Nonlinear ODE, Howto attack

[andrewr]

Nonlinear ODE, Howto attack...

Hi,

I've got a general nonlinear ODE equation that I have been solving in various situations, and I needed to make an approximate correction to -- but after the correction, an analytical solution to the new form evades me...

So I am studying it, but my math classes are quite a few years back now, and I would sure appreciate some help on figuring out how to attack it..
The subscript r denotes ordinary derivative with respect to r.
$$\varphi$$ is a function of (r), and K and k are arbitrary constants, I like them real ... but they can be made imaginary if it helps solve the problem. Also, the answer generated (Right hand side of =) can be off by a constant, or even a small deviation in r,r^-1,r^-2, but no more than that if that helps make the problem tractable.

$$K\left\{ \varphi_{r}^{2}+\frac{2\varphi_{r}}{r}+\varphi_{rr}\right\} =\frac{1}{r^{3}}$$

I am not able to find a solution by inspection. However I did find that:
$$\varphi=k*ln(r)$$ generates $$K\left\{ \frac{k^{2}}{r^{2}}+\frac{2k}{r^{2}}-\frac{k}{r^{2}}\right\} =K\left\{ \frac{k^{2}+k}{r^{2}}\right\}$$
and
$$\varphi=\frac{k}{r}$$ generates $$K\left\{ \frac{k^{2}}{r^{4}}-\frac{2k}{r^{3}}+\frac{2k}{r^{3}}\right\} =K\left\{ \frac{k^{2}}{r^{4}}\right\}$$

So that I figured the superposition of these would generate the non-linear $$r^{-3}$$ and it does:
$$\frac{k}{r}+ln\left(r\right)$$ generates $$K\left\{ \left(\left\{ \frac{k^{2}}{r^{4}}\right\} +\frac{2}{r^{2}}\right)-2\left(\frac{k}{r^{3}}\right)\right\}$$

I compared the results, and I think it is fairly easy to see that I can take advantage of a binomial expansion's predictability and do some kind of variation. It would be solvable by superposition if not for the fact that a binomial expansion produces additional terms to the squares of the components: eg:
$$(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2ab+2ac+2ad+2bc+2bd+2cd$$

In the superposition, this generates the needed $$r^{-3}$$ term as the nonlinear expansion. However, it also leaves a $$r^{-4}$$ as a side effect; If I eliminate that, by trying the obvious next negative power of r, it will generate 1 more term I have to solve...

$$\varphi=\frac{k}{r^{2}}$$ generates $$K\left\{ \frac{4k^{2}}{r^{6}}-\frac{4k}{r^{4}}+\frac{6k}{r^{4}}\right\} =K\left\{ \frac{4k^{2}}{r^{6}}+\frac{2k}{r^{4}}\right\}$$
In general:
$$\varphi=\frac{k}{r^{n}}$$ generates $$K\left\{ \frac{n{}^{2}k^{2}}{r^{2n+2}}+\frac{n\left(n-1\right)k}{r^{n+2}}\right\}$$

eg: using 1/r squared to generate a term to eliminate r**-4, works but it will add another term r**-6 that needs to be eliminated ... ending in an infinite series of corrections.

So, I'm stuck. Is there an easier way to figure out an analytical solution?

 

[gato_]

any available boundary conditions?
I only find consistent solutions for K<0. That said, if the domain of application includes r->0, I have reasons to think the solution does not apply. After some trial, I tried brute force, and it turns out that maple gives an exact solution. redefining r as $$r\equiv -Kr$$, the equation reads
$${r}^{3}{\frac {d^{2}}{d{r}^{2}}}\phi \left( r \right) +2\,{r}^{2}{ \frac {d}{dr}}\phi \left( r \right) +{r}^{3} \left( {\frac {d}{dr}} \phi \left( r \right) \right) ^{2}+1$$
for which a general solution is
$$ln[C_{1}\frac{J_{1}(2/\sqrt{r})}{Y_{1}(2/\sqrt{r})J_{0}(2/\sqrt{r})}+C_{2}\frac{Y_{1}(2/\sqrt{r})}{J_{1}(2/\sqrt{r})Y_{0}(2/\sqrt{r})}]$$
which has an infinite number of singularities as r->0. So the validity of the approximation will depend on where you are applying the solution.

