@radaar Let me know if you understand what I wrote or if you have any further questions. Using procedures as parameters in other procedures is a tricky issue in any language. Maple makes it much easier than most.

@Rouben Rostamian  

Regarding the smoothness of the IC: Yes, of course. I didn't intend for readers to infer that my technique was generally applicable. Instead, I was reacting to what I perceived to be Tom's implying that there was some choice involved for the new IC.

I'm glad that you noticed that my code computes D[1,1](u)(x,0) from the u(x,0) given in the original bounds.
 

Regarding the original problem being flawed: I understood your previous comments about this to mean that the system couldn't represent any practical physical situation. But are you indeed saying further that it's mathematically flawed: that it has no mathematical solution, however impractical?
 

Regarding union=: This is a new feature of Maple 2019 that's documented on the help page ?assignment. Let's let OP represent almost any binary infix operator (such as +, union, and, etc.). Then the expression

X OP= Y

is equivalent to 

X:= X OP  Y

I'm just guessing, but see if the VariationalCalculus package does what you want. See help page ?VariationalCalculus

@tomleslie Isn't D[2](u)(x,0) uniquely determined by plugging t=0 into the original PDE and applying the original initial conditions? Doing so, I get D[2](u)(x,0) = x^2*(x-1), like this:

restart:

#Original partial-differential-integro equation (PDIE) problem:
f:= (x^3+t^2*x^2-t^2*x+4*t*x-2*t+1)*exp(x*t)-x+1:
pdie:= diff(u(x,t),t) + diff(u(x,t),x,x) + u(x,t) + int(u(x,tau), tau= 0..t) = f;
bounds:= {u(0,t)=0, u(1,t)=0, u(x,0)=x*(x-1)}:

#Generate IC for D[2](u)(x,0):
bounds union= {simplify@isolate}(
   convert(value(eval(convert(eval(pdie, {t= 0}), Diff), bounds)), D), 
   D[2](u)(x,0)
);

#Convert original PDIE to a true PDE (i.e., without integrals): 
pde:= diff(pdie, t);

#Solve PDE system:
psol:= pdsolve(pde, bounds, numeric):
psol:-plot3d(u(x,t), x= 0..1, t= 0..0.167); 

@Joe Riel As it's currently coded, if your &@ associates, then it must associate to the right. Maple 19 has introduced right-associative neutral operators. They begin with && and otherwise are completely the same as those begining with single &.

@Joe Riel My choice for a name for a built-in identity function is `1`.

@Joe Riel Perhaps D[1$0] would be more clear. Personally, I'll use D[].

@Christian Wolinski Your effort to make it into a worksheet is appreciated.

@abbastalebi Word format is not very helpful. Plaintext or worksheet is best. Anyway, I think that I got all the necesssary code from Christian's worksheet transcription of your Question. 

So, what does parameter beta represent? Is it meant to be input to or output from the system?

Currently, the total differential order of the system is 3, and there's 1 parameter, beta. That's 4, yet you have 6 initial and boundary conditions. Those two things (the 4 and the 6) need to be equal. You could add two more output parameters.

What is your version of Maple?

@Christian Wolinski The ubiquitous line 

print(`output redirected...`); #input placeholder

is garbage automatically added by some method of copy-and-pasting from 2D Input to plaintext. It's just clutter, and there's no reason to retain it in the worksheet.

Your worksheet above is not an Answer to this Question. It's a small help, but it's not an Answer or even an attempted Answer.

@Joe Riel In this context, I'd like to use D[ ], not 1, as the left-operand identity. Anyway, vote up. I was going to post a solution using LinearAlgebra:-Generic, but I think that yours is adequate.

It's always bothered me to need to waste the time needed to evaluate Maple-level procedures x-> x or ()-> args when it could be a builtin. Even D is Maple level.

You need to post your code. There are hundreds of unrelated things that could cause that error.

Please post your input code in a form such that I can easily copy-and-paste it into a worksheet. The output of the code is just clutter in this format. You can also simply attach a worksheet.

You have a three-point problem, with conditions at -10, 0, and 10. It'll need to be solved stepwise.

Would you please define "eigenvalue" in this context? I'm familiar with eigenvalues of square matrices and and of boundary value problems, but not of solitary algebraic equations.

Regarding your Digits:= 5: It causes no problem here, because this is just presentation rather than numeric computation. But, if you're habituated to it, it's going to cause problems for you eventually: It's too low. This number affects the precision of almost all internal floating-point computations, not just the results that are displayed. It's generally not enough significant digits to be safe.

Someone will probably mention at this point that there's a setting for the number of displayed digits, which doesn't affect the internal computations. However, the implementation of this feature is so cruddy that I won't mention it further.