@bunmipopoola The manuals found these days in Maple itself are not bad at all.
In Maple look under the menu item Help/Manuals, Resources, and more/Manuals.

The manuals can also be downloaded from
http://www.maplesoft.com/documentation_center/

It used to be possible to subscribe to a post or question. This was useful when you didn't answer or comment yourself, but still wanted to follow the discussion. That feature seems left out.

Well, maybe it is just put under Comments feed. Haven't tried that.

@traiannp It appears to me that L1 appears in all the equations. Also if you remove the first 4 equations, then you still have 9 unknowns, viz.
{L1d, L1q, Ladssec, Laqssec, Lfd, Ll, R1d, R1q, Rfd}

I deleted my initial response as I realized that Carl had already said the same thing.

@bunmipopoola You can do like this
PrF; #Just recalling what PrF is
odeplot(sol,[t,rhs(PrF)],0..0.00431,labels=[t,Pr]);
#Analogously
odeplot(sol,[t,rhs(OF)],0..0.00431,labels=[t,O]);
#To plot D(P) we need to express it in terms of P, F, and T.
#Notice that in sys2 we have D(P) expressed in terms of D(T).
sys2;
#Thus we need to isolate the derivatives:
sys3:=solve(sys2,{D(F)(t),D(P)(t),D(T)(t)});
#(We might as well have done that earlier.)
#Thus D(P)(t) is given by
DP:=eval(D(P)(t),sys3);
odeplot(sol,[t,DP],0..0.00431,labels=[t,D(P)]);

Notice that double brackets in odeplot are not necessary if you have only one pair as we have.
If you have several, then you need them as in
odeplot(sol,[[t,log10(P(t))],[t,log10(T(t))]],0..0.00431,labels=[t,"log(P), log(T)"]);

@danlun If you set p to be what you have above (where multiplication signs are missing), you can correct that by using the following device:

indets(p,function);
nops(%); #6 multiplication signs are missing
q:=evalindets(p,function,s->op(0,s)*op(1,s)); #Correcting that
indets(q,function); #Output {}
factor(q);
nops(%);
type(q,`*`);

@Carl Love Here is a version which doesn't need unevaluation quotes, but is requires the sequence of variable names surrounded by a pair of double quotes:

restart;
a:=6: b:=7: c:=8:
p:=proc(s::string) unassign(parse(s)) end proc;
p("a,b,c");
a,b,c;
%;


One could even overload the existing unassign:

Unassign:=overload([
                    proc(s::string) option overload; unassign(parse(s)) end proc,
                    subsop(3=overload,eval(unassign))]
):
a:=6: b:=7: c:=8:
Unassign("a,b,c");
a,b,c;
%;
a:=6: b:=7: c:=8:
Unassign('a','b','c');
a,b,c;
%;

@Carl Love Would it be possible in Maple to make unassign not evaluate its arguments, so that uevaluation quotes were not necessary?
If you restrict unassign to a fixed number of arguments it is easy enough by just using the existing unassign:
p:=proc(x::evaln) unassign(x) end proc;
a:=7:
p(a);
a;
An attempt to extend this to a sequence of arguments using the seq-modifier:
q:=proc(x::seq(evaln)) unassign(x) end proc;
Error, `evaln` cannot be used within `seq` modifier for parameter `x` in procedure q

@acer My guess is that there is no (significant) difference between a Danish keybord and a Norwegian.
Ctrl+Alt+=  works in Maple 17.02 on a Danish keyboard.

@wzelik

Using that the matrix A is symmetric and has an LU-decomposition with no switching of rows, i.e. A = L.U, A can also be written as A = L.D1.L^T, where D1 is the diagonal matrix having the diagonal of U as its diagonal.
A will be negative definite iff all diagonal elements of U are negative. Since all your constants are positive all the elements but the last one U[n,n] are always negative. 
The last element U[n,n] is
((n-1)*hext*hint*`δx`+hext*lambda-hint*lambda)/(`δx`*((n-1)*hint*`δx`+lambda)*rho*cp);

Since `δx` = L/(n-1) this can be written

(L*hext*hint+hext*lambda-hint*lambda)*(n-1)/(L*(L*hint+lambda)*rho*cp)

and this is negative iff

L*hext*hint+hext*lambda-hint*lambda < 0.

Notice that this inequality does not contain n, thus it doesn't help to increase n.
The inequality can be rewritten as
L < lambda*(1/hext-1/hint)
For your values of the constants L, lambda, hext, and hint this inequality is not satisfied.

This part
(diff(v(r,t),r)-v(r,t)/r+beta*diff(diff(v(r,t),r),t)
- 1/r * beta*diff(v(r,t),t)
-Re*(1+lambda)*exp(1-p(r,t))*diff(v(r,t),t)-(1+lambda)*exp(1-p(r,t))*g2
starts with a parenthesis '(' , but its partner ')' is missing.
Furthermore, Re has a multiplication sign following it. It should just be Re(....).
Your IBC has sint, it should be sin(t).

@wzelik I get an error 404. File or directory not found.

@Carl Love My guess is that the intended meaning was plotting z as a function of x which is accomplished as Adri van der Meer is describing or as I commented to the asker's question.
But only the asker can tell us, what he meant. Clearly these different interpretations of the intended meaning are quite different.

It is not clear to me what you want to do: What does "after indicate to extract u(t) etc. " mean?

Change the last line to
odeplot(sol,[x(t),z(t)],0..10,numpoints=200,color=blue,legend="z",scaling=constrained);

@bunmipopoola From the picture you posted it appears that the maximal time considered is 3.1 ms, i.e. 3.1*10^(-3) s. Why do you use .0000031? Is the unit of time not seconds?
Are you sure that you have the correct equation eq1? Is the sign simply wrong?
It follows from eq1 that D(F)(t) > 0 for all time t. Thus F is increasing. If F is the amount of fuel, then that is clearly wrong, since you are not gathering wood for your wood burning stove.