I'm solving the below differential equation:

$$ 3z^2 U'(z) -2 U(z)^2 - z U(z) + 2 = 0$$

I have two boundary conditions - $U(0)=1$, $U'(0) = -\frac{1}{4}$, however it is apparent that the set of solutions to this equation, before imposing the initial conditions, satisfy the first initial condition $U(0)=1$. Looking at the differential equation this is because it can be rearranged to:

$$ [U(0)]^2 = 1 $$

So it is evident that all of the general solutions should fit this condition. What's more, through "numerical experiment", I believe that the first derivative condition should be satisfied as well (though I can't prove it from the differential equation).

I am not sure that I can set the constant of integration to be anything I want, as this changes the behaviour of the solution. However, I can simplify the solution greatly by choosing "convenient" values of this constant.

Also the resulting function is a function of $\frac{1}{z}$, so it cannot actually be evaluated at 0, so I have been using limits, which feels quite mathematically dodgy.

This is a peculiarity that I have not come up against before - how exactly am I to interpret this?

  • $\begingroup$ Is $z$ a parameter or the independent variable? $\endgroup$ – Henning Makholm Dec 22 '11 at 21:34
  • $\begingroup$ Independant variable, sorry. It's not typical notation, so I'll make it more clear. $\endgroup$ – VolatileStorm Dec 22 '11 at 21:42
  • 2
    $\begingroup$ The real issue is that $3z^2=0$ when $z=0$, so you cannot apply the fundamental existence/uniqueness theorem. In fact, your calculation (forcing $U(0)^2=1$) shows that this equation has no solution if we try to impose an initial condition where $U(0)^2\not=1$. $\endgroup$ – Bill Cook Dec 22 '11 at 22:13

I'll assume you're interested in $z\ge 0$; for $z\le 0$ you can turn everything 180° about the origin and satisfy the same ODE.

You don't necessarily have $U(0)=1$; it can also be $-1$. So the first boundary condition is not completely superfluous.

Suppose that we have such a solution, and that it is reasonably smooth -- in particular that $U'(z)$ is bounded on $[0,\epsilon]$. Then for small enough $z$, the $z^2$ term will be insignificant, so we just need to solve $-2U^2-zU+2=0$ for $U$. The quadratic formula gives $$ U\approx\frac{z\pm\sqrt{z^2+16}}{-4} \approx \frac{-1}{4}z \mp 1$$ so your second boundary condition is automatically satisfied for any otherwise well-behaved solution that is defined for $z=0$.

Imagine that we know a "true" $U(z_0)$ for some $z_0>0$. Assume that $z_0$ is "not too large" such that $U(z_0)>0$. What happens to the general solution with $f(z_0)=U(z_0)+\varepsilon$? For a fixed $z_0$, the differential equation tells us that $f'(z_0)$ must be greater or smaller than $U'(z_0)$ according to whether $\varepsilon$ is positive or negative. But that means that when we follow $f$ backwards from $z=z_0$ towards $z=0$, the absolute difference between $U$ and $f$ decreases monotonically as $z$ decreases, so $f$ also hits $f(0)=1$, and by the argument above, $f'(0)=-1/4$ too.

Thus, as you speculate, your boundary conditions is not enough to select one particular solution -- the ODE has a singularity at $(0,1,-1/4)$.

You can't have solutions for every possible value of $f(z_0)$ though; if you set $f(z_0)$ too low and extend towards $z=0$, it will blow up towards $-\infty$ instead of being attracted by the $(0,1)$ singularity. Just on the boundary between these two possibilities there ought to be a single unstable solution with $f(0)=-1$.

A way forward might be to find the unique solution for $z\le 0$ that hits $U(0)=1$, and then try to continue it analytically across the singularity. I'm not quite sure how to go about that, however, and it might not even make any particular sense in your application (whatever that is).


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