- #1
andrewr
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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.
[tex]\varphi[/tex] 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.
[tex]
K\left\{ \varphi_{r}^{2}+\frac{2\varphi_{r}}{r}+\varphi_{rr}\right\} =\frac{1}{r^{3}}
[/tex]
I am not able to find a solution by inspection. However I did find that:
[tex]\varphi=k*ln(r)[/tex] generates [tex]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\} [/tex]
and
[tex]\varphi=\frac{k}{r}[/tex] generates [tex]K\left\{ \frac{k^{2}}{r^{4}}-\frac{2k}{r^{3}}+\frac{2k}{r^{3}}\right\} =K\left\{ \frac{k^{2}}{r^{4}}\right\} [/tex]
So that I figured the superposition of these would generate the non-linear [tex]r^{-3}[/tex] and it does:
[tex]\frac{k}{r}+ln\left(r\right)[/tex] generates [tex]K\left\{ \left(\left\{ \frac{k^{2}}{r^{4}}\right\} +\frac{2}{r^{2}}\right)-2\left(\frac{k}{r^{3}}\right)\right\} [/tex]
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:
[tex]
(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2ab+2ac+2ad+2bc+2bd+2cd
[/tex]
In the superposition, this generates the needed [tex]r^{-3}[/tex] term as the nonlinear expansion. However, it also leaves a [tex]r^{-4}[/tex] 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...
[tex]\varphi=\frac{k}{r^{2}}[/tex] generates [tex]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\} [/tex]
In general:
[tex]\varphi=\frac{k}{r^{n}}[/tex] generates [tex]K\left\{ \frac{n{}^{2}k^{2}}{r^{2n+2}}+\frac{n\left(n-1\right)k}{r^{n+2}}\right\} [/tex]
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?
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.
[tex]\varphi[/tex] 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.
[tex]
K\left\{ \varphi_{r}^{2}+\frac{2\varphi_{r}}{r}+\varphi_{rr}\right\} =\frac{1}{r^{3}}
[/tex]
I am not able to find a solution by inspection. However I did find that:
[tex]\varphi=k*ln(r)[/tex] generates [tex]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\} [/tex]
and
[tex]\varphi=\frac{k}{r}[/tex] generates [tex]K\left\{ \frac{k^{2}}{r^{4}}-\frac{2k}{r^{3}}+\frac{2k}{r^{3}}\right\} =K\left\{ \frac{k^{2}}{r^{4}}\right\} [/tex]
So that I figured the superposition of these would generate the non-linear [tex]r^{-3}[/tex] and it does:
[tex]\frac{k}{r}+ln\left(r\right)[/tex] generates [tex]K\left\{ \left(\left\{ \frac{k^{2}}{r^{4}}\right\} +\frac{2}{r^{2}}\right)-2\left(\frac{k}{r^{3}}\right)\right\} [/tex]
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:
[tex]
(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2ab+2ac+2ad+2bc+2bd+2cd
[/tex]
In the superposition, this generates the needed [tex]r^{-3}[/tex] term as the nonlinear expansion. However, it also leaves a [tex]r^{-4}[/tex] 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...
[tex]\varphi=\frac{k}{r^{2}}[/tex] generates [tex]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\} [/tex]
In general:
[tex]\varphi=\frac{k}{r^{n}}[/tex] generates [tex]K\left\{ \frac{n{}^{2}k^{2}}{r^{2n+2}}+\frac{n\left(n-1\right)k}{r^{n+2}}\right\} [/tex]
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?
Last edited: