b0) local a, b, c, p1, p2, p3, p4, r, II; a := evalf(a0); b := evalf(b0); c := f(a, b); if b0-0.1e-2

  I have read the article attached in the following

I wrote the code mentioned in the article in order to find the Hirota formula This is the code that I wrote For Hirota Method But when running the program it does not work well 

Could you help me to fix the mistake and run the program

 

>  >  (1) >  >  >  (2) >  (3) >  (4) >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  >  (5) >  >  diff(diff(u(x, t), t), t)-(diff(diff(u(x, t), x), x))-3*(diff(diff(u(x, t), x), x))^2-(diff(diff(diff(diff(u(x, t), x), x), x), x)) = 0 Error, (in getFnumer) expected 1 st argument be type `+`.   >  >

 

Download Hirota_to_send.mw

Let Q be a quadrilateral which has a inscribed circle and a circumscribed circle. Show that the centers of these 2 circles and the point of intersection of the diagonals
of the quadrilateral are aligned.

restart; unprotect(D, O);

with(plots); with(LinearAlgebra);
Soit Q un quadrilatère qui possède un cercle inscrit et un cercle circonscrit. Montrer que les centres de ces 2 cercles et le point d'intersection des diagonales
du quadrilatère sont alignés.
_EnvHorizontalName := 'x';

_EnvVerticalName := 'y';

Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end proc;
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end proc;
EQ := proc (M, N) local eq; eq := (y-M[2])/(x-M[1]) = (N[2]-M[2])/(N[1]-M[1]) end proc;
Pour un quadrilatère A'B'C'D' circonscrit autour d'un cercle avec points de contact A, B, C, D, les droites A'C', B'D', AC, BD sont concourantes
varphi1 := (1/3)*Pi; varphi2 := varphi1+(1/4)*Pi+.1; varphi3 := varphi2+3*Pi*(1/5); varphi4 := 2*Pi-varphi1-varphi2-varphi3; R := 5;
                    varphi4 := -0.514159263
O := [0, 0];
for i to 4 do M || i := [R*cos(varphi || i), R*sin(varphi || i)]; OM || i := plot([O, M || i], color = blue, linestyle = 3); cfOM || i := -1/(diff(solve(EQ(O, M || i), y), x)); eq || i := y = cfOM || i*x+R*sin(varphi || i)-cfOM || i*R*cos(varphi || i); T || i := plot(cfOM || i*x+R*sin(varphi || i)-cfOM || i*R*cos(varphi || i), x = -10 .. 10, color = green) end do;

for i to 4 do j := `mod`(i+1, 4); if j = 0 then j := 4 end if; sol || i := evalf(op(solve([eq || i, eq || j], [x, y]))); P || i := [subs(sol || i, x), subs(sol || i, y)] end do;
for i to 4 do Q || i := geometry:-point(Q || i, P || i[1], P || i[2]) end do;
geometry:-AreConcyclic(Q1, Q2, Q3, Q4, 'cond');
                             false
AC := plot([M1, M3], color = blue);
BD := plot([M2, M4], color = blue);
diago1 := plot([P1, P3], color = coral); diago2 := plot([P2, P4], color = coral);
quadri := plot([seq(M || i, i = 1 .. 4), M1], color = black);
cer := plot([R*cos(t), R*sin(t), t = 0 .. 2*Pi], color = blue);
Points1 := pointplot([seq(M || i[], i = 1 .. 4)], symbol = solidcircle, color = [blue], symbolsize = 15);
Points2 := pointplot([seq(P || i[], i = 1 .. 4)], symbol = solidcircle, color = [green], symbolsize = 15);
Tt1 := plots:-textplot([[M1[], "A"], [M2[], "B"], [M3[], "C"], [M4[], "D"]], font = [times, 10], align = {below, right});
Tt2 := plots:-textplot([[P1[], "A'"], [P2[], "B'"], [P3[], "C'"], [P4[], "D'"], [O[], "O"]], font = [times, 15], align = {above, right});
display([cer, quadri, seq(OM || i, i = 1 .. 4), seq(T || i, i = 1 .. 4), AC, BD, diago1, diago2, Points1, Points2, Tt1, Tt2], axes = none, view = [-9 .. 7, -12 .. 6], scaling = constrained, size = [500, 500]);

Hello, 

For a few days Maple crashs everytime i try to use the command "plot3d()". 

