Fractional differential equation Does someone know how to solve this fractional differential equation? $$a\frac{d^2}{dx^2}u(x)+b\frac{d^\frac{1}{k}}{dx^\frac{1}{k}}u(x)+cu(x)=0$$ assuming $(a,b,c) =const$ and $k$ a parameter?
Thanks in advance
 A: You can use the formula obtained via Laplace Transforms, which works as far as I know for $n \in \mathbb{Q}$
$$D^{-n}\{f(t)\} = f(t) *\frac{t^{n-1}}{\Gamma (n)} =\int_0^t \frac{{(t-u)}^{n-1}}{\Gamma(n)}f(u)du$$
So for example, setting $n= \dfrac{1}{2}$ and multiplying by $D$ you get an expression for the half derivative of $f(t)$
$$D^{1/2} f(t)= \frac{1}{{\sqrt \pi  }}\frac{d}{{dt}}\int\limits_0^t {\frac{{f\left( u \right)}}{{\sqrt {t - u} }}du} $$
So maybe you can multiply your equation by $D$ to get $D^{\frac{k+1}{k}}$ and then use the equation I give you with appropiate $n$.
A: it depends on the function $u(x)$. you can try Laplace Transform or Fourier Transform or some other approaches for sufficiently suitable functions to the chosen method.
A: Since some of the solutions (from integral transforms) are already listed I'll discuss some of the other approaches to this.
Looking at this:
$$
a\frac{d^2}{dx^2}u(x)+b\frac{d^\frac{1}{k}}{dx^\frac{1}{k}}u(x)+cu(x)=0
$$
One thing to note is that a solution is likely anylytic (since x is a real variable), so a polynomial series solution is an option.
Assume u(x) is an analytic function with series expansion (I'll use alpha in the series to avoid confusion with your a):
$$
u(x) = \sum_{i=0}^\infty \alpha_ix^i
$$
$$
u^{({{1}\over{k}})}(x) = \sum\alpha_i{{\Gamma(i+1)}\over{\Gamma(i-{{1}\over{k}}+1})}x^{i-{{1}\over{k}}}
$$
Which then gives an equation of the form:
$$
a\sum \alpha_i {\Gamma(i+1) \over \Gamma(i-2+1)}x^{i-2} + b\sum\alpha_i{{\Gamma(i+1)}\over{\Gamma(i-{{1}\over{k}}+1})}x^{i-{{1}\over{k}}} + c\sum\alpha_ix^i =0
$$
So something like:
$$
({bx^{-{{1}\over{k}}}  \over \Gamma( 1-{{1}\over{k}})} +c)\alpha_0+ ({2bx^{1-{{1}\over{k}}}  \over \Gamma( 2-{{1}\over{k}})} + cx)\alpha_1 + \sum_{i=2} \alpha_i ({a\over \Gamma(i-2+1)}+{bx^{-{{1}\over{k}}}  \over \Gamma( i-{{1}\over{k}}+1)} +{c\over\Gamma(1+i)}) \Gamma(i+1)x^i =0
$$
will give a recurrance relation of the second order for the $\alpha_i$ terms. This will be rather messy though (and will depend on your initial conditions).
Also, if you don't need an exact solution as k grows large the solution will be approximated by the solution to:
$$
a\frac{d^2}{dx^2}u(x)+(b+c)u(x)=0
$$
due to the order of differentiation approaching 0 (you can see this by taking the limit of the series representation of the fractional part for $ k \to \infty$.
and for k ~= 1:
$$
a\frac{d^2}{dx^2}u(x)+(b\frac{d}{dx}+c)u(x)=0
$$
The error will grow in factorial order from the points where these O.D.E. approximations exist, but for finding particular points (or a general form to start with variation of prarameters) they're useful.
