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I'm faced with the following nonlinear 2nd order system of ODEs: $$ \phi''(r)+\frac{4r^3-1}{r^4-r}\phi'(r)+\frac{r^2 h(r)^2+2r(r^3-1)}{(r^3-1)^2}\phi(r)=0, \\ h''(r)+\frac{2}{r}h'(r)-\frac{2r\phi(r)^2}{r^3-1}h(r)=0. $$ Here, $\phi$ and $h$ are real-valued functions of $r\in[1,\infty)$ and $h(1)=0$ and $\phi(1)$ is finite and nonzero.

I've been able to find numerical solutions of this system. I did so by assuming $h$ and $\phi$ can be written as a power series around $1$ and by imposing initial conditions at $r=1.00001$ and integrating to larger $r$.

However, I am wondering now if the solutions I found are unique, because the system is singular at the point $r=1$.

So my question is: for this specific system, is it possible that there exist singular solutions (by which I mean that they are not uniquely fixed by the initial value problem at $r=1.00001$)? Or are the solutions I can find uniquely fixed by the initial values because I impose them at $r=1.00001$ rather than at $r=1$? Note: I am only looking for solutions which in fact do behave analytically at $r=1$!

Any reference to existence theorems relevant for systems like the one above is also appreciated!

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  • $\begingroup$ The method you have used (looking for coefficients assuming analytic behavior of your functions at your origin) has limited range. Have you tried a differential system solver as exist on many systems, like ode43 on Matlab (based on Runge Kutta) in order to gain a better understanding of it ? I understand that you are looking for theoretical guarantees, but, as your system is fairly complicated, it is unlikely that good conditions will be found... $\endgroup$
    – Jean Marie
    Feb 7 '16 at 22:36
  • $\begingroup$ @JeanMarie I am looking only for solutions which should behave analytically at my origin. Does that restrict the uniqueness some more? My numerical skills are limited unfortunately, I only used Mathematica, but I will look up the method you mentioned if this can help me. $\endgroup$ Feb 8 '16 at 8:48
  • $\begingroup$ Also, theoretical guarantees of existence of singular solutions would be nice, but any statement like "it is not immediately clear that singular solutions cannot exist" also helps. $\endgroup$ Feb 8 '16 at 8:49
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I have made a Matlab implementation of your system (script below) to get approximate solutions.

a) If $h(1)=0$ (your hypothesis), function $h$ is found by the solver to be identically zero. This is one of the singular solution you are looking for... even it is rather trivial, isn't it ? Fonctions $\phi$ have are decreasing steadily, in a smooth manner.

I have thus attempted to solve the first equation by replacing $h(r)$ by $0$.

I have turned to Mathematica for solving it;

DSolve[p''[r]+(4r^3-1)p'[r]/(r(r^3 - 1))+2 r p[r]/(r^3-1) == 0, p[r], r]

gives two exact independent solutions :

$p_1[r]= \ $ Hypergeometric2F1[1/3, 2/3, 1, r^3] and

$p_2[r]= \ $ MeijerG[{{}, {1/3, 2/3}}, {{0, 0}, {}}, r^3]}}

(I never met any Meijer before !).

The general solution is thus a linear combination of these two.

BUT a major drawback is that this solution is defined only for $0<r<1$... Should there exist an analytical continuation through the complex plane ?


b) If one relaxes the constraint on $h(0)$ by letting $h(0)=0.001$ for example, the solutions for $h$ are increasing functions ; the curves of the $phi$ functions are highly oscillatory (see figures which are but a number of very different cases I have tried).


Matlab program:

function soldiffequ;

clear all;close all;

rspan = 1.00001:0.001:15; % domain for r values

ode = @(r,y) sc(r,y);

for k=2:5

y0 = [2^k;0.0001;0.;0.];% initial values for phi,h,phi',h', resp.

[r,Y] = ode45(ode, rspan, y0);%Runge Kutta solver

functions phi and h as functions of r

figure(1);hold on;plot(r,Y(:,1),’r’); % phi(r)

figure(2);hold on;plot(r,Y(:,2),’k’); % h(r)

end;

function yp = sc(r,y)

phi=y(1);h=y(2);Phi=y(3);H=y(4);

Dphi=Phi;% a name for the der. of phi

Dh=H;%a name for the der. of h

DPhi=-((4*r.^3-1)./(r.^4-r))Phi-phi.((r.^2.*h.^2+2*r*(r.^3-1))./((r.^3-1).^2));

DH=-(2./r).H+h.((2*r.*(phi.^2))./(r.^3-1));

yp = [Dphi;Dh;DPhi;DH];


Fig. 1a : case h(1)=0 : in red, functions phi for initial values 4,8,16 ,32. Functions h remain identically 0… :

Fig. 2a and 2b : Cases where h(1) is non zero (though very small : 0.0001) showing that functions phi may have a very oscillatory behaviour, according to their initial values (here 4,8,16,32),; Functions h are no longer zero : they are even rapidly increasing.

Fig. 1

Fig. 2a

Fig. 2b

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  • $\begingroup$ Thanks a lot for this! However, I am not sure I completely understand this result. Are you assuming here that $h'(1)=0$ and $\phi'(1)=0$? The numerical solutions I have found depend on the parameters $h'(1)$ and $\phi(1)$, so it might not be so surprising that putting $h'(1)=0$ yields the trivial solution for $h$. Also, it should follow from the ODEs that $3\phi'(1)=-2\phi(1)$. $\endgroup$ Feb 8 '16 at 13:36
  • $\begingroup$ This last condition follows from multiplying the first DE by $r^3-1$ and then evaluating at $r=1$, where if $h$ is analytic $h(r)\approx h'(1)(r-1)$ so that the term proportional to $h(r)^2$ vanishes (i.e. $h(r)^2/(r^3-1)=h'(1)^2(r-1)^2/(r^3-1)\rightarrow 0$. $\endgroup$ Feb 8 '16 at 13:43
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    $\begingroup$ Yes indeed, I had taken the most natural initial conditions (all derivatives equal to zero).The fact that there $3\phi'(1)=-2\phi(1)$ changes things..I will take it into account. $\endgroup$
    – Jean Marie
    Feb 8 '16 at 13:54
  • $\begingroup$ I just incorporated condition $3\phi'(1)=-2\phi(1)$ in the program : no change ; the results are the same as in fig. 1 for $\phi$ and $h$ is still identically zero. Can some similar condition be found on $h'(1)$ ? $\endgroup$
    – Jean Marie
    Feb 8 '16 at 14:11
  • $\begingroup$ When I say "the same", one must understand "have the same shape"... $\endgroup$
    – Jean Marie
    Feb 8 '16 at 14:22

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