# Bessel function solution on extremum point

#### David Shulman

##### New member
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
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
How do you go from "$$\displaystyle \frac{1}{v^2}z$$" to $$\displaystyle v^2z$$?

#### David Shulman

##### New member
of course it $$\displaystyle 1/v^2$$