On Laplace Finite Marchi Fasulo Transform Of Generalized Functions. A.M.Mahajan

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1 On Laplace Finite Marchi Fasulo Transform Of Generalized Functions A.M.Mahajan Department of Mathematics, Walchand College of Arts and Science, Solapur Abstract : In this paper the Laplace Finite Marchi Fasulo Transform F(n, s) = f(z, t)e p (z)dtdz has been extended to certain class of generalized function and an inversion and uniqueness theorem has been derived. Introduction : We construct an integral transform whose kernel is the product of the kernels of Laplace and Finite Marchi Fasulo transform. The Laplace Finite Marchi Fasulo transform F (n,s) of a function f (z,t) of two variables ( z, t ) on the domain {(z, t)/ h < z < h, 0 < t < } is then defined as M [ f (z, t) ] F(n, s) = f(z, t)e p (z)dtdz. (.) We firstly extend the transform defined by (.) to a class of generalized function by applying similar technique used by Zemanian (4 ) and then we shall prove the inversion and uniqueness theorem for the class of generalized function. Testing function spaces +, a and + (w) :- We shall denote the open set(-h, h ) ( 0, ) by I. Let +, a denotes the space of all complex valued smooth functions φ (z, t ) that are infinitely differentiable with respect to z and t on I on which the functionals λ,, defined by λ,, (φ) sup e at D Ω φ ( z, t ) 0 < t < -h < z < h assumes finite values where k, k 2 are nonnegative integers and D t =, D z = Ω (φ) = ( D ) (φ) 2) Now +, a is linear space under the pointwise addition of functions and their multiplication

2 by complex numbers. Each λ,, is a seminorm on +, a and λ a, 0, 0 is a norm, hence the countable collections λ,,, of seminorms is a countable multinorm on +, a. We assign to +, a the topology generated by λ,,, making it countably multinormed space. A sequence φ converges in +, a to φ if and only if for each pair of nonnegative integers k, k 2 λ,, (φ p φ) 0 as p and a sequence φ is a cauchy sequence in +, a if and only if λ,, (φ φ ) 0 k, k 2 = 0,,2, as p, q Lemma (.) :- The space +, a is complete and therefore it is a Frechet space. Proof : Let φ be a cauchy sequence in +, a. Then there exists a smooth function φ (z, t) such that for each pair of non negative integers k, k 2 D Ω φ (z,t) D Ω φ (z,t) as p. Moreover given any > 0 there exists N, every p, q > N e, D Ω [ φ (z, t) φ (z, t)] <, for all z. such that for Taking the limit as q, we obtain e D Ω [ φ (z, t) φ (z, t)] < 0 < t <, h < z < h.(.2) Thus as p λ,, ( φ (z, t) φ (z, t)) 0 for each k, k 2. Finally because of the uniformity of the convergence and the fact that each e D Ω [ φ (z, t)] is bounded on 0 < t <, h < z < h there exists constant C, not depending on p such that e D Ω [ φ (z, t)] < C, for all z, t. Therefore (.2) implies that e D Ω [ φ (z, t)] < C, + ε which shows that the limit function φ is a member of +, a hence +, a is complete, so it is Frechet space. The countable union space + (w) :- Let w denote either a finite real number or. We choose the monotonic sequence Since the convergence is uniform and a such that a w.

3 Then, is a sequence of testing function spaces such that,, for all p and the topology of, is stronger than the topology induced on it by, Define (w) =, A sequence φ converges in + (w) it converges in, for some p. Then with these properties + (w) is countable union space. + (w) is a linear space. Further it is complete space, since for fixed p, is complete. The dual spaces +, a and + (w) : - The dual space +, a of +, a is the collection of all continuous linear functionals on +, a. Since +, a is complete, +, a is also complete. If a c, then +,c +, a and the topology of +,c is stronger than the topology induced on it by +, a. Therefore the restriction of any member f Є +, a to +,c is in +,c. We assign the customary weak topology generated by the multinorm { ξ }, φ Є +, a to the dual space +, a, where ξ (f) = < f, φ > φ Є +, a. We denote by + (w) the dual space of + (w). since all, are complete and + (w) is complete. the dual space + (w) is also complete. The properties of testing function spaces and their duals :- ) D +, a and the convergence in D (I) implies the convergence in +, a (I). Therefore the restriction of any member of +, a (I) to D(I) is in D (I). Similarly D(I) is subspace of + (w) whatever may be the value of w. The convergence in D(I) implies the convergence in + (w) and the restriction of any member of + (w) to D(I) is a member of D (I).. The member of +, a (I) and + (w) are called distribution. 2) D(I) is dense in + (w) for every w. Therefore + (w) is subspace of D (I). 3) For each f Є +, a (I) there exists a +ve constant c and a nonnegative integer r, such that for all φ Є +, a (I) < f, φ > c max λ,, (φ) 0 k r 0 k 2 r

