1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

Edit:

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.

$\endgroup$
1
  • $\begingroup$ 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 $s \neq 0$, $I_s(0)=0$. $\endgroup$ – guanton Jul 16 '20 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.