@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);

@Mazaya Jamil For each concrete value of phi you can use fsolve:

f9:=RootOf(126*exp(I*phi)*_Z^5+(15-351*exp(I*phi))*_Z^4+(107+740*exp(I*phi))*_Z^3+(354-1110*exp(I*phi))*_Z^2+(1056*exp(I*phi)+624)*_Z+480-480*exp(I*phi));
p:=op(f9); #The polynomial in _Z
fsolve(eval(p,phi=.12345),_Z); #Example
#Here is a procedure which computes the roots for each given value of phi:
Q:=proc(phi1) if not phi1::numeric then return 'procname(_passed)' end if;
   fsolve(eval(p,phi=phi1),_Z)
end proc;
Q(8);
Q(.12345);
Q(x);
#It returns unevaluated if it receives non numeric input.
#Notice that in your case you don't need the extra argument 'complex' because the coefficients of p are imaginary.
Here the extra argument is necessary (if you want imaginary roots too):
fsolve(z^2+z+1,z,complex);
############
Plot of all 5 roots as phi varies from 0 to 2*Pi:
For efficiency add 'option remember' to Q:
Q:=proc(phi1) option remember; if not phi1::numeric then return 'procname(_passed)' end if;
   fsolve(eval(p,phi=phi1),_Z)
end proc;
plots:-complexplot([seq(Q(phi)[i],i=1..5)],phi=0..2*Pi,style=point);
#And an animation of the same plot:
plots:-animate(plots:-complexplot,[[seq(Q(phi)[i],i=1..5)],phi=0..pp,style=point,numpoints=round(pp*5+1)],pp=0..2*Pi);

Try leaving out the initial condition:
pdsolve(sys[1]);

If that is correct it is no wonder that the initial value problem cannot be solved.

Not knowing very much about MatLab I was wondering if Maple's 'break' is what you need:

?break

In the help page you find the example

L := [1, 2, "abc", "a", 7.0, infinity]:
for x in L do
    if type(x, 'string') then
        print(x);
        break;
    end if;
end do;