@Carl Love Maybe there is something to that, however
caption = typeset(a^(b^(c^d))+u^(v^(w^t)))
takes up significantly larger space but causes no problem.

Here is the desired potential function:

V := unapply(piecewise(x < x__1, psi__1(x), x < x__2, psi__2(x), psi__3(x)), x);

If you meant to ask something else, then you need to be more specific.

@student_md Essentially your entire calculation should be enclosed in a procedure.  In the attached worksheet I have made minimal modifications to your worksheet to do that.  It's likely that your calculations can be streamlined and improved significant;ly.  I haven't attempted that.

Download worksheet:   code-ver2.mw

@student_md 

In your worksheet you have all the necessary pieces but some good organization is lacking.

As a first step, instead of specifying J at the top level, see if you can write a procedure that receives J as an argument and produces the desired graphics for that particular J, as in:
frame := proc(J)
     ...
     calculations here
     ...
     return plots:-display([AA, BB]);
end

Next, generate the sequence of frames:
frames := seq(frame(J), J=0..5):

Finally, display the animation:
plots:-display([frames], insequence);

@griffgruff Your statement is clear enough although I don't understand why you asssume that del(u) is normal to the boundary.

That would be the case only when the Dirichlet data is constant.  But you say that it isn't.  If so, then your del(u) is not normal to the boundary, and therefore you cannot calculate the vector n from it.

You need to give that some additional thought.

@griffgruff You say that u is a solution of a PDE and at the same time u represents the domain's boundary. That would be the case only if the boundary is a level curve of u, that is, u(x,y) is specified as a constant on the boundary.

@Kitonum You seem to have misiterpreted the question.  The vectors n and del(u) are unrelated.  The desired answer is  n(x,y)^+ . del_u(x,y), where

n := (x,y) -> < n1(x,y), n2(x,y) >;    # assuming n1*2 + n2^2 = 1
del_u := (x,y) -> < diff(u(x,y),x), diff(u(x,y), y) >;
 

@waseem

Do:

dsol := dsolve(ODEs);

to see what the general solution looks like.  Then you may apply any boundary conditions that you like.

@ivanfanthony You have four equations in the five unknowns, S(t), I(t), E(t), J(t), R(t).  You need to supply one more equation.

Additionally, the symbols I and gamma are reserved for internal use in Maple.  You should use some other symbols instead of those.

@ilpadrino As tomleslie has pointed out, you have a system of seven equations that involve nineteen symbols {b1, b2, ca, cb, cc, jmax, s1, s2, t, t1, t2, t3, t4, t5, t6, t7, tf, v1, v2}.  The best you can hope for is to solve for seven of those symbols.  You haven't said which seven.  Say it without a "..." in your statement.

@Christopher2222 In Maple's Tools menu go to:

Tools -> Options -> Interface

Inspect the item labled "Open worksheet at startup".  Select one of the three possible choices.  Probably you want the "New, blank" choice.

After selecting it, click on "Apply Globally" .

@tomleslie A link to his worksheet is at the very top of his original post.  It's called Пример02.mws.

This looks like a bug to me.  A workaround is to solve the PDE:

diff(u(x, y), x, x)+diff(u(x, y), y, y) = a,

and then substitute Pi for a.

PS: Actually Maple is able solve the more general case where the right-hand side is an arbitrary function f(x,y).

restart;
pde := diff(u(x, y),x,x) + diff(u(x,y),y,y) = f(x,y);
pdsol := pdsolve(pde, u(x,y), series, order = 4);
pdetest(pdsol, pde);

The test result in not zero but that's alright -- what we get is the difference between f(x,y) and its Taylor series expansion. That's what we expect when we are looking at a series solution.

@ecterrab

Dear Edgardo, I certainly did not mean to criticize or belittle your wonderful contributions to Maple in general, and to the pdsolve utility in particular. If I didn't like it, I wouldn't use it, would I?

It seems to me that the offense that I have caused is due to the use of the phrase "not reliable" in the discussion. Admittedly, that phrase when taken in isolation is both unkind and very untrue. But there is context—the question asked was whether one may rely on the uniqueness of solutions returned by pdsolve. My answer was that no, we may not rely on pdsolve for that since it does not return all possible solutions. That's a statement of fact, and I believe that you don't disagree with it.

At any rate, hereby I sincerely apologize to you for any unkind language that I may have used and any offense that I may have caused. Cheers!

@Mariusz Iwaniuk The solution returned by Wolfram is the unique solution of the problem under the two assumptions that we (a) seek solutions in the positive quadrant; and (b) want a solution that tends to zero as we approach infinity.

There are infinitely many solutions to that problem if we remove those assumptions.  For instance::
    u(x,t) = A*sin(c*x)*sinh(c*t)
is a solution for any A and c.