How to get rid of imaginary values that come from Bessel $\ K_n (-x)$??

Question

In solving a 2nd order differential equation, I got a solution like $\ y= A I_1(\frac{bx}{a\ln M})+ B K_1(\frac{bx}{a\ln M})$. when $\ M<1, (\frac{bx}{a\ln M}) $ becomes negative and $\ K_1(\frac{bx}{a\ln M}) $ gives imaginary value. M is a constant. It doesnt depend on $\ x $ or $y$. M is just a ratio that can be less than $1$ too. but what does this imaginary part mean in my modeling and how can i find a real-valued answer? Can I use $\ K_n(-x)= (-1)^n K_n(x)$

Answer 1

In case of $M<0$ , I suppose that you can find the reals $X$ and $Y$ such as $$\frac{bx}{a\ln(M)}=X+iY$$ So, your problem is to find the real and imaginary parts of $I_1(X+iY)$ and $K_1(X+iY)$

On the form of series : $$I_1(X+iY)=\frac{1}{2}(X+iY)\sum_{k=0}^\infty \frac{1}{k!(k+1)!}\left(\frac{X^2-Y^2}{4}+i\frac{XY}{2}\right)^k$$ The separation of real and imaginary parts of the power terms is possible with binomial coefficients. This will leads to an huge formula with double sum. Probably that is not what you expect, so I don't continue this way.

On the form of integral : $$I_1(X+iY)=\frac{2(X+iY)}{\pi}\int_0^1\sqrt{1-t^2}\cosh\big((X+iY)t\big)dt$$ $\cosh\big((X+iY)t\big)=\cosh(Xt)\cos(Yt)+i \sinh(Xt)\sin(Yt)$

$I_1(X+iY)=\frac{2(X+iY)}{\pi}\int_0^1\sqrt{1-t^2} \big(\cosh(Xt)\cos(Yt)+i \sinh(Xt)\sin(Yt) \big)dt$ $$I_1(X+iY)=\text{Re}+i\text{ Im}$$ $$\text{Re}=\frac{2X}{\pi}\int_0^1\sqrt{1-t^2} \big(\cosh(Xt)\cos(Yt)\big)dt - \frac{2Y}{\pi}\int_0^1\sqrt{1-t^2} \big(\sinh(Xt)\sin(Yt)\big)dt$$ $$\text{Im}=\frac{2X}{\pi}\int_0^1\sqrt{1-t^2} \big(\sinh(Xt)\sin(Yt)\big)dt + \frac{2Y}{\pi}\int_0^1\sqrt{1-t^2} \big(\cosh(Xt)\cos(Yt)\big)dt$$ As far as I know, there is no closed form for these integrals. So, the above integrals are the simplest analytic form to express the real and imaginary parts.

One can proceed on the same manner for : $$K_1(X+iY)=(X+iY)\int_1^\infty \sqrt{t^2-1}\:e^{-(X+iY)t}dt $$

$K_1(X+iY)=(X+iY)\int_1^\infty \sqrt{t^2-1}\:e^{-Xt}\big(\cos(Yt)-i\sin(Yt) \big)dt $

$$K_1(X+iY)=\text{Re}+i\:\text{Im}$$

$$\text{Re}=X\int_1^\infty \sqrt{t^2-1}\:e^{-Xt}\cos(Yt)dt + Y\int_1^\infty \sqrt{t^2-1}\:e^{-Xt}\sin(Yt)dt $$ $$\text{Im}=-X\int_1^\infty \sqrt{t^2-1}\:e^{-Xt}\sin(Yt)dt + Y\int_1^\infty \sqrt{t^2-1}\:e^{-Xt}\cos(Yt)dt $$

NOTE :

In case of pure imaginary argument, the formulas are much simpler : $$I_1(iY)=i\,J_1(Y)$$ $$K_1(iY)=\frac{\pi}{2}\big(-J_1(Y)+i\,Y_1(Y) \big)$$ Where $Y_1$ is the modified Bessel function of first kind and order $1$ (Don't confuse with the real argument $Y$).