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

Anne Wacker

Anne Wacker

Answered question

2022-01-20

Solve fractional differential equation?
d2dx2u(x)+bd1kdx1ku(x)+cu(x)=0
assuming (a,b,c)=const and k a parameter?

Answer & Explanation

Kindlein6h

Kindlein6h

Beginner2022-01-20Added 27 answers

Since some of the solutions (from integral transforms) are already listed I'll discuss some of the other approaches to this.
Looking at this: ad2dx2u(x)+bd1kdx1ku(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)=i=0αixi
u(1k)(x)=αiΓ(i+1)Γ(i1k+1)xi1k
Which then gives an equation of the form:
aαiΓ(i+1)Γ(i2+1)xi2+bαiΓ(i+1)Γ(i1k+1)xi1k+cαixi=0
So something like:
(bx1kΓ(11k)+c)α0+(2bx11kΓ(21k)+cx)α1+
i=2αi(aΓ(i2+1)+bx1kΓ(i1k+1)+cΓ(1+i))Γ(i+1)xi=0
will give a recurrance relation of the second order for the αi 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:
ad2dx2u(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.
and for k∼=1:
ad2dx2u(x)+(bddx+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.

Mary Nicholson

Mary Nicholson

Beginner2022-01-21Added 38 answers

You can use the formula obtained via Laplace Transforms, which works as far as I know for nQ
Dn{f(t)}=f(t)tn1Γ(n)=0t(tu)n1Γ(n)f(u)du
So for example, setting n=12 and multiplying by D you get an expression for the half derivative of f(t)
D12f(t)=1πddt0tf(u)tudu
So maybe you can multiply your equation by D to get Dk+1k and then use the equation I give you with appropiate n.
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

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.

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