## Monotonicity of a combination of Bessel functions

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Prove that the following function is decreasing (as a function of a) for a > 0 when 0 < r < 1: $${K_2(ar)I_2(a)-I_2(ar)K_2(a)\over I_2(a)}I_2(ar).$$

The problem arose in the analysis of a model for yield stress fluids. We have numerical evidence, but I would be interested in an analytical proof.

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## 1 Answer

This function is quite interesting as it reaches a maximum in 0. This can be seen without difficulty computing the first and second derivatives. It is

$$B(a,r)=\frac{K_2(ar)I_2(a)-I_2(ar)K_2(a)}{I_2(a)}I_2(ar)$$

that gives

$$B(0,r)=\frac{1}{4}(1-r^4)>0$$

$$B'(0,r)=0$$

$$B''(0,r)=-\frac{1}{12}r^2(1-r^2)^2<0$$

for the given interval. This means that this function is decreasing for $a>0$. The problem is if there are some other points where the first derivative can become zero changing the concavity of the curve. So, the first derivative has the following involved expression

$$B'(a,r)=\frac{1}{2 I^2_2(a)}\left[I_2(ar)^2 (I_1(a)+I_3(a)) K_2(a)+\right.$$ $$I_2(a) I_2(ar) (-2 r (I_1(ar)+I_3(ar)) K_2(a)+$$ $$I_2(ar) (K_1(a)+K_3(a))+r I_2^2(a) ((I_1(ar)+I_3(ar)) K_2(ar)$$ $$\left.-I_2(ar) (K_1(ar)+K_3(ar)))\right]$$

Now we notice that $I_2$ is a monotonic increasing function that never becomes zero and $K_2$ is a monotonic decreasing function that never becomes zero for the given intervals and similarly is true for $I_1,\ I_3$ and $K_1,\ K_3$. All these functions are positive. So, it is not difficult to realize that the first derivative is a monotonic function and never hits zero again being just a balance of functions never reaching zero unless $a=0$ and having monotonic behavior, $K_i$ are decreasing functions and $I_i$ increasing functions. We also note that

$$\lim_{a\rightarrow\infty}B'(a,r)=0$$

that can be proved using the asymptotic formula for these Bessel functions. Now, combining monotonicity and positivity of these Bessel functions, starting from 0 and reaching asymptotically 0 at increasing values of the argument, they can just reach an extremum and never cross zero again. This can also be seen with a simple plot

evaluated at $r=0.1,0.3,0.5,0.7,0.9$.

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I understand that $I_2$ and $K_2$ do not become zero, but why does this imply $B'$ does not become zero? – Michael Renardy May 7 2012 at 11:37
I have improved the answer. You are doing a balance of positive terms in the first derivative. This balance can only be zero in zero where the maximum is reached. Then, Is increase exponentially while Ks go down exponentially. – Jon May 7 2012 at 13:38
I still cannot follow. B' has positive and negative contributions. You have to show the negative ones outweigh the positive ones, and I do not see how you get this out of sign of I and K and their derivatives alone. I tried to check your expressions for B'. I could get them to agree neither with B' nor with each other. Maybe there are typos. Your second expression seems to have an unbalanced parenthesis. – Michael Renardy May 7 2012 at 14:34
Ok, I finally found out where you are missing the parenthesis. I still do not follow your argument beyond the positivity and monotoncity of the I's and K's. – Michael Renardy May 7 2012 at 15:00
I should say that the derivative does not change sign and so does not cross to zero again. I will try to expand the answer to make this evident. – Jon May 8 2012 at 7:54
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