Your function f : R^3 --> R should be locally integrable, but it's not!
So, the Hardy−Littlewood maximal function makes no sense.

@Thomas Richard Sometimes the situation is reversed!

@sursumCorda So, you want an explicit solution instead of a parametric one (t3 being the parameter).
This could be very difficult (or even impossible) to obtain in general.

It would be nice to tell us how the equation was obtained.

Here is a (simpler) parametric one (Klimek G.&M. - Discovering curves and surfaces with Maple, Springer 1997.)
 

plot3d([(4+3.8*cos(y))*cos(x),
        (4+3.8*cos(y))*sin(x), 
        (cos(y)+sin(y)-1)*(1+sin(y))*log(1-Pi/10*y)+7.5*sin(y)],
      x=-Pi .. Pi, y=-Pi .. Pi, color=[sin(x+1)+cos(y), sin(y), 0.3],
      light=[85, 50,  0.6, 0.8, 0.2], orientation=[47, 55],
      style=patchnogrid, axes=none);

Your function of Pe also depends on several variables and some arbitrary functions!
How can you imagine that the limit could be computed?

@mmcdara For me, it's not about assume or assuming. 
For the same assumptions, assumptions := a>0, b>0, a*b < 1:
assuming gives the same results.
Also, in Maple 2022, assumptions := a>0, b>0, b < 1/a: 
==> all answers are true (But it seems that in 2023, one answer is FAIL !!).

So, everything seems to be about weakness!

@Carl Love  Unfortunately the assume facility is very weak here.

restart;
# _EnvTry := true;  # useless
assume(a>0, b>0, a*b<1);
is( (1 + sqrt(1-a^2*b^2))/(a*b) >0 ); # FAIL   ?
is( (1 + sqrt(1-a*b))/(a*b) >0 );     # FAIL   ?
is( (1 + sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a*b)) >0 );           # true
is( (1 + sqrt(1-a*b)) >0 );           # true

@nm The general solution of the ode  y'' + y = 0  is y = c1*sin(x) + c2*cos(x)  but also
 y = c1*(2*sin-7*cos(x)) - c2*(sin(x)-3*cos(x)).

@dharr a) A square complex matrix C has not a square root iff its Jordan form has a Jordan block having dimension > 1 and eigenvalue 0.

b) For C as in the file (40x40), M := evalf[n](MatrixPower(C, 1/2)), eps = Norm(M^2-C)  then:
n=14-->eps=0.003;  n=16-->eps=2e-4;   n=18-->eps=1e-6;  n=25--> eps=2e-13;  n=30-->eps= 2e-15.

Use odetest to check the solution. Yes, it's a bug.

restart;
ode:=[diff(x(t), t, t) + 2*diff(x(t), t)/sqrt(L*C) + x(t)/(L*C)];
s5 := dsolve(ode, [x(t)]);
odetest(s5, ode);
s6 := dsolve(ode, [x(t)], method = laplace);
odetest(s6, ode);   # bug

@Christian Wolinski The problem is that {a>0, b>0, c>0, d>0} is far for being equivalent to {e1>0, e2>0, e3>0, e4>0}.

@tomleslie 
G is the centroid of the (inscribed) quadrilateral P1P2P3P4.
E is the Euler center of  the quadrilateral P1P2P3P4  defined as the point in the intersection of the four Euler circles of the triangles P1P2P3, ..., P4P1P2. (E exists!).

Note that OG = (OP1+OP2+OP3+OP4)/4, OE = (OP1+OP2+OP3+OP4)/2,  OPk  being vectors.

@Alex0099  1. The order of summation does not matter, the summand being symmetric.

2. The proof could be very tricky. Probably here the Residue Theorem should be useful.  

@Thomas Dean  In Windows, after a restart everything is OK. Sorry, I don't use Linux.

@emendes  procedure is a type, so, its use is covered.