Bessel function solution on extremum point

[David Shulman]

Hello,

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

 

[Deleted member 4993]

Hello,

I have a heterogeneous modified Bessel equation:
[MATH]z'' +\frac{1}{r}z'-\frac{1}{v^2}z=G(r),[/MATH]and I interesting in the solution of the function on extremum points,
[MATH]z'(0)=0[/MATH]so I solve the equation:
[MATH]z''-v^2z=G(r),[/MATH]I have for homogeneous solution:
[MATH]z_h=c_1 exp⁡(v^{-1} r)+c_2 exp⁡(-v^{-1} r) [/MATH]and for a Particular solution with Variation of Parameters:
[MATH]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)[/MATH]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]

How do you go from "\(\displaystyle \frac{1}{v^2}z\)" to \(\displaystyle v^2z\)?

 

[David Shulman]

of course it [MATH]1/v^2[/MATH]