# Solve fractional differential equation? \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

Solve fractional differential equation?
$\frac{{d}^{2}}{{dx}^{2}}u\left(x\right)+b\frac{{d}^{\frac{1}{k}}}{{dx}^{\frac{1}{k}}}u\left(x\right)+cu\left(x\right)=0$
assuming $\left(a,b,c\right)=const$ and k a parameter?
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Kindlein6h

Since some of the solutions (from integral transforms) are already listed I'll discuss some of the other approaches to this.
Looking at this: $a\frac{{d}^{2}}{{dx}^{2}}u\left(x\right)+b\frac{{d}^{\frac{1}{k}}}{{dx}^{\frac{1}{k}}}u\left(x\right)+cu\left(x\right)=0$
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):
$u\left(x\right)=\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{i}{x}^{i}$
${u}^{\left(\frac{1}{k}\right)}\left(x\right)=\sum {\alpha }_{i}\frac{\Gamma \left(i+1\right)}{\Gamma \left(i-\frac{1}{k}+1\right)}{x}^{i-\frac{1}{k}}$
Which then gives an equation of the form:
$a\sum {\alpha }_{i}\frac{\Gamma \left(i+1\right)}{\Gamma \left(i-2+1\right)}{x}^{i-2}+b\sum {\alpha }_{i}\frac{\Gamma \left(i+1\right)}{\Gamma \left(i-\frac{1}{k}+1\right)}{x}^{i-\frac{1}{k}}+c\sum {\alpha }_{i}{x}^{i}=0$
So something like:
$\left(\frac{b{x}^{-\frac{1}{k}}}{\Gamma \left(1-\frac{1}{k}\right)}+c\right){\alpha }_{0}+\left(\frac{2b{x}^{1-\frac{1}{k}}}{\Gamma \left(2-\frac{1}{k}\right)}+cx\right){\alpha }_{1}+$
$\sum _{i=2}{\alpha }_{i}\left(\frac{a}{\Gamma \left(i-2+1\right)}+\frac{b{x}^{-\frac{1}{k}}}{\Gamma \left(i-\frac{1}{k}+1\right)}+\frac{c}{\Gamma \left(1+i\right)}\right)\Gamma \left(i+1\right){x}^{i}=0$
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:
$a\frac{{d}^{2}}{{dx}^{2}}u\left(x\right)+\left(b+c\right)u\left(x\right)=0$
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 $k\to \mathrm{\infty }$.
and for $k\sim =1$:
$a\frac{{d}^{2}}{{dx}^{2}}u\left(x\right)+\left(b\frac{d}{dx}+c\right)u\left(x\right)=0$
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.

###### Not exactly what you’re looking for?
Mary Nicholson
You can use the formula obtained via Laplace Transforms, which works as far as I know for $n\in Q$
${D}^{-n}\left\{f\left(t\right)\right\}=f\left(t\right)\cdot \frac{{t}^{n-1}}{\Gamma \left(n\right)}={\int }_{0}^{t}\frac{{\left(t-u\right)}^{n-1}}{\Gamma \left(n\right)}f\left(u\right)du$
So for example, setting $n=\frac{1}{2}$ and multiplying by D you get an expression for the half derivative of f(t)
${D}^{\frac{1}{2}}f\left(t\right)=\frac{1}{\sqrt{\pi }}\frac{d}{dt}{\int }_{0}^{t}\frac{f\left(u\right)}{\sqrt{t-u}}du$
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.
###### Not exactly what you’re looking for?
RizerMix
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.