Analytic solution of this, Advice.

[Clausius2]

I am looking for the analytic solution of this ODE (if it were one):

$$s^2G''+sG'-(1+s^2+s\text{coth}(s))G=-4s^2e^{-s}$$

I have solved this equation numerically, it only gives one physically realizable configuration rejecting conveniently one of the homogenous solutions. I don't have those solutions BUT I have the asymptotic behavior of $$G(s)$$, which turns out to be $$G\sim As^{\sqrt{2}}$$ and $$G(s)\sim se^{-s}}$$ for small and large $$s$$ respectively, where A is a coefficient that I have worked out by means of a linear shooting.

When writting it on Maple in OdeAdvisor, it says to me that it is a Linear ODE (easy thing to know) and with Linear Symmetries. Does this last thing have something to do with Lie Symmetries?. May this ODE be solvable employing that theory, I don't have a clue.

Thanks.

 

[Matthew Rodman]

For small s we can start with the equation

$$u^{\prime \prime} +\lambda u^{\prime}<br /> +\lambda^{\prime}u=\gamma e^{\int \eta ds}$$

and make the substitution

$$u = G(s) e^{\int \eta ds}$$

This will give us the equation

$$G^{\prime \prime}(s) + (2 \eta + \lambda) G^{\prime}(s) <br /> + (\eta^{\prime} + \eta^2 + \lambda \eta + \lambda^{\prime}) G(s) = \gamma$$which if we set

$$\lambda = \frac{1}{s} - 2 \eta$$

and

$$\gamma = -4 e^{-s}$$

gives us

$$G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s}<br /> + ( \frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2}) G(s) = -4e^{-s}$$

which we can see is similar to your equation once we divide it by $$s^2$$.

$$G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s}<br /> - ( \frac{1}{s^2} + 1 + \frac{\coth{(s)}}{s}) G(s) = -4e^{-s}$$Now the next bit is to match the $$G(s)$$ term with your equation. So,

$$\frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2} <br /> = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}$$

Hence we need to choose an $$\eta$$ that fits. If we try substituting

$$\eta = \coth{(s)} \pm \frac{\sqrt{2}}{s}$$

we get, with some fiddling,

$$\frac{2(1 - \sqrt{2})}{s} \coth{(s)} = <br /> \frac{2(1 - \sqrt{2})}{s^2}$$

or

$$\frac{2(1 + \sqrt{2})}{s} \coth{(s)} = <br /> \frac{2(1 + \sqrt{2})}{s^2}$$

So, for small s, where we can make the assumption that

$$\coth{(s)} \simeq \frac{1}{s}$$

we therefore have a solution.
The only thing left is to solve the very first equation.
By inspection we can see that it can be integrated directly to yield

$$u^{\prime} + \lambda u = \int \gamma e^{\int \eta ds} ds$$

which is straightforward to solve for u, and hence G.

For large, positive s we can use the same method, except in our choice of $$\eta$$. If we choose

$$\eta = \coth{(s)} - 2$$

and plug this into

$$\frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2} <br /> = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}$$

and rearrange, we get

$$(\frac{2}{s} + 4) \coth{(s)} = \frac{2}{s} + 4$$

which as s tends to (positive) infinity, coth(s) tends to 1, and this becomes true.

You can then solve for u and G as above.

 

[Matthew Rodman]

will get back to this later...

 

[Clausius2]

Mathew! THANKS A LOT MAN!. Keep on posting, I will keep track of every line. At first sight it seems you don't find a closed form though...it's an asymptotic behavior isn't it?.

 

[Clausius2]

Hey man. I re-did all the steps, and it pretty much works. The only problem is that finding an exact solution of:

$$\eta^2+\eta'-\eta/s=1+coth(s)/s$$

seems a pretty difficult task. It's a Riccatti equation. It would be nice to have it for closing the problem. I will mess around with some books to look for a solution (Maple doesn't say anything as always), and also I am going to see how good is your asymptotic approximation compared with the one (coming from Frobenius at s=0 and approximating $$scoth(s)\sim s$$ for large s) that I had before. Where did you find this stuff?. I mean, you have the book where this stuff can be found?.

Thanks.

 

[Matthew Rodman]

Sorry, no book - I've just been fiddling around with it, seeing what comes up.

Yeah - for the general case I can't find a solution (one that doesn't involve an approximation), but I'll keep looking.