 

[andrewr]

any available boundary conditions?
I only find consistent solutions for K<0. That said, if the domain of application includes r->0, I have reasons to think the solution does not apply. After some trial, I tried brute force, and it turns out that maple gives an exact solution. redefining r as $$r\equiv -Kr$$, the equation reads
$${r}^{3}{\frac {d^{2}}{d{r}^{2}}}\phi \left( r \right) +2\,{r}^{2}{ \frac {d}{dr}}\phi \left( r \right) +{r}^{3} \left( {\frac {d}{dr}} \phi \left( r \right) \right) ^{2}+1$$
for which a general solution is
$$ln[C_{1}\frac{J_{1}(2/\sqrt{r})}{Y_{1}(2/\sqrt{r})J_{0}(2/\sqrt{r})}+C_{2}\frac{Y_{1}(2/\sqrt{r})}{J_{1}(2/\sqrt{r})Y_{0}(2/\sqrt{r})}]$$
which has an infinite number of singularities as r->0. So the validity of the approximation will depend on where you are applying the solution.

Thanks! The application does include r near zero, but it need not go all the way to zero because that location affects the final result very little. Maple is correct, in general K is negative. The redefinition of the equation looks like a simple multiply through by r**3, so that's fine. I think the form of the solution has enough information for me to make the correction I need -- the parameter varies as the square root of r, and the two terms inside the log have a reciprocal relationship in two of the functions

As far as boundary conditions go, I actually have a solution to a slightly different equation that's too messy to post... so I think I'll just try to use what you provided here. I appreciate the help .

 

[gato_]

Well, I have to warn you that the domain does not need to go to 0 to imply a problem. Any arbitrary neighbourhood of 0 contains singularities... I think, in that case the approximation is not valid. If this has been derived from a perturbation expansion, I'm affraid the solution is telling you it breaks up, so I'd be careful. A singular perturbation approach would be better. What is this problem about?

 

[andrewr]

Well, I have to warn you that the domain does not need to go to 0 to imply a problem. Any arbitrary neighbourhood of 0 contains singularities... I think, in that case the approximation is not valid. If this has been derived from a perturbation expansion, I'm affraid the solution is telling you it breaks up, so I'd be careful. A singular perturbation approach would be better. What is this problem about?

Yes, I understand.

The problem is a reformulation of the Schrodinger Equation (concentrating on the time invariant version for now). I have been studying the idea of spin, and alternate models of what it could be. You can see a link here giving a crude idea of the equation's origin.
https://www.physicsforums.com/showpost.php?p=3152874&postcount=39"
And the 2 posts following that one have links to the most relevant literature, but by no means the whole story...

In general, one of the ways to gain understanding is to work an equation through under various circumstances to become familiar with what kind of solutions appear, and how they relate to various models. So, that is what I have been doing. The equation I am showing you is simply the Schrodinger equation with a substitution of variables that I have found useful in solving non-spin related equations numerically using some numerical methods I developed for another project some years ago.

The solutions of the equation are almost trivial in the cases where only the first and second negative powers of r are considered. But I have found quite often, when trying alternate models of "spin", that a cubic term becomes one way to propose a possible solution, yet that approach dead ended as I mentioned at the start -- in an infinite series. So, I am simply studying the equation by trial and error at the moment to allow my intuition time and material to sift through in order to gain a clearer idea of why the equation blows up. In physics, dipole moments go as the r**-3 power, and so it seems strange to me that the Schrodinger equation would have difficulties with it. The solutions of the Schrodinger equation are very similar to those in Laplace's equation in classical electrodynamics, although the solutions are real in classical electrodynamics. I have been able to find real solutions in place of eigenvectors that are imaginary because the problem appears to be very symmetrical. eg: if a solution exists, there is almost always exists another answer which is a complex conjugate of the first one -- and therefore the two of them added together appropriately can construct a third 'real' solution.

