Hello! I need to solve a system of vector equations (a system with vectors, not a linear system). The coordinates of the vectors are not given. (It is a system of kinetostatic equations, I must to define reaction forces as functions of vectors of external forces.

I'm using the differential operator D to differentiate variables and look at the structure of the equations, i.e. turn things like

D[1]u-vt+v^2=0 into

D[1,2](u)-D[2](v)*t+v*D[1](t)+2*v*D[2](v)=0

However, it is assumed that this is a commutative operator, i.e. D[1,2]=D[2,1]. Can I use an option to have it not commute, or should I look into using the DETools package?

Thanks,

Mike

restart;
with(LinearAlgebra):

C:=proc(k,M) local N,P,m,L,T,j,Ld,n,i; 
N:=2^(k-1)*M:
P[0]:= t -> 1:
P[1]:= t -> t:
for m from 2 to M-1 do
   P[m]:= unapply(expand((2*m-1)*t*P[m-1](t)/m - (m-1)*P[m-2](t)/m), t)
end do: 
L:=proc(n,m) local a,b; 
  a := (2*n-2)/2^k; 
  b := 2*n/2^k; 
  return piecewise(a <= t and t <= b, P[m](2^k*t-2*n+1)*sqrt(((m+1)/2)*2^k))
end proc:
T := Vector(N):
for j from 1 to N do T[j] := (j-1/2)/N end do; 
Ld := unapply(Vector(N),t):
for n from 1 to 2^(k-1) do 
  for m from 0 to M-1 do
i := n+2^(k-1)*m:
Ld(t)[i] := L(n,m)
     end do
end do:

return Ld(t):  
end proc:
A:=t->C(2,3);
A(0);
A(0.5);
A(0.3);

Ld is piecewise function depends on t. I want to calculate Ld(0)=A(0), Ld(0.5)=A(0.5) etc.  

What is the problem?

with(Physics);
Setup(mathematicalnotation = true):
Setup(dimension = 3, coordinates = (X = [x, y, t]), metric = -ut^2*(dx^2)+ut^2*(dy^2)+(-ux^2-uy^2+1)*(dt^2)+2*ux*ut*dx*dt+2*uy*ut*dy*dt):
              ut := Physics:-diff(u(X), t)
ux := diff(u(x, y, t), x);
                  ux := Physics:-diff(u(X), x)
uy := diff(u(x, y, t), y);
                  uy := Physics:-diff(u(X), y)
Define(S[mu, nu] = 1/(d-2)*(Ricci[mu, nu]-1/(2*d-2)*Ricci[`~alpha`, alpha]*g_[mu, nu]));
             Defined objects with tensor properties
Define(C[mu, nu, rho] = D_[nu](S[rho, mu])-D_[rho](S[nu, mu]));
             Defined objects with tensor properties
g_[];
Physics:-g_[mu, nu] = 

 
C[mu, nu, rho, nonzero];
          [Length of output exceeds limit of 1000000]
utt := diff(u(x, y, t), t, t);
               utt := Physics:-diff(u(X), t $ 2)
uxx := diff(u(x, y, t), x, x);
               uxx := Physics:-diff(u(X), x $ 2)
uyy := diff(u(x, y, t), y, y);
               uyy := Physics:-diff(u(X), y $ 2)
uxt := diff(u(x, y, t), x, t);
                uxt := Physics:-diff(u(X), t, x)
uyt := diff(u(x, y, t), y, t);
                uyt := Physics:-diff(u(X), t, y)
utt = ut*ut*(1/(-ux^2-uy^2+1))*(uxx+uyy)-2*ut*(1/(-ux^2-uy^2+1))*(ux*uxt+uy*uyt);
Basically here the cotton tensor exceeds the output limit in maple, however I am supposed to be showing how the cotton tensor vanishes, the last part of the code is me trying to input a relation of u_tt to various other derivatives of u in the hope that the cotton tensor can be simplified from this equation but nothing has changed as a result, can anyone help me? 

restart;
PDE := diff(y(x,t), x,x,x,x)+(diff(y(x, t), t,t))=0;  
# Initial/boundary conditions 
  BCs:=y(0,t) = 0, y(1,t) = 0,D[1](y)(0,t)=0,D[1](y)(1,t)=0;

  ICs:=y(x,0) =0, D[2](y)(x,0)=1 ;
  num_solution := pdsolve(PDE, {BCs,ICs}, numeric); 
  num_solution :-plot3d(x=0..1, t=0..1);

1st Question:How to plot the 2D at point x=0.5?

2nd Question: How to use PDEplot?

How to use a Lyapunov function to prove the stability of a system of differential equations and then check the stability? My initial thoughts were to use a contour plot?

