@Alfred_F 

I never said the serie was not convergent: I only said it was not summable.

The classical example of a convergent serie which is not summable is sum((-1)ⁿ / n, n=1..+oo): as its limit is -ln(2), splitting this serie as sum((-1)^2*n-1 / (2*n-1), n=1..+oo) + sum((-1)^2*n / (2*n), n=1..+oo) converges to -ln(2)/2. 

So we must be fery careful in the limit we (here Maple) get.

Before going any further, I have to say that I'm not a specialist, and that what I'm about to say may be incorrect.

Let A the set of couples (m, n) where both m and n are strictly positive integers such that m <> n.
Then you can separate A as A^> + A^<  where (m, n) belongs to A^> if m > n and  (m, n) belongs to A^< if m < n.
Rewritting S = sum(1/(m^2-n^2), (m, n) in A) as S = sum(1/(m^2-n^2), (m, n) in A^>) + sum(1/(m^2-n^2), (m, n) in A^<) gives (immediate result) S = 0.
So the limit os the serie depends on the way the summation is done
Thus depending on the way the summation is done the limits are diferent.

Download summability.mw

The limit pi²/8 is nothing but some possible limit value of S for a given arbitrary choice of a summation strategy.

@dharr @acer @Alfred_F 

The serie is not summable... which raises some questions concerning  Maple's Pi^2/8 result.

Here is a demonstration (in French): see Exercice 21, p 3 (Exercise 21) and its correction (follow the link).
Translations:

• Justifier : Prove that
• En déduire : Deduce that
• Qu'en déduire? : What conclusion can we draw?

Translations (from correction)

• La série converge compte tenu des critères usuels : Convergence comes from clasical criteria
• Par telescopage : Using a telescoping sum
• De plus : More of this
• Donc : Thus
• Puis : Next
• Cependant : But
• On en déduit... : We deduce that the family...is not summable.

Observe that the first question in Exercice 21 corresponds to @acer's inner sum comment

@Alfred_F 

What will happen when to the condition m <> n when m = +oo and n=+oo but m=n-p for any strictly posiive defined integer p?

Download Infinities.mw

@nm 

I use to use notepad++ at the office and you're right, it is a very good tool especially when you use syntax coloring and code versionning.
Unfortunately when I'm at home I use an iMac and I wasn't capable to find any "true" alternative to notepad++ for Mac OSX: are you aware os some notepad++-like for Mac OSX?

TIA

@Arya-S-AA 

I'm happy for you

want to find 7 parameters which verify 252 equations ???

numelems(eqs); 
                              252
numelems(vars);
                               7

Maybe you shoulg think twice about that, unless your et of equations has a very specific structure (for instance 245 equations are linear combinations of the remaining 7 equations, for instance)

Did you give a look to the IterativeMaps package?

What does your phrase "So we want to find a substitution that removes the time dependence from u. One way is to find the maximum and see how it moves. Here, the first solution gives what we want." mean precisely?

What kind of transformation are you talking about?

It is obvious that eq17 also writes

u(x, y, z, t) = 2*diff(L(x, y, z, t)), x$2);

# with 

L(x, y, z, t) = log(f, x, y, z, t)

So if you don't want u to depend on t you have to take f  to be for instance

f(x, y, z, t) = A(x, y, z)*B(t)

Is that what you need?

In such case have a look to  hfz_mmcdara.mw

If not forget it.

For_your_information.mw shows it worked well with Maple 2015: likely a code regression

@Kitonum 

"The variant ... more universal"

I agree but it wasn't the question.

BTW an answer coser to what the OP asked would have been

A := n -> Matrix(n, (i, j) -> a_||i||j);
 #or
A := n -> Matrix(n, (i, j) -> cat(a_, i, j))

Have a good day

@Kitonum 

Why didn't you write simply

A:=n->Matrix(n, symbol=a):

instead of 

A:=n->Matrix(n,(i,j)->a[i,j]):

?

Indeed the two definitions below seem equivalent

A:=n->Matrix(n,(i,j)->a[i,j]):
lprint(A(2))
Matrix(2, 2, {(1, 1) = a[1, 1], (1, 2) = a[1, 2], (2, 1) = a[2, 1], (2, 2) = a[2, 2]}, datatype = anything, storage = rectangular, order = Fortran_order, shape = [])

B:=n->Matrix(n, symbol=a):
lprint(B(2))
Matrix(2, 2, {(1, 1) = a[1, 1], (1, 2) = a[1, 2], (2, 1) = a[2, 1], (2, 2) = a[2, 2]}, datatype = anything, storage = rectangular, order = Fortran_order, shape = [])

@salim-barzani 

Remove the "&" character in the name of your file for it can be downloaded

and solve(%, R) returns 0 with multiplicity 2.
Explanation pde_mmcdara.mw

To complete @gkokovidis answer: as 0 is neither negative nor positve, it's worth looking to what happens when you used the Heaviside function:

restart
f := x -> Heaviside(-x)
x -> Heaviside(-x)
f~([-1, 0, 1])
                       [1, undefined, 0]
NumericEventHandler(invalid_operation = `Heaviside/EventHandler`(value_at_zero = 1)):
f~([-1, 0, 1])
                           [1, 1, 0]
NumericEventHandler(invalid_operation = `Heaviside/EventHandler`(value_at_zero = 1/2)):
f~([-1, 0, 1])
                           [   1   ]
                           [1, -, 0]
                           [   2   ]
NumericEventHandler(invalid_operation = `Heaviside/EventHandler`(value_at_zero = 0)):
f~([-1, 0, 1])
                           [1, 0, 0]
_EnvUseHeavisideAsUnitStep := true: 
f~([-1, 0, 1])
                           [1, 1, 0]

# Watch out: running _EnvUseHeavisideAsUnitStep := true:  
# makes `Heaviside/EventHandler`(value_at_zero = 0) ineffective
NumericEventHandler(invalid_operation = `Heaviside/EventHandler`(value_at_zero = 0)):
f~([-1, 0, 1])
                           [1, 1, 0]

Here are two tricks I use to ease the coding and avoid forgetting some backslashes (I used the substitute "&" in the second trick but you use any other character).
Based on @acer's answer:Sugggestions.mw