4 4) If f ( z,t ) is a locally integrable function defined on the interval I such that e f(z, t) dtdz exists then f (z, t )generates a reguler member of, (I) through the definition < f, φ > = < f, φ > = f(z, t) φ (z, t)dtdz φ Є, (I).. (.3) f (z, t ) e e φ (z, t) dtdz λ,, e f(z, t) dtdz which exists in view of our assumption. Therefore (.3) defines a functional f on +, a (I), this functional is linear. Further if converges in +, a (I) to zero then λ a, 0, 0 ( φ ) 0 so that φ < f, φ > 0 therefore f is continuous on +, a (I). Similarly if w < a, then f generates a regular distribution of + (w) through the distribution < f, φ >= where f(z, t) φ (z, t)dtdz φ Є, (I), e f(z, t) dtdz exists. This condition is satisfied if e f (z,t) bounded on o < t < for every choice of σ where σ>w. 5) For each positive integer n and Re. S a the function e p (z) is a member of +, a (I) and is also member of + (w) if a > w. Proof :- e [ D Ω e p (z) ] = ( ) s (a ) e - ( s - a ) t p (z) for t 0 h < z < h.

5 The right hand side of above equation is bounded for Re. s a for the positive eigen values a n. Thus for each k, k 2, λ,, [ e p (z)] exists which shows that e p (z) belongs to +, a (I) In the similer way we can show that for every a > w, e p (z) Є + (w). The Generalized Laplace Finite Marchi Fasulo Transform :- We call the generalized function f as Laplace Finite Marchi Fasulo transformable if it belongs to + (w) for some real number w. Let σ f be defined as, σ f = inf { w f Є + (w) } If f is Laplace Finite Marchi Fasulo transformable function, then we see that σ f s. t. f Є + (w) w < σ f Thus for a given Laplace Finite Marchi Fasulo transformable function f Є + (w) if D f denote strip of definition i.e. D f = { ( n,s ) / σ f < Re s,n +ve integer }.Then Laplace Finite Marchi Fasulo transformation F (n, s) of f (z, t) is defined by M [ f (z, t ) ] F (n, s) = < f (z, t), e p (z) > i.e, it is defined as the application of f Є + (σ f ) to kernel e p (z) Є + (σ f ) or equivalently f Є +, a to e p (z) Є +,a for any σ f < a Re. S < f ( z, t), e p (z) > Boundedless property of Generalised Laplace Finite Marchi Fasulo Transform: We show that the generalized Laplace Finite Marchi Fasulo transform F (n, s) defined as above is bounded for (n, s) Є D f. Theorem : - Let f Є +, a (I) and F (n, s) = < f (z, t), e p (z) > For (n, s) Є D f. Then F (n, s) satisfies the inequality F (n, s) C. A. Proof : - Let f Є +, a (I). Then by property (3) above there exists a non negative integer r and a positive constant C such that for Re. S a F (n, s) = < f (z, t), e p (z) > C max λ,, [ e p (z)] 0 k r 0 k r

6 C max sup e D ( D ) e p (z) 0 k r ( z,t ) Є I 0 k r = C max sup e s a e p (z) 0 k r ( z,t ) Є I 0 k r but e e p n (z) < A on 0 < t <, -h < z < h so that F (n,s) C. A max s. a 0 k r 0 k r F (n, s) C. A. P ( S (a ) ) where P ( S (a ) ) is alpolynomial that depends in general on the choice of a. Analyticity Theorem : - Theorem : -The Generalized Laplace Finite Marchi Fasulo transform is an analytic function of s proof : Let (n, s) be arbitrary but fixed point in D f and choose the real number a and r such that σ f < a Re. S -r Re. S + r let s be complex increment such that s < r. Now for s 0 by definition of F (n, s) as (, ) (,) < f (z, t), e p (z) > = < f (z, t), ψ s ( z, t ).(.4)

