The attached file assigns a value of 1 for all unknown parameters - since I have no idea what values are relevant

It then solves the PDE system, and plots the 3D solution of all independent variables over the ranges t=0..1, x=0..12. The value 12 for the upper limit of the range in x is chosen simply because there are some boundary conditions specified at x=12.

pdeSol.mw

Your remark,

how to obtain the partial derivative of u(x,t) with respect to y

makes no sense at all. Rather obviously, u(x,t) does not depend on y so the partial derivative of u(x,t) with respect to y, is identically zero. Somehow I'm guessing that this is not the answer you want - so what is the answer you want? The partial derivative of u(x,t) wrt to x, maybe? The partial derivative of u(x,t) wrt to t maybe? Something else?

when you have a product of two derivatives - which of the two do you want to "transform". How am I (or Maple) expected to "know" - should I guess?

In my original response

1. psdsolve(numeric), returns a numeris approximation to the function C(x,t)
2. You supplied a fixed value for t, ie t=T, which allowed us to define the curve C(x,T)
3. Given the curve C(x,T), it is possible to "fit" a function to this curve
4. Having fitted a function to the curve C(x,T), it is possible to differentiate this function wrt x

However in your reply, the thought process seems to go

1. psdsolve(numeric), returns a numeris approximation to the function C(x,t)
2. Supply a fixed value for x, ie x=L, which allows us to define the curve C(L.t)
3. But since L is a constant, it makes no sense to differetniate with respect to L
4. The boundary condition you cite requires you to differentiate C(x,t) with respect to x, and then substitute x=L. This is not equilvalent to evaluating C(L,t), and then differntiating wrt L, because L is a constant, so the differentiation makes no sense. As a "toy" example, consider C(x,t)=sin(x), differntiate wrt x to get cos(x), then substitute x=L to get cos(L): now try these operations the other way round C(x,t)=sin(x), substitutex=L, to get  sin(L): this is a constant. Unless your expecting to get zero as a result differentiating it with respect to anything is a bit pointless.

Now I think I know what you want to achieve, which is

1. pdsolve(numeric), returns a numeris approximation to the function C(x,t)
2. somehow(?!) differentiate this expression wrt x, then substitute x=L in the result. The differentiation is the tricky part and I'm still thinking about how it might be achieved, but I don't think it is trivial! I will comyimue to contemplate, but I'm a bit busy over the weekend

Both Kitonum and I (at the second attempt) provided the correct answer to your original question. This disagreed with the "answer" you wanted for the simple reason that the "answer" you wanted was incorrect..

You now seem to want to want t use integration by parts sometimes, for some function combinations - but not for other function combinations. Furthermore, the choice of 'parts' seems rather arbitrary. From your file cc.mw, I can easilyy come up with two possibilities for the integration of the first term -see the attached

Download cc2.mw

On what basis would I be expected to select the particular option which you want??

All of these representations of the original espression are equivalent. What is being achieved by converting them to different (but identically equivalent) forms?

and suggest that you read the following, very carefully - and if necessary explain what is incorrect

Download parts.mw

Well I've just tried it again with the following steps. It is repoducible, and I still have no idea why!!

Steps:

1. Shut down Maple
2. Start Maple
3. Load hidChar.mw using File->Open, from within Maple, and selecting hidChar.mw
4. Re-execute using !!! - error occurs
5. Shut down Maple
6. Start Maple by double-clicking by double-clicking the file hidChar.mw
7. Re-execute using !!! - error occurs
8. Remove all output from worksheet using Edit->Remove Ouput->from Worksheet
9. Mouse select the 1D input from the Maple worksheet and copy/paste (keyboard, CTRL-C, CTRL-V) to an editor (in my case, for the moment, this is Notepad++, introducing other editors might just complicate things, so I haven't tried Emacs or anything else)
10. Open new blank sheet in "worksheet" mode
11. Mouse select all text from editor and copy/paste (keyboard CTRL-C, CTRL-V) from editor to the new, blank Maple worksheet obtained at (10) above
12. Re-execute using !!! - error has disappeared
13. (Alternative to starting from (10) above. Open new blank sheet in Maple in "document" mode
14. Mouse select all text from editor and copy/paste (keyboard CTRL-C, CTRL-V) from editor to the new, blank Maple worksheet obtained at (13) above
15. Re-execute using !!! - error has disappeared

All I can conclude from the above is that somehing "nasty" got removed in the process of copying from Maple to the editor, or possibly from the editor back to Maple.

I am prepared to admit that the suggestion of a "mysterious, non-printing character", in my original response, is more-or-less equivalent to stating that I have no idea wtf is happening.

In case it matters, everything was done using Maple 2017.2 on 64bitWin7

My reason for using fdiff() in this problem was that having using cubic splines to produce the fitted curve, I was then in the position of having to generate derivatives at the points where the splines "join" (known as 'knots' maybe?). The basic diff() command failed at these points,producing 'Float(undefined)' so I reverted to fdiff().

