Hi,

In what follows you have the symbolic exact solution to the ODE system you posted, matching the ics you posted, and without specializing any of the symbolic parameters a1, a2, a3, b1, b2 or b3.

Your ODE system:

Strategy:

1) rewrite this system as an ODE system for three unknowns C1, C2, C3, plus an expression giving O1 in terms of them

2) compute the exact solution for C1, C2, C3 in terms of three arbitrary constants

3) adjust the three arbitrary constants to match your initial conditions ics for C1, C2 and C3

4) substitute this solution for C1, C2, C3 matching your initial conditions into the expression giving O1 in terms of them.

While performing these steps, do not display expressions that are too large.

Step 1), one of the equations you present as ics is actually a constraint between the functions:

The actual system to be solved then involves 4 DEs and one constraint

Rewrite now this as "one constraint and three DEs involving only C1, C2 and C3" (see casesplit )

In the above you see a system entirely equivalent to yours, but now involving 3 DEs instead of four. Remove then the constraint (that has O1 for lhs) to work just with the ODE system, and assign it to

Step 2) Compute now the exact solution to this system; note that dsolve will solve and simplify the solution - most of the time is spent in simplifying

As mentioned by others, these solutions involve large algebraic expressions

You can transform these into slightly shorter expressions using simplify/size

Before proceeding further, let's verify this solution

Step 3) Now you need to adjust the integration constants to match your ics for C1, C2 and C3

 are:

So: evaluate  at  and substitute in  in order to have a system for the Cn

Check lengths

The symbols in these ics are

Solve now for the Cn and simplify the size; check lengths and left-hand sides

At this point we have solved the problem:  is the general solution in terms of three arbitrary constants Cn and  gives the value of these three arbitrary constants that match your  for C1, C2 and C3. So evaluate  at  to get the answer you are looking for; check lengths

Step 4) To get the answer to O1, substitute these solutions for C1, C2 and C3 in the

In addition, the following three would help in compacting the expressions, but they are rather large and simplifying takes time - I am keeping the lines here in comments in case you want to try them

Hi,
Yes, in Maple 17 you can do substitutions of tensors of the form you mention, taking into account free and dummy indices the same way we do when computing with paper and pencil. You need to update your Physics library with the latest one, available for download at the Maplesoft Physics: Research & Development webpage. The command you use to do these substitutions, SubstituteTensor, is currently a Physics:-Library command, part of an upcomming more general Physics:-Substitute command, to handle also tensorial subexpressions within noncommutative products similar to what algsubs and simplify/siderels do with commutative nontensorial products.

The repeated and free indices of the lhs and rhs of (2)

The easy case

The free index in the target expression is not  but

Distinction between covariant and contravariant indices

The index  found repeated in the rhs of the substitution equation (2) apears also repeated in the following target expression

Substitute (2) in the rhs of (2) itself

The example you posted as a question

Example with the Schwarzschild metric in spherical coordinates: instead of using Setup (you can, but you do not need to), set the spacetime metric directly from the the metric command g_

The Kretschmann scalar is given by

The above concerns your fist post, and to see the Ricci tensor you can always

Now in your second post, however, you do not use the Physics package as suggested but the old 'tensor' package. The computations with the old 'tensor' are overly complicated and that package is by now obsolete.

My suggestion in the previous replies were computing this using Physics, it is really simpler.

Copying the metric info from the worksheet attached in your last reply and pasting here:

Noticing now that your coordinates  are the same introduced above with the Schwarzschild metric where  = t, you can now re-enter the metric in different ways, see Physics:-Setup  .For instance, to avoid disquisitions about the ordering of the coordinates, enter the metric directly as the square of the line element

That is all.

To compute the Kretschmann scalar for this metric, just Simplify(K) again, and no need to re-enter, re-construct or re-define the Riemann tensor or anything else.

Regarding the Schwarzschild limit when  mentioned in your worksheet, this is the metric with

(If you want to see the inversse of the metric, just enter  )

This is the value of  at q = 0 :

This is the value of the K scalar when q = 0

And this is the KScalar for  (a bit of a wallpaper even after simplifying in size)

For more information please check the help pages for g_, Ricci, Riemann and Setup.

dsolve's solution is correct:

This simplification (see ?PDEtools/casesplit) explains dsolve's solution:

You can arrive at the same result by hand. Take the first equation, and you have the solution for f(t) as a function of u(t)

Remove now f(t) from the system

Take the second equation

Substitute into the third equation

So, the solution for u(t) is

Then from (4) the solution for f(t) is 0.