7 where ψ s (z,t) = { [ e-( s + s ) t - e - s t ] - e } p (z).(.5) since ψ s (z,t) Є +,a (I) equation () has meaning. We show that ψ s (z,t) 0 in +,a (I) as s 0 but as f Є +,a (I) this implies that < f, ψ s > 0 let c denote the circle with center at s and radious r where o < r < r < Re. S a by equation (.4) D Ω ψ s (z,t) = ( ) a p (z) { [ ((s + s) e - ( s + s ) t s e s t ] s e -st } since e -st p n (z) is analytic in s using Cauchy s integral formula to the r.h.s. of the last equation we obtain D Ω ψ s (z,t) = ( ) (a ) p (z) [ 2πi s ξ e dξ ξ (s + s) ξ e dξ ξ s 2πi ξ e dξ ( ξ s) = ( ) (a ) p (z) ξ e dξ 2πi (ξ s)(ξ s s) ξ e dξ (ξ s) = ( ) k +k 2 a n 2 k2 p n (z) s 2πi ξ e dξ (ξ s s)(ξ s) now for all ξ Є C and 0 < t <, -h < z < h e D ψ Ω (z, t) = (a ) p (z) s 2π e ξ k e dξ (ξ s s)(ξ s) s 2π. k (a ) dξ ((r r)r, (a n2 ) k 2 ( ), where e at ξ e p (z)] k k being constant independent of ξ and t. The right hand side of the last equality is independent of z,t converges to zero as s o

8 this proves that Ψ s (z,t) converges to zero in +, a (I) as s o. Hence if f Є +, a (I), < f, Ψ s > o as s o. Lemma (.2) : - Let M [ f (z,t) ] = F ( n,s ), where ( n,s ) Є D f, Re. S > σ and φ (z,t) Є +,a (I) and -h < a < b < h assume that, φ (n, s) = φ(z, t) e p (z) λ dt, dz. Then for any fixed real number r with o < r < < f(u, v), e p (v) > φ ( n, s)dρ = < f(u, v), e p (v) φ ( n, s)dρ >, where s = σ + iρ and σ is fixed and σ > σ f. Proof : - We know that e -st p (z) is a member of +,a (I) and e p (v) φ ( n, s) dρ is also the member of +, a (I) is also the member of +,a (I indeed, D D e p (v) φ ( n, s) dρ e s (a ) p (v) φ ( n, s) dρ A. e -σu s (a ). 2r which exists, where A is bound for φ (n,s). p (v) both sides has sence. If φ (z,t) = o, the proof is obvious. So assume that φ (z,t) o.

9 partition the path of integration on the straight line s = σ ir to s = σ + ir into m intervals each of length and let s p = σ + i ρ p be any point in path interval θ (u, v) = e s p u by applying f ( u,v) to (.5) term by term p (v) φ n, s 2r m.. (.6) < f(u, v), θ (u, v) >= < f(u, v), e s p u p (v) > φ n, s 2r m < f ( u, v) e p (v) φ( n, s) dρ. Since < f ( u,v), e -su p (v) > φ (n,s) is a continuous function of ρ choose a such that σ f < a < σ since f Є +,a To show that To show that θ (u, v) converges in +, a (I) to e p (v) φ( n, s) dρ, We will to show that for fixed k, k 2 A m ( u,v) convsrges uniformly to zero, o < u <, -h < v < h, where A (u, v) = e D Ω [ θ ( u, v) e p (v) φ ( n, s) dρ ]. A (u, v) = e ( ) s (a ) e s p u p (v) φ( n, s) 2r m ( ) s (a ) e p (v) φ( n, s) dρ, Now e au. e -su p (v) e (a-σ) u. p (v) o as u as a < σ. So given Є > 0 we choose T so large that for all u > T

10 ENCLOSURE I e. e p (v) ε 2 [ s (a ) φ ( n, s) dρ ]. (.7) since φ (z,t) o the right hand side of (.7) is finite Hence for all u > T the magnitude of the second term on the right hand side of (.6) is bounded by Є. Morever for u > T the magnitude of first term on right hand side of (.6) is bounded by ε 2 [ s (a ) φ ( n, s) dρ ] s (a ) φ n, s 2r m. We can now Choose m 0 so large that for all m > m 0 the last expression is less than Є. Hence for all u > T and for all m > m 0 A (u, v) < Є. * Theorem ( Inversion) :- Let f Є +, a (I), and let F (n,s) be the distributional Laplace Finite Marchi Fasulo transform of f ( z,t ). defined by M [ f (z, t ) ] F (n, s) = < f (z, t ), e p (z)] > For (n,s) Є D f, in the sense of convergence in D (I), f(z, t) = lim r, m 2πi p (z) F(n, s) e ds λ, where σ is any fixed number s.t. σ > σ f. Proof : - Let φ (z,t) be an arbitray member of D (I). To show that lim r, m < 2πi p (z) F(n, s) e ds λ φ(z, t) > < f (u, v), φ (u, v) >...(.8) Let φ Є D ( I ) has support contained in [ A, B] [ a, b ], where 0 < A < B < and -h < a < b < h.

