# How many solutions?

1. Aug 19, 2007

### scotlass

Hello, this is my first post so I am not familiar with how to write maths in here.

I have a third order ode of the form

y'''+A/y=0 where y=y(x) (A is just a constant) with well defined boundary conditions.

I believe there is no analytic solution to an ode of this form, but I can (by imposing a further condition) solve this numerically for y.

My question is this; I solve this numerically in Maple and one answer pops out. Is there any way I can easily prove this is the only real, non-negative solution?

2. Aug 19, 2007

### HallsofIvy

Staff Emeritus
I can't think of any way to answer your question because I can't imagine why you would "I believe there is no analytic solution to an ode of this form". It seems to me pretty obvious that there exist a solution as long as the boundary conditions do not require y= 0. Perhaps you could explain a little more exactly what you mean.

You don't say WHAT boundary conditions you have. Whether or not a boundary value problem has a unique solution depends strongly on the boundary conditions themselves.

For example, the problem y"+ y= 0, with boundary conditions, y(0)= 0, y(1)= 0 has a unique solution. The same equation with boundary conditions y(0)= 0, $y(\pi)= 1$ has no solution. The same equation with boundary conditions y(0)= 0, $y(\pi)= 0$ has an infinite number of solutions.

Last edited: Aug 20, 2007
3. Aug 20, 2007

### scotlass

Thanks for the reply,

my bcs are y(-1)=0, y(1)=0, y'(-1)=beta1, y'(1)=(beta1^2+4A)^(1/2) and y must also satisfy (integral y dx, x=-1..1)=2/3.

I have a third order ode, 2 unknowns and 5 conditions so I have enough for a solution.

This does mean y''' is singular at x=-1, 1 but that is ok.

As far as I can see this cannot be solved analytically, but I can find one numerical solution, can I know there are no other solutions that are everywhere real and positive?

4. Aug 21, 2007

### Matthew Rodman

Hm. All I can think of is that your equation may be derived from

$$y y^{\prime \prime} + A x + B - \frac{y^{\prime 2}}{2} = 0$$

where B is a constant. But

$$y y^{\prime \prime} - \frac{y^{\prime 2}}{2} = 2 y^{\frac{3}{2}} (\sqrt{y})^{\prime \prime}$$

Therefore you can re-write your equation

$$u^{\prime \prime} + \frac{Ax + B}{2 u^{ 3}} = 0$$

with the substitution $$u = \sqrt{y}$$

...and with a suitable sub for x, you should get a standard Emden-Fowler equation. See here. (EqWorld)

Last edited: Aug 21, 2007