# Anyone ever work with a DE like this?

1. Nov 29, 2009

### Mosis

This has come up in my research, and my supervisor and I don't really know how to proceed. It reads

$$r^2\frac{d^2Y}{dr^2} + r\frac{dY}{dr} - \left(\frac{3}{2}\right)^2Y = Cr^3Y^{1/3}$$

I know the RHS is an equidimensional DE which has the nice solution $$Y = r^{\pm 3/2}$$, but I have no idea how this could help with that nasty $$Y^{1/3}$$ term sitting on the RHS.

Really, we just want to learn SOMETHING about this DE. What do the solutions look like? Does a bounded solution exist? What can we do if C is small? etc.

Any ideas?

2. Nov 29, 2009

### Mosis

wow, 46 views, no reply - is this really that uncommon?

3. Nov 29, 2009

### HallsofIvy

Staff Emeritus
The problem is that that is a seriously non-linear equation. Since solutions of non-linear equations do not add to another solution, Knowing a solution to LHS= 0 does not help. For C small, you might try a perturbation method, writing the solution as a series in powers of C: $y_0+ Cy_1+ C^2y_2+ \cdot\cdot\cdot$. Substituting that into the equation and equating like powers of C will give, for the "C0" term, the homgeneous equation as the equation for $y_0$. Then the "C1 term will gvie a non-homogeneous equation for $y_1$ with the (now known) $y_0$ on the right side, etc.

Last edited: Dec 11, 2009
4. Nov 29, 2009

### Mosis

I was thinking about that kind of perturbation scheme, but the fact that we have Y^1/3 makes that intractable i.e. what to do with $$\left(Y_0 + CY_1\right)^{1/3}$$?

5. Nov 29, 2009

### hamster143

You could express it as $$Y_0^{1/3} (1+C(Y_1/Y_0)+C^2(Y_2/Y_0)+...)^{1/3} = Y_0^{1/3}(1 + \frac{1}{3}(C(Y_1/Y_0)+C^2(Y_2/Y_0)...) - \frac{1}{9}(C(Y_1/Y_0)+C^2(Y_2/Y_0)...)^2 ...$$

because $$(1+x)^a = 1 + xa + x^2 a(a-1)/2! + x^3 a(a-1)(a-2)/3! + ..., x \ll 1$$

It will be incredibly messy beyond the first two terms, but at least you can get some results.

By the way, it would simplify your calculations if you did a substitution $$z = \ln{r}$$.

Last edited: Nov 29, 2009
6. Nov 30, 2009

### hamster143

Further thoughts.

There are two distinct cases, C>0 and C<0.

Once you rewrite your expression as r = ln z, $$Y'' = 9/4 Y + C e^{3z} Y^{1/3}$$,

it's easy to see that, as long as C>0, all nontrivial solutions are unbounded (because the second derivative is always positive, therefore the graph of the function is convex).

Every individual solution will look like a bunch of exponentials a*exp(b*z) stitched together, depending on which of the two terms on the right hand side dominates (most of the time, one of the two terms would be negligible). Draw a plot of the solution of the equation $$9/4 Y = C e^{3z} Y^{1/3}$$. Use log scale on the vertical axis. Think about the behavior of a solution of your DE that passes through a point to the left of the curve. See if you can prove (or at least convince yourself) that almost all solutions either blow up because Y crosses below zero, or diverge as exp(-1.5 z) as $$z\rightarrow -\infty$$ and as exp(4.5 z) as $$z\rightarrow +\infty$$.

If C<0, things are more complicated, and many solutions would blow up at some point because there's nothing to keep Y from going negative.

Last edited: Nov 30, 2009
7. Nov 30, 2009

### Mosis

why are you assuming Y is positive? also, exp(-r) is convex but decays as r goes to infinity. Of course, it blows up in the other direction, but in my case, Y = Y(r), and r > 0, so I'd really need some information about the first derivative to conclude all nontrivial solutions are unbounded.

8. Nov 30, 2009

### Mosis

also, I'm not sure why you suggest plotting the solution to that equation - how can that help me?

9. Nov 30, 2009

### hamster143

Because Y^(1/3) is only defined for positive Y.

Yes, exp(-r) is convex. But notice my variable change: from r that goes 0..infinity, to z that goes from minus infinity to infinity.

If you have a function whose argument goes from minus infinity to infinity and its graph is convex, it must be unbounded on either side.

Because either of two terms on the right hand side would dominate, depending on which side of the graph you are. So, as long as you're sufficiently far to the left, solutions of DE would behave as if there's no pesky cubic root term.

10. Dec 1, 2009

### trambolin

I agree with hamster143, even it does not help you in terms of the details it will give you a lot of information about the vector field. My suggestion is setup your DE in MATLAB or your favorite solver while keeping it nonlinear, and let the solutions from many (I mean many) initial conditions and get a feeling for the differential equation. You can even read off if there is a bounded solution or not.

You never know! It might act like almost linear in the region of interest. (Happened to me also)

11. Dec 2, 2009

### gato_

you can further write that using the fact that $$Y=exp(9z/2)$$ is an exact solution, and write $$Y=exp(9z/2)y(z)$$, with :
$$y''+9y'+18y-Cy^{1/3}$$
which is autonomous, and proceed by local analysis.

12. Dec 2, 2009

### gato_

you can check that the critical points are
$$y = y'=0$$, and $$y=(-C/18)^{3/2}, y'=0$$
the second stationary points leads to a non bounded solution Y(z). The jacobian at (0,0) is not defined, but we have the inequality
$$[y'^{2}/2+9y^{2}-Cy^{4/3}3/4]'=-9y'^{2}$$
So, for C<0, y=0=y' is stable, and you might have bounded solutions
* beware when simulating that in matlab, y^(1/3) does not give the proper solution. try sign(y)*abs(y)^1/3 instead.

13. Dec 11, 2009

### cheky

what do you mean by a convex graph