Bessel function solution on extremum point

David Shulman

New member
Joined
Jul 29, 2020
Messages
2
Hello,

I have a heterogeneous modified Bessel equation:
\(\displaystyle z'' +\frac{1}{r}z'-\frac{1}{v^2}z=G(r),\)
and I interesting in the solution of the function on extremum points,
\(\displaystyle z'(0)=0\)
so I solve the equation:
\(\displaystyle z''-v^2z=G(r),\)
I have for homogeneous solution:
\(\displaystyle z_h=c_1 exp⁡(v^{-1} r)+c_2 exp⁡(-v^{-1} r) \)
and for a Particular solution with Variation of Parameters:
\(\displaystyle z_p=exp⁡(v^{-1} r)∫G(r)/(2v^{-1}) exp⁡(-v^{-1} r)+exp⁡(-v^{-1} r)∫G(r)/(2v^{-1} ) exp⁡(v^{-1} r)\)
But how I can find the constant parameters in homogeneous solution? Probably I can not use my boundary condition, because I change my function...
 
Last edited:

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
21,538
Hello,

I have a heterogeneous modified Bessel equation:
\(\displaystyle z'' +\frac{1}{r}z'-\frac{1}{v^2}z=G(r),\)
and I interesting in the solution of the function on extremum points,
\(\displaystyle z'(0)=0\)
so I solve the equation:
\(\displaystyle z''-v^2z=G(r),\)
I have for homogeneous solution:
\(\displaystyle z_h=c_1 exp⁡(v^{-1} r)+c_2 exp⁡(-v^{-1} r) \)
and for a Particular solution with Variation of Parameters:
\(\displaystyle z_p=exp⁡(v^{-1} r)∫G(r)/(2v^{-1}) exp⁡(-v^{-1} r)+exp⁡(-v^{-1} r)∫G(r)/(2v^{-1} ) exp⁡(v^{-1} r)\)
But how I can find the constant parameters in homogeneous solution? Probably I can not use my boundary condition, because I change my function...
I have not worked out this problem - just thinking.

You are using the original solution (function or graph resulting from the original DE) - with an additional condition z' = 0

So your original bdy. conditions should be still applicable.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
6,402
How do you go from "\(\displaystyle \frac{1}{v^2}z\)" to \(\displaystyle v^2z\)?
 

David Shulman

New member
Joined
Jul 29, 2020
Messages
2
of course it \(\displaystyle 1/v^2\)
 
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