How to Solve a Bessel-like Equation for R(r) with Unknown BC

[member 428835]

Homework Statement

Solve $$r^2R''+2rR' +\lambda^2 r^2 R = 0$$ where $R=R(r)$. (Let's not worry about BC).

Homework Equations

Looks like something that could be transformed into Bessel's ODE but I'm unsure how.

The Attempt at a Solution

I'm not sure how to start. I typed it into Mathematica and the solution is a sum of exponentials divided by $r$, so evidently Bessel's ODE is not the right course of action. Sure, I could "guess" the solution now that I know it, but if I didn't know it what would I do?

 

[Ray Vickson]

Homework Statement

Solve $$r^2R''+2rR' +\lambda^2 r^2 R = 0$$ where $R=R(r)$. (Let's not worry about BC).

Homework Equations

Looks like something that could be transformed into Bessel's ODE but I'm unsure how.

The Attempt at a Solution

I'm not sure how to start. I typed it into Mathematica and the solution is a sum of exponentials divided by $r$, so evidently Bessel's ODE is not the right course of action. Sure, I could "guess" the solution now that I know it, but if I didn't know it what would I do?

If you note that
$$\frac{d^2}{dr^2} ( r R )= r R^{\prime \prime} + 2 R^{\prime},$$
you will soon see how to solve the DE. Of course, that does not motivate looking at the function $r R$ in the first place.

 

[member 428835]

If you note that
$$\frac{d^2}{dr^2} ( r R )= r R^{\prime \prime} + 2 R^{\prime},$$
you will soon see how to solve the DE. Of course, that does not motivate looking at the function $r R$ in the first place.

Awesome, thanks so much! But yea, what made you look for the function $r R$? Should I see something?

 

[Ray Vickson]

Awesome, thanks so much! But yea, what made you look for the function $r R$? Should I see something?

Well, of course I cheated, knowing that the solution has the form $R = f(r)/r$, so $rR = f(r)$ must satisfy some DE as well.

However, a more useful answer is based on seeing lots of examples in the past: whenever we see something like $r dR/dr$ the default trick to fall back on is to look at $d( rR)/dr$, to see what we get. Sometimes it works, and sometimes it doesn't, but it is always worth a try.

 

[member 428835]

Experience always wins! Ok, so now that I understand the solution, how would I handle the BC and IC: the initial PDE was $$\partial_t \Theta + \frac{1}{r^2}\partial_r(r^2\partial_r\Theta) = 0\\
\Theta(1,t)=Ct:C\in\mathbb{R}\\
\Theta(r,0)=0.$$

While no second BC is given, I assume the solution must be finite and converge for long times, which may help eliminate a solution later. I started by trying to homogenize the BC by introducing $\Theta = \hat\Theta+Ct$. Then the PDE becomes $$\partial_t \hat\Theta +C+ \frac{1}{r^2}\partial_r(r^2\partial_r\hat\Theta) = 0$$ which does not appear to be separable. Any ideas? I don't think the guess $\Theta(t,r)=tf(r)$ works.

 

[Ray Vickson]

Experience always wins! Ok, so now that I understand the solution, how would I handle the BC and IC: the initial PDE was $$\partial_t \Theta + \frac{1}{r^2}\partial_r(r^2\partial_r\Theta) = 0\\
\Theta(1,t)=Ct:C\in\mathbb{R}\\
\Theta(r,0)=0.$$

While no second BC is given, I assume the solution must be finite and converge for long times, which may help eliminate a solution later. I started by trying to homogenize the BC by introducing $\Theta = \hat\Theta+Ct$. Then the PDE becomes $$\partial_t \hat\Theta +C+ \frac{1}{r^2}\partial_r(r^2\partial_r\hat\Theta) = 0$$ which does not appear to be separable. Any ideas? I don't think the guess $\Theta(t,r)=tf(r)$ works.

Using the notation $F(r,t)$ instead of $\Theta(r,t)$, Maple gets a solution of the form
$$F(r,t) = F_1(r) F_2(t) -\frac{C}{6} r^2 - \frac{a}{r} - b,$$
where $C$ is the constant in the PDE, $a,b$ are some other constants and where $F_1, F_2$ satisfy ODEs
$$F_1^{\prime \prime}(r) +\frac{2}{r} F_1^{\prime}(r) -c F_1(r) =0$$
and
$$F_2^{\prime}(t) + c F_2(t) = 0. $$
Here, $c$ is another (arbitrary) constant, the same in both ODEs.

I have not checked this, and I have not asked Maple to outline the steps it takes to arrive at the solution.

 

[member 428835]

Wow, Maple does all of that? Does Mathematica? Perhaps I am using the wrong platform?

 

[Ray Vickson]

Wow, Maple does all of that? Does Mathematica? Perhaps I am using the wrong platform?

I do not have access to Mathematica, so I cannot say, but I would be very surprised if it does not. However, sometimes Maple can do things that Mathematica cannot, and sometimes vice versa. However, about 99.9% of the time they have the same capabilities.

Anyway, if you look at $G(r,t) = \Theta(r,t) + u + v r^2 + w/r$ for some constants $u,v,w$ you can look at the PDE satisfied by $G$. You will see how to choose $u,v,w$ to get a separable PDE.

 

[member 428835]

I do not have access to Mathematica, so I cannot say, but I would be very surprised if it does not. However, sometimes Maple can do things that Mathematica cannot, and sometimes vice versa. However, about 99.9% of the time they have the same capabilities.

Anyway, if you look at $G(r,t) = \Theta(r,t) + u + v r^2 + w/r$ for some constants $u,v,w$ you can look at the PDE satisfied by $G$. You will see how to choose $u,v,w$ to get a separable PDE.

We could go one step further than this, couldn't we? Taking the guess $\Theta = F_1(r)F_2(t) + \sum a_n r^n$ and forcing that the polynomial term yield only a constant yields $a_n = 0 \forall n$ except for $n=-1,0,2$, right?

Also, when I plug the $u + v r^2 + w/r$ part of the solution into the PDE a constant is left over. However, this constant does not seem to appear in the ODE of $F_1$ or $F_2$. Can you explain?

 

[Ray Vickson]

We could go one step further than this, couldn't we? Taking the guess $\Theta = F_1(r)F_2(t) + \sum a_n r^n$ and forcing that the polynomial term yield only a constant yields $a_n = 0 \forall n$ except for $n=-1,0,2$, right?

Also, when I plug the $u + v r^2 + w/r$ part of the solution into the PDE a constant is left over. However, this constant does not seem to appear in the ODE of $F_1$ or $F_2$. Can you explain?

You will need to show exactly what you mean, because I don't get that. When I substitute $\Theta(r,t) = G(r,t) +ur^2 + vr + w + p/r$ into your PDE I end up with
$$ \frac{\partial^2}{\partial r^2} G(r,t) +\frac{2}{r}\frac{\partial}{\partial r} G(r.t) + \frac{\partial}{\partial t} G(r,t) +\frac{2 v}{r} + 6u + C = 0,$$
so taking $u = -C/6, v = 0$ gives a separable PDE. The two constants $w$ and $p$ drop out, so can be used to help meet boundary conditions, for example.