I'm sorry my mathematical vocabulary is limited, my brother's the mathemetician in the family. I'm an engineer.
In any event, trying various methods of attack on the problem help expand the mind -- so long as my eyes don't glaze over...

The solution you proposed has a form that I have seen repetitively. There is a function plus its' inverse (**-1) embedded in the solution. Historically, I have found those kinds of solutions embedded in exponentials / erf(). although the precise location of the form you have shown me still eludes me.
It's a matter of time before my memory kicks out the connection, but I have to wait... :) I think, however, that Schrodinger's equation is also very similar to the heat transfer problems which gave Forier his impetus to come up with Forier series solutions. So, there is a specific area of overlap that is significant between wave and heat equations that I need to explore also.

Ciao!

 

[gato_]

So, correct me if I am wrong, you are trying to solve the schrodinger equation with a potential $$V(r)=-K/r^{3}$$, and 0 angular momentum and energy. Without making the substitution $$\psi=exp(\phi)$$, that is:
$$\frac{d}{dr}(r^{2}\frac{d\psi}{dr})+\frac{K}{r^3}\psi=0$$
for which a solution is
$$\psi=C_{1}\frac{J_{1}(\sqrt{8K/r})}{\sqrt{r}}+C_{2}\frac{Y_{1}(\sqrt{8K/r})}{\sqrt{r}}$$
Did I understand right?

 

[andrewr]

So, correct me if I am wrong, you are trying to solve the schrodinger equation with a potential $$V(r)=-K/r^{3}$$, and 0 angular momentum and energy. Without making the substitution $$\psi=exp(\phi)$$, that is:
$$\frac{d}{dr}(r^{2}\frac{d\psi}{dr})+\frac{K}{r^3}\psi=0$$
for which a solution is
$$\psi=C_{1}\frac{J_{1}(\sqrt{8K/r})}{\sqrt{r}}+C_{2}\frac{Y_{1}(\sqrt{8K/r})}{\sqrt{r}}$$
Did I understand right?

It would be equivalent to trying to solve the equation for a potential like you are mentioning, yes.
In General, though, it would be summed with other solutions -- so that the cubic potential you are seeing doesn't have to be the only term on that side of the equation. It is just the one that was causing me difficulty in solving.
The substitution of variables that I made, merely changes psi that you list from the exponential form -- to one where only the phase is used. (eg: a log of a ratio of exponentials). Normalization is not required, unless explicitly wanted and it is very simple to work with numerically. As a side effect, it makes the equation look nonlinear -- but the squared term acts like a components of a vector; where the length is a simple sum of the squares of the components. So that term actually adds with other square terms from other solutions. I'm not sure of the exact word to use for this type of combination -- it isn't linear in the mathematical sense of "going through zero", but it has the properties similar to adding lines with nonzero intercepts -- to produce another line with a different intercept -- eg: something that would be used in linear algebra.



You are right, there is zero angular momentum explicitly applied to the problem. However, I wasn't sure if placing the cubic term there would require a square law term in order to solve the equation. If it does, then the solution may in fact have angular momentum of some value related to the cube. That is why I am exploring it -- to get an idea if there is an obscure relationship. Having a zero energy potential is arbitrary, as I mentioned in the formulation of the problem it could be off by a constant if that helped solve the problem, or even a square law term.. etc..

In engineering, a typical approach to problems which are "linear" or close to it in nature is to inject a stimulus, record the response, and then try to form approximate solutions from weighted summations of the responses -- often it is called the linear network, or black box, approach. Often times, in simulations, the Laplace transform is used to translate a linear network with frequency dependent components into a space where the solutions can be found by simple algebraic manipulation. Eg: the program "Spice" is such a simulator. However, the Laplace transform actually makes the program unstable in many practical circuits which I know work in practice -- but which will not simulate correctly.
I replaced the Laplace transform by a numerical technique that provided a very stable simulator to work around the problem. I hope to develop something similar to work with the Schrodinger equation and atomic orbitals -- but I have to understand the nature of the equation first, before I can design anything really useful.