Preben Alsholm

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These are replies submitted by Preben Alsholm

You must have introduced a floating point number (i.e. a decimal number) at some stage. After that things like you describe can happen.
To point out where it happened we need to see the code. You could upload a worksheet.

The site was down for maintenance a day or so ago, but it surely didn't last for days.

Do you need to refresh the page or clear the cache in your browser?

I have been able to access the page each day for months with the exception mentioned. Needless to say, I'm not looking at it at all times though.

@kuker2008 You could use infnorm from the numapprox package. It finds the infinity norm of the expression, i.e. the supremum (or maximum) of the absolute value of the input on the given interval.
Thus there is no difference between these two:
 

numapprox:-infnorm(abs(b-a),x=-1..1);
numapprox:-infnorm(b-a,x=-1..1);

 

@Markiyan Hirnyk So Markiyan, if that is right, then how do you explain the difference between the very different results of

int(int(F,x=-infinity..infinity),y=-infinity..infinity);

and

int(F,[x=-infinity..infinity,y=-infinity..infinity]); 

where F is given by

f:=x->1/sqrt(2*Pi)*exp(-x^2/2);
F:=convert(f(x)*f(y)*x*x*abs(x+y),piecewise,x);

And is the note I quote from the help page for int in my answer false in its statement about a difference?

##### But I notice that my argument for VectorCalculus:-int using the iterated form was faulty.
Consider this:
 

restart;
int(phi(x,y),[x=-infinity..infinity,y=-infinity..infinity]);
lprint(%);

The result of lprint is the iterated form. I take that as an indication that after the "assumptions" job is done the integral is computed via an iterated integral.
That is in full agreement with the help page note.

First of all you need to use a midpoint method as the error message says, when you execute the dsolve command.
If you solve for the highest derivatives there is a denominator in the ode for H of the form eta*(....). So that will cause problems at eta=0.
Thus change that dsolve command to
dsys1 := {Eq1, Eq2, Eq3, Eq4, IC1, IC2}; dsol1 := dsolve(dsys1, numeric, output = listprocedure, method=bvp[midrich]);
or use the other midpoint method bvp[middefer].
I tried with no luck whatsoever.
Do you have any reason to believe that your system has a solution at all?

@Markiyan Hirnyk It appears that VectorCalculus:-int with the Rectangle input returns an iterated integral.
Let phi be unassigned:

VectorCalculus:-int(phi(x,y), [y, x] = Rectangle(-infinity .. infinity, -infinity .. infinity));
lprint(%);
int(int(phi(x, y), y = -infinity .. infinity), x = -infinity .. infinity)

I think this explains why VectorCalculus:-int with Rectangle works correctly for the function g. In my answer I used iterated integrals instead of double range integrals to get the correct answer.

I simply don't understand what you are talking about. Could you be more specific? Has it got to do with Maple?

@Christopher2222 Quite right; you can't.

@shakuntala In the sense I described above the ode involving only f (and with the boundary conditions with theta left out) does indeed have two solutions (at least). I only looked at the first value in B, i.e. B[1] = 1.
The other solution for f is negative and decreasing. I'm not familiar with the physics of this, but this may be an unwanted solution for those reasons alone.
If not, then there is a mathematical (or at least a numerical) problem for the total system with that solution: The numerator in the theta equation when solved for diff(theta(eta),eta) is 7*f(eta)+30:

That is bad news because 7*f(eta)+30 becomes zero in the interval 0..1.1 as seen in this graph:

@shakuntala Do you have just one solution? Why talk about dual solutions if you don't even have one?

## But to clear up matters: Do we mean the same by the expression dual solution?

In my use of the term "dual solution" it means that a given bvp problem has two solutions, as e.g. the simple algebraic equation x^2 = 4 has two solutions (-2 and 2).

@shakuntala Plotting two (or more) solutions is no problem at all. In fact it seems that your worksheet has the code for it.
Your problem (as it has been presented so far) is to find any solution at all.

Have you done that?

You wanted to impose two conditions on theta and three on f, yet the total order of the system is 3+1=4, as I pointed out. You must leave out one of your five conditions!
You can e.g. leave out theta(1.1) = 0 or you can leave out D(f)(1.1) = 0 (just examples, but in both cases dsolve gives results without problem.

@vv I tried an animation of a disk with a "background" consisting of your 4 plots (p1,p2,p3,p4) and a texplot.
It just confirms what has already been described by you.
 

restart;
with(plots):with(plottools):
p1:=plot( x^3,x=-1..1,thickness=20,color=red):
p2:=plot(-x^3,x=-1..1,thickness=20,color=blue):
p3:=display(rectangle([0.5, 1],[0.75,-1],color=green)):
p4:=display(rectangle([-1,-0.1],[1,0.1],color=yellow)):
#######################################################
pt:=textplot([0,0.5,"Happy New Year", 'font'=["times","roman",20]]):
Q:=a->display(disk([0,0],a,color="NavajoWhite"),axes=none):
plots:-animate(Q,[a],a=0..1.25,background=[p1,p2,p3,p4,pt],paraminfo=false);

You don't explain what plrPt is.
In order to transform the 2D animation to a 3D animation you would have to transform each frame.

@Christopher2222 To combine your animated field with the fixed field you can produce both and display them together:

with(DETools):
LVS := [diff(x(t),t)=x(t)*(1-y(t)),diff(y(t),t)=.3*y(t)*(x(t)-1)];
DEplot( LVS, [x(t),y(t)], t=0..12, x=0..3, y=0..2, arrows=comet,
  animatefield=true,numframes=30,dirfield=50); p1:=%:
DEplot( LVS, [x(t),y(t)], t=0..12, x=0..3, y=0..2,color=gray); p2:=%:
plots:-display(p1,p2);

But here you have just one ode. In Kitonum's revised example there were 90 odes, one for each frame.

@Christopher2222 The field does indeed move in Kitonums animation (New version).

If you add animatefield you get an animation having each frame an animation. Just like having a movie whose individual frames are movies themselves. How do you expect that to play?
The movie could be it all of Hitchcock's movies played in sequence chronologically according to release date?

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