@Preben Alsholm 

I am very grateful for your help. Thanks Alsholm and Carol.

I changed some parameters. But I failed to run the code correctly. I had a singularity on t=t0=-1 (initial time). Should I modifiy something in the code?

restart; mu := 0.1e-2; nu := .1; sigma := 1; k := 1;
ode1 := mu*(diff(u(z), z, z))/t+(z-2*sigma*(diff(1/v(z)^(1/2), z))/t^2)*(diff(u(z), z))+(1-2*sigma*(diff(1/v(z)^(1/2), z, z))/t^2)*u(z) = 0; ode2 := nu*(diff(v(z), z, z))-z*t*(diff(v(z), z))+2*t*v(z)-k*v(z)^(1/2)*u(z) = 0; ics := u(0) = 1, v(0) = 1, (D(u))(0) = 0, (D(v))(0) = 0;
sol := dsolve({ics, ode || (1 .. 2)}, numeric, parameters = [t]);
proc(x_rkf45)  ...  end;
q := proc (t) sol(parameters = [t]); sol(-2) end proc;
proc(t)  ...  end;
FindSingularity := proc (q, t::numeric) try q(t) catch "cannot evaluate the solution further left of %1, probably a singularity": return lastexception[-1] catch "cannot evaluate the solution":  end try; FAIL end proc;
FindSingularity(q, -1);
                          -0.015360725
Kind Regards. 

find_sing2.mw

@Preben Alsholm 

To clear up a bit of confusion between z and t. Actually, I tried to solve a system of PDEs through its conversion into a system of ODE: u(x,t) --> u(z) where z=|x|/t, t in [-1,0).

The new system of ODEs show still existing (t). So, I considered this (t) as a parameter, not a variable.

The objectif is to seek the blow-up (singularity as Dirac) when (t) is approaching zero.

According to your explanation  I see that it's not possible by using Maple.

Regards.

@Preben Alsholm 

I start with time t=-1<0  rather than t=0

So, Ics: u(0)=1, v(0)=1, D(u)(0)=0, D(v)(0)=0 should be at t=-1. Let's say as a parameter in initial condition.

If I don't specify the value of t in Ics, I got singularity at t=-1 which is not correct.

How can I modify the Ics so as to take into account the start time?

Regards. 

find_sing.mw

@Preben Alsholm

Actually, we have u(z,t) and v(z,t) --> u(z) and v(z). I fix time t and solve the probem. 

In addition, t is involved in the initial conditions u(z,-1)=u_0(z)=1, v(z,-1)=v_0(z)=1

T refers to the blow-up time. So, at t=T, singularity appears.

Thanks in advance.

@Carl Love 

Thanks a lot.

If it's possible I also need your help for the same system but modified according to the value of time t in [-1,0)

ode1: 0.001*diff(u(z),z,z)/t+(z-2*diff(v(z)^-1/2,z))*diff(u(z),z)+(3-2*diff(v(z)^-1/2,z$2))*u(z)=0

ode2: 0.1*diff(v(z),z,z)-z*t*diff(v(z),z)-2*t*v(z)-u(z)*v(z)^1/2=0

Ics: u(z)=1 and v(z)=1 for t=-1

So, I shouldn't have any singularity at the starting time and I have to find T as the value of time when singularity can appear.

Thanks for your time and consideration.

@tomleslie 

Actually it's a non linear PDE system which I converted to a non linear ODE system in the hope to find a solution that blows up in finite time. I try to simplfy the problem by assuming some parameters equal to one or zero and use Maple for solving the problem in its different cases.

What could you suggest for me. I mean as an ODE system.

Thanks.

@ecterrab 

Thanks for helping me to solve my non linear ODE system.

I have an error like:

Error, (in dsolve/numeric/make_proc) solutions for ODE about singular points cannot be computed if the ODE system contains global variables.

res_syst.mw

@tomleslie 

Hi, Thanks for your time and consideration.

Setting initial and boundary conditions I got an error like :

Error, (in dsolve/numeric/make_proc) solutions for ODE about singular points cannot be computed if the ODE system contains global variables

What's wrong?

Again Thank you for help.

Download res_syst.mw

@tomleslie 

Sorry, please find the file below. Other conditions on u and v that I have not been able to implement them are:

limit(u(x), x=infinity)=0:

limit(v(x), x=infinity)=0:

Thanks for your help.

res_syst1.mw

@tomleslie