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Zbl 0711.34019
Campos, L.M.B.C.
On the solution of some simple fractional differential equations. (English)
[J] Int. J. Math. Math. Sci. 13, No.3, 481-496 (1990). ISSN 0161-1712; ISSN 1687-0425/e

The aim of the author is to show that the classical elementary methods used to solve the ordinary differential equations with constant coefficients or the equations of Euler type can also be used to solve some functional equations in which the fundamental operator consists of a generalization of the classical concept of derivative, namely: $$ \frac{D\sp{\nu}F}{Dz\sp{\nu}}=\frac{\Gamma (1+\nu)}{2\pi i}\int\sp{(z+)}\sb{\infty}\frac{F(\zeta)d\zeta}{(\zeta -z)\sp{\nu +1}},\quad \nu \ne -1,-2,...\quad, $$ $$ \frac{D\sp{\nu}F}{Dz\sp{\nu}}=\frac{1}{\Gamma (- \nu)}\int\sp{z}\sb{\infty}\frac{F(x)dx}{(z-x)\sp{\nu +1}},\quad Re \nu <0. $$ Notice that $D\sp nF/Dz\sp n$ is exactly equal to $F\sp{(n)}(z)$ when $n\ge 0$ is an integer and F(z) is regular near the point z while for $n=-1,-2,..$. it becomes equal to the n-times repeated integral of F(z). The generalization made by the author relies upon the fact that $D\sp{\nu}(e\sp{az})/Dz\sp{\nu}=a\sp{\nu}e\sp{az},$ which permits to write the solution of an equation of the form $\sum A\sb mD\sp{\nu\sb m}F/Dz\sp{\nu\sb m}=0$ as $F(z)=\sum\sb{m}\sum\sb{\ell}C\sb{\ell m}z\sp{\ell}e\sp{a\sb mz}.$ Similar results concerning the equations of Euler type as well as those with forcing terms are also given.
[M.Idemen]

MSC 2000:: *34A99 General theory of ODE
34B30 Special ODE
26A33 Fractional derivatives and integrals (real functions)
45D05 Volterra integral equations
45J05 Integro-ordinary differential equations

Keywords: Euler differential equation; constant coefficients

Cited in: Zbl 0789.30030

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