S.O.S. Mathematics CyberBoard

[jinydu]

Here's a question that came up while I was working on an interesting problem that I made up for myself. How can I prove that, given the differential equation:

, where is a constant

with the initial condition



the only solution is the trivial solution:



For bonus points, show that with the differential equation

, where n is a nonzero integer

we must have (assuming that the function is defined at that point)

[mathisfun]

Hint for the first problem:

Let .

Assuming that the condition for the first problem () also applies to problem two, you can employ the above approach to dispatch the second problem as well.

[jinydu]

[mathisfun]

I apologize for the error. I meant to say: Let . Such a substitution should reduce the given equation to a Bessel equation of order zero.





Mr. Fantastic, please feel free to castigate me most severely.

[jinydu]

[mathisfun]

Because must remain bounded as r -> 0, we must discard because it has a logarithmic singularity at r = 0. Thus, we obtain:

[jinydu]

Thank you mathisfun

So essentially, the question reduces down:

Show that is finite and nonzero.

(Can you see? I'm trying to rediscover properties of the Bessel function myself...)

[mathisfun]

Correct. You can show that by solving Bessel's differential equation of order 0 in a neighborhood about the regular singular point r = 0 and using the definition of .

[jinydu]

[mathisfun]

Ordinary series methods will fail because the coefficients of y' and y are not analytic at x = 0; hence, x=0 is not a regular point. To solve the equation about x = 0, use the Frobenius method. If you are not familiar with this technique, refer to the section on ODEs near a regular singular point in your differential equations textbook.

[jinydu]

[Pwrong]

Have a look at equation 41 on http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
If m=0 and x=0, then every term in series is zero except the first. . If x=0 but m is nonzero, then every term in the series is zero, so