@lolly Do the links you just posted work when you try yourself? I still get "Bad Request".

None of your links work for me. I get "Bad Request".
In the code posted (as an image, not a good idea) you definitely should not have assignment (:=) but equality (=).

@siamak taghavi You could do like this:

VR22:=0.1178*diff(phi(x),x,x,x,x)-0.2167*diff(phi(x),x,x)+0.0156*diff(psi(x),x,x)+0.2852*phi(x)+0.0804*psi(x);
VS22:=0.3668*diff(psi(x),x,x)-0.0156*diff(phi(x),x,x)-0.8043*psi(x)-0.80400*phi(x);
bok:=evalf(dsolve({VR22=0,VS22=0}));

PHI,PSI:=op(subs(bok,[phi(x),psi(x)]));
Eqs:={eval(PHI,x=.83)=1,eval(diff(PHI,x),x=0.83)=0,eval(PHI,x=-.83)=1,eval(diff(PHI,x),x=-0.83)=0,
eval(PSI,x=.83)=1,eval(PSI,x=-0.83)=1};
C:=fsolve(Eqs,indets(%,name));
eval(bok,C);
SOL:=fnormal(evalc(%));
plot(rhs~(SOL),x=-.83..0.83);

@Markiyan Hirnyk There are (at least) 2 solutions for Q. For N=22 fsolve(Q,0) gave the negative one.
With fsolve(Q,0.002) we get the positive one.
I added a remark to my answer above.

Runge-Kutta methods are used for solving initial value problems. You have a boundary value problem.
So please elaborate.

@Preben Alsholm I tried the continuation I mentioned above, but had only success with A1=4.5.
So just a slight improvement from A=0.43 obtained without continuation.
Then I tried using the result obtained for A=0.45 by continuation from A=0.2 as an aproximate solution for the problem with A=0.5. It didn't work. But I was unpleasantly surprised to find that the continuation solution didn't even work when A=0.45! That is: I'm handing dsolve the solution, but yet it cannot find the solution with that as a start!!

sys:=eval({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1},A=0.2*(1-c)+c*0.45);
res:=dsolve(sys, [f(eta), F(eta), G(eta), H(eta), theta(eta), theta1(eta)], numeric, output = listprocedure,continuation=c); #Works
plots:-odeplot(res,[[eta,diff(f(eta),eta)],[eta,F(eta)]],0..blt,linestyle=3);
#Now dsolve is given sys with A=0.45 the approximate solution (res) found above for A = 0.45 by continuation from A=0.2, yet we get the usual error:
res2:=dsolve(eval({Eq1, Eq2, Eq3, Eq4,Eq5,Eq6,bcs1}, A = 0.45),numeric, output = listprocedure,approxsoln=res); #Does not work!!

 

spline is deprecated, use Spline from the CurveFitting package.
Maybe you could use CurveFitting:-ArrayInterpolation?

@cloz54 Cutting and pasting your module definition into my version of Notepad and saving it as a textfile. Reading it into Maple 18 I got

                     "Est?n en perspectiva"
                      "Spain means Espa?"
Actually the question marks appeared as squares.

I also tried executing the moduledefinition in Maple, saving it using
save TestModule,"F:/MapleDiverse/MaplePrimes/testmod.mpl";

and then reading it into a fresh worksheet:

read("F:/MapleDiverse/MaplePrimes/testmod.mpl");

Then I got:
with(TestModule):
test();
                     "Están en perspectiva"
                      "Spain means España"
                               0

@Axel Vogt I tried the following. All had the same result on my computer:

print("Están en perspectiva");
p:=proc() print("Están en perspectiva") end proc;
p();
m:=module() export p; p:=proc() print("Están en perspectiva") end proc; end module;
m:-p();
Q:=proc() module() export p; p:=proc() print("Están en perspectiva") end proc; end module end proc;
m:=Q();
m:-p();

I tried various interpretations of what you are saying, but have not encountered any problem. Could you give us the full code necessary for exhibiting this behavior?

@love maths You could try this:

This illustrates that the elements affected by Average are the elements inside a square:
plots:-implicitplot(abs(x-N/2)+abs(y-N/2)-N/4=0,x=0..N,y=0..N,rangeasview,gridrefine=2);
#That square has vertices (N/4,N/2),(N/2,N/4),(N/2,3*N/4),(3*N/4,N/2), which is confirmed by F which returns 0 at those points:
F(N/4,N/2),F(N/2,N/4),F(N/2,3*N/4),F(3*N/4,N/2);
#A little experimentation:
A:=curry(Average,F);
seq((A@@i)(f0),i=0..5);
#Guess: The sequence converges to a matrix of ones.
#This guess is confirmed by (but certainly not proved by)
f:=f0: f1:=Matrix(N,fill=1):
for i from 1 to 500 while LinearAlgebra:-Norm(f-f1)>1e-12 do f:=A(f) end do:
i-1;
evalf[12](f);

@mablecat I repeated the commands and made sure that I used Maple 17. Still no problem:

interface(version);
Standard Worksheet Interface, Maple 17.02, Windows 7, September 5 2013 Build ID 872941

@Carl Love I was surprised that the square brackets you have after D in the intital conditions actually work.
However, I corrected that and used your system:

restart;
sys:={diff(a[1](t),t$2) = -a[1](t), diff(a[2](t),t$2) = -a[2](t),a[1](0)=1, a[2](0)=0, D(a[1])(0)=0, D(a[2])(0)=1};
#Analytical solution
dsolve(sys);
#Info on events:
?dsolve,events
#Example of using events. These two events will trigger when a[i](t) = 0 i=1..2. The discrete variables b1 and b2 will be taking the values for t at those times.
sysE:=sys union {b1(0)=0,b2(0)=0};
SolE:=dsolve(sysE,numeric,discrete_variables=[b1(t)::float,b2(t)::float],events=[[a[1](t),b1(t)=t],[a[2](t),b2(t)=t]]);
SolE(6);
plots:-odeplot(SolE,[[t,b1(t)],[t,b2(t)]],0..6,thickness=2);

I shall try to see if I can upload anything. So here is an extremely short mw-file.

Well, I cannot do it either. MaplePrimes has temporary problems.

@sharena2 Well, you do need one more condition. Not necessarily the one I used.
If you have tried the same problem earlier and it was OK then you ought to show us that "same" problem:
It couldn't be the same, could it?