The form of Modified Bessel Function is expressed as:
$$\frac{d^2p(r)}{dr^2}+\frac{1}{r}\frac{dp(r)}{dr}-(\alpha^2+\frac{v^2}{r^2})p(r)=0$$ Where $\alpha$ and $v$ both are real numbers. Its general solution could be expressed as: $$p(r)=c_1I_v(\alpha x)+c_2K_v(\alpha x)$$
Recent I met a problem about Modified Bessel Function that I didn't understand very well. See below.
$$\frac{d^2p(r,t)}{dr^2}+\frac{1}{r}\frac{dp(r,t)}{dr}=\frac{1}{c}\frac{dp(r,t)}{dt}$$
Where $p$ is a function of $r$ and $t$. My purpose is to get the solution of $p$. In order to solve it, firstly Laplace transform with respect with $t$ is performed. Therefore, $$\frac{d^2P(r,s)}{dr^2}+\frac{1}{r}\frac{dP(r,s)}{dr}=\frac{1}{c}[s*P(r,s)-p(r,0)]$$ With the initial boundary value $p(r,0)=0$, we have $$\frac{d^2P(r,s)}{dr^2}+\frac{1}{r}\frac{dP(r,s)}{dr}=(\sqrt \frac{s}{c})^2P(r,s)$$ Based on the solution to Modified Bessel Function, we have $$P(r,s)=c_1I_0(r\sqrt\frac{s}{c})+c_2K_0(r\sqrt\frac{s}{c})$$
However, the correct solution to this PDE is $$\frac{sP^"(r,s)}{p_0}=-\frac{K_0(r\sqrt\frac{s}{c})}{K_0(a\sqrt\frac{s}{c})}$$ Where $a$ is a real number.
I don't know why there is such a big difference in my solution and the correct solution.
I will appreciate it if you can leave your valuable viewpoint or comment.
Thank you.
Kl.