I havev to calculate the intersection of three planes and need to plot them, when I try to plot them I get this error:

Error, (in plot3d) bad range arguments: (Vector(3, {(1) = -4*_t2[3], (2) = 3*_t2[3], (3) = _t2[3]})) = -8 .. 8, y = -20 .. 20

This is my code so far:

with(LinearAlgebra);
A := <1, 1, -2|3, 4, -7|-5, -8, 13>
b := <0,0,0>

x := LinearSolve(A, b)
P1 := x+3*y;
P2 := x+4*y;
P3 := -2*x-7*y;
plot3d([P1, P2, P3], x = -8 .. 8, y = -20 .. 20, plotlist = true, color = [blue, red, green]);
Error, (in plot3d) bad range arguments: (Vector(3, {(1) = -4*_t2[3], (2) = 3*_t2[3], (3) = _t2[3]})) = -8 .. 8, y = -20 .. 20

Hello,

I will buy some books for improving Maple.

Especially, I'm looking for the good books with Maple code for solving differential equations (Analytical and numerical solutions for Both Ordinary and Partial Differential equations).

Book Language: English.

Best regards.

Hello,

I am currently doing this thing:

unapply(piecewise(x>37-0.25*t and x<37+0.25*t,-(11*sin(t/2+1.145)-10)*((x-37)^2)+(-3.5*sin(t)+ 1),10),x);

(It is used with "for t from 0 to 8 do *things* end do", hence why I use "t")

The problem I have is I don't want this (over) complicated function to go over y=10, but the conditions I tried were not working that well.

I saw on the Help page that it should be possible, but my attempts were unsuccessful...

Maple 2018.2 generates wrong latex in this example. In Latex a space in command name is important. So "\tau L" is not the same as "\tauL".  Since in the later case, Latex will complain that there is no command "\tauL"

Maple generates "\tauL" in the latex, when it should be "\tau L" in the following example, so the latex fails to compile because there is no command called "\tauL" in Latex.

Here is screen shot showing the problem and the Maple command to reproduce it

restart;
interface(showassumed=0);
pde :=  diff(u(x,t),t)=k*diff(u(x,t),x$2)+(exp(-c*t)*sin(2*Pi*x/L));
ic  :=  u(x,0)=f(x);
bc  :=  D[1](u)(0,t)=0, D[1](u)(L,t)=0;
sol:=pdsolve({pde,ic,bc},u(x,t)) assuming L>0,t>0,k>0;

latex(sol)

Any chance Maplesoft could fix this?

Possible workaround for now is to use something like "\newcommand{\tauL}{\tau L}" in preamble for this specific case.

How could this be set up in Maplesim?  A vibration signal is imparted on the box and we want to observe the behavior of the sprung object inside the box.

Hello,

Assume we have the following "intervals" (I am not sure what is its formal name in Maple)

C :=[0,1/11],[1/11,1/9],[1/9, 1/7],[1/3,1/2],[1/2,1]

How can we get the "union" of these intervals? That is to say, obtain  [0,1/7],[1/3,1] 

Many thanks in advance for your comments and suggestions.

Hi,

I'm trying to plot a 3 dimensional vector equation of motion in Maple for my dissertation. The equation is as follows:

vec(v_r) = (v_b)*i + (v_g*cos(Omega*t))*j + (v_g*sin(Omega*t))*k

where Omega is the gyrofrequency, v_b is the velocity parallel with the i direction, and v_g is the perpendicular velocity of gyration. v_g and v_b are both in terms of the velocity of incidence and two angles, which are all just numerical values that I've introduced.

It should look like a helix spiralling around the i axis, but I'm fairly new to Maple and I've tried everything I can think of and I can't get it to work.

When I use plot3d it turns out looking like a sphere or a cylinder. When I use spacecurve, it doesn't work at all and I get the error message " Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct.".

Any help is greatly appreciated.

I have an ODE. I solved it numerically by Runge Kutta method.

The solution is X(t).

Now, I want to multiply X(t) by sin(y) and to plot in 3D.

How can we do it? 

question.mw

I am trying to find the number a and b (-20 < a < 20, -20 < b <20) so that two circles (x+1)^2+(y+3)^2 = 125 and (x-a)^2+(y-b)^2 = 225 cut at two points A and B and coordinates A and B are pairs of integers.
I tried
 

restart; L := []; 
for a from -20 to 20 do 
for b from -20 to 20 do 
for x from -20 to 20 do 
for y from -20 to 20 do 
if (x+1)^2+(y+3)^2 = 125 and (x-a)^2+(y-b)^2 = 225 and nops({a, b, x, y}) = 4 and x*y*a*b <> 0 then 
L := {op(L), {[a, b], [x, y]}} 
od: od: od: od:
nops(L); 
L

How to select the number a and b so that the system of equations (x+1)^2+(y+3)^2 = 125  and (x-a)^2+(y-b)^2 = 225 have two integral solutions. For example
 

restart; 
solve({(x-6)^2+(y+2)^2 = 225, (x+1)^2+(y+3)^2 = 125}, {x, y})