Write as numerator over denominator then just take the numerators, which we want equal to zero.

Let's try @rcorless's suggestion to reduce the number of variables.
All the terms in Q2 have the same dimensions. If we divide all terms by the first term, then each of the new terms is dimensionless.

Combine with the ones from Q3 and Q4 and remove duplicates. We want the "real" variables Pr, Pn and w last.

We construct a matrix K that contains the powers of allvars (rows) in each of the terms (columns)
Matrix K has 12 rows but rank 10, so we can eliminate 12-10=2 variables. The code below shows we can, for example eliminate beta and Ce (these are the ones = 1 in "rewrites").

Some technical stuff. See E. Hubert, G, Labahn, Found Comput Math 13 (2013) 479–516, doi:10.1007/s10208-013-9165-9.

or https://youtu.be/Nl2FBAbU1pE

The naming isn't optimal here so we'll ignore it and make up our own. Basically (see nondims), we can leave all variables the same except introduce R=Ce*rho0, B=beta*upsilon. (I used R instead of rho0__new, B instead of upsilon_new.) So we need to do the following substitutions to get rid of Ce and beta.

We are down to 3 unknowns and 7 parameters, but this is still too hard for solve. Trying a polynomial solve specifying that we want vars in terms of the others is still too slow. We can get solutions without backsubstitution if we let it choose the solver choose the variable order by specifying newvars as a set.

We have 8 solutions. Some are simple, e.g. tau=Cr=0.  We should check that even if they solve {Q2,Q3,Q4}, they also don't give a division by zero error for {K2,K3,K4}

Error, numeric exception: division by zero

For solution 1 we have B=0 (so either beta=0 or upsilon=0), and a range of solutions only for particular relationships involving Pn-Pr, not Pn and Pr separately. The solution is only valid when delta is not 1.

Error, numeric exception: division by zero

This one also depends on Pn-Pr

Error, numeric exception: division by zero

Cr = 0 and delta=1 is a solution but not to the originals

Error, numeric exception: division by zero

The next one has only the combination Pr-Pn=0 i.e. Pr=Pn

Error, numeric exception: division by zero

Similar to above but a more complicated relationship. For a given Pr we can get w and Pn. delta=0

Error, numeric exception: division by zero

In this case we get a solution for all of Pr,Pn,w for {Q2,Q3,Q4} but not for the original system. Close!

Error, (in simplify/normal) numeric exception: division by zero

And now the hard one, which is probably not so different from the original

And we still cannot solve this in any reasonable time.

Define trees with k branches from each node. Z here is a leaf on the tree.

Find all possibilities

Convert the Prod(...) to [...]

The LeafToSymbol routine below then substitutes the a,b,c,d for the Z's

Here is a procedure to carry out the whole thing

Move the lhs over to the rhs

It looks as thogh Maple can solve the whole problem, but it will be a nightmare sorting through the various 1368 cases (and the "otherwise" case).

A typical condition and result.

Let's simplify by collecting the complicated combinations of parameters into simpler parameters

So in simpler form we have

Aside from the cases with b=0, there are three different cases, depending on the value of , which is just the discriminant of the quadratic a*i1^2+b*i1+c. We want to know the values of i1 for which the parabola is above the i1 axis. Since a is positive, the parabola is not inverted. If  the roots are complex, and the whole parabola is above the axis. This is case 4. If   the roots are real (part of the parabola is below/on the axis) and for the relation to hold i1 must be greater than the more positive root or less than the more negative root (outside the interval between the roots) - this is case 5. Case 6 is for a<0, which is not of interest.

So we want to find when the discriminant is positive or negative

This is going to depend on the exact values of the parameters and is still a complicated problem. But if you know some extra conditions you might be able to narrow it down. For example it seems important to know if delta<1

So this is progress, but not enough to narrow it down

Unless you know you will always be using a particular range, I would not put a range here. If there is no range, then the range and values are determined when you actually need them, in this case in odeplot. The range option here doesn't apply to all methods. On the other hand if you do use range here, you can use refine = 2 in odeplot to get twice the normal number of points.

rkf45 is an automatic step size method (not the standard fixed stepsize Runge-Kutta method), so gives the default accuracy over all points plotted, with a reasonable number of points. I would specify the range here.

