The Maple share library was replaced (as of Maple 6) by the
on-line Maple Application Center. However, I can't find
IntSolve there. In any case, I think IntSolve could
handle a single integral equation (of certain types) but
not a system of integral equations.

If you give us an example of the type of system you're
trying to solve, we might have some ideas of how to solve it.

It evokes mixed feelings in me. Certainly it was full of rather useful code. However, when Maplesoft "took it over" to include in the product, we couldn't just include it as-is, we had to make sure that a minimum level of quality and consistency was met by all packages.

And that turned out to be a huge amount of work. I led that effort, and it involved work by a lot of people over several months to get all the help pages cleaned up, all the code to 'work' (it had bit-rotted a lot more than we'd expected). It does seem like Maple users did appreciate the effort though.

The Application Center was an interesting political and technological solution to the maintenance of user-shared code. If the code shipped with Maple, then Maplesoft had to ensure a minimum quality level. But if it was on the App Center, then it was up to the authors of the packages to ensure that they did not bit-rot. [The fact is that a lot of them have bit-rotted, the grand claims to backwards compatibility by Maplesoft notwithstanding.] It was also nice because it was the first foray of Maplesoft into using the web constructively, and it really was a success. It was both useful for users, gave some prestige to some avid Maple fans&developers, and was a useful marketing too. It also meant that this app center could be updated continuously instead of having to wait through release cycles.

In another 5 years, maybe I'll be able to tell the full story of the decline and rebirth of the app center. Hint: a lot of the app center consists of .mws files.

Dear Comrades,
Sorry I've been offline for a while. The equations I'm looking at aren't really integral equations in the standard sense. They involve integrals, but the integrals can be done (as opposed to standard integral equations where the mystery function is under one of the integrals).

So, I have two equations involving four quantities; alpha, beta, Omega and delta, and I want to plot the surface of beta as a function of Omega and delta (eliminating alpha leaves one equation in three variables, which defines a surface). Without having to eliminate alpha myself by hand, I want to give maple both equations and have it plot those points which satisfy both equations. My equations are complicated, so I tried a simpler example to get the syntax right. I tried to plot the circle where a cylinder cuts through a sphere. Unfortunately, using plot(eq1, eq2) command it plots both surfaces (cylinder and sphere), not just the intersection. So, do any of you know a way to give maple two equations and have it plot the points which satisfy both equations?

intersectplot won't help, though, with two equations in four variables. If the equations involve polynomials, you can
take the resultant with respect to the variable you want to eliminate. For example, consider the equations

> e1:= x^2 + y^2 + z^2 + w^2 - 1 =0 ;
> e2:= x*y^2 + y*z^2 + z*w^2 + w*x^2 = 0; 
> Res := resultant(lhs(e1),lhs(e2),w);
> plots[implicitplot3d](Res,x=-1..1,y=-1..1,z=-1..1,
    grid=[30,30,30],axes=box,style=patchnogrid,
    scaling=constrained);

If they're not polynomial (and can't be recast
in terms of polynomials), it's not so easy.

Actually there is another possibility that might make sense:
build up a sequence of cross-sections using intersectplot.
For example, using the two equations I used before:

> with(plots):
  e1:= x^2 + y^2 + z^2 + w^2 - 1 =0 ;
  e2:= x*y^2 + y*z^2 + z*w^2 + w*x^2 = 0; 
  display([seq(intersectplot(e1,e2,x=-1..1,y=-1..1,z=-1..1,
    colour=COLOR(HUE,(w+1)/2)),w=[seq(0.1*j,j=-10..10)])]);

Andrew writes:

In other words, is it possible to give an implicit finction of three variables, and have it plot two of the variables on two axes and some function of the third variable on the third axis. For the simple sphere case, imagine giving the equation x^2+y^2+z^2-1=0 and asking it to plot not (x,y,z) but rather (x,y,f(z)) where f is some function of z you specify. Does anybody know how to do this?

You can use plots[implicitplot3d] to generate a plot p of the (x,y,z) values satisfying your equation. Then, use the plottools[transform] command to transform the points to (x,y,f(z)):
plottools[transform]((x,y,z)->[x,y,f(z)])(p)

Paulina Chin
Maplesoft

Andrew, could you send the worksheet to me? It'll be easier to determine the problem if I can see the exact input you're giving to Maple. Thanks.

Paulina Chin
Maplesoft

It seems that all is well now.
One last thing, though. I've calculated the values of (alpha,beta,Omega) from one equation using plot. I then used plottransform to change these into (Omega,delta,alpha) in the one case and (Omega,delta,beta) in the other case as you suggested.

plotB=implicitplot3d(f4(alpha,beta,Omega)=0,alpha=0..1,beta=0..1,Omega=0..5,grid=[50,50,50],labels=[alpha,beta,Omega]):

plotBTrans:=transform((alpha,beta,Omega)->[Omega,2*(-beta*Omega/sqrt(2))+Omega/sqrt(2)*(alpha^2+(3*(abs(beta*Omega))^(3/2)*(IntegralA(1,0,infinity))/(4*2^(3/4))))*(1/beta),alpha])(plotB);

plotBTrans1 := (transform(proc (alpha, beta, Omega) options operator, arrow; [Omega, -2*beta*Omega/sqrt(2)+Omega*(alpha^2+(3/8)*abs(beta*Omega)^(3/2)*IntegralA(1, 0, infinity)*2^(1/4))/(sqrt(2)*beta), beta] end proc))(plotB)

This seems to be all working fine, but I wonder if it would be more efficient to take the initial values of (alpha,beta,Omega) and use plottransform only once to get a bigger array of (Omega,delta,alpha,beta) and then tell it to plot, in the one case columns 1,2 & 3 and in the other case columns 1,2 & 4. This means that all the values of delta would only be calculated once rather than twice. Is this possible?

Thanks again, everybody,
Andrew

Thanks everybody. That seems to all be working. I have another problem with a related system, but I think it's different enough to warrent starting a new thread.
Regards,
Drew