I had'nt this problem befor and I have no idea what the reason could be. It ist irrelevant what Funktion I try to visualize,  the window just get closed evertime.

I hope someone can help me.

Thank you!

Tom

The question reads as follows: 

a) Use Maple to find the linear function f(x,y) = c +mx + ny, whose graph passes through the points P=(5,1,0), Q=(-1,7,1), and R = (2,1,4). 

b) Use Maple to graph the linear function whose formula you found in part a.

c) Use Maple to find the (exact) area of the triangle having P,Q and R as vertices. 

Thank you so much. I am new to Maple and just trying to figure it all out.  

I am trying to write a program for the representation of a function F with three variables i, j, x, where i, j are non-negative integers and x is real. I have tried with different indexings like F(i,j,x), F[i](j,x), F[i,j](x), but did not get the desired result. My objective is to get F as a function of i, j, x.

Here is my maple code with detailed description of my problems in each step.

 

test1702.mw

Plot the Vector Field f(x,y,z) = (-x,y,cos z).

How do I go about plotting this? I have tried everything and nothing is working for me. 

a1 := 5; b1 := 3; a2 := 3; b2 := 4; a3 := 3; b3 := 7; eq1 := expand((y-2)^2/b1^2+(x-5)^2/a1^2 = 1); 1 2 4 13 1 2 2 eq1 := - y - - y + -- + -- x - - x = 1 9 9 9 25 5 eq2 := expand((y+2)^2/b2^2+(x+1)^2/a2^2 = 1); 1 2 1 13 1 2 2 eq2 := -- y + - y + -- + - x + - x = 1 16 4 36 9 9 Sys := {eq1, eq2}; Sol := [solve(Sys, explicit)]; L := map(proc (t) options operator, arrow; eval([x, y], t) end proc, simplify(`~`[fnormal](evalf(Sol), 9), zero)):I don't want complex roots. Tank you.

Can you please help me in telling different ways to speed up the following code. thank you.

 

thread_map.mw

Can you guess what P() produces, without executing it?

P:=proc(N:=infinity) local q,r,t,k,n,l,h, f;
q,r,t,k,n,l,h := 1,0,1,1,3,3,0:
while h<N do 
   if 4*q+r-t < n*t
   then f:=`if`(++h mod 50=0,"\n",`if`(h mod 10=0," ","")); printf("%d"||f,n);   
        q,r,t,k,n,l := 10*q,10*(r-n*t),t,k,iquo(10*(3*q+r),t)-10*n,l
   else q,r,t,k,n,l := q*k,(2*q+r)*l,t*l,k+1,iquo(q*(7*k+2)+r*l,t*l),l+2
   fi
od: NULL
end:

I hope you will like it (maybe after execution).

While using the Physics package in Maple 2018, I am facing a few hurdles. In the file tov.mw ,

1. I have defined a dual-vector with components $u_a$ in step 12, but when I am trying to compute $g(u,u)$ using SumOverRepeatedIndices  in step 15, it is not evaluating the sum.

2. I want to define a tensor $B_{ab}=\nabla_a u_b$. Do I just write Define(B{a,b}=D_[a] u[b]) ?

3. Is there any command to find the symmetric and antisymmetric components and trace of the tensor B or do we evaluate them by writing out the expressions?

Thank you.

restart; with(plots); _EnvHorizontalName := 'x'; _EnvVerticalName := 'y'; para := -2*p*x+y^2 = 0; para1 := -2*p1*x+y^2 = 0; t := y-m*x-p/(2*m) = 0; t1 := y+x/m+(1/2)*p1*m = 0; sol := op(solve([t, t1], [x, y])); eliminate({rhs(sol[1]), rhs(sol[2])}, m); for example : m := 2; p := 1; p1 := -2; PARA := implicitplot(para, x = -3 .. 3, y = -3 .. 3, color = blue); PARA1 := implicitplot(para1, x = -3 .. 3, y = -3 .. 3, color = green); Tang := implicitplot(t, x = -3 .. 3, y = -3 .. 3, color = brown); Tang1 := implicitplot(t1, x = -3 .. 3, y = -3 .. 3, color = aquamarine); NULL; #m is not constant display([PARA, PARA1, Tang, Tang1], view = [-3 .. 3, -3 .. 3], scaling = constrained);#on what curve is the vertex of the angle of 2 tangents ? Thank you.