ODE Problem: Solution for y'' + (1/x)y' - lambda y = 0 with Boundary Conditions

[member 428835]

Homework Statement

$$y'' + \frac{1}{x}y' - \lambda y = 0$$
where $x \to \infty \implies y \to 0$ and $x \to 0 \implies y' \to 0$

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??

 

[pasmith]

Homework Statement

$$y'' + \frac{1}{x}y' - \lambda y = 0$$
where $x \to \infty \implies y \to 0$ and $x \to 0 \implies y' \to 0$

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??

Multiplying by x^2 gives <br /> x^2 y&#039;&#039; + xy&#039; - (\lambda x^2)y = 0<br /> which if \lambda = -k^2 &lt; 0 can be turned into Bessel's equation of order zero by a rescaling of the independent variable; if \lambda = k^2 &gt; 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.

 

[member 428835]

Multiplying by x^2 gives <br /> x^2 y&#039;&#039; + xy&#039; - (\lambda x^2)y = 0<br /> which if \lambda = -k^2 &lt; 0 can be turned into Bessel's equation of order zero by a rescaling of the independent variable

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!

 

[Chestermiller]

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!

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

Chet

 

[pasmith]

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.

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}.

 

[member 428835]

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!

 

[member 428835]

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, what's 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!

 

[pasmith]

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?

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.

and secondly, as i mentioned earlier, Bessels equation arises from a PDE, where i have taken separation of variables.

Can you post the original PDE and boundary conditions?

if so, what's the point of solving $a_1$ since we will have more constants from the other ODE (much easier to solve)?

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.

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

It depends on the nature of the boundary conditions. If 0 &lt; x &lt; \infty then all values of \lambda &lt; 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.

 

[Chestermiller]

and secondly, as i mentioned earlier, Bessels equation arises from a PDE, where i have taken separation of variables. if so, what's 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.

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

Chet

 

[member 428835]

Can you post the original PDE and boundary conditions?

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!

thanks for both your help!

 

[member 428835]

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 $\frac{\partial r}{\partial u} = \frac{u}{r} \implies \frac{\partial f}{\partial r} = \frac{\partial f}{\partial u} \frac{r}{u}$

 

[member 428835]

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!

 

[Chestermiller]

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!

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

Chet

 

[member 428835]

thanks, i just emailed you!