# second order ODE with non-constant coefficients

I'd like to know how to solve the differential equation: $$u_{tt}+\frac{1-2s}{t}u_t-u=0$$ where s is a constant. Let $$u(0)=1$$ and $$\lim_{t \to 0+} t^{1-2s}u_t = c$$. I'm studying a fractional laplacian equation and this ODE arose after applying the Caffarelli-Silvestre extension (https://arxiv.org/pdf/math/0608640.pdf). I have only learnt how to solve second order linear ODEs with constant coefficients, so I wasn't sure how to go about solving this equation.

Let $$u=t^{s} v$$. Therefore we have $$u' = t^{s}v' + st^{s-1}v$$ and $$u'' = t^{s}v'' + 2s t^{s-1}v' + s(s-1)t^{s-2}v$$. Substituting into the ODE yields \begin{align} t^{s}v'' + 2s t^{s-1}v' + s(s-1)t^{s-2}v + \frac{1-2s}{t}(t^{s}v' + st^{s-1}v) - t^{s} v = & \, 0 \\ \implies t^2v'' + 2s tv' + s(s-1)v + \frac{1-2s}{t}(t^{2}v' + stv) - t^2v = & \, 0\\ \implies t^2v'' + tv' -(s^2+t^2)v = & \, 0. \end{align} This is the modified Bessel equation. To get to the classic Bessel equation we make the transformation $$t\to i x$$. Therefore $$\frac{d}{dt}=-i \frac{d}{dx}$$ and thus $$x^2 v'' + x v' + (x^2-s^2)v=0.$$ This is Bessel's equation https://en.wikipedia.org/wiki/Bessel_function.
In response to the comment on initial conditions I am expanding my answer a bit. Let us consider the solution $$u(t) =t^s\left ( C_1 I_s(t) + C_2 K_s(t) \right )$$ where $$I_s$$ and $$K_s$$ are modified Bessel functions. Assuming $$s> 0$$ (or $$\Re (s) > 0$$ for the complex case) the condition $$u(0)=1$$ leads to $$1 = \lim_{t\to 0^+} t^s\left ( C_1 I_s(t) + C_2 K_s(t) \right ) = C_2 \lim_{t\to 0^+}t^s K_s(t).$$ This still works as $$K_s(t)=O(t^{-s})$$ as $$t\to 0$$. I think (but you should check) that we get $$\lim_{t\to 0^+} t^s K_s(t) = 2^{s-1}\Gamma (s).$$ Thus we find that $$C_2 = \frac{1}{\Gamma(s)} 2^{1-s}$$. The case $$s=0$$ needs to be considered separately. I will leave it to you to solve for the other condition.
Note: I checked the other condition and I think both can only be satisfied if $$0 \le s \le 1$$. If this is an unacceptable limitation then you may want to check both my work, and your work in deriving the equation and IC's. A physical interpretation of what $$s$$ is may help.
• I see, however, wouldn't this contradict the initial condition $u(0)=1$? We would have $u(t)=t^s I_s(t)$ where $I_s$ is a modified Bessel function. For all $s \neq 0$, $I_s(0)=0$. – guanton Jul 16 '20 at 19:39