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CE Adamson

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Anybody have any suggestions on how to attack the problem?

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- Thread starter CE Adamson
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- #1

CE Adamson

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Anybody have any suggestions on how to attack the problem?

- #2

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Peter J. Olver, has some of the most comprehensive texts on the subject. See "Symmetry Groups of Differential Equations" by him.

Specifically a symmetry transformation which maps the equation to itself will map solutions to solutions. By the looks of your k/t solution it looks like you have a scaling symmetry: [itex] f\mapsto \alpha f, t\mapsto \alpha t[/itex]. It maps the equation to an equivalent one and if you divide the whole ODE by [itex] t[/itex] this scaling transformation will map the equation exactly to itself. Often insights into other symmetries can be found in the original application. Look for natural symmetries to your geometry problem.

I'll play with this a bit and see if I can spot anything helpful.

[edit:] This looks a little familiar. You playing with GR metrics? [itex] f = r[/itex]?

- #3

romsofia

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l.

[edit:] This looks a little familiar. You playing with GR metrics? [itex] f = r[/itex]?

This looks like some Ricci tensor DEQs

- #4

CE Adamson

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... The only way I know of to tackle such equations in general is to look for symmetries...

Hmmm... interesting. Not sure if it's what you intended, but gets me thinking there might be a coordinate transform that would result in a DEQ that's easier to solve even if it looks more complicated. (Yes, this is from a GR metric I've been playing with in my spare time :) ) Though I'm not entirely hopeful as I think the form above is about as simple as it gets...

- #5

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The existence of the scaling symmetry I mentioned should mean you can reduce the order by 1. If my review gets to the point where I remember how to implement that I'll post it. It should be something along the lines of the change of dependent variable of [itex] g(t) = f(t)/t[/itex].

Hmm...

[tex] f'' + \frac{2}{t} f' + \frac{f'^2}{t(1-f/t)} + \frac{f'f}{t^2(1-f/t)}=0[/tex] times [itex]t[/itex]

becomes

[tex] tf'' + 2 f' + \frac{f'^2}{1-f/t} + \frac{f'f}{t(1-f/t)}=0[/tex]

which is invariant under ##f\to af, t\to at##. Now applying the change of var:[itex] f = tg, f' = g+tg', f'' = 2g' + tg''[/itex]

gives us

[tex]2tg'+t^2g'' + 2g+2tg' + \frac{g^2+2tgg'+t^2g'^2}{1-g} + \frac{(g+tg')g}{(1-g)}=0[/tex]

or

[tex]t^2g'' + 2g+4tg' + \frac{(g+tg')^2}{1-g} + \frac{(g+tg')g}{(1-g)}=0[/tex]

It's still nastily non-linear but notice then you can replace the derivative operator with ## D = t\frac{d}{dt}## or equivalently make the change of independent variable: ## t = e^\tau## wherein

[tex]\frac{d}{d\tau} = \frac{dt}{d\tau} \frac{d}{dt} = t \frac{d}{dt}[/tex]

carefully (I about skipped this part!) we need also to work out the 2nd derivative:

[tex]\frac{d^2}{d\tau^2} = \frac{d}{d\tau}[t\frac{d}{dt}] = t\frac{d}{dt}[t\frac{d}{dt}] = t\frac{d}{dt}+t^2\frac{d^2}{dt^2}[/tex]

Thence using the notation ##\dot{u}=\frac{du}{d\tau}## the equation becomes:

[tex]\ddot{g}-tg' + 2g+4tg' + \frac{(g+tg')^2}{1-g} + \frac{(g+tg')g}{(1-g)}=0[/tex]

or

[tex]\ddot{g}+ 2g+3\dot{g} + \frac{(g+\dot{g})^2}{1-g} + \frac{(g+\dot{g})g}{1-g}=0[/tex]

So now there's no explicit occurrence of the independent var. From there a lot of algebraic possibilities should be popping out at you.

[tex](1-g)(\ddot{g}+ 2g+3\dot{g}) + (g+\dot{g})^2+ g(g+\dot{g})=0[/tex]

We should now be able to write it as a first order system (separable?) system. ##\dot{g}= -h## (minus sign so I don't have to put many in below).

[tex]\begin{array}{c} \dot{h}= 2g+3h + \frac{(g+h)^2}{1-g} + \frac{(g+h)g}{1-g}\\ \dot{g}=-h\end{array}[/tex]

I can feel an integration just out of reach but I'm tired and under the influence of a bit of bourbon. Play with this and see what you get. And crucially check my work as I well may have made at least 1 error here. But again, the 1 parameter scaling symmetry of the original equation should indicate one level of reduction here. I'll meditate further.

- #6

CE Adamson

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$$ ( \ddot{g}+ \dot{g} ) + 2(\dot{g} +g) + \frac{(\dot{g} +g)^2}{1-g} + \frac{g(\dot{g} +g)}{1-g}=0 $$

Looks so tantalizingly close :) Trying to get to ## h = \dot{g} + g## and ## \dot{h} = \ddot{g} + \dot{g}##...

However, I'm not sure I follow about the scaling symmetry. If I just make the straightforward substitution ##g = af, u=at## in the original equation, I get

$$

\frac{g''}{a} + \frac{2}{u}g' + \frac{g'^2}{a \left( 1 - \frac{g}{u} \right) u } + \frac{g'g}{\left(1- \frac{g}{u} \right) u^2} = 0

$$

which is obviously equivalent but has somewhat different form than the original. Can you explain what the "invariance" means?

- #7

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[tex] \frac{(af'/a)(af)}{(1-\frac{af}{at})(at)^2} = \frac{1}{a} \frac{f'f}{(1-\frac{f}{t})t^2}[/tex]

As it stands you don't get exactly the same equation, you get a ##1/a## multiple of the original equation which is equivalent. But if you simply rewrite the original equation times ##t## (or times ##f##) then you'll note that

The scaling symmetry generator is: ##\Xi=x\partial_x + t\partial_t## where ##x=f(t)## is a solution. The trick is to find new dependent and independent vars ##y,s## such that in this new set of coordinates ##\Xi = \partial_s##... or so the books say. Hmmm... in terms of polynomial terms you can think of scaling operator as the degree operator in that ##\Xi ax^p = p\cdot ax^p## and ##\Xi bt^q = q\cdot bt^q##. In particular ##\Xi(x/t) = 0## and ##\Xi \sqrt{xt} = \sqrt{xt}##.

So I think the transformation: [itex] s=\sqrt{xt}, y=x/t[/itex] should have the effect of transforming the equation to one without any explicit occurrence of ##y## and thus it will become a 1st order equation in ##u=y'##.

I'm working here on very shaky memory and you should read up on the method and check my assertions. I'm not inclined to work this out myself in the near future but may get curious again and throw some graphite at it.

But do a search on Symmetry method for differential equations and you'll find no shortage of references. There are some good pdf's you can download out there but I am very uncertain about their license and so refrain from posting them, or links to them here.

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