Stuck at Transforming a Differential Equation: x(d^x X/dx^2) + dX/dx + lambda*X = 0

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mario99

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I was trying to solve a partial differential equation. I did all the basic steps easily to get this ordinary differential equation.

[imath]\displaystyle x\frac{d^2X}{dx^2} + \frac{dX}{dx} + \lambda X = 0[/imath]

This is not one of the equations that I am familiar with. Therefore, I had to cheat and I looked at the final solution and discovered that this equation can be transformed to the following Bessel equation.

[imath]\displaystyle r^2\frac{d^2X}{dr^2} + r\frac{dX}{dr} + \lambda r^2X = 0[/imath]

I have told topsquark before that the difficult part in solving a differential equation is how to transform it to one that you know how to solve. Can you suggest a substitution for [imath]x[/imath]? Let us see if it works!

FYI: The solution to the first equation is [imath]X(x)[/imath] and to the second equation is [imath]X(r)[/imath].
 
mario99 said:
I was trying to solve a partial differential equation. I did all the basic steps easily to get this ordinary differential equation.

[imath]\displaystyle x\frac{d^2X}{dx^2} + \frac{dX}{dx} + \lambda X = 0[/imath]

This is not one of the equations that I am familiar with. Therefore, I had to cheat and I looked at the final solution and discovered that this equation can be transformed to the following Bessel equation.

[imath]\displaystyle r^2\frac{d^2X}{dr^2} + r\frac{dX}{dr} + \lambda r^2X = 0[/imath]

I have told topsquark before that the difficult part in solving a differential equation is how to transform it to one that you know how to solve. Can you suggest a substitution for [imath]x[/imath]? Let us see if it works!

FYI: The solution to the first equation is [imath]X(x)[/imath] and to the second equation is [imath]X(r)[/imath].
Click to expand...
This is a coordinate transformation problem. Try [imath]x = \dfrac{1}{2} r^2[/imath]. (Note: this does give an extra factor of 2 on the X term... that doesn't really alter the solution method.)[imath]^*[/imath]

This is one of those things that is a function of experience. Now that you've seen it, you know what to do if you see it again.

As far as how I knew to use [imath]x = \dfrac{1}{2} r^2[/imath], let's work backward.

We want to make
[imath]x X^{\prime \prime}(x) + X^{\prime}(x) + \lambda X(x) = 0[/imath]

into
[imath]X^{\prime \prime}(r) + \dfrac{1}{r} X^{\prime}(r) + \lambda X(r) = 0[/imath]

So try
[imath]\dfrac{d}{dx} = \dfrac{1}{r} \dfrac{d}{dr}[/imath]

and see what transformation that gives. Obviously, this might not work. But it's a place to start.

This was a relatively easy one: if it's more complicated, inspired guesswork (or even better, Google!) is going to be necessary.

[imath]^*[/imath] Notice that we have a potential domain problem. If x is allowed to be negative in the original problem, this transformation will not give us a good solution for that part of the domain. Fortunately, we also get a Bessel equation if we use [imath]x = -\dfrac{1}{2} r^2[/imath].

-Dan
 
I was about to ask you how did you choose to guess [imath]x = \frac{1}{2}r^2[/imath], but then when I looked at this,

topsquark said:
We want to make
[imath]x X^{\prime \prime}(x) + X^{\prime}(x) + \lambda X(x) = 0[/imath]

into
[imath]X^{\prime \prime}(r) + \dfrac{1}{r} X^{\prime}(r) + \lambda X(r) = 0[/imath]

So try
[imath]\dfrac{d}{dx} = \dfrac{1}{r} \dfrac{d}{dr}[/imath]
Click to expand...
I thought that it was a brilliant hint to take what you need from the transformed equation itself.

[imath]\dfrac{d}{dx} = \dfrac{1}{r} \dfrac{d}{dr}[/imath]


[imath]\dfrac{d^2}{dx^2} = \dfrac{d}{dx}\left(\dfrac{1}{r} \dfrac{d}{dr}\right) = \left(\dfrac{1}{r}\dfrac{d^2}{dr^2} - \dfrac{1}{r^2}\dfrac{d}{dr}\right)\left(\dfrac{1}{r}\right) = \left(\dfrac{1}{r^2}\dfrac{d^2}{dr^2} - \dfrac{1}{r^3}\dfrac{d}{dr}\right)[/imath]


[imath]\dfrac{1}{2}r^2\left(\dfrac{1}{r^2}\dfrac{d^2X}{dr^2} - \dfrac{1}{r^3}\dfrac{dX}{dr}\right) + \dfrac{1}{r} \dfrac{dX}{dr}+ \lambda X = 0[/imath]


[imath]\left(\dfrac{d^2X}{dr^2} - \dfrac{1}{r}\dfrac{dX}{dr}\right) + \dfrac{2}{r} \dfrac{dX}{dr}+ 2\lambda X = 0[/imath]


[imath]\dfrac{d^2X}{dr^2} + \dfrac{1}{r} \dfrac{dX}{dr}+ 2\lambda X = 0[/imath]


[imath]r^2\dfrac{d^2X}{dr^2} + r\dfrac{dX}{dr}+ 2\lambda r^2X = 0[/imath]

This equation needs a second transformation to look like the required one, but fortunately, I am familiar with these extra constants in the 3rd term.

The solution will be [imath]X(r) = J_0(\sqrt{2\lambda} r)[/imath].

[imath]X(x) = J_0(\sqrt{2\lambda} \sqrt{2x}) = J_0(2\sqrt{\lambda x})[/imath]

The solution in the book is [imath]J_0(2\alpha_n\sqrt{x})[/imath] which means that our substitution [imath]x = \dfrac{1}{2}r^2[/imath] helped us to get the correct solution as [imath]\lambda_n = \alpha^2_n[/imath], however, it is not the straightforward substitution.


Thank you topsquark for the help. What you have done was a brilliant idea.
 
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