Rouben Rostamian

MaplePrimes Activity


These are replies submitted by Rouben Rostamian

@nm  In this problem the initial and boundary conditions are artificially designed/fine-tuned to yield a simple analytical solution.  The slightest change, for instance replacing the 12 by 13, will destroy the balance and the simple solution will be lost.  In the general case the solution would be expressed as an infinite sum—a Fourier series.

As to your question on how I thought of the "+" form of the solution, I looked at the boundary conditions and saw that the two ends move together, one end moving as −12 t2, and the other lagging by 1 relative to the other.  That led me to think of a rigid translational motion.  Since the initial condition is given as x4, the natural candidate would be x4 − 12 t2.  I tried that in my head, and saw that it works.

That gave me the idea of the HINT=`+` in Maple, and that yielded the same result.

This does not make the HINT=`+` option a useful choice for such problems.  That choice works in this case by "accident" because of the artificially fine-tuned conditions.

@assma Did you try saving it as EPS?

@assma I don't have Maple 18 to test, but it is likely that it should be able to handle the saving of a 3D figure as EPS.

Note, however, that the EPS format cannot handle transparency.  If your figure has transparent parts, then you will have to save it in the JPG format which is not the best thing since it is a lossy format and results in somewhat blurry images.

@josephrajck The equation and the boundary conditions are correct, but Initial conditions are missing.  Additionally, you can't get a numerical solution if you leave the coefficient "a" and the length "l" inspecified.  Here I will take both of these to be 1.  You may change then as needed.

restart;

pde := diff(y(x,t),t,t) + diff(y(x,t),x$4) = 0;

diff(diff(y(x, t), t), t)+diff(diff(diff(diff(y(x, t), x), x), x), x) = 0

bc := y(0,t)=0,
      D[1](y)(0,t)=0,
      D[1,1](y)(1,t)=0,
      D[1,1,1](y)(1,t)=0;

y(0, t) = 0, (D[1](y))(0, t) = 0, (D[1, 1](y))(1, t) = 0, (D[1, 1, 1](y))(1, t) = 0

ic := y(x,0) = x^2,
      D[2](y)(x,0)=0;

y(x, 0) = x^2, (D[2](y))(x, 0) = 0

sol := pdsolve(pde, {bc, ic}, numeric);

_m140082965461792

sol:-animate(t=5.23, frames=100, thickness=12, title="time = %f");

Download mw.mw

@vv Uh, OK, I see it now and I agree.  Also thanks to Preben for his explanations.

 

@Preben Alsholm Aren't all variables at the top level global by default?
restart;
type(a, `global`);          #true
type(whatever, `global`);   #true
type(_Z, `global`);         #true

I don't understand what it is so special about singling out _Z as being global.

 

@Preben Alsholm The error message that you have shown does not complain about "global".  It complains about "protected".  It may be a global variable, but that's beside the point.

In fact _Z is not unique in that respect.  Try any of:

_a := 12;
_b := 12;
_c := 12;

In general it's good to work under the assumption that all identifiers beginning with underscore are reserved for Maple's internal use, and avoid messing around with them.  Even if some of such identifiers are unprotected now, they may become unavailable in future releases.

This is exactly parallel to the C programming language where identifiers beginning with underscore are reserved for the compiler's internal use.  One may assign to them at one's own risk.

 

I haven't checked all of the calculations in that manuscript, however just

spot-checking it reveals some errors which weakens my trust in the rest

of its statements.

 

First, I have already noted the F(t, x, y) versus F(t, x, y, z) confusion,

but perhaps that can be attributed to a careless copy-paste error.  It is

clear that all F(t, x, y, z) and G(t, x, y, z) are supposed to be F(t, x, y)

and "G(t,x,y)."

 

Second, each of the two phase portraits in Figure 1 on page 161 show

an equilibrium near x = .5, y = 0.6e-1.  That is inconsistent with the

given data as we see in the calculations below. Again, that may be due

to typographical errors in the article but there is no way for us to tell.

 

Of course, it is possible that I am mistaken, in which case I would like

to be corrected.

