Can this first order non-linear ODE be solved for all values of r?

[LAHLH]

Hi,

I have the differential equation \left(\frac{df}{dr}\right)^2-\frac{1}{r-1}\left(1+\frac{1}{4r^3}\right)=0, does anyone know how to attack this? (I'm led to believe this is only possible if r>1, not sure why not r<1, although I can see it does blow up at r=1)

thanks

 

[JJacquelin]

The integral (below) involves Eliptic Integrals.
If it is a physical problem, better use numerical computation.

 

[LAHLH]

The integral (below) involves Eliptic Integrals.
If it is a physical problem, better use numerical computation.

It is a physical problem, did you manage to solve it in Maple/Mathematica under some set of assumptions in terms of Elliptic functions? I couldn't get Maple to do anything with it.

Is there any obvious reason this integral is impossible for r<1 without actually doing it?

 

[JJacquelin]

Since we have to integrate a function of the form sqrt[P(x)/Q(x)] where P(x) and Q(x) are polynomials of degre 4 at the most, it is known that the result will be a combination of several Elliptic Integrals and usual functions.
Theoreticaly, the anylitical solving is possible, but probably very arduous.
Mathematical package for formal calculus will probably give a big and complicated formula as a result, with Elliptic functions appearing in the formula.
In the scope of physical problem, all ends with numbers. So, I suppose that it would be simpler to carry out the computation by direct numerical integration, than deriving first a complicated formula and then using this formula to compute the numerical results.

 

[LAHLH]

I mean I don't really need the explicit functions or anything all the author claims is that it can't be solved (or at least doesn't make sense physically, maybe it will give complex or infinite answers or something in the r<1 region), so I'm just trying to see that really.

 

[HallsofIvy]

Yes, that is the point. Your equation is the same as
\left(\frac{df}{dr}\right)^2= \frac{1}{r-1}\left(1+ \frac{1}{4r^3}\right)

If r< 1 then
\frac{1}{r- 1}\left(1+ \frac{1}{4r^3}\right)
is negative so that, if it is equal to the square of df/dr, dr/dr must be imaginary. If we are interpreting f(r) in some physical sense that cannot happen.

 

[JJacquelin]

In the scope of physical problem, there are other ways to obtain approximative results with small deviation (but not mathematically exact).
Depending on the order of magnitude of r , the function to be integrated might be replaced by another approximative function simpler to integrate.
Another way consists in using series development of the function, with a sufficient number of terms, depending on the range of integration.

 

[JJacquelin]

Your equation, given in your first message, implies r>1.
May be, in the case r<1, the correct physical model is different than the model in the case r>1. So, the equations might be different in both cases.

 

[JJacquelin]

although I can see it does blow up at r=1

That is not true.
Close to r=1, then f(r) is equivalent to (sqrt[5(r-1)]/4)+constant.
sqrt(5(r-1))/4 doesn't blow up, but on the contrary tends to 0.

 

[LAHLH]

Yes, that is the point. Your equation is the same as
\left(\frac{df}{dr}\right)^2= \frac{1}{r-1}\left(1+ \frac{1}{4r^3}\right)

If r< 1 then
\frac{1}{r- 1}\left(1+ \frac{1}{4r^3}\right)
is negative so that, if it is equal to the square of df/dr, dr/dr must be imaginary. If we are interpreting f(r) in some physical sense that cannot happen.

Oh yes, of course. Thanks alot

 

[gato_]

Actually, maple does give a solution:

f \left( r \right) =1/4\,\ln \left( r \right) +5/4\,\ln \left( r-1<br /> \right) \left( r-1 \right) -r+5/4-1/4\,r\ln \left( r \right) -1/8\,<br /> {r}^{-1}+{\it \_C1}\,r+{\it \_C2}<br />

The equation is autonomous , so you can reduce it to a quadrature. I haven't checked it myself, though

 

[LAHLH]

Actually, maple does give a solution:

f \left( r \right) =1/4\,\ln \left( r \right) +5/4\,\ln \left( r-1<br /> \right) \left( r-1 \right) -r+5/4-1/4\,r\ln \left( r \right) -1/8\,<br /> {r}^{-1}+{\it \_C1}\,r+{\it \_C2}<br />

The equation is autonomous , so you can reduce it to a quadrature. I haven't checked it myself, though

How did you get Maple to give that?

 

[gato_]

Sorry. I read d^{2}f/dx^{2} instead of (df/dx)^{2}