I accept that using a 'spline' fit may be an overall bad fitting strategy.  The curve to be fitted is "smoothly-varying" and more-or-less parabolic, so a low-order (continuous) polynomial may be a better choice of fitting model. Such a choiice would allow the subsequent use of diff(), rather than fdiff(). But for this particular example, I'd be surprised if there were much difference

Don't still have the original, but I have managed to recreate. It does show the error occurring in 1D, but I can't swear that it is exactly what I had before, because both then and now I was doing a lot of cut+paste/reformatiing/whatever

Download hidChar.mw

One difference on this occasion is that if I open a new sheet in "worksheet" mode, and copy/paste from the above(document mode) to this new sheet, the error "goes away".

I'm nearly certain that last time I did this, the error remained in the worksheet mode (which is why I did the extra steps of going to/from an external editor)

You - that is YOU torabi - are generating two expressions in three unknowns and then trying to solve. This is obviously impossible. When you stop generating two expressions in three unknowns, I will look at this again. Until then, you are on your own, and I guarantee you will get absolutely nowhere

1a. Your original question referred to evalr(1/INTERVAL(-1 .. 1) which is a real problem with a real range. So far as I am aware, a real range includes its endpoints - so the denominator does not have " absolute value less than 1", which makes the rest of your response at 1a earlier, a bit meaningless

1b. This was more of a curiosity for me, and I'm not sure why evalr(1/INTERVAL(-1..x,x..1)) seems to "work", when  evalr( 1/INTERVAL(-1 .. 1)) onbiously does not. Note however that the help for evalr() states "ranges written INTERVAL(a..b) with a<=b" (my emphasis), So INTERVAL(-1..x,x..1) specifies two conditions -1<=x , x<=1. Since I owuld expect Maple to apply these two conditions consistently, then x does have to be between -1 and 1

2. Nowhere in your original question do you mention evalrc(), or the fact that you are dealing with complex quantitities. Since I assumed that you were dealing with real quantities, I tried to interpret your problem in that context.

3 Don't propose to comment on your remark (3) because I cannot reproduce the result. evalr(1/INTERVAL(x, -1..1)) returns INTERVAL(1/INTERVAL(x)). I'm assuming that the discrepancy between my result and your is a Maple version issue. (I'm running 2017.1)

as the Laplace tranform of a periodic function - derivation given here

https://proofwiki.org/wiki/Laplace_Transform_of_Periodic_Functions

Several issues

1. You are using 'pi' in various places. This is simply a Greek letter. If you want the quantity which relates circumference to diameter, then you need to be using 'Pi', so I changed these occurrences
2. Both of the odes in your worksheet are second-order, so only two boundary condition are required (or can be satisfied) - so I arbitrarily deleted one.
3. The attached now gernates solution procedures for both of the odes in your worksheet

Download odeProb.mw

as in my original response. try

 a:=0.6364562590
evalf[2](a);
evalf[3](a);

change the number in the square brackets depending on how many significant figures you want

using the big green up-arrow in the MaplePrimes edit toolbar.

Clarify what exactly your problem is - because the statement

the infint  summation  maple cant evaluated it  when my variables  are more than one decimal number 

doesn't make much sense

OP's original code, converted to 1D input just for clarity, is actually

sol1:=dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2 = 0,
                        diff(diff(diff(g(eta), eta), eta), eta)+f(eta)*(diff(diff(g(eta), eta), eta))+2*g(eta)*(diff(diff(f(eta), eta), eta))-3*(diff(g(eta), eta))*(diff(f(eta), eta)) = 0,
                      (diff(diff(g(eta), eta), eta))*(diff(diff(g(eta), eta), eta))-g(eta)*(diff(diff(diff(g(eta), eta), eta), eta))*f(0) = 1,
                      (D(f))(0) = 1, (D(f))(5) = 0, g(0) = 1, (D(g))(0) = -1, (D(g))(5) = 0], numeric, method = bvp
                    );

Note the part I have hihlighted above in 'red'. You appear to have interpreted this as

sol1:=dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2 = 0,
                        diff(diff(diff(g(eta), eta), eta), eta)+f(eta)*(diff(diff(g(eta), eta), eta))+2*g(eta)*(diff(diff(f(eta), eta), eta))-3*(diff(g(eta), eta))*(diff(f(eta), eta)) = 0,
                      (diff(diff(g(eta), eta), eta))*(diff(diff(g(eta), eta), eta))-g(eta)*(diff(diff(diff(g(eta), eta), eta), eta)),
                      f(0) = 1, (D(f))(0) = 1, (D(f))(5) = 0, g(0) = 1, (D(g))(0) = -1, (D(g))(5) = 0], numeric, method = bvp
                    );

which I am sure you agree, are two entirely different things.

The former (ie what the OP actually posted) seems to be more difficult to solve - I haven't succeeded so far. With your interpretation (ie the latter of the two above), then the solution is much simpler