@mehdi jafari It used to be that if you answered a question or if you made a comment you were automatically signed up for email notification, i.e. subscription by default.

@samiyare Looking at the results obtained for NBT=1 (actually I think I looked at NBT=0.7) I saw that basically the graphs of phi, u, and T look like parabolas. So that is how I got those. You don't have to include the derivatives, as Maple can compute these, but I did initially since I was experimenting with some code of my own, which requires that the system be rewritten as a first order system. If the derivatives are included I assume those are used even if they are not really derivatives of the given approximations. For U and UF I used simple linear approximations.

About the case:

Phiavg:=0.1:
EPSILONE:=0.5:
Nr:=30:
lambda:=0.1;
Ha:=5;
NBT=0.2

I tried with those values, and with nothing else changed. I got no problem for NBT=0.2. In fact no problem from NBT=1 down to NBT=0.08. The first problem occurred at NBT=0.07.

I notice that the following device works:
latex(subs(erf=Erf,r));
and for the other example:
latex(subs(arctanh=Artanh,r));

The error comes up also with this simple version:
latex(erf(2*x));

but not with
latex(erf(x));
Same with arctanh.

The problem may be in `latex/print`, which is different from the procedure with the same name in Maple 17, where there is no such problem.
You may try
`latex/erf`(2*x);
p:=`latex/print`(erf(2*x)); #In Maple 17 p is a symbol, in Maple 18 a sequence
whattype~([p]); #Notice the type `*`
Compare with
p:=`latex/print`(sin(2*x)); #In Maple 17 and 18 p is a sequence
whattype~([p]);

I submitted an SCR about latex(erf(2*x)).

@Muhammad Ali The following two constructions both work. One uses piecewise, the other Heaviside:

pde:=diff(f(t,x),t)=diff(f(t,x),x,x)*piecewise(x>=1,1,0)+piecewise(x<1,1,0)*f(t,x);
#The alternative:
pde:=diff(f(t,x),t)=diff(f(t,x),x,x)*Heaviside(x-1)+Heaviside(1-x)*f(t,x);
res:=pdsolve(pde,{f(0,x)=piecewise(x>=1,(x-1)^2*x,0),f(t,0)=0,f(t,2)=2},numeric);
res:-plot3d(t=0..1);
p:=res:-value();
p(1,.5);
p(1,1.5);

Notice that f(t,x) does not appear inside piecewise or Heaviside.

@Muhammad Ali I'm not sure that I understand your example, because if f(t,x) is required to be 0 for x <= ln(b/K) then f(t,x) is known in that region, so that is that! In the remaning region x<ln(b/K) you must solve
diff(f(t,x),t)=diff(f(t,x),x,x) subject to initial and boundary conditions. The boundary condition at x=ln(b/K) ought to be chosen to be zero so that at least you have continuity.
For ln(b/K)=1 here is a simple example:
pde:=diff(f(t,x),t)=diff(f(t,x),x,x);
res:=pdsolve(pde,{f(0,x)=(x-1)^2*x,f(t,1)=0,f(t,2)=2},numeric);
res:-plot3d(t=0..1); p1:=%:
plots:-display(p1,plot3d(0,x=0..1,t=0..1));

@samiyare Notice that the results reported by LSSolve corresponds to the smallest values of a[1] and a[2], not to the last ones computed.

@samiyare I tried a different continuation but otherwise using the same loop as before:
N_bt:=cc*NBT+(1-cc)*1;#Works down to NBT=0.37
I also tried without continuation, i.e. removing the option continuation=cc and simply setting
N_bt:=NBT;
That worked down to NBT=0.41.

Your system is linear in x and y, so certainly you can trust solve to have found all solutions of this simple system. What does t have to do with this system?

@abscissa If you have several equilibrium points for a given set of concrete values of the parameters (i.e. in your case A, B, C, D, E, F, G, H, S, T, U, V, W, Y, x) then you may find that defining J as a function of the variables a, b, c, d, e, f, g, h, s, t, u, v may be convenient.
In two steps for clarity:
J1 := VectorCalculus:-Jacobian([eq_1, eq_2, eq_3, eq_4, eq_5, eq_6, eq_7, eq_8, eq_9, eq_10, eq_11, eq_12], [a, b, c, d, e, f, g, h, s, t, u, v]);
J:=unapply(J1,[a, b, c, d, e, f, g, h, s, t, u, v]):
#Now you can use J as a function as in
pt:=seq(1..12);
J(pt);

@pedromneto If there is a way to solve this system by using the laplace transform, I would like to hear about it!

@RemonA What is the capital X appearing in the error message as an argument to sin and cos?

We need more detail than that to give a useful answer.

@roussea18u 
I notice several problems among which are:
1. F is differentiated w.r.t. the variable in the form diff(F,t). This makes sense only if F is an expression in t, not a function. But you assign to F(0) which means that F is supposed to be a function. Also what is F?
Presumably it is globally given and is the right hand side of the ode? Or is it the F appearing on page 10 of the paper in the link. In that case it is a function of two variables?
2. Now if t is time and is used as above, then you cannot assign t to be a vector. Also that assignment is strange. What is t:=vector(i,1) supposed to do? i is unassigned at that point. (Actually you have =, not :=, but then nothing is being assigned to, but I assume that was the intention. There are more cases  of equality signs which should be assignments).
3. You need Newmark:=proc(...) local ... : Notice := and no semicolon.

My suggestion actually is to give us the differential equation including initial conditions. The we can let Maple's dsolve do the problem.

You provide an image. It would be so much easier for us to analyze your code if you gave us the code as text here in MaplePrimes or if you uploaded a worksheet with the code.

A somewhat less trivial example (Volterra-Lotka model):
sys:=diff(x(t),t)=x(t)*(1-y(t)),diff(y(t),t)=-y(t)*(1-x(t));
res:=dsolve({sys,x(0)=1,y(0)=1/2},numeric,output=Array([seq(.1*i,i=0..100)]));
A:=res[2,1];
plot(A[..,2..3],caption="Plot in phase space",labels=[x,y]);
plot([A[..,[1,2]],A[..,[1,3]]],labels=[t,"x,y"]);
ExportMatrix("F:/testMatrix.txt",A);