# Ode problem

1. May 16, 2014

### joshmccraney

1. The problem statement, all variables and given/known data
$$y'' + \frac{1}{x}y' - \lambda y = 0$$
where $x \to \infty \implies y \to 0$ and $x \to 0 \implies y' \to 0$

3. The attempt at a solution
to begin, this was initially a pde, and i've applied separation of variables. to solve this ODE, it seems i cannot assume $y=e^{rx}$ since the $x^{-1}$ term is present. i've thought of a series solution like $\Sigma_{-\infty}^{\infty} A_n x^n$ but the boundary conditions are bringing me to a stop.

any advice would be great! how would you solve this??

2. May 16, 2014

### pasmith

Multiplying by $x^2$ gives $$x^2 y'' + xy' - (\lambda x^2)y = 0$$ which if $\lambda = -k^2 < 0$ can be turned into Bessel's equation of order zero by a rescaling of the independent variable; if $\lambda = k^2 > 0$ the ODE can be turned into the modified Bessel's equation of order zero by the same rescaling.

Don't forget the case $\lambda = 0$.

3. May 16, 2014

### joshmccraney

thanks! could you help me with the rescaling? i'm new to this and not sure how. i was thinking of letting $x = z / \lambda ^{1/2}$ but this doesn't look like it will help me beyond the $y$ term.

thanks so much for the help!

4. May 16, 2014

### Staff: Mentor

It should help you on all three terms. Don't forget, λ is a constant.

Chet

5. May 16, 2014

### pasmith

If $z = kx$, then the chain rule gives $\dfrac{dy}{dx} = k \dfrac{dy}{dz}$. But $x = z/k$, so $x\dfrac{dy}{dx} = z\dfrac{dy}{dz}$.

6. May 16, 2014

### joshmccraney

oh my gosh, of course! $\frac{dy}{dz} = \frac{dy}{dx} \frac{dx}{dz} \implies \frac{dy}{dz} / \frac{dx}{dz} = \frac{dy}{dx}$

thanks to both of you! (i made a multiplication error). sorry, i should at least triple check before I ask for more help.

thanks!

7. May 17, 2014

### joshmccraney

i did have another question for you both, if you don't mind?

it seems obvious that the Bessel function of first and second kind on order of zero solve the bessel equation, and it also seems, even with the independent variable change, my boundary conditions stay the same: $z \to \infty \implies y \to 0$ and $z \to 0 \implies y' \to 0$

my question is, if $y(z) = a_1 J_0 (z) + a_2 Y_0 (z)$ where $J_0 (z)$ and $Y_0(z)$ are bessel functions of first and second kind, respectively, then to satisfy $z \to 0 \implies y' \to 0$ we need to set $a_2 = 0$ from the natural logarithm in $Y_0$; is this correct?

if so, we now have $$y(z) = a_1J_0(z)$$ where $z \to \infty \implies y \to 0$ governs $a_1$.

i have two questions left (assuming the above is correct). firstly, the bessel function does converge (by the alternating series test i believe), but what does it converge to? wouldn't i need to know this to fully solve?

and secondly, as i mentioned earlier, Bessels equation arises from a PDE, where i have taken separation of variables. if so, whats the point of solving $a_1$ since we will have more constants from the other ODE (much easier to solve)?

this being said, i'm still real curious on how to match $a_1$ to the b.c.

thanks so much!

8. May 17, 2014

### pasmith

Both $J_0(x)$ and $Y_0(x)$ are oscillatory with the amplitude decaying so that $J_0(x) \to 0$ and $Y_0(x) \to 0$ as $x \to \infty$. However $Y_0 \to -\infty$ as $x \to 0$, and so cannot be used if x = 0 is in the region you are interested in.

Can you post the original PDE and boundary conditions?

You cannot determine $a_1$ from the eigenvalue problem; a constant multiple of an eigenfunction is also an eigenfunction. It is simplest to take $a_1 = 1$.

It depends on the nature of the boundary conditions. If $0 < x < \infty$ then all values of $\lambda < 0$ are possible and one must use the Hankel transform. This assumes that the eigenvalue problems in the other independent variables don't restrict the possible values of $\lambda$.

9. May 17, 2014

### Staff: Mentor

You don't solve for a1 separately. You combine it with the constant from the other ODE and then apply the BCs.

Chet

10. May 17, 2014

### joshmccraney

sure! it is $$\alpha \frac{\partial f}{\partial x} = \frac{1}{x}\frac{\partial}{\partial u}(x\frac{\partial f}{\partial x}) + \frac{\partial^2 f}{\partial u^2}$$ where $\alpha$ is a constant. this is subject to $$\lim_{x \to \infty}f(x,u) = 0$$ $$\lim_{u \to \infty}f(x,u) = 0$$ $$\lim_{\substack{{u \to 0 \\ x \to 0}}}-4 \pi \frac{1}{u} (x^2+u^2)^{3/2} C \frac{\partial f}{\partial u} = W$$ $$\lim_{x \to 0}\frac{\partial f}{\partial x}= 0$$

and $C$ and $W$ are constants. i've broken this down by separation of variables. now that i can solve the $x$ portion, i think i can get it all. i'll try it tonight (i can't right now) and post what i get!

Last edited: May 17, 2014
11. May 17, 2014

### joshmccraney

i should comment here. this boundary condition: $$\lim_{\substack{{u \to 0 \\ x \to 0}}}-4 \pi \frac{1}{u} (x^2+u^2)^{3/2} C \frac{\partial f}{\partial u} = W$$ initially looked like $$\lim_{r \to 0}-4 \pi r^2 C \frac{\partial f}{\partial r}= W$$ such that $r := \sqrt {u^2 + x^2}$. i changed the boundary condition as shown, which i believe still holds. but, please check me here:
$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial u}$$ and $\frac{\partial r}{\partial u} = \frac{u}{r} \implies \frac{\partial f}{\partial r} = \frac{\partial f}{\partial u} \frac{r}{u}$

Last edited: May 17, 2014
12. May 19, 2014

### joshmccraney

if either of you are interested, i have formally written the initial problem in latex and turned it into a pdf. it's probably a little too big to post on pf, but if you have an email and are willing to check it out i can forward the .tex and pdf to you.

either way, thanks so much for the help!

13. May 19, 2014

### Staff: Mentor

Sure. I'll send you a PM with my email address.

Chet

14. May 19, 2014

### joshmccraney

thanks, i just emailed you!