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$Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...$

804 Ž. Tomovski et al. Proof Our demonstration of Theorem 5 is based upon the Laplace transform method. Indeed, if we make use of the notational convention depicted in (1.7) and the Laplace transform formula (1.6), by applying the operator L to both the sides of (3.7), it is easily seen that Downloaded By: [Srivastava, Hari M.] At: 18:19 27 October 2010 Furthermore, since and s β 1(α 1 −1) s β 2(α 2 −1) F (s) Y (s) = ac 1 + bc as α 1 + bs α 2 + . 2 + c as α 1 + bs α 2 + c as α 1 + bs α 2 + c s β i(α i −1) = 1 ( s β i (α i −1) ) ⎛ ⎝ 1 as α 1 + bs α 2 + c b s α 2 + c b 1 + a b [ 1 ∞∑ = L b F (s) as α 1 + bs α 2 + c = 1 b = L m=0 ∞∑ m=0 [ 1 b ( s α 1 s α 2 + c b ⎞ ) ⎠ = 1 b ∞∑ m=0 ( (−1) m a ) m x (α 2 −α 1 )m+α 2 +β i (1−α i )−1 b · E m+1 α 2 ,(α 2 −α 1 )m+α 2 +β i (1−α i ) ( − a b ) m ( s α 1m ( s α 2 + c b ( − a ) m s α 1m+β i α i −β i ( b s α 2 + c b ( − c b xα 2) ] (s) (i = 1, 2) ) m+1 F (p) ) [ 1 = L b ( ( · x (α 2−α 1 )m+α 2 −1 E m+1 α 2 ,(α 2 −α 1 )m+α 2 − c b xα 2 ∞∑ m=0 ∞∑ m=0 ( − a ) m b ) ∗ f (x) ) m+1 )] (s) ( (−1) m a ) m ( )( E m+1 α b 2 ,(α 2 −α 1 )m+α 2 ,− c ;0+f − c 2) ] b b xα (s) in terms of the Laplace convolution, by applying the inverse Laplace transform, we get the solution (3.9) asserted by Theorem 5. 3.2. An Application of Theorem 5 The next problem is to solve the fractional differential equation (3.7) in the space of Lebesgue integrable functions y ∈ L(0, ∞) when α 1 = α 2 = α and β 1 ̸= β 2 , that is, the following fractional differential equation: ( ) ( ) a D α,β 1 0+ y (x) + b D α,β 2 0+ y (x) + cy (x) = f (x) (3.10) under the initial conditions given by ( ) I (1−β i)(1−α) 0+ y (0+) = c i (i = 1, 2). (3.11) We now assume, without loss of generality, that β 2 ≦ β 1 . If c 1 < ∞, then c 2 = 0 unless β 1 = β 2 .
Integral Transforms and Special Functions 805 Corollary 1 The fractional differential equation (3.10) under the initial conditions (3.11) has its solution in the space L (0, ∞) given by ( ac1 y (x) = a + b ( bc2 + a + b ( 1 + a + b ) ( x β 1+α(1−β 1 )−1 E α,β1 +α(1−β 1 ) − c ) a + b xα ) ( x β 2+α(1−β 2 )−1 E α,β2 +α(1−β 2 ) − c ) a + b xα ) ( ) Eα,1,− 1 c ;0+f (x) . (3.12) a+b Downloaded By: [Srivastava, Hari M.] At: 18:19 27 October 2010 Proof Our proof of Corollary 1 is much akin to that of Theorem 5. We choose to omit the details involved. 3.3. A Fractional Differential Equation Related to the Process of Dielectric Relaxation Let 0
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