The 0..3 solution is stored so when we do 0..30, the solver starts actually starts from 3.

Look at a small section of the solution

Use another method. rosenbrock is default for stiff equations, and can be obtained by stiff=true or method=rosenbrock

The help pages have more

This version assumes wolf speed = goat speed and each is constant. It also assumes the wolf's velocity vector is always directed at the goat.

Take the origin to be the location of the stake fixing one end of the rope, which is of length1, and take the goat's initial position as . Without loss of generality, we take the common speed to be 1.
Let  be the wolf coordinates and  be the goat coordinates.
We assume the wolf never "second guesses" the goat's position so that the wolf's velocity vector is always pointed directly at the goat. i.e., the tangent to the wolf's trajectory points at the goat. The tangent vector is given by and therefore  for some function . Therefore we have the two odes

And the speeds are each 1.

And the goat moves on the unit circle.

Let the initial position of the wolf be . Then the vector from here to the goat is , so.

But the squares of these sum to 1 so , or  For , the initial velocity of the goat is downwards (presumably, though we could assume otherwise).

Sample start point for the wolf for numerical solution.

Numerical solution

Red = wolf, blue = goat.

Separation vs time - looks like the wolf catches the goat in an asymptotic sense.

Note that the motion of the goat is predetermined by its speed and initial condition since it must maintain constant speed, i.e., it can't turn around.

In terms of an analytical solution, Maple can reduce the system to a single ode in  but not solve it. There is no result if the initial conditions are included.

Use the interactive feature of dsolve to sample different initial conditions and record cases where the separation at  is less than 0.05.

A point here indicates the wolf (effectively) catches the goat if starting from this point. It seems the wolf always catches the goat

First one of result 2

The following gives lambda[0] differing by a sign from in the paper (aleph^2-1) instead of 1-aleph^2 in the numerator

Case of aleph = 1

Case of aleph = 0 - here the coefficient goes to zero so the whole thing does.

Just looking at the JacobiCN part, it goes to cos as expected.

Error, (in series/function) unable to compute series

Note the following is imaginary.

Go back to the ode

This form with sn is close, but not sure about the right signs for assumptions to get this result without symbolic

Circles k1 and k2 have radii 1 and sqrt(2), centred WLOG at the origin.

"From the outside, five congruent circles k3 are placed tangent to k1, each with a radius r yet to be calculated."
WLOG take centre of one k3 circle at

Find the intersection points p1 and p2 between circles k2 and k3

Find the angle p1-o-k3c

This angle needs to be 2*Pi/10, and we can solve for r.

Warning, solve may be ignoring assumptions on the input variables.

Update these objects to reflect the calculated value of r.

I changed the part in red - correct?

Check it looks correct.

This needs to be linearized with respect to theta. I'll just throw away second and higher order terms in theta. We can't use coeff on theta(x,t), so will do it in jet notation.

But we still have half-integral powers of Q, which don't appear in Eq, 11. So there is definitely something wrong something I'm doing here.
If I manually remove any terms with half integral powers of Q, I'm left with something that is still not right - there are the right number of terms but there is no term with coefficient 18

So I'll just proceed with the above incorrect result

We assume theta has the form

We substitute this in P3 and collect the coefficients of the two exp terms

Set up these as matrix equations in the unknowns alpha[1] and alpha[2]

There is a nontrivial solution to these equations iff the determinant is zero. The determinant is a quartic in w.

And we solve for w to define the dispersion relationship eq 13. The coefficients under the square roots aren't correct, but presumably would be if P3 were correct. The first two solutions seem to be the ones of interest.

w is real when the expression under the square root is positive