Rouben Rostamian

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These are questions asked by Rouben Rostamian

How to calculate flux in the heat equati...

Asked: Rouben Rostamian 8541 Product: Maple

September 02 2019

3 4

I solve a boundary value problem for a linear system of first order PDEs
in the unknowns and Ultimately, I am interested in the graph
of the function but numerical artifacts distort that graph badly by
imposing spurious oscillation.

Is there a way to maneuver the calculations to obtain the graph of
without the oscillation?

restart;

>	pde1 := diff(u(x,t),t)=diff(v(x,t),x);

>	pde2 := diff(u(x,t),x)=v(x,t);

>	ibc := u(x,0)=1, u(0,t)=0, u(1,t)=0;

>	dsol := pdsolve({pde1,pde2}, {ibc}, numeric, spacestep=0.01, timestep=0.01);

>	dsol:-value(output=listprocedure): my_u := eval(u(x,t), %): my_v := eval(v(x,t), %%):

>	plot3d(my_u(x,t), x=0..1, t=0..0.5);

>	plot3d(my_v(x,t), x=0..1, t=0..0.5);

The oscillations are numerical artifacts. Can they be avoided?

>	plot(my_v(0,t), t=0..0.5);

A side comment: By eliminating between the two PDEs we see that
satisfies the standard heat equation, thus is the flux. The expression
expresses the heat flux at the boundary, and that's what I am after.

Download calculating-flux-in-the-heat-equation.mw

How to update a matrix in a proc?...

Asked: Rouben Rostamian 8541 Product: Maple

August 05 2019

2 2

I expect this proc to return a matrix of all ones but it returns a matrix of all zeros. Why does it do that?

doit := proc()  
  local i, j, M;                                     
  M := Matrix(3,3);                    
  for i from 1 to 3 do  
    for j from 1 to 3 do  
      M[i][j] := 1;                                      
    end do;                             
  end do;              
  return M;  
end proc:                                           

doit();                            
                                 [0    0    0]
                                 [           ]
                                 [0    0    0]
                                 [           ]
                                 [0    0    0]

Unexpected solution to PDE...

Asked: Rouben Rostamian 8541 Product: Maple 2019

July 28 2019

1 1

We solve Laplace's equation in the domain
in polar coordinates subject to prescribed Dirichlet data.

Maple produces a solution in the form of an infinite sum,
but that solution fails to satisfy the boundary condition
on the domain's outer arc. Is this a bug or am I missing
something?

restart;

>	kernelopts(version);

>	with(plots):

>	pde := diff(u(r,t),r,r) + diff(u(r,t),r)/r + diff(u(r,t),t,t)/r^2 = 0;

diff(diff(u(r, t), r), r)+(diff(u(r, t), r))/r+(diff(diff(u(r, t), t), t))/r^2 = 0

>	a, b, c, d := 1, 2, Pi/6, Pi/2;

>	bc := u(r,c)=c, u(r,d)=0, u(a,t)=0, u(b,t)=t;

We plot the boundary data on the domain's outer arc:

>	p1 := plots:-spacecurve([bcos(t), bsin(t), t], t=c..d, color=red, thickness=5);

Solve the PDE:

>	pdsol := pdsolve({pde, bc});

Truncate the infinite sum at 20 terms, and plot the result:

>	eval(rhs(pdsol), infinity=20): value(%): p2 := plot3d([rcos(t), rsin(t), %], r=a..b, t=c..d);

Here is the combined plot of the solution and the boundary condition.
We see that the proposed solution completely misses the boundary condition.

>	plots:-display([p1,p2], orientation=[25,72,0]);

Download mw.mw

What is the asymptotic value of the root...

Asked: Rouben Rostamian 8541 Product: Maple

June 16 2019

2 6

I am interested in the root of the equation ((2*n-1)*x + 2*n + 1)*x^n = 1 - x, where 0 < x < 1 and n is a large positive integer. I believe that the root converges to 1 as n goes to infinity. How does one obtain an asymptotic estimate of the root for large n?

How to simplify this expression?...

Asked: Rouben Rostamian 8541 Product: Maple

February 10 2019

2 1

I don't know how to simplify the expression A/B in the worksheet below. I know that it should simplify to exp(I*Pi/3). How do I lead Maple to discover that result without telling it the solution ahead of the time? All variables other than A and B are real.