 

restart;

The system of differential equations is defined in equation (1.2) in terms

of the functions F and G:

F := r*x*(1-x/k) - beta*x*y/(a+x^2);
G := mu*beta*x*y/(a+x^2) - d*x - eta*x*y;

r*x*(1-x/k)-beta*x*y/(x^2+a)

 

mu*beta*x*y/(x^2+a)-d*x-eta*x*y

(1)

The numerical values of the coefficients are specified on page 159:

params := r=0.05, a=0.8, mu=0.8, d=0.24,
          eta=0.01, beta=0.6, k=1.6;

r = 0.5e-1, a = .8, mu = .8, d = .24, eta = 0.1e-1, beta = .6, k = 1.6

(2)

The equilibria are the roots of the system F = 0, G = 0:

sys := eval({F=0, G=0}, {params});

{0.5e-1*x*(1-.6250000000*x)-.6*x*y/(x^2+.8) = 0, .48*x*y/(x^2+.8)-.24*x-0.1e-1*x*y = 0}

(3)

Figure 1 on page 161 indicates that there is an equilibrium at around "(x=0. 5, y=0.06)".

However we may verify in a number of ways that there are no equilibria near

that point at all.  One (but not the only) way is through Maple's fsolve() which fails to

find a root in the specified region:

fsolve(sys, {x=0.1..1.0,y=0.01..0.10});

fsolve({0.5e-1*x*(1-.6250000000*x)-.6*x*y/(x^2+.8) = 0, .48*x*y/(x^2+.8)-.24*x-0.1e-1*x*y = 0}, {x, y}, {x = .1 .. 1.0, y = 0.1e-1 .. .10})

(4)
 

 

Download mw2.mw

@torabi In your differential equations we have F(t,x,y) but in the discretized equations we have F(t,x,y,z).  What is z? Something isn't quite right there.

Additionally, in order to do actual computations, we need to know the functions F and G. What are they?

 

@kainmuth Let w := s -> 1/abs(s), and then try:

int(w(x-y), x=0..1, y=0..1);

The result is infinity.  Consequently, your E is always infinity, regardless of the choice of f.  There is no hope of minimizing E.

@torabi

Here is the general solution of your problem:

restart;

de := (diff(r*(diff(u(r), r)), r))/r = 0:

dsolve(de);

u(r) = _C2*ln(r)+_C1

You see that the graph of the solution, regardless of any boundary conditions, would

be a scaled and translated copy of the graph of the logarithm function.  But the graph

of the logarithm function is monotonically increasing, that is, it does not go up

and down.  Therefore it is impossible for the solution to start at a zero value at

some r, become nonzero, and then return to zero again.

Maple can solve the wave equation in one-dimensional space, but as far as I know it cannot handle the two-dimensional case, so you are out of luck here.

If you have the knowledge and ambition, you may set yourself the goal of solving the problem numerically with your own implementation of a finite element algorithm in Maple.  That will take some time and effort.

The commercial software Comsol Multiphysics provides ready-to-use finite element solvers for all sorts of PDEs.  If you have access to it, you may want to give it a try although it will take a few days just to learn the interface.  If you don't have Comsol, probably you cannot afford it since it is quite expensive.

I have no idea whether Matlab or Mathematica can solve your problem.  Ask someone who knows about them.  Good luck!

 

In the definition of E you have f(y).  What is y?  Perhaps you need a double integral that integrates on both x and y?

Additionally, what is w?

 

@deniscr OK, that's good.  Almost.  I see that you are entering your code in Maple's 2D input mode.  That's very much error-prone.  As is, you have two different "r" symbols in your worksheet.  They look the same but they are not the same.  Because of that if you try
simplify(U^+ . C . U - M) assuming positive;
you don't get zero.

To see where the problem lies, convert your worksheet to 1D input.  One way of doing this is through the Maple menus:
Edit -> Select All
Format -> Convert To -> 1-D Math Input

After you do that, look at your definition of the matrix M to see the problem.

To avoid such problems, consider making 1-D input the default for your worksheets.  Look at how acer and vv do it.

That said, I am puzzled why vv's calculations do not capture your solution U.  This requires a closer look at how Maple solves systems.

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