# Apparently this 2nd-order ODE has 3 solutions?

1. Sep 30, 2004

### Farina

Apparently this 2nd-order ODE has 3 solutions??

The following apparently has 3 solultions:

$$\frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})$$

where:

u = 1/r
m = mass
l = angular momentum

One of the solutions is:

$$r=r_0e^{k\theta} \text{ where } \theta \text { varies logarithmically with time}$$

Apparently there are also 2 additional solutions (depending on the value of the constant $$\alpha$$)
that could be in the form of:

$$r=Ae^{\sqrt{\alpha x}}+Be^{-\sqrt{\alpha x}} \text{ or }$$
$$r=A\theta + B \text{ or }$$
$$r=Asin({\sqrt{\alpha x}})+Bcos({\sqrt{\alpha x}})$$

So, knowing:

$$\frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})$$

and

$$r=r_0e^{k\theta}$$

How does one specifically determine the equations of the additional solutions?

Thanks!

2. Sep 30, 2004

### NateTG

I saw your other post, and really wasn't able to understand the equation. What function is $$f$$?

Something like:

$$\frac {d^2}{d\theta^2}u(\theta)^2+ u(\theta) = -\frac {1}{ml^2u(\theta)^2}$$

Makes more sense, but if you plug in $$u(\theta)=\cos \theta$$ which represents a solution to your original problem you get:
$$-\cos \theta + \cos \theta = k \sec(\theta)^2$$
or
$$0 = k \sec(\theta)^2$$
which is...problematic.

Perhaps you mean that $$f(u^{-1})=f(r)=-\frac{1}{r^3}=-u^3$$
then your eqation simplifies to:
$$\frac {d^2}{d\theta^2}u(\theta)^2+ u(\theta) = \frac {u(\theta)}{ml^2}$$
or
$$\frac{d^2}{d\theta^2}u(\theta)^2 + (1-\frac{1}{ml^2}) u(\theta)=0$$
which has solutions of the form
$$u(\theta)=k_1e^{i(\theta)\sqrt{1-\frac{1}{ml^2}}}$$
so
$$r(\theta)=k e^{-i(\theta) \sqrt{(1-\frac{1}{ml^2}}}$$
It's relatively easy to see that if $$1-\frac{1}{ml^2} < 0$$ the functions are not orbits.

3. Oct 3, 2004

### Farina

I think you've addressed this NateTG, but I still
have some digesting to do...

The following is basically Newton's DE of motion using two-dimensional polar coordinates instead of three:

$$\frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})$$

where:

u = 1/r
m = mass
l = angular momentum
$$f(u^{-1}) \text { = the central force}$$

And, yes, l = angular momentum per unit mass (thanks ehild!).

There is a fair amount of derivation provided in the text. The DE is labled as "the differential equation of the orbit of a particle moving under a central force."

Setting u = 1/r you thus have:

$$u={r_0}^{-1}e^{-k\theta}$$

Referring back to the DE and using

$$\text{u, } \frac {d^2u}{d\theta^2}$$

then simply re-arranging and solving algebraically
for $$f(u^{-1} )$$ you get

$$f(r)=-(k^2+1)ml^2\cdot \frac {1}{r^3}$$

which shows that the central field corresponds to an inverse-cube force.

I'm stuck at the proposition of having to find two other types of possible orbits (and their equations). I was thinking it was a simple matter of expressing the solution to the DE in basic alternative forms -- but maybe not??

4. Oct 3, 2004

### e(ho0n3

If the diff. equation is of the form au'' + bu' + cu = 0, then shouldn't it only have two possible solutions? Unless of course you are cosidering u = 0 a solution.

5. Oct 3, 2004

### cj

I was thinking the same thing, but I also seem to
remember that DE of this form have a bunch of solutions
based on the sign of a combination of constants.

I guess I need to refresh my working knowledge of DE.

6. Oct 4, 2004

### ReyChiquito

Your problem is incomplete. You need two boundary conditions in order to set a unique solution for the ode. Also you need to consider under wich circumstances au'' + bu' + cu = 0 + (boundary conditions) has a solution and if this solution is unique. (Existence and Uniqueness Theory).

In the case that au'' + bu' + cu = 0 + (boundary conditions) satisfies the uniqueness and existence theorem, the there is only two l.i. solutions to the ode.

On relation to 1 If you already founded a solution, use the reduction of order method to find the second solution ie
$$y_{2}(x)=v(x)y_{1}(x)$$

Last edited: Oct 4, 2004
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