I wish to know about the existing methods to solve a fractional order algebraic equation. I am not asking about solution of fractional order differential equations. To make things clear, the problem is described below:
How to solve an equation of the form:
a0yn+a1yn1+a2yn2+.....+amynm=0
where n, n1,n2...nm have non-integer values and
a0,a1,a2...am are real constants
I have achieved the solution to some example fractional-order equations using Mathematica. But, I need to learn the underlying mathematics.
Please suggest!
If all n, n1,n2,....nm are rational numbers, try subsitution Y=y1/r, where r is chosen so that all numbers r*n, r*n1, r*n2,......,r*nm are integers.
Hello,
it depends on whether the powers n0, n1, n2, ... all are rational (see Vladimir's answer) or more generally there exist a change of variable Z=\phi(y) such that the original equation can be transformed into a polynomial one in the variable Z.
In other case I think only a numerical approach would be valid.
Gabriel
Hello,
Many books is devoted to a rapidly developing area on the research of the qualitative theory of fractional differential equations. that is self-contained and unified in presentation, and provides readers the necessary background material required to go further into the subject and explore the rich research literature.
The tools used include many classical and modern nonlinear analysis methods such as fixed point theory, measure of non compactness method, topological degree method, the Picard operators technique, critical point theory and semigroups theory. Based on research work carried out by the authors and other experts during the last years.
Dear Audai, it seems that you have not read well the question. This is NOT about fractional differential equations.
---------------------------------------------------
As Vladimir says, when the exponents are rational, the problem reduces to that of the roots of "normal" integer-valued polynomials. It is interesting to note that the orders of those polynomials tend to be much larger than the original exponents and give rise to many roots (solutions). For example, the equation a*y^(2/3) + b*y^(1/5) = 0, by setting Y = y^(1/15), transforms to Y^3*(a*Y^7 + b) which has in C eight roots (Y=0, plus the seven ones for the second factor). Since each of the non-zero roots for Y gives 15 complex roots for y, arranged around a circle, the total number of C solutions of the original equation (for generic values a, b) is 106.
This seems to indicate (by thinking in terms of the limit of rational approximations of the irrational exponent values) that for irrational exponents the roots form a continuum of some kind - probably continuous lines in the complex plane, or maybe even something much more complicated. It is worth investigating :-) If somebody would like to give it a try with me, let me know.
Deep Shekhar Acharya ,
In addition to the previous helpful answers, I have the following note to add:
It is well known that a polynomial P(x) = anxⁿ+an-1xⁿ⁻¹+...a₂x²+a₁x¹+a₀
where an, an-1,..a₂, a₁, a₀ are real numbers, has n roots counted with their multiplicities. In general, if n > 4, there are no formulas to find the exact solutions.
In case:
f(x) = an(1)xⁿ⁽¹⁾ +an(2)xⁿ⁽²⁾+...an(m-1)xn(m-1)+an(m)xn(m) +an(0)
where an(1) , an(2),..an(m-1), an(m),an(0) are real numbers and
n(1),n(2),..,n(m-1),n(m) are rational numbers.
Without loss of generality, we assume that
n(i)=(pi/qi), gcd(pi , qi)=1 are rational numbers
such that n(1) > n(2)>...> n(m-1) > n(m).
We can prove the following theorem:
Theorem: The function
f(x) = an(1)xⁿ⁽¹⁾ +an(2)xⁿ⁽²⁾+...an(m-1)xn(m-1)+an(m)xn(m) +an(0) = 0
has
(1) d2n(1) non trivial solutions if an(0) is not zero.
(2) d2(n(1)- n(m)) non trivial solutions if an(0) = 0 and an(m) is not zero.
(3) d2(n(1) - n(r)) non trivial solutions if an(0) =an(m) =an(m-1) =..an(r-1) = 0
and an(r) is not zero. where d = gcd(q1,q2,...qn).
Proof: Elementary.
Best regards
- Badges
- Science topic
- Similar topics
- Mathematical Sciences
- Algebra