11 2πi p (z) F(n, s) e ds λ is locally integrable function on I. Equation (.7 ) can be written without limit notation as 2πi [ p (z) F(n, s) e ds ] φ(z, t) dt dz λ substituting s = σ + iρ ds = idρ we get 2π [ p (z) λ F(n, s) e dρ ] φ(z, t) dt dz. We interchange the order of integration as φ ( z,t) has bounded support and integrand is continuous function of z, t and ρ. Therefore last expression takes the form 2π F(n, s) φ(z, t) e p (z) λ dt dz dρ = 2π < f(u, v) e p (v) > φ(z, t) e p (z) λ dt dz dρ = < f (u, v), 2π e p (v) φ(z, t) e p (z) λ dt dz dρ > the order of integration for repeated integrals can be changed because again φ(z, t) is of bounded support and the integrand is continuous function of ( t,z, ρ ), we obtain < f (u, v), 2π [ e p (v) φ(z, t) e p (z) λ dt dz dρ > = < f (u, v), 2π φ(z, t) p (z) p (v) λ dt dz e () e () dρ >

12 = < f (u, v), 2π φ(z, t) T e () dt dz e () e () i(t u) >, where T = p (z) p (v) λ = < f (u, v), π sin r (t u) () φ(z, t) T e (t u) dtdz > < f ( u,v), φ ( u, v ) > as r, m. The Uniqueness Theorem : - Let M [ f (z,t) ] = F ( n,s ) and M [ g (z,t)] = G ( n,s ) for all ( n,s ) Є D f and ( n,s ) Є D g respectively and if F ( n,s ) = G ( n,s ) for ( n,s ) Є D f Λ D g 0, then f = g in sense of equality in D ( I ). Proof :- In the sense of convergence in D ( I ) and in view of inversion theorem f(z, t) = lim r, m 2πi p (z) F(n, s) e ds, λ the right hand side of this equation becomes lim r, m 2πi p (z) G(n, s) e ds λ which by inversion theorem equal to g (z,t) Hence f = g. Lemma 3 : - if ψ (z,t) Є D (I) then

13 π e T sin r(t u) t u [e () ψ (z, t) ψ (u, v) ] dtdz converges to zero uniformly an 0 < u <, as r,m where the support of ψ (z, t) is contained in [ A,B ] [ a, b ] O < A < B <, -h < a < b < h Proof : - Let us devide the interval ( 0, ) ( -h,h) in to six disjoint sets. ( ) ( 0,A ) U ( B, ) ( -h, h ) ( 2 ) ( A, B ) ( b, h ) ( 3 ) ( A, B ) ( -h, a ) ( 4 ) ( 0, A ) ( a, b ) ( 5 ) ( B, ) ( a, b ) ( 6 ) ( A, B ) ( a, b ) Case I : - Proof : - for ( u, v) Є ( o, A) U ( B, ) ( -h, h ) ψ (u, v) = 0 since ψ (u, v) has support contained in [ A, B ] [ a, b ] therefore e π T sin r(t u) t u e () ψ (z, t) ψ (u, v) dtdz = e π T sin r(t u) t u e () ψ (z, t) dtdz By lemma 2 as r this integral reduces to = e π T ψ (z, v) dz we shall now show that for fixed u and -h < z < h

14 Lim e m π T ψ (z, v) dz converges to zero uniformly for all z. since ψ (z, v) Є D ( I ) where I { (z, v) / -h< z < h,u is fixed } since e T ψ (z, v) dz bounded on 0 < v < the last integral converges to zero Uniformly by lemma 2 as m in the same way we prove the remaining cases. Case 6:- proof we show that the given equation is also true for the interval [ A, B ] [ a, b ] the given equation can be written as, e π T sin r(t u) t u e () ψ (z, t) dtdz e π T sin r(t u) t u ψ (u, v) dtdz this tends to e T ψ (z, u)dz e T ψ (v, u)dz as r (.8) by lemma 2 therefore (.8) can be written as e T [ψ (z, u) ψ (v, u)]dz but by lemma (2) this converges to zero as m for fixed u combining all the cases we get proof of lemma

