Y''=y^2-exp[y], Y=y(x)

johnk

22nd October 2009 - 10:17 PM

Please,
I have to found
a point x=x_o
where the solution y(x) of equation:
y''=y^2-exp[y], y=y(x).
is oscillatory

I have think the rule (y''/y)<0...

thanks...
johnk.

rpenner

23rd October 2009 - 12:02 AM

I think you problem is incorrectly stated.

Your problem is in the form y'' = f(y) and has no dependence on x. Therefore no x is special in the general solution.

Your one-trick pony is here. Can we solve it using our new trick for y'' =f(y) ?

(d^2y/dx^2) = d/dy [(1/2)(dx/dy)^-2] = y^2 - e^y

(dx/dy)^-2 = 2 ∫ y^2 - e^y dy = (2/3) y^3 - 2 e^y + A

± (dx/dy) = 1/sqrt((2/3) y^3 - 2 e^y + A)

± (∫ 1 dx) = ± (x + C) = ∫ 1/sqrt((2/3) y^3 - 2 e^y + A) dy

Which is not going to happen. Time to get more descriptive.

There is a certain number p, p is the only real solution of p^2 = e^p and is about -7/10.

y = p is the only solution of y'' = y^2 - e^y = 0 so it is a fixed point when y = 0.

And for y> p, y'' < 0 and for y < p, y'' >0, and is strictly increasing in magnitude.

Thus y''/(y-p) < 0 when y is not p.

Thus p is the only possible stable asymptotic value of y. But it is not stable, and y oscillates everywhere but exactly at p.

Trout

23rd October 2009 - 03:24 PM

QUOTE (rpenner+Oct 23 2009, 12:02 AM)

I think you problem is incorrectly stated.

Your problem is in the form y'' = f(y) and has no dependence on x. Therefore no x is special in the general solution.

Your one-trick pony is here.

You really like this trick :-)
(I do, too)

johnk

28th October 2009 - 06:39 AM

An other explanation is:
First expand with Taylor,
y''=f(p)+f'(p)y+f''(p)y^2/2!+...
suggest |y|<<1,
y''=f(p)+f''(p)y
y''-f(p)-f''(p)y=0
with
-f(p)=0, and -f''(p)<0
first equation gives,
f(p)=0
p^2-e^(2p)=0.
p=-7/10

johnk

28th October 2009 - 06:44 AM

any suggestion of solving:
(k*p)^2-e^(2p)=0,
I know there are
1) one solution for k<e
2) tree solution for k>e
but how?

Trout

28th October 2009 - 07:06 AM

QUOTE (johnk+Oct 28 2009, 06:44 AM)

any suggestion of solving:
(k*p)^2-e^(2p)=0,
I know there are
1) one solution for k<e
2) tree solution for k>e
but how?

This equation is transcendental, so there is no closed solution.
Let f(p)=(k*p)^2-e^(2p)
f'(p)=2k^2p-2e^(2p)
Use the hint that between two solutions of the equation f'(p)=0 there is at most one solution for f(p)=0.

Trout

28th October 2009 - 05:45 PM

QUOTE (Trout+Oct 28 2009, 07:06 AM)

One more hint:

f(p)->+oo for p->-oo
f(p)->-oo for p->+oo
f'(p)->-oo for p->-oo
f'(p)->-oo for p->+oo
f''(p)=0 has one root only

How many roots does f'(p)=0 have? What does this tell you about the roots of f(p)=0?

Trout

28th October 2009 - 10:58 PM

QUOTE (johnk+Oct 28 2009, 06:44 AM)

any suggestion of solving:
(k*p)^2-e^(2p)=0,
I know there are
1) one solution for k<e
2) tree solution for k>e
but how?

I don't think that this is correct.

PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.

To quit out of "lo-fi" mode and return to the regular forums, please click here.