Because implicitplot allows the range for the second variable to depend on the first, but not vice versa.

> display(plottools[reflect](%, [0,0],[1,1]),labels=[r,x]);

> display(plottools[reflect](%, [0,0],[1,1]),labels=[r,x]);

Since you wrote: dx/dt = rx - sin(x), I guess you're looking for the equilibrium solutions of this differential equation, which (for any r) are the solutions x of r*x - sin(x) = 0.  You also may want to distinguish between stable and unstable equilibria: the equilibrium solution r of the differential equation dx/dt = f(x) is stable if f'(x) < 0, and unstable if f'(x) > 0.  In the following, the stable equilibria are green and the unstable ones are red.
 

> f := (r, x) -> r*x - sin(x);
   with(plots):
   display([implicitplot(f(r,x),x = -10 .. 10, r=-2 .. cos(x), colour=green, axes=box, gridrefine=3, crossingrefine=3),
            implicitplot(f(r,x),x = -10 .. 10, r=cos(x) .. 2,  colour=red,   axes=box, gridrefine=3, crossingrefine=3)]); 

Since you wrote: dx/dt = rx - sin(x), I guess you're looking for the equilibrium solutions of this differential equation, which (for any r) are the solutions x of r*x - sin(x) = 0.  You also may want to distinguish between stable and unstable equilibria: the equilibrium solution r of the differential equation dx/dt = f(x) is stable if f'(x) < 0, and unstable if f'(x) > 0.  In the following, the stable equilibria are green and the unstable ones are red.
 

> f := (r, x) -> r*x - sin(x);
   with(plots):
   display([implicitplot(f(r,x),x = -10 .. 10, r=-2 .. cos(x), colour=green, axes=box, gridrefine=3, crossingrefine=3),
            implicitplot(f(r,x),x = -10 .. 10, r=cos(x) .. 2,  colour=red,   axes=box, gridrefine=3, crossingrefine=3)]); 

What do you mean by "the vector field for" a function?  And what is the parameter for the animation?

What do you mean by "the vector field for" a function?  And what is the parameter for the animation?

By singularity I didn't mean it necessarily goes to infinity.  "Collapse of the wave" or formation of a shock would be enough to defeat the numerical solver, I think.

By singularity I didn't mean it necessarily goes to infinity.  "Collapse of the wave" or formation of a shock would be enough to defeat the numerical solver, I think.

Something like this?

> M:=pdsolve(diff(u(x,t),t)=diff(u(x,t),x,x),
     {u(x,0)=10, u(0,t)=0, u(1,t)=0},numeric, timestep=0.01);

> M:-plot3d(u(x,t),x=0..1,t=0..0.5,axes=box,shading=zhue,
     orientation=[90,180]);

Something like this?

> M:=pdsolve(diff(u(x,t),t)=diff(u(x,t),x,x),
     {u(x,0)=10, u(0,t)=0, u(1,t)=0},numeric, timestep=0.01);

> M:-plot3d(u(x,t),x=0..1,t=0..0.5,axes=box,shading=zhue,
     orientation=[90,180]);

I tried this:

> de:= diff(diff(y(x),x),x) = .1239104454e-3*y(x)^4-938417.2491
     +428020.2082*(.6+.874e-1*(y(x)-295)^(1/6))^2*(y(x)-295);

> ude:= simplify(eval(de, y(x) = 295 + u(x)^6)) assuming u(x) > 0;

> dsolve([ude, u(0) = (328-295)^(1/6), D(u)(0.5)=0], numeric);

Error, (in dsolve/numeric/bvp) Newton iteration is not converging
 

But this worked:

> dsolve([ude, u(0)=(328-295)^(1/6), D(u)(0.5)=0],numeric,
     initmesh=1000);

  proc(x_bvp) ... end proc
 

It seems there is a "boundary layer" with very rapid change near x=0, and y is very close to 295 everywhere else.
 

> plots[odeplot](%,[x,295+u(x)^6],x=0..0.05, labels=[x,y(x)]);

With time[real] you get the time as you would measure it on a clock.  This may be influenced by whatever other things your computer is doing besides running Maple. 
CPU time, as measured by time without the [real], may be a better indication of the performance of your procedure.

With time[real] you get the time as you would measure it on a clock.  This may be influenced by whatever other things your computer is doing besides running Maple. 
CPU time, as measured by time without the [real], may be a better indication of the performance of your procedure.

In particular I'm thinking of problems introduced by implicit multiplication:
x y means x*y, but xy is a single name; x(y+z) is the function x applied to y+z, while
sin (x) is sin*x.  See e.g.
www.mapleprimes.com/forum/how-do-i-get-rid-of-unable-to-match-delimiters-error-message (starting with Doug Meade's first comment) and
www.mapleprimes.com/forum/solvingvariable.