15 Application :- Consider a solid circular cylinder whose axis is coincident with z axis defined by -h z h and 0 r d, where d is the radius. Consider the heat conduction problem with symmetry with respect to z axis. i.e, 2 u r u = z 2 (.9) where k = where k is conductibility of medium ρ is calarific capacity assumed to be ρ constant with initial and boundry condition, u ( r,h,t) + k u ( r,-h,t) + k 2 u (r, z, o) = 0 ( r,h,t) = θ a = constant 0 < r < d, t >0 ( r,-h,t) = θ b = constant 0 < r < d, t >0 u ( ξ, z,t) = f ( z,t) -h < z < z, t > 0 o < ξ < d u ( d, z,t) = g ( z,t) t > 0 We shall now have the generalized solution u ( r, z,t) in the space, applying the Laplace Finite Marchi Fasulo transform to equation (.9). and taking into account the boundary condition we get u r + r u r + u z e p (z)dt dz = r u t e p (z)dt dz Since u = u ( r, z, t) U ( r, n, p) r + r U( r, n, p) r + u z e p (z) dt dz = k e [ t p (z) u (r, z, t)dz ]dt

16 U r + r U r + η p a U = k [ e p (z)u ( r, z, t) dz ] p e p (z) u( r, z, t) dz ] dt But u ( r,z,0) =0 and after substituting the limits and by the property that U r p (z)dz = p (h) α β u ( r, z, t) + α u z - () [ β 2 u ( r, z, t) + α 2 ] z = -h and using the condition, substituting β = β 2 =, α = k α 2 = k 2 η = [ () - () ] = constant. Hence, we get. 2 U r a n 2 U = U 2 U r 2 + let α 2 - ( a n 2 + ) U = = ( a n 2 + ) then, 2 U r α2 υ =. The solution of this differential equation is, υ ( r, n, p) = A I 0 (αr ) + B k 0 (αr ) + pα 2..(.0) where I 0 (αr ) and k 0 (αr ) are the modified Bessel functions of first and second kind of order zero respectively. As r tends to zero, k 0 (αr ) tends to infinity. But υ ( r, n, p) remains finite,hence B=0. Then equation becomes U ( r, n, p) = A I 0 (αr ) + Since u (ξ, z, t) = f ( z, t ) pα 2.(.)

17 U (ξ, n, p) = F ( n, p ) U ( r, n, p) = F ( n, p ) = A I 0 (αξ ) + pα 2 by putting r = ξ A = (, ) ( ) pα 2, putting this value of A in (.) U( r, n, p) = [ (, ) ( ) ] I 0 (αr ) + pα 2..(.2) Applying the inverse Laplace finite Marchi Fasulo transform to (.2), we get u(r, z, t) = lim r, m 2πi p (z) λ { Now for each φ (z,t) Є D (I) it can be shown that n F (n, p ) pα I I (αξ ) (αr) + η pα } e dp. < u( r, z, t), φ (z, t) = 2πi p (z) λ [ F (n, p) η pα { ] I I (αξ ) (αr) + η pα} φ (z, t) dp dt dz. So that u( r,z, t) as a conventional function is obtained as u(r, z, t) = 2πi n= p n (z) λ n σ+i σ i [ F (n, p) η pα 2 ] { I 0 (αr) + I 0 (αξ ) η pα2} dp. (.3) We show that (.3) is the solution since F(n,p) is bounded on the domain [ A, B ] [ a, b ]. it can also be shown that the series obtained by applying D, D, D seperatly under the summation and under integral sign converges uniformly on domain D. To verify the boundry condition we shall show that for each φ Є D (I) u Є +, a (w) < u( r,z, t) φ (z,t) > < f( z, t) φ (z,t) > p n (z) λ n { [ (, ) η ] ( ) I 0 (α ξ ) + p α 2 }dp φ (z,t) dt dz = p (z) n λ n F (n,p) φ (z,t) dt dz

18 = < f (z,t), φ (z,t) > < u(ξ,z, t), φ (z,t) > = Similarly we can show that < u (d,t,z), φ (z,t) > = < g ( z,t), φ (z,t) > REFERENCES [] K.W.KHOBRAGADE An Inverse Transient Thermoelastic Problem Of A Thin Annular Disc Applied Mathematics 6(2006) [ 2] S. R. PATEL Inverse problems of Transient Heat conduction with Radiation, The mathematics Education (97) vol. No. 4. [3 ] A.H. ZEMANIAN Generalised Integral Transformation Interscience Pub. New York, 968